Binomial Theorem

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Theorem

Integral Index

Let $X$ be one of the set of numbers $\N, \Z, \Q, \R, \C$.


Let $x, y \in X$.

Then:

$\displaystyle \forall n \in \Z_+: \left({x+y}\right)^n = \sum_{k=0}^n {n\choose k}x^{n-k}y^k$


Ring Theory

Let $\left({R, +, \odot}\right)$ be a ringoid such that $\left({R, \odot}\right)$ is a commutative semigroup.


Let $n \in \Z: n \ge 2$. Then:

$\displaystyle \forall x, y \in R: \odot^n \left({x + y}\right) = \odot^n x + \sum_{k=1}^{n-1} \binom n k \left({\odot^{n-k} x}\right) \odot \left({\odot^k y}\right) + \odot^n y$

where $\displaystyle \binom n k = \frac {n!} {k! \left({n-k}\right)!}$ (see Binomial Coefficient).


If $\left({R, \odot}\right)$ has an identity element, then:

$\displaystyle \forall x, y \in R: \odot^n \left({x + y}\right) = \sum_{k=0}^n \binom n k \left({\odot^{n-k} x}\right) \odot \left({\odot^k y}\right)$


General Binomial Theorem

Let $\alpha \in \R$ be a real number.

Let $x \in \R$ be a real number such that $\left|{x}\right| < 1$.


Then:

$\displaystyle \left({1 + x}\right)^\alpha = \sum_{n=0}^\infty \frac {\prod \limits_{k=0}^{n-1} \left({\alpha - k}\right)} {n!} x^n$


That is:

$\displaystyle \left({1 + x}\right)^\alpha = 1 + \alpha x + \frac {\alpha \left({\alpha - 1}\right)} {2!} x^2 + \frac {\alpha \left({\alpha - 1}\right) \left({\alpha - 2}\right)} {3!} x^3 + \cdots$


Proof

Proof for Integral Index

Base Case

For $n = 0$ we have:

$\displaystyle \left({x+y}\right)^0 = 1 = {0\choose 0}x^{0-0}y^0 = \sum_{k=0}^0 {0\choose k}x^{0-k}y^k$

Therefore the base case holds.


Inductive Hypothesis

This is our inductive hypothesis:

$\displaystyle \left({x+y}\right)^j = \sum_{k=0}^j {j\choose k}x^{j-k}y^k$ for all $j \ge 1$


Inductive Step

This is our inductive step:

\(\displaystyle \) \(\displaystyle \left({x+y}\right)^{n+1}\) \(=\) \(\displaystyle \left({x+y}\right) \left({x+y}\right)^n\) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle x \sum_{k=0}^n {n \choose k} x^{n-k}y^k + y \sum_{k=0}^n {n \choose k} x^{n-k} y^k\) \(\displaystyle \)          by the inductive hypothesis          
\(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \sum_{k=0}^n {n \choose k} x^{n+1-k}y^k + \sum_{k=0}^n {n \choose k} x^{n-k} y^{k+1}\) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle {n \choose 0} x^{n+1} + \sum_{k=1}^n {n \choose k} x^{n+1-k} y^k + {n \choose n} y^{n+1} + \sum_{k=0}^{n-1} {n \choose k} x^{n-k} y^{k+1}\) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle x^{n+1} + y^{n+1} + \sum_{k=1}^n {n \choose k} x^{n+1-k} y^k + \sum_{k=0}^{n-1} {n \choose k} x^{n-k} y^{k+1}\) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle {n+1 \choose 0} x^{n+1} + {n+1 \choose n+1} y^{n+1} + \sum_{k=1}^n {n \choose k} x^{n+1-k} y^k + \sum_{k=1}^n {n \choose k-1} x^{n + 1 - k} y^k\) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle {n+1 \choose 0} x^{n+1} + {n+1 \choose n+1} y^{n+1} + \sum_{k=1}^n \left({{n \choose k} + {n \choose k-1}}\right) x^{n + 1 - k} y^k\) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle {n+1 \choose 0} x^{n+1} + {n+1 \choose n+1} y^{n+1} + \sum_{k=1}^n {n+1 \choose k} x^{n + 1 - k} y^k\) \(\displaystyle \)          Pascal's Rule          
\(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \sum_{k=0}^{n+1} {n+1 \choose k}x^{n + 1 - k} y^k\) \(\displaystyle \)                    

And so we are done by the Principle of Mathematical Induction.

$\blacksquare$


Proof for Ring Theory

The proof for the Ring Theory version follows the same strategy.


Also see


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