Binomial theorem: Difference between revisions
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In [[elementary algebra]], the '''binomial theorem''' is the identity that states that for any non-negative [[integer]] ''n'', | {{subpages}} | ||
In [[elementary algebra]], the '''binomial theorem''' or the binomial expansion is a mechanism by which expressions of the form <math>(x + y)^n</math> can be expanded. It is the identity that states that for any non-negative [[integer]] ''n'', | |||
: <math> (x + y)^n = \sum_{k=0}^n {n \choose k} x^k y^{n-k}, </math> | : <math> (x + y)^n = \sum_{k=0}^n {n \choose k} x^k y^{n-k}, </math> | ||
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where | where | ||
: <math> {n \choose k} = \frac{n!}{k!(n - k)!}. </math> | : <math> {n \choose k} = \frac{n!}{k!(n - k)!} </math> | ||
is a [[binomial coefficient]]. Another useful way of stating it is the following: | |||
<math>(x + y)^n = {n \choose 0} x^n + {n \choose 1} x^{n-1} y + {n \choose 2} x^{n-2} y^2 + \ldots + {n \choose n} y^n</math> | |||
===Pascal's triangle=== | |||
An alternate way to find the binomial coefficients is by using [[Pascal's triange]]. The triangle is built from apex down, starting with the number one alone on a row. Each number is equal to the sum of the two numbers directly above it. | |||
n=0 1 | |||
n=1 1 1 | |||
n=2 1 2 1 | |||
n=3 1 3 3 1 | |||
n=4 1 4 6 4 1 | |||
n=5 1 5 10 10 5 1 | |||
Thus, the binomial coefficients for the expression <math>(x + y)^4</math> are 1, 3, 6, 4, and 1. | |||
==Proof== | |||
One way to prove this identity is by [[mathematical induction]]. | |||
'''Base case''': n = 0 | |||
: <math> (x + y)^0 = \sum_{k=0}^0 {0 \choose k} x^{0-k} y^k = 1 </math> | |||
'''Induction case''': Now suppose that it is true for n : <math> (x + y)^n = \sum_{k=0}^n {n \choose k} x^{n-k} y^k, </math> and prove it for n + 1. | |||
:<math> (x+y)^{n+1} = (x+y)(x+y)^n \,</math> | |||
::::<math> = (x+y) \sum_{k=0}^n {n \choose k} x^{n-k} y^k \,</math> | |||
::::<math> = \sum_{k=0}^n {n \choose k} x^{n+1-k} y^k + \sum_{j=0}^n {n \choose j} x^{n-j} y^{j+1} \,</math> | |||
::::<math> = \sum_{k=0}^n {n \choose k} x^{n+1-k} y^k + \sum_{j=0}^n {n \choose {(j+1) -1}} x^{n-j} y^{j+1} \,</math> | |||
::::<math> = \sum_{k=0}^n {n \choose k} x^{n+1-k} y^k + \sum_{k=1}^{n+1} {n \choose {k -1}} x^{n+1-k} y^k \,</math> | |||
::::<math> = \sum_{k=0}^{n+1} {n \choose k} x^{n+1-k} y^k - {n \choose {n+1}} x^0 y^{n+1}+ \sum_{k=0}^{n+1} {n \choose {k -1}} x^{n+1-k} y^k - {n \choose {-1}}x^{n+1} y^0 \,</math> | |||
::::<math> = \sum_{k=0}^{n+1} \left[ {n \choose k} + {n \choose {k -1}} \right] x^{n+1-k} y^k \,</math> | |||
::::<math> = \sum_{k=0}^{n+1} {{n+1} \choose k} x^{n+1-k} y^k, </math> | |||
and the proof is complete. | |||
== Examples == | |||
These are the expansions from 0 to 6. | |||
<math> \begin{align} | |||
(x + y)^0 &= 1 \\ | |||
(x + y)^1 &= x + y \\ | |||
(x + y)^2 &= x^2 + 2xy + y^2 \\ | |||
(x + y)^3 &= x^3 + 3x^2 y + 3xy^2 + y^3 \\ | |||
(x + y)^4 &= x^4 + 4x^3 y + 6x^2 y^2 + 4xy^3 + y^4 \\ | |||
(x + y)^5 &= x^5 + 5x^4 y + 10x^3 y^2 + 6x^2 y^3 + y^5 \\ | |||
(x + y)^6 &= x^6 + 6x^5 y + 15x^4 y^2 + 20x^3 y^3 + 15 x^2 y^4 + 6xy^5 + y^6 | |||
\end{align} </math> | |||
== Newton's binomial theorem == | |||
There is also '''Newton's binomial theorem''', proved by [[Isaac Newton]], that goes beyond elementary algebra into mathematical analysis, which expands the same sum (''x'' + ''y'')<sup>''n''</sup> as an infinite series when ''n'' is not an integer or is not positive. | There is also '''Newton's binomial theorem''', proved by [[Isaac Newton]], that goes beyond elementary algebra into mathematical analysis, which expands the same sum (''x'' + ''y'')<sup>''n''</sup> as an infinite series when ''n'' is not an integer or is not positive. |
Revision as of 07:34, 9 August 2010
In elementary algebra, the binomial theorem or the binomial expansion is a mechanism by which expressions of the form can be expanded. It is the identity that states that for any non-negative integer n,
where
is a binomial coefficient. Another useful way of stating it is the following:
Pascal's triangle
An alternate way to find the binomial coefficients is by using Pascal's triange. The triangle is built from apex down, starting with the number one alone on a row. Each number is equal to the sum of the two numbers directly above it.
n=0 1 n=1 1 1 n=2 1 2 1 n=3 1 3 3 1 n=4 1 4 6 4 1 n=5 1 5 10 10 5 1
Thus, the binomial coefficients for the expression are 1, 3, 6, 4, and 1.
Proof
One way to prove this identity is by mathematical induction.
Base case: n = 0
Induction case: Now suppose that it is true for n : and prove it for n + 1.
and the proof is complete.
Examples
These are the expansions from 0 to 6.
Newton's binomial theorem
There is also Newton's binomial theorem, proved by Isaac Newton, that goes beyond elementary algebra into mathematical analysis, which expands the same sum (x + y)n as an infinite series when n is not an integer or is not positive.