# Partial fraction decomposition

• Aug 16th 2009, 04:25 AM
transgalactic
Partial fraction decomposition
$\displaystyle \frac{1}{x^3(x^2+1)}=\frac{ax^2+bx+c}{x^3}+\frac{d x+e}{x^2}\\$
$\displaystyle ax^4+bx^3+cx^2+dx^4+ex^3=x^4(a+d)+x^3(b+e)+cx^2=1\ \$
$\displaystyle c=0\\$
$\displaystyle b+e=0\\$
$\displaystyle a+d=0\\$
i got 4 variables with 2 equations
what to do??
• Aug 16th 2009, 04:56 AM
songoku
Hi transgalactic

Your partial fraction is wrong :)

$\displaystyle \frac{1}{x^3(x^2+1)}\neq\frac{ax^2+bx+c}{x^3}+\fra c{dx+e}{x^2}\\$ because if we put the denominator of RHS together, it won't become $\displaystyle x^3(x^2+1)$ but $\displaystyle x^3$ (different from LHS)

It should be :
$\displaystyle \frac{1}{x^3(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+\fr ac{C}{x^3}+\frac{Dx+E}{x^2+1}$
• Aug 16th 2009, 05:00 AM
skeeter
Quote:

Originally Posted by transgalactic
$\displaystyle \frac{1}{x^3(x^2+1)}=\frac{ax^2+bx+c}{x^3}+\frac{d x+e}{x^2}\\$
$\displaystyle ax^4+bx^3+cx^2+dx^4+ex^3=x^4(a+d)+x^3(b+e)+cx^2=1\ \$
$\displaystyle c=0\\$
$\displaystyle b+e=0\\$
$\displaystyle a+d=0\\$
i got 4 variables with 2 equations
what to do??

$\displaystyle \frac{1}{x^3(x^2+1)} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x^3} + \frac{dx+e}{x^2+1}$

$\displaystyle 1 = ax^2(x^2+1) + bx(x^2+1) + c(x^2+1) + (dx+e)x^3$

$\displaystyle 1 = (a+d)x^4 + (b+e)x^3 +(a+c)x^2 + bx + c$

equating coefficients ...

$\displaystyle a+d = 0$

$\displaystyle b+e = 0$

$\displaystyle a+c = 0$

$\displaystyle b = 0$

$\displaystyle c = 1$

from the above equations ...

$\displaystyle a = -1$ , $\displaystyle b = 0$ , $\displaystyle c = 1$ , $\displaystyle d = 1$ , $\displaystyle e = 0$

$\displaystyle \frac{1}{x^3(x^2+1)} = -\frac{1}{x} + \frac{1}{x^3} + \frac{x}{x^2+1}$
• Aug 16th 2009, 05:00 AM
transgalactic
can you give me a link to the manual of this stuff
in order to know thurely all the laws
• Aug 16th 2009, 05:04 AM
skeeter
Quote:

Originally Posted by transgalactic
can you gove me a link to this stuff
in order to know thurely all the laws

partial fraction decomposition - Google Search
• Aug 16th 2009, 05:07 AM
transgalactic
thanks