Newton polynomial interpolation

Today, we will learn about polynomial interpolation.

Suppose we have the following polynomial $$P(x) = 2x^2 - 3x + 3$$

Let us calculate the values of the polynomial $P(x)$ for a few values of $x$ as follows $$P(1) = 2 - 3 + 3 = 2,$$ $$P(2) = 8 - 6 + 3 = 5,$$ $$P(3) = 18 - 9 + 3 = 12, \dots$$

Now, suppose that we are given the following information $$P(1) = 2, ~~P(2) = 5, ~~P(3) = 12,$$ can we reconstruct the polynomial $P(x)$?

The answer is, yes, we can. An interpolation formula enables us to reconstruct the polynomial $P(x)$ based on its values. Today, we will learn about Newton's interpolation formula, and in the next post, we will cover Lagrange's interpolation.

1 = 2012 = 2013

In previous posts, we learned about mathematical induction and we used induction to solve some problems. We can see that mathematical induction is a useful technique in problem solving. Today, we will consider two induction proofs that lead to a wrong result that $$1 = 2012 = 2013$$ Let us know if you can identify the wrong steps in the proofs.

Binomial identity

As we are learning about mathematical induction, in this post, we are going to use induction to prove the following:

• formula for the Pascal's triangle $$p_{n,k} = {n \choose k} = \frac{n!}{k! (n-k)!}$$
• the binomial identity $$(x+y)^n = x^n + {n \choose 1} x^{n-1} y + {n \choose 2} x^{n-2} y^2 + \dots + {n \choose {n-2}} x^{2} y^{n-2} + {n \choose {n-1}} x y^{n-1} + y^n$$

Mathematical induction III

Today, we will solve some more problems using mathematical induction.

Problem 7. Observe that $$\cos 2 \alpha = 2 \cos^2 \alpha - 1$$

Prove that we can write $\cos n\alpha$ as a polynomial of $\cos \alpha$.

Mathematical induction II

Today we will use induction to solve some more problems. 

Problem 4. Prove that $$1 \times 2 \times 3 + 2 \times 3 \times 4 + \dots + n (n+1)(n+2) = \frac{1}{4} n(n+1)(n+2)(n+3).$$

Mathematical induction

Today, we will learn about mathematical induction. We usually use induction to prove a certain statement to be correct for all natural numbers.

Let us use $P(n)$ to denote a statement that involves a natural number $n$. To prove that $P(n)$ is correct for all natural number $n$, an induction proof will have the following steps

Step 1: is called the initial step. We will prove that the statement $P(n)$ is correct for the case $n=0$.

Step 2: is called the induction step. This is the most important step. In this step,

• we assume that for any $0 \leq n \leq k$, the statement $P(n)$ is correct;
• with this assumption, we will prove that the statement $P(n)$ is also correct for the case $n=k+1$.

With these two steps, by the mathematical induction principle, we conclude that the statement $P(n)$ must be correct for all natural number $n$.

Pascal's triangle

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

Pascal's triangle

Pythagorean triples

In geometry, there is a well known theorem, called the Pythagorean Theorem, which says that in a right triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides. 

Pythagorean Theorem: $BC^2 = AB^2 + AC^2$

That is why we call the equation $$x^2 + y^2 = z^2$$ the Pythagorean equation and its solution $(x,y,z)$ is called a Pythagorean triple. Of course, we only consider integer solutions.

Today, we will solve the Pythagorean equation and show that this equation has an infinite number of solutions.

Wilson's Theorem

Today, we will study Wilson's Theorem - a theorem concerning prime numbers. Wilson's theorem says that if $p$ is a prime number then the number $(p-1)! + 1$ will be divisible by $p$.

Here, the notation $n!$ denotes $$n! = 1 \times 2 \times 3 \times \dots \times n.$$

For example,

• $1! + 1 = 2$ is divisible by $2$
• $2! + 1 = 3$  is divisible by $3$
• $4! + 1 = 25$  is divisible by $5$
• $6! + 1 = 721$  is divisible by $7$

Some problems on prime numbers

In previous posts, we have learned about prime numbers, and we know from Euclid's Theorem that there exists an infinite number of primes. In this post, we continue to look at prime numbers. Leading mathematicians like Fermat, Euler, Gauss were all fascinated by prime numbers. There are many problems about primes that even the statements are simple, but they still remain unsolved even to this day.

Euclid's theorem on prime numbers

Continuing with our story about prime numbers, today we will prove that there exists an infinite number of primes. This is called the Euclid's theorem on prime numbers. This theorem has a very simple proof but it is probably one of the most beautiful proofs ever in mathematics.

Prime numbers

Today we will learn about prime numbers - a basic building block of arithmetic.

A prime number is a natural number greater than 1 and has no divisors other than 1 and itself. For example, the numbers 2, 3, 5, 7, 11, 13 are prime numbers. The number 9 is not a prime number because it is divisible by 3. The number 2012 is not a prime number because it is divisible by 2.