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Difference between revisions of "Partial fraction decomposition"

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== Examples ==
 
== Examples ==
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One of the uses of decomposition by partial fractions transforms an otherwise difficult (or possibly infinite!) sum to [[telescope]].
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Partial fraction decomposition has several common uses. It allows for much easier integration of rational functions, allowing one to integrate a complicated rational function term by term. Additionally, fraction decomposition can also be used to transform difficult summations into more manageable ones. If used in certain cases, the sum even telescopes into a simple arithmetic problem.
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Consider the sum
 
Consider the sum
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<math>\sum_{n=1}^{99}\frac{1}{n(n+1)}=\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{99*100} </math>.
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<math>\sum_{n=1}^{99}\frac{1}{n(n+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{99\cdot 100} </math>.
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We can rewrite the general term <math>\frac{1}{n(n+1)}=\frac{a}{n}+\frac{b}{n+1}</math> for some constants a and b (I wanted A and B but it wouldn't "parse").  Multiplying both sides of the equation by n(n+1) and solving by substituting n=1, -1, we find that a=1 b=-1.  Hence, the sum can be rewritten <math>\sum_{n=1}^{99}\frac{1}{n(n+1)}=\sum_{n=1}^{99}\left(\frac{1}{n}+\frac{-1}{n+1}\right)=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{99}+\frac{1}{100}\right)=1-\frac{1}{100}=\frac{99}{100}</math>.
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We can rewrite the general term <math>\frac{1}{n(n+1)}=\frac{a}{n}+\frac{b}{n+1}</math> for some constants <math>a</math> and <math>b</math>.  Multiplying both sides of the equation by <math>n(n+1)</math> and solving by substituting <math>n=1, -1</math>, we find that <math>a=1,\  b=-1</math>.  Hence, the sum can be rewritten <math>\sum_{n=1}^{99}\frac{1}{n(n+1)}=\sum_{n=1}^{99}\left(\frac{1}{n}+\frac{-1}{n+1}\right)=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\ldots +\left(\frac{1}{99}-\frac{1}{100}\right)=1-\frac{1}{100}=\frac{99}{100}</math>.
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Partial fraction decomposition has several common uses. It allows for much easier integration of rational functions, allowing one to integrate a complicated rational function term by term. Additionally,  it can be used in summations, causing sums to often telescope or have a much easier form which can be expressed with a closed form. It has several other minor uses, as well.
 

Revision as of 13:56, 16 February 2007

Any rational function of the form $\frac{P(x)}{Q(x)}$ maybe written as a sum of simpler rational functions.

To find the decomposition of a rational function, first perform the long division operation on it. This transforms the function into one of the form $\frac{P(x)}{Q(x)}=S(x) + \frac{R(x)}{Q(x)}$, where $R(x)$ is the remainder term and $\deg {{R}(x)} \leq \deg {{Q}(x)}$.

Next, for every factor $(a_nx^n+a_{n-1}x^{n-1}+\ldots +a_0)^m$ in the factorization of $Q(x)$, introduce the terms

$\frac{A_1x^{n-1}+B_1x^{n-2}+\ldots+Z_1}{a_nx^n+a_{n-1}x^{n-1}+\ldots+a_0}+\frac{A_2x^{n-1}+B_2x^{n-2}+\ldots+Z_2}{(a_nx^n+a_{n-1}x^{n-1}+\ldots+a_0)^2}+\ldots+\frac{A_mx^{n-1}+B_mx^{n-2}+\ldots+Z_m}{(a_nx^n+a_{n-1}x^{n-1}+\ldots+a_0)^m}$

(Note that the variable $Z_i$ has no relation to being the 26th letter in the alphabet.)

Next, take the sum of every term introduced above and equate it to $\frac{R(x)}{Q(x)}$, and solve for the variables $A_i, B_i, \ldots$. Once you solve for all the variables, then you will have the partial fraction decomposition of $\frac{R(x)}{Q(x)}$.


Examples

Partial fraction decomposition has several common uses. It allows for much easier integration of rational functions, allowing one to integrate a complicated rational function term by term. Additionally, fraction decomposition can also be used to transform difficult summations into more manageable ones. If used in certain cases, the sum even telescopes into a simple arithmetic problem.

Consider the sum $\sum_{n=1}^{99}\frac{1}{n(n+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{99\cdot 100}$. We can rewrite the general term $\frac{1}{n(n+1)}=\frac{a}{n}+\frac{b}{n+1}$ for some constants $a$ and $b$. Multiplying both sides of the equation by $n(n+1)$ and solving by substituting $n=1, -1$, we find that $a=1,\ b=-1$. Hence, the sum can be rewritten $\sum_{n=1}^{99}\frac{1}{n(n+1)}=\sum_{n=1}^{99}\left(\frac{1}{n}+\frac{-1}{n+1}\right)=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\ldots +\left(\frac{1}{99}-\frac{1}{100}\right)=1-\frac{1}{100}=\frac{99}{100}$.

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