Formulas and Theorems

Binomial Expansion

For any two real numbers \(x\) and \(y\), and a whole number \(n\) as the power, the binomial expansion, also named binomial theorem, describes the algebraic expansion of powers of a binomial as below

(1)\[\begin{split} \begin{aligned} (x+y)^n & = \begin{pmatrix} n \\ 0 \end{pmatrix} x^{n} y^0 + \begin{pmatrix} n \\ 1 \end{pmatrix} x^{n-1} y + \begin{pmatrix} n \\ 2 \end{pmatrix} x^{n-2} y^2 + \ldots + \begin{pmatrix} n \\ n-1 \end{pmatrix} x y^{n-1} + \begin{pmatrix} n \\ n \end{pmatrix} x^0 y^{n} \\ & = \sum_{k=0}^{n} \begin{pmatrix} n \\ k \end{pmatrix} x^{n-k} y^k, \end{aligned} \end{split}\]

where

(2)\[\begin{split} \begin{pmatrix} n \\ k \end{pmatrix} = \dfrac{n!}{k!(n-k)!} =\dfrac{n(n-1)(n-2)\cdots(n-k+1)}{k!} \end{split}\]

is called the binomial coefficient.

Substituting \(x=1\) into (1), we get

(3)\[ (1+y)^{n}=1+ny+\frac{n(n-1)}{2!} y^{2}+\frac{n(n-1)(n-2)}{3!}y^{3}+\cdots +ny^{n-1}+y^{n}. \]

Newton’s generalised binomial theorem

Instead of strictly using a whole number \(n\) as the power in the binomial formula (1), we can loose the condition to introduce any real number \(s\) as the power. First we replace the whole number \(n\) with a real number \(s\) to get the generalised binomial coefficient

(4)\[\begin{split} \begin{pmatrix} s \\ k \end{pmatrix} =\dfrac{s(s-1)(s-2)\cdots(s-k+1)}{k!}, \end{split}\]

then we can get the generalised bionomial expansion

(5)\[\begin{split} \begin{aligned} (x+y)^{s}&=\sum _{k=0}^{\infty }{s \choose k}x^{s-k}y^{k}\\ &=x^{s}y^{0}+s x^{s-1}y+{\frac {s(s-1)}{2!}}x^{s-2}y^{2}+{\frac {s(s-1)(s-2)}{3!}}x^{s-3}y^{3}+\cdots . \end{aligned} \end{split}\]

Substituting \(x=1\) into (5), we get

(6)\[ (1+y)^s = 1 + s y + \frac{s(s-1)}{2!}y^2+\frac{s(s-1)(s-2)}{3!}y^3 + \cdots \]

Mean Value Theorem

Theorem 1 (Mean Value Theorem)

Let \(f=f(x)\), \(x \in \R\) be continuous over the closed interval \([a, b]\) and differentiable over the open interval \((a, b)\). Then, there exists at least one point \(\xi\in (a,b)\) such that

\[ f'(\xi)=\frac{f(b)-f(a)}{b-a} \]