tag:blogger.com,1999:blog-83568752095479654202024-03-06T05:57:07.833+11:00Math GardenMath for the curiousUnknownnoreply@blogger.comBlogger93125tag:blogger.com,1999:blog-8356875209547965420.post-31931495516988030912016-12-20T09:53:00.002+11:002016-12-22T10:06:43.720+11:00Merry Christmas and Happy New Year<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgvnHfRAug2eHG1X_otDWbXIU5KA0dhyj_b56q-ZzxRdt59pTY3IThXBDYNaH8jGyB-HayswwzGJthGhygy6ARY28jGxDwVelYIx1c9Q_dLj_kaRYJ1k1RCfU4B_r2H90N3TLacLFMfM03m/s320/vuontoan_mathgarden_MotherMary.jpg" width="219" /></div>
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<img border="0" height="226" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdj4H2B0G4hzNyjQn9vLYA-g4knchD4BYXj8Ff8WdenOr65HGQQgPo2nG-sTQ-fUqLgf6R4is_gkheeJbOajlfBTYdLYs6vwK3VDVhLnO7yNa5Qb1rRGa5oIBHo2vfCrsB1jnHcAFMYDjj/s320/vuontoan_mathgarden_vianney.jpg" width="320" /></div>
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<img border="0" height="276" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgMdLXt1OXQxdgqf_WmPLXuaJdyu_nSbfc5tWvgeemzX_TKdNv3VNxPOqtDGchu_CyYx3Tmlf3Tai9IztHK6kPJX4cWo47tU2ryJLiESucxqKVjfyRtVcTsaLbNtejT_N0_-MHGZJwggKBh/s320/vuontoan_mathgarden_lfh.jpg" width="320" /></div>
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<blockquote class="tr_bq">
<span style="color: purple;"><br /><b>New Year Puzzle:</b> In a <i>magic square</i>, the sum of numbers on each row, each column, and each diagonal is the same.
Fill in the numbers in the following squares to make them become magic squares.</span>
</blockquote>
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<img border="0" height="186" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgB3VKq2_Abj214qlXUYkyccJb1PC7SRdDhOcaFKDeulyE5Fg29kMtW0ZbqXheTSxw7pAkR-QczI-pPKlnKJhK-mkNpIgr4m7d4oyVi0tad_FdkrEtB9BYHz8xPqRrb6saRFG4KpxgevnsL/s400/vuontoan_mathgarden_y2017.png" width="400" />
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<img border="0" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj63QxpOgKHrvhw5RUpXsxpjhC5e1ZHINq0jnqFWSjogqozssePhv9BENpAlBf9ABqz1xngYs-BicrN8eWci0DTNZlQD7IURMX-qLJQNlE3WlOHoNpsHIONs276RwcmHq8pDp4Bv6FvMclJ/s400/vuontoan_mathgarden_3king.jpg" width="282" />
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<img border="0" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhBEn5BKbb912rSSY79ANbEqYA-hPKjH-8wXHhgQa_VoAZD4P7-5BBbY-0lGhcm6odT3qDN_50EDCEMGRhZ-NXVrDJ-x1CpSHyBwq1PMckbjKRWcBdJnGWkeCo7X8F_XDeCzp_a_VfB7yq0/s320/vuontoan_mathgarden_rudolph.jpg" width="226" /></div>
Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-78245572074516454392016-10-13T11:12:00.002+11:002016-10-13T11:43:49.302+11:00Which is smaller? which is larger?<table align="center">
<tbody>
<tr><td><img border="0" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhc6_GXoyXYAldvSHkonZf0FnR2FCg9m6kN__3pRqC23jtl5oyApNZsvms1UQ9xHBlonx4d_hUnLtIVMDE4kIIIkMRFGnx-2-iCDcFTyrwDMOpf0o2czCzeFsfKhtGItvIJ2Twyl9iHoLg/s400/vianney_fatherday16_7f.jpg" width="142" />
</td>
<td><img border="0" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhFGWO5DJoXzViujYxzX47-qumLZKE6LasMeVnYZd2k9T8OasskHp5q11PjK8VhhzKdX1L29vgF9TXAjn27L-lkdI-7bu-TCH02UH3DOBeaU0VxV2D9OdPuFBPjjzYg0_Csd5PBPnnIwZmk/s400/maryanna_fatherday16_3.jpg" width="133" />
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<br />
<span style="color: purple; font-size: x-large;">T</span>oday let us look at the following two tricky questions:<br />
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<span style="color: #3d85c6; font-size: large;"><b>Question 1: </b></span><span style="color: #6fa8dc; font-size: large;">Among these two numbers </span><span style="color: #3d85c6; font-size: large;">4444 x 55555</span><span style="color: #6fa8dc; font-size: large;"> and </span><span style="color: #3d85c6; font-size: large;">5555 x 44444</span><span style="color: #6fa8dc; font-size: large;">, which one is smaller? which one is larger?</span><br />
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<span style="color: #3d85c6; font-size: large;"><b><br /></b></span> <span style="color: #3d85c6; font-size: large;"><span style="color: #3d85c6; font-size: large;"><b>Question 2: </b></span><span style="color: #6fa8dc; font-size: large;">How about </span><span style="color: #3d85c6; font-size: large;">5000 x 5000</span><span style="color: #6fa8dc; font-size: large;"> and </span><span style="color: #3d85c6; font-size: large;">4999 x 5001</span><span style="color: #6fa8dc; font-size: large;">, which one is smaller? which one is larger?</span></span><br />
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<a name='more'></a><br />
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The answer can be found below.<br />
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<img border="0" height="276" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgx6KNYA17iUVztwVUCy34rRbnXrat08k4ZJl9gOPWB93-KokJnEsHiJLAKb16knM-e34vOEHCAIKvCUZqJxazufnu4kYAxLbceuXIPspeNHV9mMvGhsWFxF_Zd4qgDD5aUpSIUkB0GLKCA/s400/vianney_fatherday16.jpg" width="400" /></div>
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<span style="color: #3d85c6; font-size: x-large;">Answer.</span><br />
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<span style="color: purple;"><b>Question 1: </b>Among these two numbers $4444 \times 55555$ and $5555 \times 44444$, which one is smaller? which one is larger?</span><br />
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We can do a survey and ask people around us to see how many people actually think $4444 \times 55555$ is the larger number, and how many people think otherwise.<br />
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It is likely that people would pick 4444 x 55555 as the larger one, because 55555 seems to be much bigger than 44444.<br />
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The answer is, these two numbers are actually equal to each other!<br />
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The reason is<br />
$$4444 = 4 \times 1111$$ $$55555 = 5 \times 11111$$ so $$4444 \times 55555 = 4 \times 1111 \times 5 \times 11111$$<br />
Similarly,<br />
$$5555 \times 44444 = 5 \times 1111 \times 4 \times 11111$$<br />
So they are equal!<br />
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<b style="color: purple;">Question 2: </b><span style="color: purple;">Among these two numbers $5000 \times 5000$ and $4999 \times 5001$, which one is smaller? which one is larger?</span><br />
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This is hard to decide! We can see that 4999 is one less than 5000, whereas, 5001 is one more than 5000. So which one is larger, $5000 \times 5000$ or $4999 \times 5001$?<br />
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The purpose of this question is to introduce the following beautiful algebraic identity.<br />
<br />
<blockquote class="tr_bq">
<span style="color: purple;"><b>Difference of two squares identity</b>: $$a^2 - b^2 = (a + b)(a - b)$$
</span></blockquote>
This is probably one of the simplest and most elegant identities in algebra. In the next post, we will learn more about this identity.<br />
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We will use the identity as follows:<br />
$$4999 = 5000 - 1$$ $$5001 = 5000 + 1$$ so $$4999 \times 5001 = (5000 - 1)(5000 + 1)$$ $$ = 5000 ^ 2 - 1^2 = 5000 \times 5000 - 1$$<br />
This shows that $4999 \times 5001$ is smaller than $5000 \times 5000$.<br />
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Let us stop here for now. In the next post, we will learn some surprising and beautiful facts about the "<b style="color: purple;">difference of two squares"</b> identity. Hope to see you again then.<br />
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<i>Homework. </i><br />
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1. Among these two numbers 1212 x 343434 and 3434 x 121212, which one is smaller? which one is larger?<br />
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2. How about 4999 x 5001 and 4998 x 5002, which one is smaller? which one is larger?<br />
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3. 44 x 45 and 43 x 46, which one is smaller? which one is larger?<br />
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<img border="0" height="217" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjkh6rPQqfJ8Z7uJWnAlrRHkLyS9KsD0UoAsJte0nn1a01tidUsqO6_ts-QkiodHLviC_PMeq-bqJkRMdw-h9yct_cJ0ndN9nL8W6AwmUw5Kd_2RVc6MIaZMHuhyphenhyphenYVyUthfYGCqltX1YXwb/s320/maryanna_fatherday16_5b.jpg" width="320" />
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Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-28844034601478023252016-05-02T10:50:00.000+10:002016-05-02T11:06:51.454+10:00Euler's magic square<br />
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<img border="0" height="262" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgR2Y9XMfBeSVLhMcNeozCP7VUJEDjX2gPx7B7uk4Dg8rjwbo7cCJZ64UUAwX2vfQj8_a5Tk2SfxkQl9CHCXsdNgQI_iq2_A30s4RwTd_ACoS7a4g2tS5Zg4GM2ka2a15bv5FNyUaz9V5Mt/s400/vuontoan_mathgarden_eye.jpg" width="400" /></div>
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<span style="color: purple; font-size: large;">Can you spot the hidden magic in this square?</span></div>
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<span style="color: purple; font-size: large;"><br /></span></div>
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<img border="0" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhErbl-sQHMDDJVIef374QzkIP01NXZmlxOufhUBCOSOKWVVXSP3p_YjPVXyPwPaKZKu-AhDHE8oDFRkGbHvi21n_8zmCvrMmGUWeQqIEliim-g4SjprUEjtCsbKbWc28_5pwzEphLbfHN_/s400/vuontoan_mathgarden_euler_magic.png" width="395" /></div>
<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-83682043253062878782016-04-25T17:23:00.000+10:002016-07-11T21:44:29.352+10:00Vianney's awesome square!<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="312" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgrWzMpfNlM2owmKhfa7zoGjijXWL_2OHLZ_S2S2yEnGUVbgRGxMkziQC4_4T4RisEaqqezpXKr0knugSkyIAfcrOsJZFXrAG9zqukQrrYn3OVvt4vRnYZpyPM-f5E6Fz6tTinl7Jh_4gY/s640/vianney_square.png" width="640" /></div>
The reasons that this square is so awesomistic are:<br />
<ul>
<li>each entry is a square number: $$1=1^2, \quad25=5^2, \quad 49=7^2$$</li>
<li>the sum of numbers on each row, each column, and each diagonal is the same.</li>
</ul>
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<img border="0" height="260" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEix5QixKiHXaWJl0L1veZmYpHLf8GPIyLrSNGVfbhhKufwPgvzy5a9C5TOJneWmcCjxPmXfbkoYhygumkU4rXIp9yDvY9HK4edgP3GyadsqAnxyP-HueHBLPMBiqRf_pb4v0qsXJWF8Kek/s320/mathgarden_vuontoan_StJohnVianney.jpg" width="320" /></div>
<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-26552476501576023662016-03-16T21:04:00.001+11:002016-03-16T23:40:38.055+11:00Radian<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="285" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgR5Yq0aJ7odzGjWah-xSMgTl_XpQ_Ug2yr8apENKf-aGwuqBAU0CdnasyMYOSyl5CbXkkPbIVHtx5mGYaBMZEEDdKm2OuuK7TMD-aJFGu5yQUDDsyxk-YN6pH_XgxuOHjgai5Zyp6kqdsw/s400/vuontoan_mathgarden_vianney_circle.jpg" width="400" /></div>
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<span style="color: purple; font-size: x-large;">O</span>n this $\pi$ day occasion, we will learn about the concept of radian.<br />
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<a name='more'></a><br />
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<span style="color: #0b5394; font-size: x-large;">Radian</span><br />
<span style="color: #0b5394; font-size: x-large;"><br /></span>
In our normal life, when we talk about angles, we often use the degree. For example, a right angle is equal to 90 degrees, an angle of an equilateral triangle is 60 degrees, a straight angle is $180^{o}$.
However, for all trigonometric functions in mathematics, such as sin(x), cos(x), etc., the angle $x$ is always understood as in radian.<br />
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So what is the radian?<br />
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To understand angle measurement in radians, we draw a unit circle. Unit circle is a circle with radius equal to 1.
We know that <a href="http://mathgardenblog.blogspot.com/2016/03/pi-2016.html" target="_blank">by definition</a>,
$\pi$ is the length of the semicircle.<br />
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<img border="0" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiBZnVfoUB_A_nwkYX3z248ukBQPw7qR2p3vmM0ZUFPFpA-7HXkrP2HoWi-FlXEKzVYhYkdUNlqtqbAeKsSfvOJpjJNFWKjCLCp49m5I-ibZ_uDkXQPHASYWxzDjjl0lJd8RQg9JXyVApxp/s320/vuontoan_mathgarden_radian1.png" width="310" /></div>
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An angle's measurement in radians is equal to the length of the corresponding arc of the unit circle.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><img border="0" height="309" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgmiud2r52plTYGYkkIP01B9IhREW7mFBhuLljBF1TBEdmmK0D9gPdCEfISdww11veycWT6xfknsT5hIIUAaWB8AMcuwzvoNmfmsNdK64zgIatAIVeKyD4JZ8hlRC_n6Nj7u3ZCnZIW1QNV/s320/vuontoan_mathgarden_radian4.png" width="320" /></td></tr>
<tr><td style="text-align: center;">In radians, $x$ is the length of the arc</td></tr>
</tbody></table>
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Let us look at an example of a right angle.
A right angle is subtended by a quarter of the circle, and the length of a quarter of the circle is $\frac{\pi}{2}$, so a right angle is equal to $\frac{\pi}{2}$ in radians.<br />
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<img border="0" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhER7SELjvABV2pjBghJFKZ62mIa89zpSK1rS_jyUrcr-_RRKU3_VxQQuK6OPsyxxngrRUsMbvBzoNBh3poB_BLzdqm-VLo-DYNBrPb9BdjzAmCECfIrDCuTHl1QcZzqP3xCk-MZSV1dwmN/s320/vuontoan_mathgarden_radian5.png" width="318" /></div>
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A straight angle ($180^{o}$) occupies half of the circle, and the length of a half of the circle is $\pi$, so a straight angle is equal to $\pi$ in radians.<br />
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<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQuk4HTnMAY4fqLARXncocSkusCreLKc4sZhMM3oX0_WFCFG5PBF9UdFFuu6BJtWSPn-7gtRCmO0zkgLToy2OD-k7TOPvw9OcHy19rVm3TAwzFRFCB8gMI0USVRJyI3cNkNfDIpxnZN03v/s320/vuontoan_mathgarden_radian6.png" width="306" /></div>
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We can easily remember the translation from degrees to radians by the following correspondence
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<blockquote class="tr_bq">
<span style="color: purple;">straight angle 180 degrees $\to$ semicircle $\to ~~ \pi$
</span></blockquote>
Common angles are
<br />
$$180^{o} ~~\to ~~ \pi$$
$$360^{o} ~~\to ~~ 2\pi$$
$$90^{o} ~~\to ~~ \frac{\pi}{2}$$
$$45^{o} ~~\to ~~ \frac{\pi}{4}$$
$$60^{o} ~~\to ~~ \frac{\pi}{3}$$
$$30^{o} ~~\to ~~ \frac{\pi}{6}$$
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Let us stop here for now.
We will return to our series on mathematical identities in our next post.
Hope to see you there.<br />
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<br />
<i>Homework.</i><br />
<br />
In this homework section, we will prove <a href="http://mathgardenblog.blogspot.com/2016/03/pi-2016.html" target="_blank">the Vieta formula</a> of the number $\pi$ that we mention in the <a href="http://mathgardenblog.blogspot.com/2016/03/pi-2016.html" target="_blank">previous post</a>
$$ \frac{2}{\pi} = \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}} \cdots $$
In the following figure, we can see that $ZA = sin(x)$ is a line segment which is shorter than the circle arc $ZI = x$
$$
sin(x) < x
$$
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<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="309" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgvCw7tXlC8UX3LxAWtsC04sqvvsNr7XV6XIjsluUsNeavBNZ2nOjuKq-NwkVyz-8GNcThCm9qQV-GjBxEaPcRAxvUt6Zc3F1IQquXJxSXdGpESTyUmqqTRC6bIUT-G8qEnMOeqLaK-lM_R/s320/vuontoan_mathgarden_radian7.png" width="320" /></div>
Especially, when the angle $x$ is small, $sin(x)$ is getting closer and closer to $x$.
We will use this fact to prove the Vieta formula.
<br />
<br />
1. Use the trigonometric cos formula for <a href="http://mathgardenblog.blogspot.com/2014/02/moivre-trigonometric-multiple-angle-formula.html" target="_blank">double angle</a>
$$cos(2x) = 2 cos^2(x) - 1$$
to show that
$$
cos \frac{\pi}{4} = \sqrt{\frac{1}{2}}
$$
$$
cos \frac{\pi}{8} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}
$$
$$
cos \frac{\pi}{16} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}}
$$
Thus,
$$
\sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}} =
cos \frac{\pi}{4} \cdot
cos \frac{\pi}{8} \cdot
cos \frac{\pi}{16}
$$
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2. Use the trigonometric sin formula for <a href="http://mathgardenblog.blogspot.com/2014/02/moivre-trigonometric-multiple-angle-formula.html" target="_blank">double angle</a>
$$sin(2x) = 2 sin(x) ~ cos(x)$$
to show that
$$
cos \frac{\pi}{4} \cdot
cos \frac{\pi}{8} \cdot
cos \frac{\pi}{16}
=
\frac{\frac{1}{8}}{sin \frac{\pi}{16} }
=
\frac{2}{\pi} \cdot
\frac{\frac{\pi}{16}}{sin \frac{\pi}{16} }
$$
<br />
<br />
3. As we have discussed above, since the angle $\frac{\pi}{16}$ is very small,
$$
sin \frac{\pi}{16} \approx \frac{\pi}{16}$$
and
$$
cos \frac{\pi}{4} \cdot
cos \frac{\pi}{8} \cdot
cos \frac{\pi}{16}
\approx
\frac{2}{\pi}
$$
<br />
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4. In general, prove that
$$
cos \frac{\pi}{4} \cdot
cos \frac{\pi}{8} \cdots
cos \frac{\pi}{2^n}
=
\frac{2}{\pi} \cdot
\frac{\frac{\pi}{2^n}}{sin \frac{\pi}{2^n} }
$$
and
$$
\lim_{n \to \infty} cos \frac{\pi}{4} \cdot
cos \frac{\pi}{8} \cdots
cos \frac{\pi}{2^n} = \frac{2}{\pi}
$$
This is the Vieta formula of the number $\pi$:
$$\sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}} \cdots = \frac{2}{\pi}$$ Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-64512172313002322102016-03-13T00:05:00.000+11:002016-03-13T00:05:13.345+11:00Pi day<div class="separator" style="clear: both; text-align: center;">
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$\pi$ day is celebrated every year on March 14 because $\pi \approx 3.14$.<br />
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The number $\pi$ is ultimately associated with the circle.
By definition, if we draw a circle of radius 1, then $\pi$ is the length of a semicircle.<br />
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In the figure below, if we go from A $\to$ B $\to$ C $\to$ D by straight lines then the length of the trip is 3,
but if we go by the circle then the trip length is $\pi$,<br />
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so $\pi$ is a tiny bit bigger than 3:<br />
$$\pi > \approx 3$$<br />
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There is an easy way to remember the decimal value of $\pi$. First, we write the three pairs of odd numbers as follows<br />
$$<br />
11~33~55<br />
$$<br />
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Now cut the above number into two halves<br />
$$<br />
113~~\mid~~355<br />
$$<br />
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If we take the bigger number divide by the smaller number, we will have<br />
$$ \frac{355}{113} = {\bf 3.141592}~92... $$<br />
Whereas, $$\pi= {\bf 3.141592}~65...$$<br />
So with this method, $\frac{355}{113}$, we can derive and remember the first 6 decimal digits of $\pi$.<br />
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To celebrate this year's $\pi$ day, let us enjoy this beautiful identity due to the mathematician Viete:<br />
$$ \frac{2}{\pi} = \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}} \cdots $$<br />
Happy $\pi$ day everyone!!!<br />
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<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-70081509746039146582016-02-12T13:07:00.001+11:002016-02-12T13:14:02.900+11:00Pythagorean Identity<div class="separator" style="clear: both; text-align: center;">
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<span style="color: purple; font-size: x-large;">T</span>his year we will start a blog series about mathematical identities. Let us begin today with the Pythagorean Identity $$
a^2 + b^2 = c^2
$$
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<span style="color: purple; font-size: x-large;">I</span>n mathematics, the Pythagorean theorem is probably the most well known theorem. Undoubtedly, anyone who has studied geometry would remember this famous theorem.<br />
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The Pythagorean theorem asserts that, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgI86S-V9FdPBqh2p0xWAk3JI0kscvcTRxSjRPMeTxdmrb7IP3Yyn9g9Jo5vnYFoC5O-ytrLnoMMOAza9spFgHroN6u1omw04CrmRQDP8YzDaOnWDwhzCKoQI6KF-h62jU4JtpNEQhl796E/s242/py2.png" /></td></tr>
<tr><td style="text-align: center;">Pythagorean theorem: $a^2 + b^2 = c^2$.</td></tr>
</tbody></table>
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<span style="color: purple; font-size: x-large;">A</span>t school, we learn about the Pythagorean theorem around year 8. After introducing the theorem, most teachers would probably use the following example:<br />
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That is why most of us would remember the Pythagorean identity:
$$
3^2 + 4^2 = 5^2
$$
$$
9 + 16 = 25
$$
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<span style="color: purple; font-size: x-large;">M</span>e and a classmate of mine once challenged one another to write down as many as possible the Pythagorean triples like
$$3^2 + 4^2 = 5^2,$$
in 5 minutes, whoever got the most identities would win.<br />
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Both of us finished the writing in much less than 5 minutes and looking at one another, each of us was sure that we would win.<br />
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We had a good laugh when the papers were opened, we both got the same idea $$30^2 + 40^2 = 50^2$$
$$300^2 + 400^2 = 500^2$$
$$3000^2 + 4000^2 = 5000^2$$
$$\vdots$$
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So we changed the rule. The new rule was that the identities of this type $$
(3n)^2 + (4n)^2 = (5n)^2
$$
were forbidden.
With this new rule, the game was harder!<br />
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Now let us pause here for 5 minutes, would you like to have a try to see how many Pythagorean equations you can come up with?!<br />
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<span style="color: purple; font-size: x-large;">L</span>et us now get back to the Pythagorean equation $$
a^2 + b^2 = c^2
$$ how can we find $a$, $b$ and $c$ ?!<br />
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Looking at $$3^2 + 4^2 = 5^2$$ the first thing we may try is to check if $4^2 + 5^2$ is equal to $6^2$ or not?
$$
4^2 + 5^2 = 16 + 25 = 41
$$
$$
6^2 = 36
$$
So they are not equal $$
4^2 + 5^2 \neq 6^2
$$<br />
(4, 5, 6) is not a Pythagorean triple.<br />
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Now, don't lose heart! We can try $b=n$ and $c = n+1$ and see what the value of $a$ would be.<br />
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We have
$$
a^2 + n^2 = (n+1)^2 = n^2 + 2n + 1
$$
so
$$
a^2 = 2n + 1
$$
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<span style="color: orange; font-size: x-large;">A</span>h-hah!!!, $a$ is an odd number, $a = 2k + 1$, and we have<br />
$$
a^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2n +1
$$
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So $n = 2k^2 + 2k$.
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With this, we have found general formula for $a$, $b$, $c$:
$$
a = 2k + 1
$$
$$
b = n = 2k^2 + 2k
$$
$$
c = n+1 = 2k^2 + 2k + 1
$$
and we obtain the Pythagorean identity
$$
(2k + 1)^2 + (2k^2 + 2k)^2 = (2k^2 + 2k + 1)^2
$$
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With $k = 1$ we have the familiar
$$
3^2 + 4^2 = 5^2
$$
With $k=2$, we have
$$
5^2 + 12^2 = 13^2
$$
$$
25 + 144 = 169
$$
And with $k=3$, we get
$$
7^2 + 24^2 = 25^2
$$
$$
49 + 576 = 625
$$<br />
Fantastic !!!<br />
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Let us stop here for now. Next time we will learn more mathematical identities.
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In the meantime, you can read about the Pythagorean triples <a href="http://mathgardenblog.blogspot.com/2012/08/pythagorean-triple.html" target="_blank">here</a>.
The Pythagorean theorem is proved <a href="http://mathgardenblog.blogspot.com/2013/06/Garfield-Pythagorean-Theorem.html" target="_blank">here</a> and <a href="http://mathgardenblog.blogspot.com/2012/05/similar-triangles.html" target="_blank">here</a>.
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Hope to see you again soon.
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<i>Homework.
</i><br />
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1. Find other ways to generate Pythagorean identities.
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2. Go to <a href="http://google.com/">google.com</a> to search and read about the Pythagorean theorem.
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<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-66656946965122931402015-12-23T17:25:00.000+11:002015-12-23T17:25:08.311+11:00Merry Christmas and Happy New Year<br />
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<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-71354176577763243992015-11-02T18:54:00.000+11:002015-11-05T20:50:07.038+11:00Fibonacci numbers and continued fractions<div class="separator" style="clear: both; text-align: center;">
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<span style="color: purple; font-size: x-large;">F</span>ibonacci is probably the most well known sequence in mathematics. Today, we will see how Fibonacci numbers can be used to construct beautiful patterns called <i><span style="color: purple;">continued fractions</span></i>.<br />
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<span style="color: purple; font-size: x-large;">F</span>ibonacci numbers are the following $$0, ~1, ~1, ~2, ~3, ~5, ~8, ~13, ~21, ~34, ~55, ~89, ~144, \dots$$ We can easily see the pattern of this number sequence. Each Fibonacci number is equal to the sum of the previous two numbers.<br />
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Mathematically, we can write the formula as follows $$F_0 = 0, ~F_1 = 1, ~~F_n = F_{n-1} + F_{n-2}$$<br />
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Now, let us consider the Fibonacci fractions $\frac{F_n}{F_{n-1}}$.<br />
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First, $$\frac{F_2}{F_1} = \frac{1}{1}$$<br />
Next, $$\frac{F_3}{F_2} = \frac{2}{1} = 1 + \frac{1}{1}$$<br />
$$\frac{F_4}{F_3} = \frac{3}{2} = 1 + \frac{1}{2} = 1 + \cfrac{1}{\frac{2}{1}} = 1 + \cfrac{1}{1 + \cfrac{1}{1}}$$<br />
$$\frac{F_5}{F_4} = \frac{5}{3} = 1 + \frac{2}{3} = 1 + \cfrac{1}{\frac{3}{2}} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1}}}$$<br />
That is so cool, isn't it?!<br />
$$\frac{F_6}{F_5} = \frac{8}{5} = 1 + \frac{3}{5} = 1 + \cfrac{1}{\frac{5}{3}} = 1 + \cfrac{1}{\frac{F_5}{F_4}} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1}}}}$$<br />
So we can follow the pattern and derive the following beautiful <span style="color: purple;">continued fractions</span><br />
$$\frac{F_7}{F_6} = \frac{13}{8} = 1 + \cfrac{1}{\frac{F_6}{F_5}} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1}}}}}$$<br />
$$\frac{F_8}{F_7} = \frac{21}{13} = 1 + \cfrac{1}{\frac{F_7}{F_6}} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1}}}}}}$$<br />
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Let us stop here for now. You can read more about Fibonacci numbers <a href="/2013/02/fibonacci1.html">here</a>. Hope to see you again next time.<br />
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<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-44281160084611542062015-10-01T17:49:00.000+10:002015-10-01T17:57:45.950+10:00Divide a line segment by compass<div class="separator" style="clear: both; text-align: center;">
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<span style="color: purple; font-size: x-large;">U</span>sing compass and straightedge, we can construct the midpoint of a line segment, and we can easily divide a line segment into, say three equal parts. The question is, is it possible to do these constructions <b>with just the compass</b>.<br />
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The answer is "yes"! Indeed, it is possible to construct the midpoint of a line segment, and it is possible divide a line segment into a number of equal parts by using the <b>compass alone</b>. How amazing is that?! Today, we will look at these constructions!<br />
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<span style="color: #0b5394; font-size: large;">Basic compass-and-straightedge constructions</span><br />
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First, let us recall the <a href="http://mathgardenblog.blogspot.com/2013/10/basic-compass-straightedge-construction.html" target="_blank">basic compass-and-straightedge constructions</a>.
In our <a href="http://mathgardenblog.blogspot.com/2013/10/basic-compass-straightedge-construction.html" target="_blank">previous post</a>, we learn about these constructions.
These constructions are assumed as obvious and can be employed in more complex construction problems:<br />
<ul>
<li><span style="color: purple;">Construction of the perpendicular bisector of a line segment and the midpoint</span></li>
</ul>
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<ul>
<li><span style="color: purple;">Through a point, construct a line perpendicular to a given line</span></li>
</ul>
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<ul>
<li><span style="color: purple;">Through a point, construct a line parallel to a given line</span></li>
</ul>
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<ul>
<li><span style="color: purple;">Construction of the angle bisector</span></li>
</ul>
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhjLw4PPSUgfnCr4uEuPx6pEs6kUl1SuPJDCQ0BqVOQdzP1pUMLuck6VIel5AaYeKe6pFrFprsmmYOpylL9moPZT_RzQzGZ7oVKi7zpsHJjwXRN8WVy-w8q1BHWYtP4N_oFy3Ttl4_U2U8r/s1600/vuontoan_mathgarden_basic4.png" /></div>
<br />
<ul>
<li><span style="color: purple;">Construction of an equal angle to a given angle</span></li>
</ul>
<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="181" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj-N0VE-4Y3_YdMLWmx6I2msCWGkFamasjE6mexel8DLzdSyU-55PXN4BDG0zbt9c2AiVCKQTEbEJ86V15T7XRILsa09_WmP17hxndkC4EOcWWNsIpNvVaNKRWTl9azNwLDNkU80mXnYJI7/s400/vuontoan-mathgarden-ruler-compas6.png" width="400" /></div>
<br />
<ul>
<li><span style="color: purple;">Through a point, construct a tangent line to a circle</span></li>
</ul>
<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="171" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjPPCZlvcPSUQAoM5eAHUA_s4-2Cyr2UuWdGepMWsCO1IwGDGNUbpyFSyHPC4kq3kJwrrvaSoWGKue6R51JQlwse6GZmBSFx1eooIu5taQDT_BPOGLga1ujTLIMi_5mUt09Wc8JRl-Trsco/s320/vuontoan_mathgarden_basic5.png" width="320" /></div>
<br />
<br />
You can read more about these <a href="http://mathgardenblog.blogspot.com/2013/10/basic-compass-straightedge-construction.html" target="_blank">basic compass-and-straightedge constructions</a>
<a href="http://mathgardenblog.blogspot.com/2013/10/basic-compass-straightedge-construction.html" target="_blank">here</a>.
<br />
<br />
<span style="color: purple; font-size: x-large;">A</span>
simple compass-and-straightedge construction problem is how to divide a line segment into a number of equal parts. For instance, suppose we are given a line segment $AB$, using compass and straightedge, how can we divide $AB$ into three equal parts?<br />
<br />
<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="222" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJs4O3XlaBPuMn2yTv4ZBN01hP15WN9-6fUe985KY8lR0WIG7wKBHwX652-tRtrcWBuj2ZkFHv6ItqKXuY58xezflkGsvHd6eojZQbBvdBQUzSwzB4W6xzy3X0LCvlALmVGKAzUmdXZpQb/s320/vuontoan_mathgarden_divide3.png" width="320" /></div>
The solution is quite trivial:
through A construct an arbitrary ray, and on this ray, use compass to subsequently construct arbitrary points $C_1$, $C_2$, $C_3$
so that $AC_1=C_1C_2=C_2C_3$. Connect $BC_3$.
Then through $C_1$ and $C_2$, respectively,
construct the two lines parallel to $BC_3$ which meet $AB$ at $D_1$ and $D_2$.
Finally, we obtain $AD_1 = D_1D_2=D_2B$.<br />
<br />
<br />
<br />
We now move on to consider constructions that use compass but not the straightedge.
<br />
<br />
<span style="color: #0b5394; font-size: large;">Construction by compass alone</span><br />
<div class="separator" style="clear: both; text-align: center;">
<br /></div>
<span style="color: purple; font-size: x-large;">T</span>here is a <a href="http://mathgardenblog.blogspot.com/2015/07/Mohr-Mascheroni.html" target="_blank">special theorem</a> in mathematics called the <a href="http://mathgardenblog.blogspot.com/2015/07/Mohr-Mascheroni.html" target="_blank">Mohr-Mascheroni Theorem</a>. This theorem states that any geometric construction which can be done with compass and straightedge can also be done with compass alone. That means that the use of straightedge is actually not necessary.
<br />
<blockquote class="tr_bq">
<span style="color: purple;"><b>Mohr-Mascheroni Theorem.</b> Any point can be constructed by compass and straightedge can be constructed by compass alone.</span></blockquote>
By <a href="http://mathgardenblog.blogspot.com/2015/07/Mohr-Mascheroni.html" target="_blank">Mohr-Mascheroni Theorem</a>,
it is then possible to use compass alone to multiply and divide a line segment. <br />
<br />
<br />
<b><span style="color: purple;">Multiply a line segment by compass alone</span></b><br />
<br />
<blockquote class="tr_bq">
<span style="color: purple;"><b>Problem.</b>
Given two points A and B, only using compass, construct a point C on AB so that $$AC = AB \times 3.$$</span></blockquote>
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjUpitLRc9i5N82d0IAIpcVaQ8as33VLQzZBKjDWQFt3iWNltgvvXF6JJr0AlX6X0TXL_FglQ8T7Gjclj3fSPAj4GjkGRqVGnGsTnQbbJZolE3n3J-otgeMdiPiogssIloGcmHfC0tB0Jeh/s1600/vuontoan_mathgarden_triangle1.png" /></div>
<br />
If we construct two equal circles with radius AB and center at A and B, respectively, then they will intersect at two points X and Y. We obtain two <b>equilateral triangles</b> ABX and ABY. If we keep doing like this then we can construct a <b>lattice of equilateral triangles</b>. Thus, we can multiply a line segment easily.
<br />
<br />
<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="212" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgU6IJ7jwe_uQKXHnk5D1GAPFwnhRf5qa4LXhNfi3fjjJ_pTyxtdR2S_ukR5yDOfIRYdVVuN6MKPGrXMBrWTu4ihW5WD-4nhGPNvbkAPvlx39O80qUnf7UahtBlK9TwZAgirb7Rt4abgOGZ/s640/vuontoan_mathgarden_triangle2.png" width="640" /></div>
<div>
<br /></div>
<br />
<div>
</div>
<br />
<div style="-webkit-text-stroke-width: 0px; color: black; font-family: 'Times New Roman'; font-size: medium; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: auto; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 1; word-spacing: 0px;">
<div style="margin: 0px;">
<b><span style="color: purple;"><br /></span></b><b><span style="color: purple;">Divide a line segment by compass alone</span></b></div>
<blockquote class="tr_bq">
<div style="margin: 0px;">
<span style="color: purple;"><b>Problem.</b>
Given two points A and B, only using compass, construct a point D on AB so that $$AD = AB / 3.$$
</span></div>
</blockquote>
<div>
<div class="separator" style="clear: both; margin: 0px; text-align: center;">
<img border="0" height="56" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8GIpRoupAsVePDP8gBrJ0v9jirDtX9gGXyfowUYhsngZwNG6D0kozB_ETCVgpFKc_hjtjSwxs6IE2s6uI9e65anDZ2IYmF4SNEatyZQdpMvAD4I6OjEUw0K2IjBMxoWQ8XqlzBHXaK7J_/s640/vuontoan_mathgarden_triangle2b.png" style="cursor: move;" width="640" /></div>
<div style="margin: 0px;">
<br /></div>
</div>
<div style="margin: 0px;">
We can see that the two points C and D are kind of inverse of one another. On one hand, we have $$AC = AB \times 3$$ and on the other hand, we have $$AD = AB / 3$$</div>
<div style="margin: 0px;">
This is called the <i><span style="color: purple;">inversion operation</span></i>. We will learn more about this inversion operation in the future. </div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
Because of this inversion, we have $$AC \times AD = AB^2$$</div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
In geometry, inversion equation like the above is often derived from <a href="http://mathgardenblog.blogspot.com/2012/05/similar-triangles.html" target="_blank">similar triangles</a>.</div>
<div class="separator" style="clear: both; margin: 0px; text-align: center;">
<img border="0" height="206" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEia3ggVLRBxdLAjchvDKQO29D19qqCA6tJz4rOkAcUSHQTBkS65PtM15ofyaoX7rnKApOQhyphenhyphenFbLNaTCp_sXi0X2fWw5vDMQzpHy0LtjtE4Rt5nbFX5VT0IuTeyKHUhwTBGdjHXQ2jq_w_Ky/s400/vuontoan_mathgarden_triangle2c.png" style="cursor: move;" width="400" /></div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
If P is an arbitrary point such that AP = AB then the two triangles ADP and APC are <a href="http://mathgardenblog.blogspot.com/2012/05/similar-triangles.html" target="_blank">similar triangles</a>.
This is because the two triangles share the same angle A and they have equal side ratios $$\frac{AD}{AP} = \frac{AP}{AC}.$$</div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
Since P is quite arbitrary, we will choose a convenient position for P. We will choose P so that the two triangles ADP and APC become isosceles triangles. That is $$PA = PD, ~~~~~~ CP = CA$$</div>
<div style="margin: 0px;">
<br /></div>
<div class="separator" style="clear: both; margin: 0px; text-align: center;">
<img border="0" height="220" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhUvOjN7zajoMTcGlOOJcZ-rvQJME4LuAR2xADZszZWNR0XBVfZqu8sEOHgqGe74XLfHMfXgl8x-I_eVYUnsa4nfjYOClkWXM61jCjg5zgnP-sw4CJAOMfUc0y9UL4BIyyIIgUdVbjC1WGk/s400/vuontoan_mathgarden_triangle2d.png" style="cursor: move;" width="400" /></div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
Since $AP = AB$ and $CP = CA$, the point P is an intersection point of the circle centered at A with radius AB and the circle centered at C with radius CA. It means that we can construct this point P by compass alone.</div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
Furthermore, since PD = PA = AB, the point D lies on the circle centered at P with radius equal to AB. By symmetry, we can construct the point D. The construction is as follows:</div>
<div style="margin: 0px;">
<br /></div>
<ul>
<li>Construct the circle centered at A with radius AB and the circle centered at C with radius CA. The two circles intersect at P and Q.</li>
</ul>
<ul>
<li>Construct the circle centered at P with radius equal to AB and the circle centered at Q with radius equal to AB. The two circles intersect at A and D.</li>
</ul>
<div style="margin: 0px;">
<br /></div>
<div class="separator" style="clear: both; margin: 0px; text-align: center;">
<img border="0" height="237" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj8Lzs0eyLs3KCF7ALn5PH1SVIVJl0MnB-Nq7TWMNid_K6-2Mbvt5dwfEdeMl5niaL8QDB67pkt7jbakv0Os8zgOCSKgs0CWn5qCE_kautnPsswSTWrnmkG2W5wd_atmrDqW2-RkghaeNOe/s400/vuontoan_mathgarden_triangle2e.png" style="cursor: move;" width="400" /></div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
So by using compass alone we indeed can construct the point D to divide the line segment into three equal parts.
Let us stop here for now. Hope to see you again next time.
</div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
<i>Homework.</i></div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
1. Given three points A, B, and C, <i>only using compass</i>, construct the circumcircle of the triangle ABC.</div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
2. Given three points A, B, and C, <i>only using compass</i>, construct the incircle of the triangle ABC.</div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
3. Go to <a href="http://google.com/">google.com</a> and search about the geometric inversion operation.</div>
<div style="margin: 0px;">
<br /></div>
<div style="margin: 0px;">
<br /></div>
<div class="separator" style="clear: both; margin: 0px; text-align: center;">
<img border="0" height="288" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh_lCBgrZzsOWoqn4H5LzdDTbU1AL1fa5t6fuHQEKtSOdwbEbZEIAuumrfxVKFT1GiRR0mw08nh7CrH6in9hH-5QBuF669Vhm5iAygbzUe4Vt0aaVRNdw4JeXXK80w-3oI5zLegHfQQznLy/s400/james13.png" style="cursor: move;" width="400" /></div>
<div style="margin: 0px;">
<br /></div>
</div>
Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-72384961309858598092015-07-10T16:48:00.000+10:002015-11-06T12:56:09.137+11:00Construction by compass alone<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="287" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjSIDCqWUU0xTIxIg1UJNqsxGEdECJgx3Hn6rzzjl_GXc7Ev_FGLBRRgN0iVcPXr-seBLJ2Iw8fi-3XcTBF0o2d7qGmGWzASUU4pQQUoU8VP_id4gZL43vT0BWieqAqfFjNq5JoHj74-dUd/s400/james7.png" width="400" /></div>
<br />
<span style="color: purple; font-size: x-large;">N</span>ormally in geometric construction problems, we use <a href="http://mathgardenblog.blogspot.com/2013/10/basic-compass-straightedge-construction.html" target="_blank">straightedge and compass</a>. Today, we will look at an unusual type of construction problems where we are only allowed to use compass.<br />
<br />
<br />
<a name='more'></a><br />
<br />
<span style="color: purple; font-size: x-large;">T</span>here is a special theorem in mathematics called the<span style="color: purple;"><b> Mohr-Mascheroni Theorem</b></span>. This theorem states that any geometric construction which can be done with compass and straightedge can also be done with <i>compass alone</i>. That means that the use of straightedge is actually not necessary.<br />
<br />
For a long time, this theorem was known as <i>Mascheroni's theorem</i> after the Italian mathematician Lorenzo Mascheroni. Mascheroni proved this theorem in his book "Geometria del Compasso" which was published in 1797.<br />
<br />
However, the first person who discovered the theorem is actually Georg Mohr, a Danish mathematician. In more than 100 year earlier than Mascheroni, Mohr proved this theorem in a book titled "Euclides Danicus" which he published in 1672.<br />
<div>
<br /></div>
<div>
Mohr's book, <i>Euclides Danicus</i>, was forgotten until it was miraculously found in a bookstore in Copenhagen in 1928. And since then, to acknowledge the contribution from the mathematician Mohr, we have a theorem named Mohr-Mascheroni!</div>
<blockquote class="tr_bq">
<span style="color: purple;"><b>Mohr-Mascheroni Theorem.</b> Any point can be constructed by compass and straightedge can be constructed by <i>compass alone</i>.</span></blockquote>
<div>
<br /></div>
<div>
<span style="color: purple; font-size: x-large;">L</span>et us look at <a href="http://mathgardenblog.blogspot.com/2014/11/construction-algorithm.html" target="_blank">two simple construction problems</a>:<br />
<ul>
<li>multiple a line segment, and</li>
<li>divide a line segment into a number of equal parts. </li>
</ul>
<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td><img border="0" height="58" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh1rk8dn06TSKpt8jopY3Os3c48WoxHGXESTWKSU2blM4V09H-66Cw_62HdmBejVSEY_jAWE4A-FXY9sJ-GL7bjP0OWOTnsRwM5MNV0Dfbx8rQhz7gvmsQq7V-pRkN4hyI7uFxfA2Z5HI54/s1600/vuontoan_mathgarden_ca1.png" style="margin-left: auto; margin-right: auto;" width="400" /></td></tr>
<tr><td class="tr-caption" style="font-size: 12.8000001907349px;">given AB, by compass and straightedge, it is easy to construct AC = 3 AB</td></tr>
</tbody></table>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td><img border="0" height="245" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjh1lseVdkESCI1o-X82zEmosXO47_oSGpIKzpQA-7Sg20kTQ3fzH8-lthtY7LnkbCnYnxpjTL3HoLuatJAzlv8-haFELrcJrw35xvEC88NMGVykbJiGoQXN0SPcA-XGA19N7liuynrgWB0/s1600/vuontoan-mathgarden-ruler-compas7.png" style="margin-left: auto; margin-right: auto;" width="400" /></td></tr>
<tr><td class="tr-caption" style="font-size: 12.8000001907349px;">given AB, by compass and straightedge, it is easy to divide AB into 5 equals segments</td></tr>
</tbody></table>
<br />
<br />
<br />
By <i>Mohr-Mascheroni Theorem</i>, it is possible to use <i>compass alone</i> to multiply and divide a line segment. Today, we will solve the first problem and we leave the second problem for our next post.<br />
<br />
<br />
<br />
<span style="color: #0b5394; font-size: large;">Multiply a line segment by compass alone</span><br />
<br />
<blockquote class="tr_bq">
<span style="color: purple;"><b>Problem.</b> Given two points A and B, <i>only using compass</i>, construct a point C on AB so that AC = 3 AB.</span></blockquote>
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjUpitLRc9i5N82d0IAIpcVaQ8as33VLQzZBKjDWQFt3iWNltgvvXF6JJr0AlX6X0TXL_FglQ8T7Gjclj3fSPAj4GjkGRqVGnGsTnQbbJZolE3n3J-otgeMdiPiogssIloGcmHfC0tB0Jeh/s1600/vuontoan_mathgarden_triangle1.png" /></div>
<br />
If we construct two equal circles with radius AB and center at A and B, respectively, then they will intersect at two points X and Y. We obtain two equilateral triangles ABX and ABY. If we keep doing like this then we can construct a lattice of equilateral triangles. Thus, we can solve the aforementioned construction problem.</div>
<div>
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="212" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgU6IJ7jwe_uQKXHnk5D1GAPFwnhRf5qa4LXhNfi3fjjJ_pTyxtdR2S_ukR5yDOfIRYdVVuN6MKPGrXMBrWTu4ihW5WD-4nhGPNvbkAPvlx39O80qUnf7UahtBlK9TwZAgirb7Rt4abgOGZ/s640/vuontoan_mathgarden_triangle2.png" width="640" /></div>
<div>
<br /></div>
<div>
<br />
Let us stop here for now. Hope to see you again next time.<br />
<br />
<br />
<br />
<i>Homework.</i><br />
<br />
1. Given two points A and B, <i>only using compass</i>, construct the midpoint of AB.<br />
<br />
2. Given two points A and B, <i>only using compass</i>, construct two points $M_1$ and $M_2$ on AB, so that $A M_1 = M_1 M_2 = M_2 B$.<br />
<br />
3. Given three points A, B, and C, <i>only using compass</i>, construct the center of the circumcircle of the triangle ABC.<br />
<br />
<br />
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="434" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_zwP9TZ7iI7FDjpq2fG2R2zEwhUDhL6KCg7X2Ews2b4484HJ_Vr-hddU9rOw78hbo6qeCB8jTnCd9UfOd-xZ6OMqXV5xWofBifY0APnREtOMB2VlJ-vebTsZZPDQDTJRoPdt0fOHuXKY/s640/vuontoan_mathgarden_triangle.jpg" width="640" /></div>
<div>
<br /></div>
Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-66628916783866186702015-05-09T20:32:00.000+10:002015-05-09T20:32:24.869+10:00Triangles<div class="separator" style="clear: both; text-align: center;">
<img border="0" height="434" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_zwP9TZ7iI7FDjpq2fG2R2zEwhUDhL6KCg7X2Ews2b4484HJ_Vr-hddU9rOw78hbo6qeCB8jTnCd9UfOd-xZ6OMqXV5xWofBifY0APnREtOMB2VlJ-vebTsZZPDQDTJRoPdt0fOHuXKY/s640/vuontoan_mathgarden_triangle.jpg" width="640" /></div>
<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-43512912290488593422015-05-02T13:59:00.000+10:002015-05-02T14:00:13.773+10:00Star of David theorem<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjrTXFUYb-EEdHO1RFkkw_pkrlTpQB7n6WUBIYr6eVnwxcEIxg3yjlL_B6Yti4lgF1hLO7iaWfBHHDDT_X-Tj6spwa2J1924z9Ns4vp6r_Udsz0uVcut4E6XmdnwbJMTrlnsNKW2vekktRX/s1600/vuontoan_mathgarden_james_starofdavid.jpg" height="320" width="308" /></div>
<br />
<span style="color: purple; font-size: large;">T</span>oday we will learn about a very beautiful theorem in geometry -- the <b><span style="color: purple;">Star of David theorem</span></b>. This theorem is a consequence of the <a href="http://mathgardenblog.blogspot.com/2013/06/hexagrammum-mysticum1.html" target="_blank"><span style="color: purple;"><b>Pascal hexagon theorem</b></span></a> and the <a href="http://mathgardenblog.blogspot.com/2013/08/hexagrammum-mysticum3.html" target="_blank"><span style="color: purple;"><b>Pappus theorem</b></span></a>.<br />
<br />
<br />
<a name='more'></a><div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjuuQizDv2E16aVi75Pus2RV0YfzZwL_nggVa-T_peCovIjyPyArDHotH8_Mam0_7EAl3QMwvTp-6XCogS0MtXNZ5JetQbhJWBemD6OmG92GGDCnDIYescAq67IUqTDeV8NWOUyP5Zn9KCp/s1600/vuontoan_mathgarden_davidstar1d.png" height="320" width="319" /></div>
<blockquote class="tr_bq">
<span style="color: purple;"><b>Star of David theorem. </b>Given two triangles $abc$ and $xyz$ inscribed in a same circle and six intersection points: $$p = ab \cap yz, ~~s = bc \cap zx, ~~q = ca \cap yz, ~~r = bc \cap xy, ~~1 = ab \cap xy, ~~3 = ca \cap zx.$$ Then the three lines $ps$, $qr$ and $13$ must meet at a common point.</span></blockquote>
<br />
<span style="color: purple; font-size: large;">T</span>o show that the three lines $ps$, $qr$, $13$ meet at a common point, we let $ps \cap qr = 2$, and prove that the three points $1$, $2$ and $3$ lie on the same line.<br />
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEikDtX1bceFCR0sJ8w3eRKiFBB-id0mS4-SE07GWIGXC4LD6lvY1eAT_uU980fq0MkIZdxqViKV5j0lM0QEm15f8-lulvYl27p_a_8yX_BNpPYRyXJfqjkZjXUqJxqRgPcePxOfAvYpY7E8/s1600/vuontoan_mathgarden_davidstar2d.png" height="320" width="318" /></div>
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<br />
Observing the above figure, we can see that there are a lot of features look very similar to the <a href="http://mathgardenblog.blogspot.com/2013/06/hexagrammum-mysticum1.html" target="_blank"><span style="color: purple;"><b>Pascal hexagon theorem</b></span></a> and the <a href="http://mathgardenblog.blogspot.com/2013/08/hexagrammum-mysticum3.html" target="_blank"><span style="color: purple;"><b>Pappus theorem</b></span></a>.<br />
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhpmBzZIBfoN7HpNNTt9Uxk4lQ8rmlioI6jxJmI_fhpSZajsosikOeX2MmGw7-6XKpHJNGNNMJb3FY60U00EIy52-vxLfIniPFTflCICu1wr_ee2EPVyTzgtN9SQtCxGgizD2oB-aE9dlBd/s1600/vuontoan_mathgarden_pascal_pappus.png" height="297" width="640" /></div>
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<br />
Thus, we will use the <a href="http://mathgardenblog.blogspot.com/2013/06/hexagrammum-mysticum1.html" target="_blank"><span style="color: purple;"><b>Pascal hexagon theorem</b></span></a> and the <a href="http://mathgardenblog.blogspot.com/2013/08/hexagrammum-mysticum3.html" target="_blank"><span style="color: purple;"><b>Pappus theorem</b></span></a> to prove the <b><span style="color: purple;">Star of David theorem</span></b>.<br />
<br />
Draw the following intersection points: $$4 = yc \cap bz, ~~ 5 = pc \cap rz, ~~ 6 = ys \cap bq.$$<br />
We will prove that the six points $1$, $2$, $3$, $4$, $5$ and $6$ are collinear.<br />
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi1Hy0_WwzfjX4RpIxY_h5CUCLJvUjCXrksz0EDk7j8nxv67oLtMN2fVOrFOeK6GlFOQlZcC9kXNsE-vTY1GXQ8WC0-PiPRxb1fsJJi3CeCkS1MUSLiA0aRRGByhsBO3-6jAR2zeDEJIaEN/s1600/vuontoan_mathgarden_davidstar3d.png" height="320" width="318" /></div>
<br />
Here is a sketch of a proof:<br />
<ul>
<li>The three points $1$, $4$, $3$ are collinear by the <a href="http://mathgardenblog.blogspot.com/2013/06/hexagrammum-mysticum1.html" target="_blank"><span style="color: purple;"><b>Pascal hexagon theorem</b></span></a></li>
<li>The three points $1$, $4$, $5$ are collinear by the <a href="http://mathgardenblog.blogspot.com/2013/08/hexagrammum-mysticum3.html" target="_blank"><span style="color: purple;"><b>Pappus theorem</b></span></a></li>
<li>The three points $3$, $5$, $2$ are collinear by the <a href="http://mathgardenblog.blogspot.com/2013/08/hexagrammum-mysticum3.html" target="_blank"><span style="color: purple;"><b>Pappus theorem</b></span></a></li>
<li>The three points $1$, $2$, $6$ are collinear by the <a href="http://mathgardenblog.blogspot.com/2013/08/hexagrammum-mysticum3.html" target="_blank"><span style="color: purple;"><b>Pappus theorem</b></span></a></li>
<li>The three points $3$, $4$, $6$ are collinear by the <a href="http://mathgardenblog.blogspot.com/2013/08/hexagrammum-mysticum3.html" target="_blank"><span style="color: purple;"><b>Pappus theorem</b></span></a></li>
</ul>
<span style="color: purple; font-size: large;">W</span>e will see the proof more clearly in the following figures:<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3G_IhQZKVBoO-fhFTUFPrIKoxV_C9HUEbTuBqn_h9kN5qhyphenhyphen8EOs5aYQNGEk56M6EMHC4GfdL_0zIWUH_3iVC7j7sWGfDNspYeur28A3kA4bjKWJHLBDo-53Q4mFtdaKlp44MJVnIN4zDi/s1600/vuontoan_mathgarden_davidstar4d.png" height="320" width="318" /></td></tr>
<tr><td class="tr-caption" style="font-size: 12.8000001907349px;"><span style="font-size: xx-small; text-align: start;">$1$, $4$, $3$ are collinear by the <a href="http://mathgardenblog.blogspot.com/2013/06/hexagrammum-mysticum1.html" target="_blank"><span style="color: purple;"><b>Pascal hexagon theorem</b></span></a></span></td></tr>
</tbody></table>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgLvRvn-tOZmlM4dzp-njRG2FIUjETQP7APM69-fqGs6snpp8yVEPYX2LhrNsfxb2pkHd6XAIuoOxdde16N_WQIb1OynkNdF2fP31ql1m8uwvCETQ6eEIR8YjZ_K-npIflMUMcWEkevY8pa/s1600/vuontoan_mathgarden_davidstar5d.png" height="320" width="318" /></td></tr>
<tr><td class="tr-caption" style="font-size: 12.8000001907349px;"><span style="font-size: xx-small; text-align: start;">$1$, $4$, $5$ are collinear by the <a href="http://mathgardenblog.blogspot.com/2013/08/hexagrammum-mysticum3.html" target="_blank"><span style="color: purple;"><b>Pappus theorem</b></span></a></span></td></tr>
</tbody></table>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi32WgiM5CGwFCHKXCbYQ6n4p-f8qutJJaQd04JPA7FMAkngn103XpdXRW-IuFULNwLOkkpwmgEWi4LBRD7rXJcZBDTvpRHzZhmtvguSUXbDW97TndWCaCUpQjFQSJuOKSaqKTb4fCmkuof/s1600/vuontoan_mathgarden_davidstar6d.png" height="320" width="318" /></td></tr>
<tr><td class="tr-caption" style="font-size: 12.8000001907349px;"><span style="font-size: xx-small; text-align: start;">$3$, $5$, $2$ are collinear by the <a href="http://mathgardenblog.blogspot.com/2013/08/hexagrammum-mysticum3.html" target="_blank"><span style="color: purple;"><b>Pappus theorem</b></span></a></span></td></tr>
</tbody></table>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9CZHCa6rUz8-1iffDt7aaBETQb26i643SD2sr-NzXTzfiMEjYea93TIFlF5YpVXy_-HV5uJ90EQKNeRUPOVE7E2amkUH-rPLnB0P4equjD8yOGs5N55NiqzmiRmdmBRBI51PaQ54HThW5/s1600/vuontoan_mathgarden_davidstar7d.png" height="320" width="318" /></td></tr>
<tr><td class="tr-caption" style="font-size: 12.8000001907349px;"><span style="font-size: xx-small; text-align: start;">$1$, $2$, $6$ are collinear by the <a href="http://mathgardenblog.blogspot.com/2013/08/hexagrammum-mysticum3.html" target="_blank"><span style="color: purple;"><b>Pappus theorem</b></span></a></span></td></tr>
</tbody></table>
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1xXqPFPFf9bNCFShyphenhyphendeQKnZAiCCRivKpkp1ESH17Ds-iDaGFrfsxHSUBXSqkas0P937p_lu_V2sKJzDBxFUzANRb3IasFz3u9K8hUqcA5iMpjsL1s85yDVzDRzsw8i7mnIWCmPwsUfua8/s1600/vuontoan_mathgarden_davidstar8d.png" height="320" width="318" /></td></tr>
<tr><td class="tr-caption" style="font-size: 12.8000001907349px;"><span style="font-size: xx-small; text-align: start;">$3$, $4$, $6$ are collinear by the <a href="http://mathgardenblog.blogspot.com/2013/08/hexagrammum-mysticum3.html" target="_blank"><span style="color: purple;"><b>Pappus theorem</b></span></a></span></td></tr>
</tbody></table>
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<br />
<span style="color: #0b5394; font-size: large;"><br /></span><span style="color: #0b5394; font-size: large;">The Star of David theorem for the conics</span><br />
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<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiat8kh2gAhGC9zfnAGJvkHb5bwCIrOmcjilj75EOe64pf4yUD10vClJ3RwKGo2C1E6k1RD6s5XfSyXq54ckZNC7xhpbV9wcjTYbdfNYsqrkz0rza7_Ln9l9A5NH7-OzjBIqYId8Dol11Ee/s1600/vuontoan_mathgarden_star_david_sand.jpg" height="225" width="400" /></div>
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<span style="color: purple; font-size: large;">W</span>e know that the <a href="http://mathgardenblog.blogspot.com/2013/06/hexagrammum-mysticum1.html" target="_blank"><span style="color: purple;"><b>Pascal hexagon theorem</b></span></a> holds for all types of conics and the <b><span style="color: purple;">Star of David theorem </span></b>is a consequence of the <a href="http://mathgardenblog.blogspot.com/2013/06/hexagrammum-mysticum1.html" target="_blank"><span style="color: purple;"><b>Pascal hexagon theorem</b></span></a>, therefore, the <b><span style="color: purple;">Star of David theorem </span></b>also holds for conics. It means that, instead of choosing two triangles on a circle, we can choose two triangles on an ellipse, or on a parabola, or on a hyperbola, and the theorem is still true!<br />
<div>
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Here is a picture of an ellipse:</div>
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjsOcLqnFwTTybZ72xgDj7K7ZyPl6pjOBFx6sVtRd9IriF5WfcP1byQyafGNb35r0xF32jzUIB14JF5xGXJwBpTC3kLiAIk_rRF4DIeIjf3QjfJekfpFjRFmOtPk1uk6aL09l8K8a5jGjYs/s1600/vuontoan_mathgarden_davidstar_ellipse_d.png" height="277" width="400" /></div>
<br />
Here is a parabola:<br />
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj-XTJgfpYa30GLLiRIlFQ31uIDHFTwZ8SxCEH5G78HHIsWqRAD7YUH73NjBExb3pjQ6wiIsp1iIBdhrN8WXSB-4o0XxmYxIkSgvPX4TUqjRYN4XTumGiEmys1vSDoemgHeE2ucmtxPwLlS/s1600/vuontoan_mathgarden_davidstar_parad.png" height="197" width="320" /></div>
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<br />
And here is a hyperbola:<br />
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgLoGsJJNBpE1HYR3b665Og9Fv3UHNmeqJyZfjEGuUKTB6tVSzCxlZxYV29MSeYecnlrLpMJyepi7vtPF_MIKmK39e4OFFoMuRMPiTjIQVkzZ30HGmNvTn9kLRr2T5M4nwass_6KrmP0GWl/s1600/vuontoan_mathgarden_davidstar_hyd.png" height="282" width="400" /></div>
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<span style="color: purple; font-size: large;">L</span>et us stop here for now. Hope to see you again in our next post.<br />
<br />
<br />
<br />
<i>Homework.</i><br />
<br />
1. Draw the <b><span style="color: purple;">Star of David theorem</span></b> for different configurations by choosing various positions for the two triangles.<br />
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhE-4r5InMedSOB4u5wDlbPpG9qF7iSQIE9dqP8L78NxL3Cu8JPiVhVp8YrqdAoDg2hE5XCQHOBiL50S4DAnoKa3jtB9h9eqYtBeMWUVvgHgiby-UWnl8Hy_SeCSVj0seO__8Ftp8ZdHfI/s1600/vuontoan_mathgarden_starofdavid.png" height="202" width="400" /></div>
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<br />
2. Write a full proof for the <b><span style="color: purple;">Star of David theorem</span></b> based on our proof sketch above.<br />
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3. Go to <a href="http://google.com/">google.com</a> and search for other applications of the <a href="http://mathgardenblog.blogspot.com/2013/06/hexagrammum-mysticum1.html" target="_blank"><span style="color: purple;"><b>Pascal hexagon theorem</b></span></a>.<br />
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<br />
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<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0sXNxOzQu71H2TUD_xwAeaKAIciQzNkfuDRY2sKwpEmy5KTtW1fxW6s5-fc6RMzr9nSVoYXWvphFUorzwcxUFJPpP82SpmSTQYBgMM2ZyQXxadFZd2eEau7HVT-CvJkphElx4MFAgX9Y/s1600/francis_hope_faith_love.jpg" height="298" width="320" /></div>
Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-46372190883784398172015-04-25T19:54:00.000+10:002015-04-25T19:59:46.870+10:00Chessboard and pyramid<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhtWjfkyZ_QB9v15gP43FB-xZz-S1TM3cXIAJHV5NtWUr4zA1j_BFhsfIcAgmMk1TwI0kq3EStjn45J3jRSVJ81TIrCR76aGVXLDgQ2S5z5eqortvPi4Wx-PJ5yPzgS-RWaJh84ppAmunF7/s1600/vuontoan_mathgarden_chess.png" height="350" width="640" /></div>
<br />
<span style="color: purple; font-size: large;">H</span>ave you heard of the <i>legend of the chessboard</i>? The story goes like this. There was a wise man who invented the chess game and introduced it to a king.
The king loved the game so much that he told the wise man that he could choose anything for a reward.
The wise man then pointed to the chessboard and asked for <i><span style="color: purple;">1 grain of wheat</span></i> for the first chess square,
<i><span style="color: purple;">2 grains of wheat</span></i> for the second square,
<i><span style="color: purple;">4 grains of wheat</span></i> for the third square,
<i><span style="color: purple;">8 grains of wheat</span></i> for the fourth square, and repeat this doubling pattern. This sounds like a very little reward but at the end the king didn't have enough wheat to reward the wise man.<br />
<br />
<br />
<span style="color: purple; font-size: large;">T</span>oday, we will calculate to see how much wheat the wise man actually requested. Is it a lot? Is it little? I'll give you a clue, it's HUGE!<br />
<br />
To see how big this reward is, we will calculate the number of pyramids that we could build up using all this wheat.<br />
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<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMxrgdpZwA-FD2-ZPX0xMpH93wvNHJPjYAV77ktPUGL1U9WwW6jAotIXrSUwf4d3MFhPL0UNg83-tDBimHrzClvpchEk11a4AfrcPfRSCPfxDnURz-h8Qm4gn0pxf990yysEn7x_cyaiRn/s1600/vuontoan_mathgarden_pyramids2.png" height="320" width="640" /></div>
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<a name='more'></a><br />
<span style="color: #3d85c6; font-size: large;">Step 1. Find the total number of grains</span><br />
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A chessboard has $8 \times 8 = 64$ squares and the wise man asked for his reward the following:<br />
<ul>
<li>Square 1: number of grains $=1$</li>
<li>Square 2: number of grains $=2$</li>
<li>Square 3: number of grains $= 4=2^2$</li>
<li>Square 4: number of grains $= 8=2^3$</li>
<li>Square 5: number of grains $=16=2^4$</li>
<li>Square $n$: number of grains $=2^{n-1}$</li>
<li>Square 63: number of grains $=2^{62}$</li>
<li>Square 64: number of grains $=2^{63}$</li>
</ul>
<br />
So the total number of grains is $$S = 1 + 2 + 4 + 8 + 2^4 + \dots + 2^{62} + 2^{63}$$<br />
<br />
We will simplify the formula $S$ as follows. Multiply both sides of the formula $S$ by 2, we have $$2S = 2 + 4 + 8 + 16 + 2^5 + \dots + 2^{63} + 2^{64}$$<br />
Subtract it to the original formula $$S = 1 + 2 + 4 + 8 + 2^4 + \dots + 2^{63}$$ we obtain $$S = 2^{64} - 1$$<br />
<br />
<br />
<span style="color: #3d85c6; font-size: large;">Step 2. Estimate the volume of the wheat</span><br />
<br />
We have calculated that the total number of grains is $S = 2^{64} - 1$. For simplicity, we will give the wise man one more grain of wheat to make $S = 2^{64}$. We will assume that a grain of wheat has a volume of <i>2 cubic millimetres</i> so that the total volume of all the wheat is $V = 2^{65}$ cubic millimetres.<br />
<br />
But how can we estimate the number $2^{65}$?<br />
<br />
In computer science, we have<br />
<ul>
<li>1 kilobyte = $2^{10}$ bytes $\approx$ 1000 bytes,</li>
<li>1 megabyte = $2^{10}$ kilobytes $\approx$ 1000 kilobytes,</li>
<li>1 gigabyte = $2^{10}$ megabytes $\approx$ 1000 megabytes,</li>
<li>1 terabyte = $2^{10}$ gigabytes $\approx$ 1000 gigabytes.</li>
</ul>
<br />
Thus, we have made the following approximation:
$$2^{10} = 1024 >\approx 1000$$<br />
<br />
Now we will use the same approximation $2^{10} \approx 1000$ to estimate the volume $V = 2^{65}$ (cubic millimetres):
$$V = 2^{65} = 2^{5} \times 2^{60} = 32 \times 2^{60} >\approx 32 \times 1000^{6}$$<br />
<span style="color: purple;">Therefore, $V \approx 32 \times 1000^{6}$ cubic millimetres $=32 \times 1000^{3}$ cubic metres.</span><br />
<br />
<br />
The volume of all the wheat that the wise man asked for his reward is<br />
<div style="text-align: center;">
<span style="color: purple;"><b>$V\approx$ 32 billion cubic metres!</b></span></div>
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<br />
<br />
<br />
<span style="color: #3d85c6; font-size: large;">Step 3. Estimate the volume of a pyramid</span><br />
<br />
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgNovZNZ0-lll41MxX-TUInETRbAUtuBzuOw_xQZmT3UtfxeCEpk87Xs2HkYlK1H2ioSyhKqIOn47WTlAjLWGmwcIusNs72il7VAhlhhIMwucnHeU3e6FPfEoBBOgjvJhYVc2BpdzPy2UqY/s1600/vuontoan_mathgarden_pyramids1.png" height="282" width="640" /></div>
<br />
<span style="color: purple; font-size: large;">T</span>he pyramids in the above picture are the Giza pyramids. They are located in the outskirts of Cairo - capital of Egypt.
The pyramids are the burial tombs of Egypt pharaohs.
The pyramid on the far side is the largest pyramid, 147 m tall, tomb of a pharaoh named Khufu.
It was constructed first around 2550BC.
The middle pyramid, 144 m tall, 2520 BC, was the tomb of a pharaoh named Khafre.
The one in the front is the smallest, 65 m tall, tomb of a pharaoh named Menkaure, constructed around 2490BC.<br />
<br />
We have a formula to calculate the volume of a pyramid. If a pyramid has a height of $h$ and its base is a square of side length $a$
then its volume is $$V = \frac{1}{3} a^2 h$$
<br />
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjlE53D_CsC5K_bVRJyNNDnGGStMsz7qxvaXU8ILZgT7qCx_Dzjia5SDez7Wh7ydmr5jSedVwAihAYo0FjKi_xQCyhmitXPtcYIKsSa7KE2pLudw8rW9bHEP7aZGJSTvhUBjt229MyRhzI1/s1600/vuontoan_mathgarden_pyramid_volume.png" height="187" width="400" /></div>
<br />
Apply this formula, we have:<br />
<div style="text-align: center;">
<table align="center" border="1"><tbody>
<tr><td></td><td>height h</td><td>base length a</td><td>volume V</td></tr>
<tr><td>Pyramid of Khufu</td><td>147 m</td><td>230 m</td><td>2,592,100 cubic metres</td></tr>
<tr><td>Pyramid of Khafre</td><td>144 m</td><td>215 m</td><td>2,218,800 cubic metres</td></tr>
<tr><td>Pyramid of Menkaure</td><td>65 m</td><td>105 m</td><td>238,875 cubic metres</td></tr>
</tbody></table>
</div>
<br />
The largest pyramid among these three, the pyramid of Khufu, has a volume of approximately 2.6 million cubic metres.<br />
<br />
<br />
<br />
<span style="color: #3d85c6; font-size: large;">Step 4. Find the number of pyramids</span><br />
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Now we are ready to answer the question that we put forth in the beginning.
We have estimated the volume of the wheat that the wise man requested: <span style="color: purple;">32 billion cubic metres</span>.
We have calculated the volume of the largest Giza pyramid: <span style="color: purple;">2.6 million cubic metres</span>.
If we could use all this wheat to form pyramids then the number of wheat pyramids would be $$\frac{32 \times 1000^3}{2.6 \times 1000^2} > \approx 12000$$<br />
So the wise man had requested a gift of <span style="color: purple;">more than 12 thousand Giza pyramids of wheat</span>. No wonder why the king did not have enough wheat to reward the wise man!!!<br />
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<span style="color: purple; font-size: large;">L</span>et us stop here for now. The above pictures of Giza pyramids were taken from <a href="https://maps.google.com/" target="_blank">Google Maps</a>. The picture of a chessboard at the top of our post is a screenshot from a movies by Cristóbal Vila -
"<a href="https://www.youtube.com/watch?v=oVthC6neqVc" target="_blank">Inspirations</a>" - follow the link below to watch it on YouTube:<br />
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<a href="https://www.youtube.com/watch?v=oVthC6neqVc"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEidJhWkyJJgVyNd9rNau-TZeYZ0Y1UF28pmky9t6eZb1FARkBixNDJHqxr81IlCOCkI7YoyHVDKUmd7Yna_VNm_4iQ5caS2RVzlfyu5eOpWr5IulM31pAhw7HbDKvio1oXNDsx0q8C_SCFL/s1600/youtube_small.png" /></a></div>
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<i>Homework.</i><br />
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1. Go to <a href="http://google.com/">google.com</a> and find out how large the moon surface is.
Is it possible to cover the whole surface of the moon by the wise man's wheat?
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2. Simplify the sum $$1 + \frac{1}{2} + \frac{1}{4} + + \frac{1}{8} + \dots + + \frac{1}{2^{64}}$$<br />
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3. Simplify the sum $$1 + 3 + 9 + 27 + \dots + 3^{64}$$<br />
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<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgHr320YPQm2p_pfIMDSBWbQm81gZ72ygsBpQbqofl6iEnbawFKOqOKMKR9Pai_GdYvLj191ClZ88ha2MfoilMageBqaCfnvNmnnyHq-5d-zCW353x40qiybr7S-c4qG0ARCXTpDYj2Ze8T/s1600/vuontoan_mathgarden_vianney_chess.jpg" height="488" width="640" /></div>
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Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-73608365370440447302015-04-21T12:16:00.000+10:002015-04-21T12:16:49.006+10:00Is 0.9999999... equal to 1?<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfBWk-lRwoIKao8NJDUBchWOUonW6aa9VB-Hp-kYJ22rKXv4F0gPx46Ve5RahZH4n5YFFZq1_j1wsYLFoLROvTWL8Pup0BOsnU0pU1SNbE8o2UYsw6I9QDOHo7Qx_QDvdy5dUt04XcXSAu/s1600/vuontoan_mathgarden_matthew_math_symbol.jpg" height="242" width="400" /></div>
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<div style="text-align: center;">
<span style="color: purple; font-size: x-large;">Is $\frac{1}{3}$ equal to 0.3333333... ?!</span></div>
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<span style="color: purple; font-size: x-large;"><br /></span></div>
Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-83604084064020586012015-04-04T17:20:00.000+11:002015-04-04T17:23:01.494+11:00Sum of reciprocal squares<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj2mlTqgckApGS3svw0QhCvDS0inT1UVJo5aFGCbLi3uC9jYTpWfV82jSoAdwHVSqExlGJdhKTFD2WiEqzZvHDf2HoVS5ZDQOPYNJvtCRLbRSiqQdcEPeACLmCpiLM7D2LMoBwPwTeA2Dfj/s1600/vuontoan_mathgarden_francis_euler.jpg" height="283" width="400" /></div>
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<span style="color: purple; font-size: large;">T</span>oday we will look at a very fascinating proof due to Euler for the following identity: $$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \dots = \frac{\pi^2}{6}$$<br />
The mathematician Euler formulated this proof in 1734 when he was 28 year old.<br />
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<a name='more'></a><br />
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<span style="color: purple; font-size: large;">W</span>e will present Euler's proof step by step.<br />
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<span style="color: #0b5394; font-size: large;">Step 1. Using <a href="http://mathgardenblog.blogspot.com/2015/04/Taylor-series.html" target="_blank">Taylor series expansion</a> for $f(x) = \sin(x)$</span><br />
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We have<br />
$$f(x) = \sin(x) ~\Rightarrow ~ f(0) = 0$$ $$f'(x) = \cos(x) ~\Rightarrow ~ f'(0) = 1$$ $$f''(x) = -\sin(x) ~\Rightarrow ~ f''(0) = 0$$ $$f^{(3)}(x) = -\cos(x) ~\Rightarrow ~ f^{(3)}(0) = -1$$ $$f^{(4)}(x) = \sin(x) ~\Rightarrow ~ f^{(4)}(0) = 0$$ $$f^{(5)}(x) = \cos(x) ~\Rightarrow ~ f^{(5)}(0) = 1$$ $$f^{(6)}(x) = -\sin(x) ~\Rightarrow ~ f^{(6)}(0) = 0$$ $$f^{(7)}(x) = -\cos(x) ~\Rightarrow ~ f^{(7)}(0) = -1$$<br />
So the <a href="http://mathgardenblog.blogspot.com/2015/04/Taylor-series.html" target="_blank">Taylor series</a> for the function $f(x) = \sin(x)$ is as follows:<br />
$$f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 + \dots + \frac{f^{(n)}(0)}{n!} x^n + \dots$$<br />
$$\sin(x) = x - \frac{1}{3!} x^3 + \frac{1}{5!} x^5 - \frac{1}{7!} x^7 + \dots$$<br />
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<blockquote class="tr_bq">
<span style="color: purple;">$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$</span></blockquote>
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<span style="color: #0b5394; font-size: large;"><br /></span><span style="color: #0b5394; font-size: large;">Step 2. Express $\sin(x)$ as product of linear factors</span><br />
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Since the equation $f(x) = \sin(x) = 0$ has the following roots $$x=0, ~x = \pm \pi, ~x = \pm 2 \pi, ~x = \pm 3 \pi, \dots$$<br />
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we have<br />
$$\sin(x) = C x (x - \pi)(x + \pi)(x - 2\pi)(x + 2 \pi)(x - 3\pi)(x + 3\pi) \dots$$<br />
$$\sin(x) = C x (x^2 - \pi^2)(x^2 - 2^2 \pi^2)(x^2 - 3^2 \pi^2)(x^2 - 4^2 \pi^2) \dots$$<br />
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<span style="color: #0b5394; font-size: large;">Step 3. Combining step 1 and step 2</span><br />
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Comparing the two representations of $\sin(x)$ in step 1 and step 2, we have<br />
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$ $$= C x (x^2 - \pi^2)(x^2 - 2^2 \pi^2)(x^2 - 3^2 \pi^2)(x^2 - 4^2 \pi^2) \dots$$<br />
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So<br />
$$1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \dots$$ $$= C (x^2 - \pi^2)(x^2 - 2^2 \pi^2)(x^2 - 3^2 \pi^2)(x^2 - 4^2 \pi^2) \dots$$<br />
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Substitute $x^2$ by $x$, we have<br />
$$1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!} + \dots$$ $$= C (x - \pi^2)(x - 2^2 \pi^2)(x - 3^2 \pi^2)(x - 4^2 \pi^2) \dots$$<br />
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<span style="color: #0b5394; font-size: large;">Step 4. Normalization to get rid of the constant factor $C$</span><br />
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If we have an equation of the form<br />
$$f(x) = C g_1(x) g_2(x) g_3(x) \dots$$<br />
we can get rid of the constant factor $C$ by normalization as follows<br />
$$\frac{f(x)}{f(0)} = \frac{g_1(x)}{g_1(0)} \frac{g_2(x)}{g_2(0)} \frac{g_3(x)}{g_3(0)} \dots$$<br />
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With the equation that we have in step 3<br />
$$1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!} + \dots$$ $$= C (x - \pi^2)(x - 2^2 \pi^2)(x - 3^2 \pi^2)(x - 4^2 \pi^2) \dots$$<br />
by normalization, we obtain<br />
$$1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!} + \dots$$ $$= \frac{(x - \pi^2)}{-\pi^2} \frac{(x - 2^2 \pi^2)}{-2^2 \pi^2} \frac{(x - 3^2 \pi^2)}{-3^2 \pi^2} \frac{(x - 4^2 \pi^2)}{-4^2 \pi^2} \dots$$ $$= (1 - \frac{x}{\pi^2}) (1 - \frac{x}{2^2 \pi^2}) (1 - \frac{x}{3^2 \pi^2}) (1 - \frac{x}{4^2 \pi^2})\dots$$<br />
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<span style="color: #0b5394; font-size: large;"><br /></span><span style="color: #0b5394; font-size: large;">Step 5. Use Vieta formula</span><br />
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Comparing the coefficients of $x$ in both sides of the equation<br />
$$1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!} + \dots$$ $$= (1 - \frac{x}{\pi^2}) (1 - \frac{x}{2^2 \pi^2}) (1 - \frac{x}{3^2 \pi^2}) (1 - \frac{x}{4^2 \pi^2})\dots$$<br />
we have $$- \frac{x}{3!} = - \frac{x}{\pi^2} - \frac{x}{2^2 \pi^2} - \frac{x}{3^2 \pi^2} - \frac{x}{4^2 \pi^2} - \dots $$<br />
Therefore, $$\frac{1}{3!} = \frac{1}{\pi^2} + \frac{1}{2^2 \pi^2} + \frac{1}{3^2 \pi^2} + \frac{1}{4^2 \pi^2} + \dots $$<br />
Thus, we obtain<br />
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots =\frac{\pi^2}{3!}= \frac{\pi^2}{6}$$<br />
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<span style="color: purple; font-size: large;">L</span>et us stop here for now. In the homework section, you can try Euler's method with the function $f(x) = \cos(x)$ to derive another proof of the above identity. Hope to see you again next time.<br />
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<i><br /></i><i><br /></i><i><br /></i><i>Homework.</i><br />
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1. Show that the Taylor series for $f(x) = \cos(x)$ is as follows $$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$$<br />
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2. Using the fact that the equation $\cos(x) = 0$ has the following roots $$x = \pm \frac{\pi}{2}, ~x = \pm \frac{3 \pi}{2}, ~x = \pm \frac{5 \pi}{2}, \dots$$<br />
show that<br />
$$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots = (1 - \frac{4 x^2}{\pi^2})(1 - \frac{4 x^2}{3^2 \pi^2})(1 - \frac{4 x^2}{5^2 \pi^2}) \dots$$<br />
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3. Prove that<br />
$$\frac{1 }{1^2} + \frac{1 }{3^2} + \frac{1 }{5^2} + \dots = \frac{\pi^2}{8} $$<br />
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4. Prove that<br />
$$\frac{1 }{1^2} + \frac{1 }{3^2} + \frac{1 }{5^2} + \dots = \frac{3}{4} \left( \frac{1 }{1^2} + \frac{1 }{2^2} + \frac{1 }{3^2} + \frac{1 }{4^2} + \dots \right)$$<br />
thus deriving<br />
$$ \frac{1 }{1^2} + \frac{1 }{2^2} + \frac{1 }{3^2} + \frac{1 }{4^2} + \dots = \frac{\pi^2}{6}$$<br />
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5. Go to <a href="http://google.com/">google.com</a> and search about the Basel problem and the Riemann zeta function.<br />
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Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-80468915725168643482015-04-03T20:19:00.000+11:002015-04-03T20:19:49.310+11:00Taylor series<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg_PtHPNft0biVgJ92obVXdR8rVvTFR2zCq8BldZHJnptFMZgYNUjJsH-hQkGRB2q-PpAeeaVC27X3My27Oz5l6sVgraXHIolBv63Ai0wKJ3jg-tnf7UQgKUhnoWZKa_GGk4fXzpkBhZsrJ/s1600/vuontoan_mathgarden_james_art.jpg" height="309" width="320" /></div>
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<span style="color: purple; font-size: large;">T</span>o celebrate the <a href="http://mathgardenblog.blogspot.com/2015/03/pi-2015.html" target="_blank">$\pi$ day</a>, in our previous post, we were introduced to a very beautiful identity due to Euler<br />
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \dots = \frac{\pi^2}{6}$$<br />
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The mathematician Euler had an <i>intriguing method</i> to derive this identity. Euler's method employed the Taylor series, so today we will learn about Taylor series, and in the next post, we will look at Euler's technique.<br />
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<a name='more'></a><br />
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<span style="color: #0b5394; font-size: large;">An example about quadratic polynomial</span><br />
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<span style="color: purple; font-size: large;">B</span>efore learning about the Taylor series, let us look at the following little quiz question:<br />
<blockquote class="tr_bq">
<span style="color: purple;">Given a quadratic polynomial $$p(x) = a x^2 + b x + c$$ Express the coefficient </span><span style="color: purple;">$c$ in terms of the function $p$.</span></blockquote>
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Have you got an answer?<br />
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The answer is: $$c = p(0)$$<br />
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<span style="color: purple; font-size: large;">T</span>he next question is<br />
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<blockquote class="tr_bq">
<span style="color: purple;">Given a quadratic polynomial $$p(x) = a x^2 + b x + c$$</span><br />
<span style="color: purple;">Express the coefficient $b$ in terms of the function $p$.</span><br />
<span style="color: purple;">(<i>Hint: take the derivative of $p$.</i>)</span></blockquote>
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Take the derivative of $p$ we have $$p'(x) = 2 a x + b$$<br />
So $$b = p'(0)$$<br />
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By now you probably know the formula for the coefficient $a$?!<br />
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Taking the second derivative, we have $$p''(x) = 2a$$<br />
Thus, $$c = p(0), ~~ b = p'(0), ~~ a = \frac{1}{2} p''(0).$$<br />
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Substitute these formulas back into the quadratic polynomial, we obtain $$p(x) = a x^2 + bx + c=\frac{1}{2} p''(0) x^2 + p'(0) x + p(0)$$<br />
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<span style="color: #0b5394; font-size: large;">An example about quintic polynomial</span><br />
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<span style="color: purple; font-size: large;">N</span>ow, suppose we have a quintic polynomial $$p(x) = a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0$$<br />
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We can follow similar steps to derive formula for the coefficients $a_i$.<br />
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First of all, we have $$a_0 = p(0)$$<br />
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Taking the first derivative, we have $$p'(x) = 5 a_5 x^4 + 4 a_4 x^3 + 3 a_3 x^2 + 2 a_2 x + a_1$$<br />
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So $$a_1 = p'(0)$$<br />
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Taking the second derivative, we have $$p''(x) = 5 \times 4 \, a_5 x^3 + 4 \times 3 \, a_4 x^2 + 3 \times 2 \, a_3 x + 2 a_2$$<br />
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So $$a_2 = \frac{p''(0)}{2}$$<br />
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Taking the third derivative, we have $$p'''(x) = 5 \times 4 \times 3 \, a_5 x^2 + 4 \times 3 \times 2 \, a_4 x + 3 \times 2 \times 1 \, a_3$$<br />
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Thus, $$a_3 = \frac{p'''(0)}{3 \times 2 \times 1} = \frac{p'''(0)}{3!}$$<br />
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<span style="color: purple;">(The notation $n!$ reads "$n$ factorial", $n!= 1 \times 2 \times \dots \times (n-1) \times n$)</span><br />
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Keep taking the derivative, we have $$p''''(x) = 5 \times 4 \times 3 \times 2 \, a_5 x + 4 \times 3 \times 2 \times 1 \, a_4$$<br />
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So $$a_4 = \frac{p''''(0)}{4 \times 3 \times 2 \times 1} = \frac{p^{(4)}(0)}{4!}$$<br />
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<span style="color: purple;">(We will write $p^{(n)}(x)$ to denote the $n$th derivative of $p(x)$)</span><br />
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Finally, taking the fifth derivative, we have $$p^{(5)}(x) = 5 \times 4 \times 3 \times 2 \times 1 \, a_5$$<br />
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So $$a_5 = \frac{p^{(5)}(0)}{5!}$$<br />
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We can see that the general formula is $$a_n = \frac{p^{(n)}(0)}{n!}$$<br />
Therefore, $$p(x) = a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0$$ $$= \frac{p^{(5)}(0)}{5!} x^5 + \frac{p^{(4)}(0)}{4!} x^4 + \frac{p^{(3)}(0)}{3!} x^3 + \frac{p''(0)}{2!} x^2 + p'(0) x + p(0)$$ $$= p(0) + p'(0) x + \frac{p''(0)}{2!} x^2 + \frac{p^{(3)}(0)}{3!} x^3 + \frac{p^{(4)}(0)}{4!} x^4 + \frac{p^{(5)}(0)}{5!} x^5$$<br />
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<span style="color: #0b5394; font-size: large;">Taylor series</span><br />
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<span style="color: purple; font-size: large;">N</span>ow, we are ready to learn about the Taylor series.<br />
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<blockquote class="tr_bq">
<span style="color: purple;"><b>The Taylor series</b> of a function $f(x)$ is:</span><span style="color: purple;">$$f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 + \dots + \frac{f^{(n)}(0)}{n!} x^n + \dots$$</span></blockquote>
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If you ever forget about this formula, just work it out with a quadratic polynomial or a cubic polynomial then you will remember the general formula.<br />
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Let us now calculate the Taylor series for the function $e^x$.<br />
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<span style="color: #0b5394; font-size: large;">The natural exponential function $e^x$</span><br />
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<span style="color: purple; font-size: large;">T</span>here are two most important constants in mathematics, the first is $\pi \approx 3.14$ and the second is the Euler constant $e \approx 2.72$.<br />
We have $$\lim_{n \to \infty}{\left( 1 + \frac{1}{n} \right)^n} = e$$<br />
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The natural exponential function $f(x) = e^x$ is a very special function because its derivative is equal to itself. Therefore, no matter how many times you take the derivative, the answer is always $e^x$: $$f(x) = f'(x) = f''(x) = f^{(3)}(x) = f^{(4)}(x) = f^{(5)}(x) = \dots = e^x$$<br />
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So $$f(0) = f'(0) = f''(0) = f^{(3)}(0) = f^{(4)}(0) = f^{(5)}(0) = \dots = e^0 = 1$$<br />
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The Taylor series for $f(x) = e^x$ is: $$f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 + \dots + \frac{f^{(n)}(0)}{n!} x^n + \dots$$ $$e^x = 1 + x + \frac{1}{2!} x^2 + \frac{1}{3!} x^3 + \frac{1}{4!} x^4 + \dots + \frac{1}{n!} x^n + \dots$$<br />
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<blockquote class="tr_bq">
<span style="color: purple;">$$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots + \frac{x^n}{n!} + \dots$$</span></blockquote>
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Take $x = \pm 1$, we obtain the following two beautiful identities: $$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \dots $$ $$\frac{1}{e} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots $$<br />
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<span style="color: purple; font-size: large;">L</span>et us stop here for now. Next time, we will look at Euler's technique to derive the identity<br />
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \dots = \frac{\pi^2}{6}$$<br />
Hope to see you again then.<br />
<br />
<i><br /></i><i><br /></i><i><br /></i><i>Homework.</i><br />
<br />
1. Use the Taylor series for function $g(x)$ $$g(x) = g(0) + g'(0) x + \frac{g''(0)}{2!} x^2 + \frac{g^{(3)}(0)}{3!} x^3 + \frac{g^{(4)}(0)}{4!} x^4 + \dots + \frac{g^{(n)}(0)}{n!} x^n + \dots$$<br />
and take $f(x+a) = g(x)$, prove that $$f(x+a) = f(a) + f'(a) x + \frac{f''(a)}{2!} x^2 + \frac{f^{(3)}(a)}{3!} x^3 + \frac{f^{(4)}(a)}{4!} x^4 + \dots + \frac{f^{(n)}(a)}{n!} x^n + \dots$$<br />
From here, derive the following formula<br />
$$f(x) = f(a) + f'(a) (x-a) + \frac{f''(a)}{2!} (x-a)^2 + \frac{f^{(3)}(a)}{3!} (x-a)^3 + \frac{f^{(4)}(a)}{4!} (x-a)^4 + \dots + \frac{f^{(n)}(a)}{n!} (x-a)^n + \dots$$<br />
This is called the Taylor series for $f(x)$ at the point $x =a$.<br />
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2. Show that the Taylor series for $\sin(x)$ is $$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \dots$$<br />
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3. Show that the Taylor series for $\cos(x)$ is $$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \dots$$<br />
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4. Calculate the Taylor series for $f(x) = \log_{e}(1 + x)$.<br />
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5. Calculate the Taylor series for $f(x) = \sqrt{x + 1}$.<br />
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6. Go to <a href="http://google.com/">google.com</a> and search for the Euler constant $e$, read about how Euler constant is used to calculate bank interest.<br />
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<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-48982142491495152812015-03-07T15:43:00.000+11:002015-03-07T15:43:40.670+11:00Pi day<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgstbFd6cU40Q8FDs4wBvTdBAVHu_5tGFbpUO_DEaAKWMDmwcf7Ea3VE-sleRibD0dn88EAE0Uox6f67WGdpBYZOVj48D-GtgDGx0_YJnBp6SaozBvhDSX8m6W2mFMFCxA0ovlwbcSur4Q/s1600/vuontoan_mathgarden_pi_f.jpg" height="223" width="320" /></div>
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<span style="color: purple;"><b><i>$\pi$ day</i></b></span> is celebrated every year on March 14 because $\pi \approx 3.14$.<br />
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiTBn-SMTjSi3pTLaznuduIS_D-MB28J0Og_LLoBp4-mLQf_B4Vyh8IOX-CHHk-zu1OFsH1yWqOFI1tE2BvOJjt9_uiTrRHz9AmIeZW3xWhWLth6dpmdTcM1cq6-CmXfiFAUrKDZDa3SdY2/s1600/vuontoan_mathgarden_pi1c.png" height="199" width="200" /></div>
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If we draw a circle of radius 1, then $\pi$ is the length of a <b>semicircle</b>.<br />
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The length of the whole circle is denoted by $\tau$, thus, $$\tau = 2 \pi \approx 6.28$$<br />
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In the figure below, the perimeter of the hexagon is 6, so the circumference of the circle is a little bit greater than 6:<br />
$$\tau > \approx 6$$<br />
so $\pi$ is a tiny bit bigger than 3:<br />
$$\pi > \approx 3$$<br />
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi1-81SKs36ISvRISV4J7w4cViOoO2SAxUGt4f_mNM9J115m4ublaZ_sP3JQb7FG7YHYeZ2ifJ59T_OYH00rU0thXaZJJHSrav7AV_AN6qTozuxz9I7957W8GWEgq_Kp1bgNqAGwHWfQL_c/s1600/vuontoan_mathgarden_pi1.png" height="197" width="200" /></div>
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgmZw9VWlRO758obHzxvW0TrLBotjtRTNOI7CV0lCZn7l68OnwrYROW_UFfszfAMoftzf8RwI-heUAMAR76NMA_0MNJQgO2dfIPKAVjVcDaLjRe6l4MMM02OMwY5ykqMwAwLFSDCtBc-JQ/s1600/vuontoan_mathgarden_pi_m.jpg" height="206" width="320" /></div>
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To celebrate this year's $\pi$ day, let us enjoy this beautiful identity about $\pi$ due to the mathematician Euler:<br />
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \dots = \frac{\pi^2}{6}$$<br />
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Happy $\pi$ day everyone!!!<br />
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZFgb-jSoN-6IaH8Wh7FEv8AIpq24D741zkzTNafG8MkGlUMRUuRIEmOA5xeRLhLIDapWTJCpvE27ezfvGQ_KiGyt10xSla8KossUl4zVyckDlG2GJWRG9bfVSBUp2vrOcnSn1I01SuyAP/s1600/vuontoan_mathgarden_pi_v2.jpg" height="245" width="320" /></div>
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<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-5746906676031056312015-01-10T16:19:00.001+11:002015-02-03T13:13:47.557+11:00When I look in your eyes<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2RJQExCThKyZCazYzJo1zWB-SPqPYv1juiMvTeI12iUFQdq8Mm9U4iOUnAsLuBXaEMTSFvW5JWnfcM6Pdxq2_KF-63wKz8KWAnAOmsZSVpN4Wbu8gIl0Uu90CnLKIPrOyUS-3rarwONBe/s1600/james_MBHT.jpg" height="301" width="400" /></div>
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<span style="color: purple; font-size: x-large;">W</span>hen people say to each other something like "<i>there is a universe hidden in your eyes</i>" or "<i>I'm lost in your eyes, </i><i>my spirits rise to the skies</i>", they are <b>not</b> exaggerating. There is actually a mathematical truth in these sayings!<br />
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Today, we will prove the following mathematical fact:<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzhE6DYnlOFyN0dHKC_GcMTbuTZUTDafPfevwsfxTgjPC8LPstcn8j55w1VIf9tQ2z-oGhIlSM_FgKXAb6teXuSDCZoCVb3qCZlveIcxbO5a_tqp45owE4Ohl0BugVs0S7mIwQZ1QsK0On/s1600/vuontoan_mathgarden_eye.png" height="190" width="320" /></td></tr>
<tr><td style="text-align: center;"><span style="text-align: start;"><i>there exists a polygon in the human eye whose perimeter is equal to the circumference of the earth</i></span></td></tr>
</tbody></table>
<blockquote class="tr_bq">
<span style="color: purple;"><i>In a circle of diameter 1 cm (about the size of the iris of the human eye), there exists a polygon whose perimeter is equal to the circumference of the earth (about 40,000 km)!</i></span></blockquote>
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We will construct that polygon as follows. First, in the middle of the circle, we draw a small <b>equilateral triangle</b> whose perimeter is equal to 1 cm.<br />
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqpq3la4UUSQAVlSmZpYqnE64nMIo-Ir8rBzs_4mWoF_8T6zVYZd_k3FvryssRl3ZrATTbhzsrCdFfKW-CcFZAlzlvH-vgsgjvaCAH3NnhzEOj21xR9Xtz20dOTNS_wNOnc1m38pS_5Dij/s1600/vuontoan_mathgarden_eye2.png" height="190" width="320" /></div>
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Next, we divide each side of the triangle into <i>three equal segments</i>. Use the middle segment as a base, construct an equilateral triangle which points <i>outward</i>. What we obtain is a new polygon that has 12 sides.<br />
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEipau0_s4sd0RW2RZ1I2SE74X0q0xZscVXO5K5uNqO8zEmehamWxJlcLlRLJ-hS6YiIJMKfOBCA0NZb3r21H0eoVItI9Ym9ghK7gEpH4FfV4CkK_hZVrYWTFQoAGu276d3EbN9UfgzI1iEB/s1600/vuontoan-mathgarden-k1-2b.png" height="180" width="400" /></div>
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Again, we divide each side of the new polygon into three equal segments and use the middle segment to construct an equilateral triangle, we obtain a new polygon with 48 sides.<br />
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGPFuz7YFA5a_sxR8j-0Ji81qTwP-lAIMkIZsxT7GXXEsaLUJSBMCwp9RcokjonfV7ETMQ0OE5OfaO6JD5FLaTl2_8hKufrjWFx2d5vWiPKZjDwujgJljqhAKqFtthVPEacmeGyvqO94ie/s1600/vuontoan-mathgarden-k1-3.png" height="250" width="640" /></div>
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Keep doing like this, at each step, we divide each side of the polygon into three equal segments and use the middle segment to construct an equilateral triangle. What we obtain is a polygon that has the shape of a snowflake. In mathematics, we call it <span style="color: purple;"><b><i>Koch's snowflake</i></b></span> because this construction was originally described by the mathematician Koch in 1904.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiYt_17DDL0FMGfZXP7e3kSqmXeqhQPqNhhbQvGCX-rGewajf5YLCsyA-_lM3oiGK4bVXg_P9D-Wdv0_4u8KH082ZELv-YcOMHJvX3729nCKEbrPjOoIlbyx33AF4adRhleEx_mUIDTajMt/s1600/vuontoan-mathgarden-k1-6.png" height="442" width="640" /></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: xx-small; text-align: start;">Koch's snowflake</span></td></tr>
</tbody></table>
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We will show that, with this construction, after a number of steps, we will obtain a Koch's snowflake polygon whose perimeter is <i>longer than the circumference of the earth</i>!<br />
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Indeed, at each step of the construction, the perimeter of the polygon is increased by a factor of $\frac{4}{3}$.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXnb0fEbk66kD7MRRL751IbY4HRzENqoJkS0obrL0Ew0oCdITRmyfy7RwLRfl_-Tv7HIVgIhPLiZHm9F0Mu7soz3gWKVK6-TDoyBFbImVtMe1HTF-Pdd9HuM2VQkU6rULwyAtGDTnjq0VB/s1600/vuontoan-mathgarden-k1-2c.png" height="218" width="400" /></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: xx-small; text-align: start;">the perimeter of the polygon is increased by a factor of $\frac{4}{3}$</span></td></tr>
</tbody></table>
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Since we start with an equilateral triangle with perimeter $1$ cm,<br />
<ul>
<li>after the first step, the polygon with 12 sides has perimeter equal to $\frac{4}{3} \approx 1.3$ cm</li>
</ul>
<ul>
<li>after the second step, the polygon with 48 sides has perimeter equal to $\frac{4}{3} \times \frac{4}{3} \approx 1.7$ cm</li>
</ul>
<ul>
<li>in the next step, the perimeter is $\frac{4}{3} \times \frac{4}{3} \times \frac{4}{3} \approx 2.3$ cm</li>
</ul>
<ul>
<li>after $n$ steps, we obtain a Koch's snowflake polygon with perimeter equal to $\left( \frac{4}{3} \right)^n$ cm</li>
</ul>
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We make the following two observations:<br />
<ul>
<li>After $10^{11}$ construction steps, the perimeter of the Koch's snowflake polygon is longer than the circumference of the earth
$$\left( \frac{4}{3} \right)^n cm > 100,000 ~km \mbox{ when we choose } n = 10^{11}$$</li>
</ul>
<ul>
<li>No matter how many construction steps we have performed, the Koch's snowflake polygons always lie inside the following triangle $ABC$. So we can be sure that the Koch's snowflake polygons never grow outside of the human eye!</li>
</ul>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjjpr0LtE89qacYgEYFn4fP4PjASzX-SfT5d5YKqRsEGhau2bOABsL-U2osrffzMOavmmJMPV08oRCjw_cY-HAB0CU2Bz757uBoqEfo_3pf3I5ntMHPeSH59WIpOYQ20HIySMs17rDPpvGL/s1600/vuontoan-mathgarden-bound-koch.png" height="272" width="640" /></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: xx-small; text-align: start;">Koch's snowflake polygons always lie inside triangle $ABC$</span></td></tr>
</tbody></table>
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We have now proved the following surprising fact<br />
<blockquote class="tr_bq">
<span style="color: purple;"><i>Inside a circle of diameter 1 cm (about the size of the iris of the human eye), there exists a polygon whose perimeter is longer than the circumference of the earth (about 40,000 km), longer than the circumference of the sun (about 4,400,000 km), and even longer than the circumference of the whole universe (if the universe is bounded)!!!</i></span></blockquote>
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<span style="color: purple; font-size: x-large;">L</span>et us stop here for now. Hope to see you again next time.<br />
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<div style="text-align: center;">
<a href="http://www.youtube.com/watch?v=bTrSGfDgVmQ">http://www.youtube.com/watch?v=bTrSGfDgVmQ</a></div>
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<a href="http://www.youtube.com/watch?v=bTrSGfDgVmQ"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEidJhWkyJJgVyNd9rNau-TZeYZ0Y1UF28pmky9t6eZb1FARkBixNDJHqxr81IlCOCkI7YoyHVDKUmd7Yna_VNm_4iQ5caS2RVzlfyu5eOpWr5IulM31pAhw7HbDKvio1oXNDsx0q8C_SCFL/s1600/youtube_small.png" /></a></div>
<blockquote class="tr_bq">
<i><span style="color: purple;">When I look in your eyes</span></i><br />
<i><span style="color: purple;">I see the wisdom of the world in your eyes</span></i><br />
<i><span style="color: purple;">I see the sadness of a thousand goodbyes</span></i><br />
<i><span style="color: purple;">When I look in your eyes</span></i> </blockquote>
<blockquote class="tr_bq">
<i><span style="color: purple;">And it is no surprise</span></i><br />
<i><span style="color: purple;">To see the softness of the moon in your eyes</span></i><br />
<i><span style="color: purple;">The gentle sparkle of the stars in the skies</span></i><br />
<i><span style="color: purple;">When I look in your eyes</span></i> </blockquote>
<blockquote class="tr_bq">
<i><span style="color: purple;">In your eyes</span></i><br />
<i><span style="color: purple;">I see the deepness of the sea</span></i><br />
<i><span style="color: purple;">I see the deepness of the love</span></i><br />
<i><span style="color: purple;">The love I feel you feel for me</span></i> </blockquote>
<blockquote class="tr_bq">
<i><span style="color: purple;">Autumn comes, summer dies</span></i><br />
<i><span style="color: purple;">I see the passing of the years in your eyes</span></i><br />
<i><span style="color: purple;">And when we part there'll be not tears, no goodbyes</span></i><br />
<i><span style="color: purple;">I'll just look into your eyes</span></i></blockquote>
<blockquote class="tr_bq">
<i><span style="color: purple;">Those eyes, so wise, so warm, so real</span></i><br />
<i><span style="color: purple;">How I love the world your eyes reveal</span></i> </blockquote>
<blockquote class="tr_bq">
<div style="text-align: left;">
<b><i><span style="color: purple;">Leslie Bricusse</span></i></b></div>
</blockquote>
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<i><br /></i>
<i>Homework.</i><br />
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1. Use the <a href="http://mathgardenblog.blogspot.com/2012/09/pascal-triangle.html" target="_blank">binomial identity</a>, to prove that for any $x > 0$,<br />
$$(1 + x)^n > 1 + n x$$<br />
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Then show that the perimeter of the Koch's snowflake polygon obtained after $n$ construction steps satisfies<br />
$$\left( \frac{4}{3} \right)^n > \frac{n}{10}$$<br />
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And with $n = 10^{11}$, show that<br />
$$\left( \frac{4}{3} \right)^n cm > 100,000 ~km$$<br />
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2. Use <a href="http://mathgardenblog.blogspot.com/2012/09/induction1.html" target="_blank">induction</a> to prove that the Koch's snowflake polygons always lie within the triangle $ABC$ as follows.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjjpr0LtE89qacYgEYFn4fP4PjASzX-SfT5d5YKqRsEGhau2bOABsL-U2osrffzMOavmmJMPV08oRCjw_cY-HAB0CU2Bz757uBoqEfo_3pf3I5ntMHPeSH59WIpOYQ20HIySMs17rDPpvGL/s1600/vuontoan-mathgarden-bound-koch.png" height="272" width="640" /></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><br /></td></tr>
</tbody></table>
3. How many sides does the Koch's snowflake polygon have after $n$ construction steps?<br />
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4. After $n$ construction steps, what is the area of the Koch's snowflake polygon? Show that the area is finite.<br />
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<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-71504542692643599962014-12-15T10:25:00.000+11:002015-03-13T07:11:57.427+11:00Merry Christmas and Happy New Year<br />
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjR3KR0Dl5rw03sVFzQgRUMbSE91-m7BGIodh5u2Lh-v0kjGIdSpvFMLjFKTr1ta_3Zi0mhAr1EfNwK8T-_eh724MOQCzmWDI21Lpv9tEuJTzXxCk6NsxHkYDhETH2yZf5SAfStTogJ00o/s1600/james_mother_Mary.jpg" height="640" width="457" /></div>
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0sXNxOzQu71H2TUD_xwAeaKAIciQzNkfuDRY2sKwpEmy5KTtW1fxW6s5-fc6RMzr9nSVoYXWvphFUorzwcxUFJPpP82SpmSTQYBgMM2ZyQXxadFZd2eEau7HVT-CvJkphElx4MFAgX9Y/s1600/francis_hope_faith_love.jpg" height="596" width="640" /></div>
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhk5BOTPq6t3JtYsEw7jRzBizlwhviPjBZiG_Hs8imvT7Dcj8_V1aeLg63t5hzTJ8q-cdZyGfLGYMLYbnEsPOF6Xe0YaFD11q2R7YPMHexmOC8Sd_4q4x7Hi2dfKTAMWmgLdgoZojeOwhc/s1600/vuontoan_mathgarden_newyear2015.png" height="282" width="640" /></div>
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6GQWb5O2G6kCJFBNWV3aIO6j9yvYHxgOCQ0c5STlC73xtb6KLNcb_LWp0AHnsXG6qG6BKir5XI_bZkh9gd7oq8Qd7RlQdhVYdqbMlv49IwQAhYQ5PuZYgbXtQBHbEomxYTviBn7VFm_4/s1600/matthew_Christmas.jpg" height="640" width="464" /></div>
<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-68163328458233160132014-11-17T12:36:00.001+11:002014-12-17T09:36:53.402+11:00Gauss' trigonometric identities for heptadecagon<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiYYKF66GaJ9WZyqP5OFQGMLzpojfE2ppmGa_Jpddel8cwZnAsnwcIp_Gq7NVr_jYAAJLHdPsCvBYhGLqp8r9frviwL1IQBwWNyNW6-XP-qc45wbes7a2-YMhk7gc1vEu6oh31YS2FEGxI/s1600/james_face.jpg" height="640" width="444" /></div>
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<span style="color: purple; font-size: x-large;">T</span>oday, we write down Gauss' magical trigonometric identities for regular heptadecagon.<br />
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<a name='more'></a><br /></div>
$$\cos{\frac{\pi}{17}} = \frac{1}{16} \left( 1 - \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2 \sqrt{17 + 3 \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2 \sqrt{34 + 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{2 \pi}{17}} = \frac{1}{16} \left( -1 + \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2 \sqrt{17 + 3 \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2 \sqrt{34 + 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{3 \pi}{17}} = \frac{1}{16} \left( 1 + \sqrt{17} + \sqrt{34 + 2 \sqrt{17}} + 2 \sqrt{17 - 3 \sqrt{17} + \sqrt{34 + 2 \sqrt{17}} - 2 \sqrt{34 - 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{4 \pi}{17}} = \frac{1}{16} \left( -1 + \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} + 2 \sqrt{17 + 3 \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2 \sqrt{34 + 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{5 \pi}{17}} = \frac{1}{16} \left( 1 + \sqrt{17} + \sqrt{34 + 2 \sqrt{17}} - 2 \sqrt{17 - 3 \sqrt{17} + \sqrt{34 + 2 \sqrt{17}} - 2 \sqrt{34 - 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{6 \pi}{17}} = \frac{1}{16} \left( -1 - \sqrt{17} + \sqrt{34 + 2 \sqrt{17}} + 2 \sqrt{17 - 3 \sqrt{17} - \sqrt{34 + 2 \sqrt{17}} + 2 \sqrt{34 - 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{7 \pi}{17}} = \frac{1}{16} \left( 1 + \sqrt{17} - \sqrt{34 + 2 \sqrt{17}} + 2 \sqrt{17 - 3 \sqrt{17} - \sqrt{34 + 2 \sqrt{17}} + 2 \sqrt{34 - 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{8 \pi}{17}} = \frac{1}{16} \left( -1 + \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} - 2 \sqrt{17 + 3 \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2 \sqrt{34 + 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{9 \pi}{17}} = \frac{1}{16} \left( 1 - \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} + 2 \sqrt{17 + 3 \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2 \sqrt{34 + 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{10 \pi}{17}} = \frac{1}{16} \left( -1 - \sqrt{17} + \sqrt{34 + 2 \sqrt{17}} - 2 \sqrt{17 - 3 \sqrt{17} - \sqrt{34 + 2 \sqrt{17}} + 2 \sqrt{34 - 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{11 \pi}{17}} = \frac{1}{16} \left( 1 + \sqrt{17} - \sqrt{34 + 2 \sqrt{17}} - 2 \sqrt{17 - 3 \sqrt{17} - \sqrt{34 + 2 \sqrt{17}} + 2 \sqrt{34 - 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{12 \pi}{17}} = \frac{1}{16} \left( -1 - \sqrt{17} - \sqrt{34 + 2 \sqrt{17}} + 2 \sqrt{17 - 3 \sqrt{17} + \sqrt{34 + 2 \sqrt{17}} - 2 \sqrt{34 - 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{13 \pi}{17}} = \frac{1}{16} \left( 1 - \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} - 2 \sqrt{17 + 3 \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2 \sqrt{34 + 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{14 \pi}{17}} = \frac{1}{16} \left( -1 - \sqrt{17} - \sqrt{34 + 2 \sqrt{17}} - 2 \sqrt{17 - 3 \sqrt{17} + \sqrt{34 + 2 \sqrt{17}} - 2 \sqrt{34 - 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{15 \pi}{17}} = \frac{1}{16} \left( 1 - \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2 \sqrt{17 + 3 \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2 \sqrt{34 + 2 \sqrt{17}}} \right)$$<br />
$$\cos{\frac{16 \pi}{17}} = \frac{1}{16} \left( -1 + \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2 \sqrt{17 + 3 \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2 \sqrt{34 + 2 \sqrt{17}}} \right)$$<br />
<br />
If you are curious and want to know Gauss' method to derive these formulas, please follow our next few posts.<br />
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiyEehvr9YZ5yOBaoFeH0jeLZHNxN5pnTPyrNz8nxHfngh2Ibz9SzbT6e8ZuD7_RhRhag8ivzRw3fS7-CH8jgrqJvTgp7KV2oyCk8alcAo-KcVQSMlviehPN7RJyXcqbr85MaOS91QQKSHG/s1600/20141101_113141.jpg" height="260" width="400" /></div>
<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-73107644441627120232014-11-17T10:10:00.001+11:002014-11-17T10:10:10.214+11:00James' geometry art<div class="separator" style="clear: both; text-align: center;">
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiconHI850uKZyiLJ-YV5TUOH9Pu0a1KvWSoXYcX2GvIFwAylt9vMDjWC0bkhAYImOjCvRxvaQkNcK0Fx-R5ILdiGlggTNk2AQY2SUJIHwhJBsuNOyUZL-SLtJ2pVFce8mMFGU6xH2UW_8/s1600/james_geometry.png" height="508" width="640" /></div>
<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-46522411973653850672014-11-12T13:44:00.000+11:002014-11-12T13:51:24.445+11:00Construction algorithm<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhju8gn0J2Cpn9rXiLTnUZc96cjN61nbh5FCFZIXHZNp1zygb_9tWsKegjsFc3nB5IaCKkhoK_Mk2xQimImyZCeBMsyOYPc3Q4pD2X01I6ek_2JeeJnsUS7aRXvt2jAv2-b3nYLImlovNY/s1600/matthew1.jpg" height="268" width="400" /></div>
<br />
<span style="color: purple; font-size: x-large;">I</span>n our previous post, we have learnt about <i>Similar Triangles</i> and the <i><span style="color: purple;"><a href="http://mathgardenblog.blogspot.com/2014/08/right-triangle-altitude-theorem.html" target="_blank">Right Triangle Altitude Theorem</a></span></i>. Today, continuing our journey in the garden of geometry, we want to find answer to the following question<br />
<blockquote class="tr_bq">
<span style="color: purple;">Given a line segment of length $r$, by compass and straightedge, what kind of shapes can we construct? </span></blockquote>
<br />
<a name='more'></a><br />
<br />
<span style="color: purple; font-size: x-large;">T</span>here are two simple cases that we can see straightaway.<br />
<ul>
<li>We can construct a line segment whose length is a multiple of a given length, and</li>
<li>We can divide a given line segment into a number of equal parts. </li>
</ul>
<br />
Here are some examples<br />
<ul>
<li>Given a line segment $AB=r$, by compass and straightedge, we can easily construct a line segment $AC = 3r$.</li>
</ul>
<ul>
</ul>
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh1rk8dn06TSKpt8jopY3Os3c48WoxHGXESTWKSU2blM4V09H-66Cw_62HdmBejVSEY_jAWE4A-FXY9sJ-GL7bjP0OWOTnsRwM5MNV0Dfbx8rQhz7gvmsQq7V-pRkN4hyI7uFxfA2Z5HI54/s1600/vuontoan_mathgarden_ca1.png" height="58" width="400" /></div>
<br />
<ul>
<li>Given a line segment $AB=r$. Construct an arbitrary ray $Ax$ and on this ray, construct $C_1$, $C_2$, $C_3$, $C_4$, $C_5$ such that $$A C_1 = C_1 C_2 = C_2 C_3 = C_3 C_4 = C_4 C_5.$$ Through $C_1$, $C_2$, $C_3$, $C_4$, construct four lines parallel to $B C_5$ which intersect with $AB$ at $D_1$, $D_2$, $D_3$, $D_4$. Then we have $$A D_1 = D_1 D_2 = D_2 D_3 = D_3 D_4 = D_4 B = \frac{r}{5}$$</li>
</ul>
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjh1lseVdkESCI1o-X82zEmosXO47_oSGpIKzpQA-7Sg20kTQ3fzH8-lthtY7LnkbCnYnxpjTL3HoLuatJAzlv8-haFELrcJrw35xvEC88NMGVykbJiGoQXN0SPcA-XGA19N7liuynrgWB0/s1600/vuontoan-mathgarden-ruler-compas7.png" height="245" width="400" /></div>
<br />
<ul>
<li>Given a line segment $AB=r$, first construct $AC = 3r$ and then divide $AC$ into 5 equal parts then we have a line segment of length $\frac{3}{5} r$.</li>
</ul>
<br />
From these simple examples we deduce the following conclusion<br />
<blockquote class="tr_bq">
<span style="color: purple;">Given a line segment of length $r$, and $\frac{p}{q}$ is a rational number, then by compass and straightedge, we can construct a line segment of length $\frac{p}{q} r$.</span></blockquote>
<br />
<span style="color: purple; font-size: x-large;"><br /></span>
<span style="color: purple; font-size: x-large;">L</span>et us now look at two examples which are a bit more sophisticated: construction of a <a href="http://mathgardenblog.blogspot.com/2014/02/pentagon-construction.html" target="_blank">regular pentagon</a> and construction of a <a href="http://mathgardenblog.blogspot.com/2014/03/regular-polygon-construction.html" target="_blank">regular polygon with 17 sides</a>.<br />
<br />
To construct a regular pentagon, first we construct a circle and pick a point $P_0$ on it. If we can construct the point $H$ then from $H$ we can construct $P_1$, and from $P_1$ we can construct other vertices of the pentagon.<br />
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Suppose that $r$ is the radius of the circle, we have the following trigonometric identity<br />
$$\angle P_0 O P_1 = \frac{2 \pi}{5}$$ $$OH = r \cos{\frac{2 \pi}{5}} = r \frac{\sqrt{5}-1}{4} $$<br />
<br />
So, in order to construct the pentagon, we need to construct the line segment $OH$ of length $\frac{\sqrt{5}-1}{4} r$.<br />
<br />
<br />
Similarly, in order to construct the regular 17-gon in the picture below, we need to construct the line segment $OH = r \cos{\frac{2 \pi}{17}}$.<br />
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiYfGuMpI3H0zhmNz55RVntb9he4lMNenHCxHljHbTKFG2432nNicGKSej6m7sTE_gvypFJ5-Ceg_rpvl98gKWqdSQGHAWD3m2JPjocHYYvXl83ThsHmvnkUrsyJr9yXBAXLd5Rf7zkRPTJ/s1600/vuontoan_mathgarden_ca3.png" height="390" width="400" /></div>
<br />
In 1796, at the age of 19, the mathematician Gauss made a breakthrough in solving the <a href="http://mathgardenblog.blogspot.com/2014/03/regular-polygon-construction.html" target="_blank">regular polygon construction problem</a>. Using modulo arithmetic, Gauss calculated<br />
$$\cos{\frac{2 \pi}{17}} = \frac{1}{16} \left( -1 + \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2 \sqrt{17 + 3 \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2 \sqrt{34 + 2 \sqrt{17}}} \right)$$<br />
and showed that a regular 17-gon is constructible.<br />
<br />
With the above two examples, have you spotted out any pattern yet?<br />
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<br />
<span style="color: #0b5394; font-size: large;"><br /></span>
<span style="color: #0b5394; font-size: large;">Constructible numbers</span><br />
<br />
<span style="color: purple; font-size: x-large;">L</span>et us now define the concept of "<span style="color: purple;">constructible numbers</span>"<br />
<blockquote class="tr_bq">
<span style="color: purple;">A number $\alpha$ is called a <b>constructible number</b> if $\alpha$ is obtained from integers by addition, subtraction, multiplication, division and taking square root.</span></blockquote>
<br />
For example,<br />
$$\frac{3}{5} \mbox{ is a constructible number,} $$<br />
$$\cos \frac{2 \pi}{5}=\frac{\sqrt{5}-1}{4} \mbox{ is a constructible number,} $$<br />
$$\frac{\sqrt{\sqrt{2} + 5 \sqrt{3} - 2}}{\sqrt{\frac{2}{5}} + 1} \mbox{ is a constructible number.} $$<br />
<br />
We have the following theorem<br />
<blockquote class="tr_bq">
<span style="color: purple;"><b>Fundamental theorem of compass-and-straightedge construction. </b>Given a line segment of length $r$, and $\alpha$ is a <b>constructible number</b>, then by compass and straightedge, we can construct a line segment of length $\alpha \, r$.</span></blockquote>
<br />
We will prove this theorem by showing a construction algorithm.<br />
<br />
<br />
<br />
<span style="color: #0b5394; font-size: large;">Construction algorithm</span><br />
<br />
<span style="color: purple; font-size: x-large;">G</span>iven three line segments of length $r$, $a r$ and $b r$. Using compass and straightedge, we can construct<br />
<br />
<ul>
<li>a line segment of length $(a+b)r$</li>
</ul>
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjp7BPHeq_IO6PJsIihxL3CyeXZduxhFczw2HLz1z6TaLivxPYtJLDWY8RayxzfETc3KSqn4doqWjpkafZ2_nti5qTJbioN-5sANCeAr2CJqKs_fZpdqL6LRBA7GhW7FJk-EAlc_2QeZtAh/s1600/vuontoan_mathgarden_ca4a.png" height="80" width="400" /></div>
<br />
<br />
<ul>
<li>a line segment of length $(a−b)r$</li>
</ul>
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqmE388V6pH2_Hpf2bqhLekWDvqojXMs3kvqnB5MFTjZROwcUA-lgmBOP6-TDsOSWqxywmLNWXGhmYTQXc7P4J3pMuwlsUp8Cl-mMnF-3qZ7QrN_jrZD4waB-S4UbSbBjpEYOqBuiR-iUs/s1600/vuontoan_mathgarden_ca4.png" height="91" width="320" /></div>
<br />
<ul>
<li>a line segment of length $(ab) r$ </li>
</ul>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi55rZsKACpWTAHh1EHMDadKrOiKeeUPWYArAgdd3q5LT7v1gLTkK-Vb3VyY9zPZ9x1D1sFYO_6JfO4aq7VWqLGpKTyj8SReTughxSn3vihdHGHbyhBCLsm5UUTHHAcuHgV8XsjrjVTLORl/s1600/vuontoan_mathgarden_ca5b.png" height="166" style="margin-left: auto; margin-right: auto;" width="320" /></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: small; text-align: left;">$AB$ is parallel to $CD$: $\frac{OD}{OC} = \frac{OA}{OB} = a \to OD = abr$</span></td></tr>
</tbody></table>
<br />
<ul>
<li>a line segment of length $(a/b) r$ </li>
</ul>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh9ZyCdYHyzWtmrnF4aeKh-ujyuWQJY9_rlt1upHEKunZ61n1u1NR0uveNOBNpcY8_XOvXK17n_KxwDjA2WCYuMky2NURtUggGHtehQmElYHwdKlwE7iE4nnmouTejB92Mb6Dx_TwM0zRL8/s1600/vuontoan_mathgarden_ca5a.png" height="166" style="margin-left: auto; margin-right: auto;" width="320" /></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: small; text-align: left;">$AB$ is parallel to $CD$: $\frac{OD}{OC} = \frac{OA}{OB} = \frac{a}{b} \to OD = \frac{a}{b}r$</span></td></tr>
</tbody></table>
<br />
<ul>
<li>a line segment of length $(\sqrt{ab}) r$ </li>
</ul>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg3IMhV14D1Z7bu7J2QUb08pVyoKGUkpIwGZSTWr18-I4CzOPbxMd_wzFKpmRrY3mUwt43bo3iYRuqQ4CbLb0LSwSNnOvbeArTn9GG-N54mJyIBUzRetKhEUEXrs5ab1funviLoKeHzfpIp/s1600/vuontoan_mathgarden_ca6.png" height="182" style="margin-left: auto; margin-right: auto;" width="320" /></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: small; text-align: left;">by the </span><i style="font-size: medium; text-align: left;"><span style="color: purple;"><a href="http://mathgardenblog.blogspot.com/2014/08/right-triangle-altitude-theorem.html" target="_blank">Right Triangle Altitude Theorem</a></span></i><span style="font-size: small; text-align: left;">, $HC^2 = HA \times HB = ab r^2 \to HC = \sqrt{ab}r$</span></td></tr>
</tbody></table>
<br />
<br />
With the above construction algorithm, if $\alpha$ is a <i>constructible number</i> then from a line segment of length $r$, we can show step by step how to construct a line segment of length $\alpha r$.<br />
<br />
<br />
<span style="color: purple; font-size: x-large;">L</span>et us stop here for now. Next time, we will learn about <i>Fermat's little theorem</i>, <i>modulo cycle</i> and many other exciting stuff and then we will learn how to derive Gauss trigonometric identity<br />
$$\cos{\frac{2 \pi}{17}} = \frac{1}{16} \left( -1 + \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2 \sqrt{17 + 3 \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2 \sqrt{34 + 2 \sqrt{17}}} \right)$$<br />
<br />
Hope to see you again then!<br />
<br />
<br />
<i>Homework.</i><br />
<br />
1. Prove that $$\cos \frac{2 \pi}{5}=\frac{\sqrt{5}-1}{4}$$<br />
<br />
2. Use <a href="http://mathgardenblog.blogspot.com/2013/01/complex2.html" target="_blank">Moivre formula</a> to find a polynomial of integer coefficients such that $\cos \frac{2 \pi}{17}$ is one of its roots.<br />
<br />
3. <i>We will use the notation $(PQ)$ to denote the line $PQ$, and notation $(P, PQ)$ to denote the circle centre at $P$ and of radius $PQ$.</i><br />
<br />
<b>Fundamental theorem of compass-and-straightedge construction.</b><br />
<br />
Given a line segment $AB = r$. A point $M$ on the plane is said to be <span style="color: purple;"><b>constructible</b></span> from $AB$ if there exists a sequence of points $X_1$, $X_2$, ..., $X_n$ such that<br />
<ul>
<li>$X_1 = A$, $X_2 = B$ and $X_n = M$</li>
<li>For any $3 \leq i \leq n$, the point $X_i$ is either the intersection point of two lines $(X_{k_1} X_{k_2})$ and $(X_{k_3} X_{k_4})$, or an intersection point of the line $(X_{k_1} X_{k_2})$ with the circle $(X_{k_3}, X_{k_3} X_{k_4})$, or an intersection point of the circle $(X_{k_1}, X_{k_1} X_{k_2})$ with the circle $(X_{k_3}, X_{k_3} X_{k_4})$, for some $1\leq k_1, k_2, k_3, k_4 < i$.</li>
</ul>
<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhle2DIN0lcNt-h-mxsOa3lJIm1kc8hdyNzlP2xFHw_aKZIiKGvwgNq_dthOp7KNbyTJSB5uPbCA11u5g-N5p0krdWZQOnQwMlSluJg7IU3zyuudcEfxSKnPuuwLX3yziU321LbQQnbsf0/s1600/vuontoan_mathgarden_ca7.png" height="239" width="320" /></div>
<br />
Draw a Cartesian coordinate system $Axy$ where the point $B$ is on $Ax$. Each point $M$ on the plane will have a coordinate $M(x,y)$.<br />
<br />
Prove that a point $M(x,y)$ is <i>constructible</i> from $AB$ <b>if and only if </b>$x = \alpha_1 r$ and $y = \alpha_2 r$ where $\alpha_1$ and $\alpha_2$ are <i>constructible numbers</i>.<br />
<br />
4. Write about one of your favourite <a href="http://mathgardenblog.blogspot.com/2013/10/basic-compass-straightedge-construction.html" target="_blank">construction problems</a>. Use <a href="http://google.com/">google.com</a> to learn more about that problem.<br />
<br />
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8t1XOzleL9edTwqKbkXS0HVfn7YMGT8e31kdxU4CbzH_GIEd3J71GOpLB0T-eBvbBmVpFanqkKTBYr660Sb4Znl1_zL5J8XpqwZqtRLAvvAiVqfB5_rb06wVe3gkawlJ-UtU4Q2yNwLDl/s1600/wolf.jpg" height="480" width="640" /></div>
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<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-19620447376366491732014-08-21T09:16:00.001+10:002014-09-26T11:06:54.264+10:00Right Triangle Altitude Theorem<div class="separator" style="clear: both; text-align: center;">
<br class="Apple-interchange-newline" /><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgC_p2WRyD2X8GBQxSUr6-53jomcpUwCW5IXrd32mhTcFDvRlxY2sy9LM6VNxBQxJvR6iTH15APZPDm8KdwrQ8dgxiB9zAlsQofFrAhw70bsiTi3IO3JCoFnH11rUs9RuogIvMXO-qjRcs/s1600/matthewptg.png" height="276" width="400" /></div>
<div>
<br /></div>
<span style="color: purple; font-size: x-large;">T</span>oday we will learn about <i><span style="color: purple;">Right Triangle Altitude Theorem</span></i> and use it to derive <span style="color: purple;"><i>Pythagorean Theorem</i></span>.<br />
<div>
<br /></div>
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<span style="color: #0b5394; font-size: large;">Right Triangle Altitude Theorem</span><br />
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<span style="color: purple; font-size: x-large;">W</span>e know that two similar triangles have three pairs of <i>equal angles</i> and three pairs of <i>proportional sides</i>. If someone asks you what your favourite example of similar triangles is, what would you say?<br />
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For me, it has to be the <i><span style="color: purple;">Right Triangle Altitude Theorem</span></i>. The theorem is constructed as follows.<br />
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First, we draw a right triangle $ABC$ ($\angle B$ is the right angle). Next, we draw the altitude $BH$ and divide the triangle $ABC$ into two smaller right triangles $BHA$ and $BHC$. Do you see that these two smaller triangles, $BHA$ and $BHC$, are similar to the original triangle $ABC$?<br />
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We can draw the picture on a paper and use scissor to cut out the shapes as follows.<br />
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Let us look at the big triangle $ABC$ and the smaller triangle $AHB$. We can see right away that they have a pair of equal angles $\angle B = \angle H = 90^{o}$. In addition to that, they also share a common angle $A$. So these two triangles are similar.<br />
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The equal ratio between the sides of the two triangles $ABC$ and $AHB$: $$\frac{{\bf AB}}{{\bf AH}} = \frac{BC}{HB} = \frac{{\bf AC}}{{\bf AB}}$$ gives us this identity $$AB^2 = AH \times AC$$<br />
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Again, the equal ratio between the sides of the two triangles $ABC$ and $BHC$: $$\frac{AB}{BH} = \frac{{\bf BC}}{{\bf HC}} = \frac{{\bf AC}}{{\bf BC}}$$ gives us $$CB^2 = CH \times CA$$<br />
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Finally, the ratio for the two triangles $AHB$ and $BHC$ $$\frac{{\bf AH}}{{\bf BH}} = \frac{{\bf HB}}{{\bf HC}} = \frac{AB}{BC}$$ gives us the third identity $$HB^2 = HA \times HC$$<br />
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The <i><span style="color: purple;">Right Triangle Altitude Theorem</span></i> consists of these three identities. Let us call them "t<i>he left identity</i>", "<i>the right identity</i>" and "<i>the middle identity</i>".<br />
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<blockquote class="tr_bq">
<span style="color: purple;"><b>Right Triangle Altitude Theorem:</b></span></blockquote>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj2joo15KLp6hn_rKbmtfRu3BCA_PRtlVN4BcJRRtuL8W_JjfmfW82JQ87NXwp7pU-QvLIbCz8tFxhBrmssiqeLXj57ELEb41uJ34j029Osabq8lExufl6w2ZexlOXmp7qkDgbedhzmxynX/s1600/vuontoan_mathgarden_dunghinh5.png" height="272" width="640" /></td></tr>
<tr><td class="tr-caption" style="font-size: 13px;"><i style="font-size: medium; text-align: start;">left identity:</i> $AB^2 = AH \times AC$<br />
<i style="font-size: medium; text-align: start;">right identity:</i> $CB^2 = CH \times CA$<br />
<i style="font-size: medium; text-align: start;">and middle identity:</i> $HB^2 = HA \times HC$</td></tr>
</tbody></table>
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<span style="color: #0b5394; font-size: large;"><br /></span><span style="color: #0b5394; font-size: large;">Pythagorean Theorem</span><br />
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We will now use the <i><span style="color: purple;">Right Triangle Altitude Theorem</span></i> to give a proof of the <span style="color: purple;"><i>Pythagorean Theorem</i></span>.<br />
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<blockquote class="tr_bq">
<span style="color: purple;"><b>Pythagorean Theorem:</b> In a right triangle $ABC$ with the right angle $B$ we have $$AB^2 + BC^2 = AC^2$$</span></blockquote>
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<i style="color: purple;">Pythagorean Theorem </i>says that the two squares $ABXY$ and $BCPQ$ have a total area equal to the big square $CAIJ$.<br />
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<li>Using the <i><span style="color: purple;">left identity</span></i> $AB^2 = AH \times AC = AH \times AI$, we can see that the <i><span style="color: #8e7cc3;">square</span></i> $ABXY$ has the same area as the <span style="color: #8e7cc3;"><i>rectangle </i></span>$AHMI$.</li>
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<li>The <i><span style="color: purple;">right identity</span></i> $CB^2 = CH \times CA = CH \times CJ$ shows that the <span style="color: #3d85c6;"><i>square</i></span> $BCPQ$ has the same area as the <span style="color: #3d85c6;"><i>rectangle </i></span>$CHMJ$.</li>
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So indeed, the <i>big square</i> $CAIJ$ has area equal to the sum of <i>two smaller squares</i> $ABXY$ and $BCPQ$, and we have obtained the <span style="color: purple;"><i>Pythagorean Theorem</i></span>.<br />
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<span style="color: purple; font-size: x-large;">L</span>et us stop here for now. In the next post, we will explore Gauss' construction of a regular 17-polygon. We will see the reason why the <span style="color: purple;"><i>Right Triangle Altitude Theorem</i></span> makes it possible for this construction. Hope to see you again then.<br />
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<i>Homework.</i><br />
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1. Prove that if two triangles have two pairs of equal angles then all their three pairs of angles are equal.<br />
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2. Write about your favourite example of similar triangles.<br />
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3. Use trigonometry to prove the <span style="color: purple;"><i>Right Triangle Altitude Theorem</i></span>.<br />
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4. Given three line segments of length $r$, $r a$ and $r b$. Use compass and straightedge to construct<br />
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<li>a line segment of length $r (a+b)$</li>
<li>a line segment of length $r (a-b)$</li>
<li>a line segment of length $r (ab)$</li>
<li>a line segment of length $r (a/b)$</li>
<li>a line segment of length $r \sqrt{ab}$</li>
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5. Given a line segment of length $r$. Using compass and straightedge, what kind of line segments can we construct?<br />
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6. Go to <a href="http://google.com/">google.com</a> and search about compass and straightedge construction.<br />
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<br />Unknownnoreply@blogger.comtag:blogger.com,1999:blog-8356875209547965420.post-91668516605403618752014-07-19T13:19:00.000+10:002014-09-26T10:53:59.448+10:00Go Bugs Go!<div class="separator" style="clear: both; text-align: center;">
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<span style="color: purple; font-size: x-large;">I</span>n our previous post, we learned how to <a href="http://mathgardenblog.blogspot.com/2014/06/construct-15gon.html" target="_blank">construct a regular polygon with 15 sides</a> using <a href="http://mathgardenblog.blogspot.com/2013/10/basic-compass-straightedge-construction.html" target="_blank">compass and straightedge</a>. The interesting part of the construction is that it is based on the integer solutions of the Diophantine equation $$3x + 5y = 1.$$<br />
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<span style="color: purple; font-size: x-large;">W</span>e have also learned about the <a href="http://mathgardenblog.blogspot.com/2014/04/measuring-liquid-puzzle.html" target="_blank">measuring liquid puzzle</a> - <i>"how to measure out exactly 1 liter of water using a 3-liter jug and a 5-liter jug."</i> Again, we showed that the answers to the puzzle correspond to integer solutions of the equation $$3x + 5y = 1.$$<br />
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<span style="color: #0b5394; font-size: large;"><br /></span><span style="color: purple; font-size: x-large;">T</span>oday, let us consider a third problem that is related to the equation $$3x + 5y = 1.$$<br />
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We will state the problem and pose some questions for the reader to think about.<br />
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<blockquote class="tr_bq">
<span style="color: purple;"><b>A game of running bug.</b> The rule of the game is, each time we can move the bug either <b>3 steps</b> or <b>5 steps</b> to the <b>left</b> or to the <b>right</b>.</span></blockquote>
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<i><span style="color: purple; font-size: x-large;">Q</span>uestions are:</i><br />
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<li>How can we move the bug from a point $A$ to a point next to it? Is there a relationship between the moves and the equation $3x+5y=1$?</li>
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<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMa4ONWhmno8Uzgd0sr4ff_0-Zugk1MPtYc8XhxPTm19EQ2JglK65pkQWVYN4GgNV1YBFWRiCrUrXgAWefQ_X_CVgs30cMkbAujAGJEmCzdRxqos-kwDg1-mUHqcKz6LbYwMfZfslJk3kU/s1600/vuontoan_mathgarden_bug3.png" height="98" width="640" /></div>
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<li>Prove that, from a point $A$, after a number of times, we can move the bug to an arbitrary point $B$ on the line.</li>
</ul>
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<li>Instead of letting the bug running on a line, we change the game so that the bug is moving on a circle. Is there a relationship between this game and the <a href="http://mathgardenblog.blogspot.com/2014/06/construct-15gon.html" target="_blank">construction problem</a>?</li>
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Happy thinking and see you again in our next post!</div>
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