Difference between revisions of "Binomial Theorem"

Revision as of 01:07, 11 December 2009

The Binomial Theorem states that for real or complex $a$ , $b$ , and non-negative integer $n$ ,

$(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is a binomial coefficient. This result has a nice combinatorial proof: $(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}$ . Repeatedly using the distributive property, we see that for a term $a^m b^{n-m}$ , we must choose $m$ of the $n$ terms to contribute an $a$ to the term, and then each of the other $n-m$ terms of the product must contribute a $b$ . Thus, the coefficient of $a^m b^{n-m}$ is $\binom{m}{n}$ . Extending this to all possible values of $m$ from $0$ to $n$ , we see that $(a+b)^n = \sum_{k=0}^{n}{\binom{n}{k}}\cdot a^k\cdot b^{n-k}$ .

Generalizations

The Binomial Theorem was generalized by Isaac Newton, who used an infinite series to allow for complex exponents: For any real or complex $a$ , $b$ , and $r$ ,

$(a+b)^r = \sum_{k=0}^{\infty}\binom{r}{k}a^{r-k}b^k$ .

Proof

Consider the function $f(b)=(a+b)^r$ for constants $a,r$ . It is easy to see that $\frac{d^k}{db^k}f=r(r-1)\cdots(r-k+1)(a+b)^{r-k}$ . Then, we have $\frac{d^k}{db^k}f(0)=r(r-1)\cdots(r-k+1)a^{r-k}$ . So, the Taylor series for $f(b)$ centered at $0$ is

$(a+b)^k=\sum_{k=0}^\infty \frac{r(r-1)\cdots(r-k+1)a^{r-k}b^k}{k!}=\sum_{k=0}^\infty \binom{r}{k}a^{r-k}b^k.$

Usage

Many factorizations involve complicated polynomials with binomial coefficients. For example, if a contest problem involved the polynomial $x^5+4x^4+6x^3+4x^2+x$ , one could factor it as such: $x(x^4+4x^3+6x^2+4x+1)=x(x+1)^{4}$ . It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients.

In addition, the expansion of a polynomial such as $(x+y)^n$ will have coefficients corresponding to the $nth$ row of Pascal's Triangle. For example, $(x+1)^5$ = $x^5+5x^4+10x^3+10x^2+5x+1$ , and the integers $1$ , $5$ , $10$ , $10$ , $5$ , and $1$ make up the 5th row of Pascal's Triangle.

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 ==Usage==
 Many [[factoring | factorizations]] involve complicated [[polynomial]]s with [[binomial coefficient]]s. For example, if a contest problem involved the polynomial <math>x^5+4x^4+6x^3+4x^2+x</math>, one could factor it as such: <math> x(x^4+4x^3+6x^2+4x+1)=x(x+1)^{4}</math>. It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients.
+In addition, the expansion of a polynomial such as <math>(x+y)^n</math> will have coefficients corresponding to the <math>nth</math> row of [[Pascal's Triangle]]. For example, <math>(x+1)^5</math> = <math>x^5+5x^4+10x^3+10x^2+5x+1</math>, and the integers <math>1</math>, <math>5</math>, <math>10</math>, <math>10</math>, <math>5</math>, and <math>1</math> make up the 5th row of Pascal's Triangle.
 ==See also==

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Difference between revisions of "Binomial Theorem"

Revision as of 01:07, 11 December 2009

Contents

Generalizations

Proof

Usage

See also