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Difference between revisions of "Binomial Theorem"

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<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>
  
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This may be easily shown for the [[integer]]s: <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>.
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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>.
  
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==Generalization==
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==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>,
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<center><math>(a+b)^r = \sum_{k=0}^{\infty}\binom{r}{k}a^{r-k}b^k</math></center>
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<center><math>(a+b)^r = \sum_{k=0}^{\infty}\binom{r}{k}a^{r-k}b^k</math>.</center>
  
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'''Proof'''
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===Proof===
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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> is  
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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  
  
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<math>(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</math>.
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<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 $a$, $b$, 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.

See also

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