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Difference between revisions of "Binomial Theorem"
m (Reverted edits by Marianasinta (talk) to last revision by Orange quail 9)
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===Proof via Induction===
 
===Proof via Induction===
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Given the constants <math>a,b,n</math> are all natural numbers, it's clear to see that <math>(a+b)^{1} = a+b</math>. Assuming that <math>a+b)^{n} = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}</math>, <cmath>(a+b)^{n+1} = (\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k})(a+b)</cmath>
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Given the constants <math>a,b,n</math> are all natural numbers, it's clear to see that <math>(a+b)^{1} = a+b</math>. Assuming that <math>(a+b)^{n} = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}</math>, <cmath>(a+b)^{n+1} = (\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k})(a+b)</cmath>
 
<cmath>=(\binom{n}{0}a^{n}b^{0} + \binom{n}{1}a^{n-1}b^{1} + \binom{n}{2}a^{n-2}b^{2}+\cdots+\binom{n}{n}a^{0}b^{n})(a+b)</cmath>
 
<cmath>=(\binom{n}{0}a^{n}b^{0} + \binom{n}{1}a^{n-1}b^{1} + \binom{n}{2}a^{n-2}b^{2}+\cdots+\binom{n}{n}a^{0}b^{n})(a+b)</cmath>
 
<cmath>=(\binom{n}{0}a^{n+1}b^{0} + \binom{n}{1}a^{n}b^{1} + \binom{n}{2}a^{n-1}b^{2}+\cdots+\binom{n}{n}a^{1}b^{n})
 
<cmath>=(\binom{n}{0}a^{n+1}b^{0} + \binom{n}{1}a^{n}b^{1} + \binom{n}{2}a^{n-1}b^{2}+\cdots+\binom{n}{n}a^{1}b^{n})
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Therefore, if the theorem holds under <math>n+1</math>, it must be valid.
 
Therefore, if the theorem holds under <math>n+1</math>, it must be valid.
 
(Note that <math>\binom{n}{m} + \binom{n}{m+1} = \binom{n+1}{m+1} </math> for <math>m\leq n</math>)
 
(Note that <math>\binom{n}{m} + \binom{n}{m+1} = \binom{n+1}{m+1} </math> for <math>m\leq n</math>)
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===Proof using calculus===
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The [[Taylor series]] for <math>e^x</math> is <cmath>\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots</cmath> for all <math>x</math>.
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Since <math>e^{a+b} = e^ae^b</math>, and power series for the same function are termwise equal, the series at <math>x = a + b</math> is the [[Generating function#Convolutions|convolution]] of the series at <math>x = a</math> and <math>x = b</math>. Examining the degree-<math>n</math> term of each, <cmath>\frac{(a+b)^n}{n!} = \sum_{k=0}^{n} \left( \frac{a^k}{k!} \right) \left( \frac{b^{n-k}}{(n-k)!} \right),</cmath> which simplifies to <cmath>(a+b)^n = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!}a^kb^{n-k}</cmath> for all [[Natural number|natural numbers]] <math>n</math>.
  
 
==Generalizations==
 
==Generalizations==
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==Usage==
 
==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.
 
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.
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==See also==
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*[[Combinatorics]]
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*[[Multinomial Theorem]]
 +
 +
[[Category:Theorems]]
 +
[[Category:Combinatorics]]
 +
[[Category:Algebra]]

Latest revision as of 12:11, 20 February 2024

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. In other words, the coefficients when $(a + b)^n$ is expanded and like terms are collected are the same as the entries in the $n$th row of Pascal's Triangle.

For example, $(a + b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5$, with coefficients $1 = \binom{5}{0}$, $5 = \binom{5}{1}$, $10 = \binom{5}{2}$, etc.

Proof

There are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of mathematical induction. The Binomial Theorem also has a nice combinatorial proof:

We can write $(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}$. 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 the number of ways to choose $m$ objects from a set of size $n$, or $\binom{n}{m}$. Extending this to all possible values of $m$ from $0$ to $n$, we see that $(a+b)^n = \sum_{m=0}^{n}{\binom{n}{m}}\cdot a^m\cdot b^{n-m}$, as claimed.

Similarly, the coefficients of $(x+y)^n$ will be the entries of the $n^\text{th}$ row of Pascal's Triangle. This is explained further in the Counting and Probability textbook [AoPS].

Proof via Induction

Given the constants $a,b,n$ are all natural numbers, it's clear to see that $(a+b)^{1} = a+b$. Assuming that $(a+b)^{n} = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$, \[(a+b)^{n+1} = (\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k})(a+b)\] \[=(\binom{n}{0}a^{n}b^{0} + \binom{n}{1}a^{n-1}b^{1} + \binom{n}{2}a^{n-2}b^{2}+\cdots+\binom{n}{n}a^{0}b^{n})(a+b)\] \[=(\binom{n}{0}a^{n+1}b^{0} + \binom{n}{1}a^{n}b^{1} + \binom{n}{2}a^{n-1}b^{2}+\cdots+\binom{n}{n}a^{1}b^{n}) + (\binom{n}{0}a^{n}b^{1} + \binom{n}{1}a^{n-1}b^{2} + \binom{n}{2}a^{n-2}b^{3}+\cdots+\binom{n}{n}a^{0}b^{n+1})\] \[=(\binom{n}{0}a^{n+1}b^{0} + (\binom{n}{0}+\binom{n}{1})(a^{n}b^{1}) + (\binom{n}{1}+\binom{n}{2})(a^{n-1}b^{2})+\cdots+(\binom{n}{n-1}+\binom{n}{n})(a^{1}b^{n})+\binom{n}{n}a^{0}b^{n+1})\] \[=\binom{n+1}{0}a^{n+1}b^{0} + \binom{n+1}{1}a^{n}b^{1} + \binom{n+1}{2}a^{n-1}b^{2}+\cdots+\binom{n+1}{n}a^{1}b^{n} + \binom{n+1}{n+1}a^{0}b^{n+1}\] \[=\sum_{k=0}^{n+1}\binom{n+1}{k}a^{(n+1)-k}b^{k}\] Therefore, if the theorem holds under $n+1$, it must be valid. (Note that $\binom{n}{m} + \binom{n}{m+1} = \binom{n+1}{m+1}$ for $m\leq n$)

Proof using calculus

The Taylor series for $e^x$ is \[\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots\] for all $x$.

Since $e^{a+b} = e^ae^b$, and power series for the same function are termwise equal, the series at $x = a + b$ is the convolution of the series at $x = a$ and $x = b$. Examining the degree-$n$ term of each, \[\frac{(a+b)^n}{n!} = \sum_{k=0}^{n} \left( \frac{a^k}{k!} \right) \left( \frac{b^{n-k}}{(n-k)!} \right),\] which simplifies to \[(a+b)^n = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!}a^kb^{n-k}\] for all natural numbers $n$.

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

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

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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