Difference between revisions of "Binomial Theorem"
(→Generalization: Added Proof) |
|||
Line 2: | Line 2: | ||
<center><math>(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k</math></center> | <center><math>(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k</math></center> | ||
− | This | + | where <math>\binom{n}{k} = \frac{n!}{k!(n-k)!}</math> is a [[binomial coefficient]]. This result has a nice [[combinatorial proof]]: <math>(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}</math>. Repeatedly using the [[distributive property]], we see that for a term <math>a^m b^{n-m}</math>, we must choose <math>m</math> of the <math>n</math> terms to contribute an <math>a</math> to the term, and then each of the other <math>n-m</math> terms of the product must contribute a <math>b</math>. Thus, the coefficient of <math>a^m b^{n-m}</math> is <math>\binom{m}{n}</math>. Extending this to all possible values of <math>m</math> from <math>0</math> to <math>n</math>, we see that <math>(a+b)^n = \sum_{k=0}^{n}{\binom{n}{k}}\cdot a^k\cdot b^{n-k}</math>. |
− | == | + | ==Generalizations== |
The Binomial Theorem was generalized by [[Isaac Newton]], who used an [[infinite]] [[series]] to allow for complex [[exponent]]s: For any [[real]] or [[complex]] <math>a</math>, <math>b</math>, and <math>r</math>, | The Binomial Theorem was generalized by [[Isaac Newton]], who used an [[infinite]] [[series]] to allow for complex [[exponent]]s: For any [[real]] or [[complex]] <math>a</math>, <math>b</math>, and <math>r</math>, | ||
− | <center><math>(a+b)^r = \sum_{k=0}^{\infty}\binom{r}{k}a^{r-k}b^k</math></center> | + | <center><math>(a+b)^r = \sum_{k=0}^{\infty}\binom{r}{k}a^{r-k}b^k</math>.</center> |
− | + | ===Proof=== | |
− | Consider the function <math>f(b)=(a+b)^r</math> for constants <math>a,r</math>. It is easy to see that <math>\frac{d^k}{db^k}f=r(r-1)\cdots(r-k+1)(a+b)^{r-k}</math>. Then, we have <math>\frac{d^k}{db^k}f(0)=r(r-1)\cdots(r-k+1)a^{r-k}</math>. So, the Taylor | + | Consider the function <math>f(b)=(a+b)^r</math> for constants <math>a,r</math>. It is easy to see that <math>\frac{d^k}{db^k}f=r(r-1)\cdots(r-k+1)(a+b)^{r-k}</math>. Then, we have <math>\frac{d^k}{db^k}f(0)=r(r-1)\cdots(r-k+1)a^{r-k}</math>. So, the [[Taylor series]] for <math>f(b)</math> centered at <math>0</math> is |
− | < | + | <cmath>(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.</cmath> |
==Usage== | ==Usage== |
Revision as of 13:30, 6 August 2008
The Binomial Theorem states that for real or complex , , and non-negative integer ,
where is a binomial coefficient. This result has a nice combinatorial proof: . Repeatedly using the distributive property, we see that for a term , we must choose of the terms to contribute an to the term, and then each of the other terms of the product must contribute a . Thus, the coefficient of is . Extending this to all possible values of from to , we see that .
Contents
[hide]Generalizations
The Binomial Theorem was generalized by Isaac Newton, who used an infinite series to allow for complex exponents: For any real or complex , , and ,
Proof
Consider the function for constants . It is easy to see that . Then, we have . So, the Taylor series for centered at is
Usage
Many factorizations involve complicated polynomials with binomial coefficients. For example, if a contest problem involved the polynomial , one could factor it as such: . It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients.