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How could one apply partial fraction decomposition to $$\left(\frac{s}{s^2+4}\right)^2$$

I tried to separate by doing $$\frac{A}{s^2+4} + \frac{B}{s^2+4}$$ and I got strange solution $A+B=0$ and $A+B=1$.

graydad
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dsaca
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1 Answers1

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You have assumed the wrong form that the partial fraction will take. Try $$\frac{s^2}{(s^2+4)^2} = \frac{As+B}{s^2+4} + \frac{Cs+D}{(s^2+4)^2}$$ Partial fractions with squares in the denominator always seemed kind of weird to me, but this should get you the result you are after!

graydad
  • 14,077
  • is there any general equation to do partial fraction? If i need to do partial fraction on, let say, 1/(x^20(x+1)), how to I do it? – dsaca May 08 '15 at 16:49
  • @dsaca I think I could write out the general equation but it would be nasty looking. It would probably just be best to look at a few examples to get a feel for what the forms will look like. You probably won't encounter anything past a cube in the denominator in any comprehensive calculus class. As for $\frac{1}{x^{20}(x+1)}$ I believe you would assume $$\frac{1}{x^{20}(x+1)} = \frac{A}{x+1}+\frac{B_{19}x^{19}+B_{18}x^{18}+\dots+B_2x^2+B_1x+B_0}{x^{20}}$$ where $B_i$ is some constant on $x^i$ – graydad May 08 '15 at 16:58
  • Thanks. But back to the question, I got A=C=0, B=1 and D=-1 which does not seem very correct. Are you sure the for above is correct? – dsaca May 08 '15 at 17:05
  • sorry Dshould be -4 – dsaca May 08 '15 at 17:09
  • When I did it I got $A = 0, B=1, C=0$ and $D = -4$ and it worked out for me – graydad May 08 '15 at 17:09
  • @dsaca looks like you did it right to me, since $$\begin{align}\frac{1}{s^2+4}-\frac{4}{(s^2+4)^2} = \frac{1}{(s^2+4)}\frac{s^2+4}{s^2+4}-\frac{4}{(s^2+4)^2} \ = \frac{s^2+4}{(s^2+4)^2}-\frac{4}{(s^2+4)^2} \ =\frac{s^2+4-4}{(s^2+4)^2} \ = \frac{s^2}{(s^2+4)^2} \end{align}$$ yielding the desired result – graydad May 08 '15 at 17:20