Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Partial fraction decomposition"

m (→‎Examples)
m
 
Line 1: Line 1:
−
Any [[rational function]] of the form <math>\frac{P(x)}{Q(x)}</math> maybe written as a sum of simpler rational functions.  
+
Any [[rational function]] of the form <math>\frac{P(x)}{Q(x)}</math> may be 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 <math>\frac{P(x)}{Q(x)}=S(x) + \frac{R(x)}{Q(x)}</math>, where  <math>R(x)</math> is the remainder term and <math>\deg {{R}(x)} \leq \deg {{Q}(x)}</math>.
 
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 <math>\frac{P(x)}{Q(x)}=S(x) + \frac{R(x)}{Q(x)}</math>, where  <math>R(x)</math> is the remainder term and <math>\deg {{R}(x)} \leq \deg {{Q}(x)}</math>.
  
−
Next, for every factor <math>(a_nx^n+a_{n-1}x^{n-1}+\ldots +a_0)^m</math> in the factorization of <math>Q(x)</math>, introduce the terms  
+
Next, for every factor <math>(a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0)^m</math> in the factorization of <math>Q(x)</math>, introduce the terms  
  
−
<math>\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}</math>
+
<math>\frac{A_1x^{n-1}+B_1x^{n-2}+\cdots+Z_1}{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0}+\frac{A_2x^{n-1}+B_2x^{n-2}+\cdots+Z_2}{(a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0)^2}</math>
 +
<math>+\cdots+\frac{A_mx^{n-1}+B_mx^{n-2}+\cdots+Z_m}{(a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0)^m}</math>
  
 
(Note that the variable <math>Z_i</math> has no relation to being the 26th letter in the alphabet.)
 
(Note that the variable <math>Z_i</math> has no relation to being the 26th letter in the alphabet.)
Line 16: Line 17:
  
 
Consider the sum
 
Consider the sum
−
<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>.
+
<math>\sum_{n=1}^{99}\frac{1}{n(n+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{99\cdot 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 <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>.
+
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 as <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)+\cdots +\left(\frac{1}{99}-\frac{1}{100}\right)</math>
 +
<math>=1-\frac{1}{100}=\frac{99}{100}</math>.

Latest revision as of 01:02, 13 February 2009

Any rational function of the form $\frac{P(x)}{Q(x)}$ may be 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 $(a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0)^m$ in the factorization of $Q(x)$, introduce the terms

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

(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 $\sum_{n=1}^{99}\frac{1}{n(n+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{99\cdot 100}$. 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 as $\sum_{n=1}^{99}\frac{1}{n(n+1)}=\sum_{n=1}^{99}\left(\frac{1}{n}+\frac{-1}{n+1}\right)=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots +\left(\frac{1}{99}-\frac{1}{100}\right)$ $=1-\frac{1}{100}=\frac{99}{100}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Partial_fraction_decomposition&oldid=30184"