Sequence - Part 6

Today we will learn about difference operator and use it to prove a fundamental theorem for linear recurrence equations.

A fundamental theorem for linear recurrence equations. Suppose that the characteristic equation can be factored as $$f(x) = a_k x^k + a_{k-1} x^{k-1} + \dots + a_0 = (x - z)^j (b_s x^s + b_{s-1} x^{s-1} + \dots + b_0)$$ and $$f_n = p(n)~z^n,$$ where $p(n)$ is a polynomial of degree less than $j$. Then the sequence $f_n$ satisfies the recurrence equation 
$$a_k f_{n} + a_{k-1} f_{n-1} + \dots + a_1 f_{n-k+1} + a_0 f_{n-k} = 0.$$

Sequence - Part 5

Today we will go through some examples. The first type of examples involves solving a linear recurrence equation to derive a general formula for a sequence. The second type of examples does the reverse process, in which, we are given a formula of a sequence and asked to determine its recurrence equation.

Sequence - Part 4

Today we will learn how to solve a linear recurrence equation and derive a general formula for a sequence. This method will work for all cases, including the case when the characteristic equation has multiple roots.  

Sequence - Part 3

This is the third part of our series on sequences. There will be no new material in this post. The main purpose of this post is to go through some examples so we can get used to the method of solving recurrence equations. If you have not read the previous parts of our series, please read them here: Part 1, Part 2.

Sequence - Part 2

This is the second part of our series on sequences. Please have a careful read of Part 1 before you continue to this post. Today, we will introduce some more terminologies about sequences and we will present a general method to solve a linear recurrence equation.

Sequence - Part 1

Today we will begin our first post of a series on recurrence sequences. The purpose of this series is to show you how to derive a general formula for a sequence that satisfies a linear recurrence equation. We start with some basic operations on sequences. We will learn about addition of sequences and multiplication of a sequence with a number.

A proof of Wilson's Theorem using polynomial interpolation

In the last posts, we have learned about polynomial interpolation through two formulas, the Newton's interpolation formula and the Lagrange's interpolation formula. Both formulas can be used to prove the Wilson's theorem.

In this post, we will present a proof of the Wilson's theorem based on the Newton's interpolation formula. The other proof, that is, proving Wilson's theorem based on the Lagrange's interpolation formula, we leave it as a homework exercise.

Wilson's Theorem is a well known theorem in number theory. It says that if $p$ is prime number then the number $(p−1)!+1$ is a multiple of $p$.

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 $$(x+y)^n = x^n + {n \choose 1} x^{n-1} y + {n \choose 2} x^{n-2} y^2 + \dots + {n \choose {n-2}} x^{2} y^{n-2} + {n \choose {n-1}} x y^{n-1} + y^n$$

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 $7$