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Difference between revisions of "Partial fraction decomposition"
m (proofreading)
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== Examples ==
 
== Examples ==
−
One of the uses of decomposition by partial fractions transform an otherwise difficult (or possibly infinite!) sum to [[telescope]].
+
One of the uses of decomposition by partial fractions transforms an otherwise difficult (or possibly infinite!) sum to [[telescope]].
 
Consider the sum
 
Consider the sum
−
<math>\sum_{n=1}^{99}\frac{1}{n(n+1)}=\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{99*100} </math>
+
<math>\sum_{n=1}^{99}\frac{1}{n(n+1)}=\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{99*100} </math>.
−
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>
+
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>.
  
 
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.
 
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 17:38, 19 June 2006

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

One of the uses of decomposition by partial fractions transforms an otherwise difficult (or possibly infinite!) sum to telescope. Consider the sum $\sum_{n=1}^{99}\frac{1}{n(n+1)}=\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{99*100}$. We can rewrite the general term $\frac{1}{n(n+1)}=\frac{a}{n}+\frac{b}{n+1}$ for some constants a and b (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 $\sum_{n=1}^{99}\frac{1}{n(n+1)}=\sum_{n=1}^{99}\left(\frac{1}{n}+\frac{-1}{n+1}\right)=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{99}+\frac{1}{100}\right)=1-\frac{1}{100}=\frac{99}{100}$.

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.

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