Trigonometric multiple-angle formulas

In our previous post, we show a compass-and-straightedge construction of a regular pentagon based on the following trigonometric formula $$\cos{\frac{\pi}{5}} = \frac{1 + \sqrt{5}}{4}.$$
We derive this formula of $\cos{\frac{\pi}{5}}$ by observing that $\cos{\frac{2 \pi}{5}} = -\cos{\frac{3 \pi}{5}}$ and then applying the trigonometric formulas for double angle and triple angle:
$$\cos{2 x} = 2 \cos^2{x} - 1,$$ $$\cos{3 x} = 4 \cos^3{x} - 3 \cos{x}$$ to set up a cubic equation for $\cos{\frac{\pi}{5}}$.

It seems a good occasion now for us to learn about trigonometric multiple-angle formulas. In this post, we will show how to derive formulas for $\sin{nx}$, $\cos{nx}$, $\tan{nx}$ and $\cot{nx}$ using de Moivre's identity of the complex numbers.

Fibonacci sequence and Pascal triangle

In previous post, we used the result of a tile matching puzzle to prove an identity for Fibonacci sequence. Today, we will continue on and prove another identity. We will show that $${2011 \choose 0} + {2010 \choose 1} + {2009 \choose 2}+ {2008 \choose 3}+ \dots + {1007 \choose 1004}+ {1006 \choose 1005} = F_{2012},$$ $${2012 \choose 0} + {2011 \choose 1} + {2010 \choose 2}+ {2009 \choose 3}+ \dots + {1007 \choose 1005}+ {1006 \choose 1006} = F_{2013}.$$
In general, we have the following identity $$\sum_{v+u=n}{v \choose u} = F_{n+1}.$$ Through this identity, we will see an interesting connection between the Fibonacci sequence and the Pascal triangle.

Fibonacci identities

In previous post, we have learnt about Fibonacci sequence and a tile matching puzzle. Today, we will use the result of the tile matching puzzle to prove the following identity on Fibonacci sequence $$F_{n+m+2} = F_{n+1} F_{m+1} + F_{n+1} F_m + F_n F_{m+1}.$$

Like an art painting, a mathematical problem reflects different beauties if we view it at different angles. Take, for example, the Fibonacci identity that we are considering, we can prove this identity by different methods. One method is to use the formula $$F_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right]$$ to calculate the two sides of the identity and make them equal by straightforward algebraic manipulations. Another method is to make use of the tile matching puzzle that we learned in the previous post. I hope that you will find this "combinatorial" method more interesting.

Fibonacci numbers and a tile matching puzzle

Today we will learn about Fibonacci sequence and a tile matching puzzle. We will soon see that the tile matching puzzle seems to be unrelated to the Fibonacci sequence, but it turns out that the solution to the puzzle is, in fact, the Fibonacci sequence!

Let us first introduce the Fibonacci sequence. The Fibonacci sequence $\{F_n\}$ is determined by the following rule: $$F_0 = 0, ~F_1 = 1, ~F_{n+1} = F_n + F_{n-1}.$$

Thus, $$F_0 = 0, ~F_1 = 1, ~F_2 = 1, ~F_3 = 2, ~F_4 = 3, ~F_5 = 5, ~F_6 = 8, ~F_7 = 13, ~F_8 = 21, \dots$$

de Moivre's formula

In previous post we have learned about complex numbers. Today, we will learn about the trigonometric form of a complex number and the famous de Moivre's formula.

Power sum and Wolstenholme's theorem

In the last post, we have learned how to derive formulas for the power summation $$S_k(n) = 1^k + 2^k + 3^k + \dots + n^k.$$

Today we will investigate some divisibility properties of the power summation $S_k(n)$. We will prove that if $p$ is a prime number and $k$ is not divisible by $p-1$ then $$S_k(p-1) = 1^k + 2^k + 3^k + \dots + (p-1)^k = 0 \pmod{p}.$$

We will also investigate the following summation $$S_{-k}(n) = \frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + \dots + \frac{1}{n^k}.$$

There is a theorem in number theory concerning this summation, it is called Wolstenholme's theorem.

Wolstenholme's Theorem. If $p$ is a prime number $>3$ then $$S_{-1}(p-1) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{p-1} ~=_{Q} ~0 \pmod{p^2} $$ and $$S_{-2}(p-1) = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{(p-1)^2} ~=_{Q} ~0 \pmod{p}.$$

We will prove a general result, that is, if $p$ is a prime number and $k$ is not divisible by $p-1$ then $$S_{-k}(p-1) = \frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + \dots + \frac{1}{(p-1)^k} ~=_{Q} ~0 \pmod{p}.$$

Power sum

Today we will learn about how to derive the formula for power summation $$S_k(n) = 1^k + 2^k + 3^k + \dots + n^k .$$


Our main tool is the following binomial identity
$$(x+y)^k = x^k + {k \choose 1} x^{k-1} y + {k \choose 2} x^{k-2} y^2 + {k \choose 3} x^{k-3} y^3 + \dots + {k \choose {k-1}} x y^{k-1} + y^k.$$

Pascal's triangle

Today, we will look at a famous number pattern, the Pascal's number triangle.

Pascal's triangle