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I don't know why, but for some reason I cannot solve the following partial fraction decomposition no matter how much I try.

$$\frac{1}{(v-1)^2(v+1)^2}$$

When decomposing that to $\frac{1}{(v-1)^2(v+1)^2} = \frac{A_1}{v-1} + \frac{A_2}{(v-1)^2} + \frac{A_3}{v+1} + \frac{A_4}{(v+1)^2}$ I did figure out that $A_2$ and $A_4$ are $\frac{1}{4}$, but I cannot for the love of me figure out $A_1$ and $A_3$.

Sorry if this sounds like a stupid question, but I'm really stuck here and don't know how to proceed.

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Hint: I got $$(A+C)v^3+v^2(A+B-C+D)+v(A+2B-C-2D)-A+B+C+D=1$$

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  • $\begingroup$ Thank you very much! This helped a lot $\endgroup$ – pavus Jul 14 '18 at 14:19
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Reducing to the common denominator,

$$(A_1(v-1)+A_2)(v+1)^2+(A_3(v+1)+A_4)(v-1)^2=1.$$

With $v=1$, $4A_2=1$.

With $v=-1$, $4A_4=1$.

With $v=0$, $-A_1+A_2+A_3+A_4=1$.

With $v=\infty$, $A_1+A_3=0.$

Hence

$$\frac14\left(-\frac1{v-1}+\frac1{(v-1)^2}+\frac1{v+1}+\frac1{(v+1)^2}\right).$$

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  • $\begingroup$ Thank you, I figured it out! $\endgroup$ – pavus Jul 14 '18 at 14:18
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$$\frac{1}{(v-1)^2(v+1)^2}=\frac{A}{(v-1)}+\frac{B}{(v-1)^2}+\frac{C}{(v+1)}+\frac{D}{(v+1)^2}$$ If you multiply both sides by $(v-1)^2(v+1)^2$, you get that $$1=A(v-1)(v+1)^2+B(v+1)^2+C(v-1)^2(v+1)+D(v-1)^2$$ And if you expand everything, and compare the coefficients of $1,v,v^2,v^3$ on both sides, you will get a system of linear equations for $A,B,C,D$.

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    $\begingroup$ Instead of expanding everything, it's usually much faster to substitute a few values of $v$ ($-1,1$,etc.) $\endgroup$ – Théophile Jul 14 '18 at 14:10
  • $\begingroup$ @Théophile Yes, you can do that too. $\endgroup$ – Botond Jul 14 '18 at 19:33
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An easy way to do this particular one is $$\begin{align} \frac{1}{(v-1)(v+1)}&=\frac12\left(\frac{1}{v-1}-\frac{1}{v+1}\right)\\ \frac{1}{(v-1)^2(v+1)^2}&=\frac14\left(\frac{1}{(v-1)^2}+\frac{1}{(v+1)^2}-\frac{2}{(v-1)(v+1)}\right)\\ \frac{1}{(v-1)^2(v+1)^2}&=\frac{1}{4(v-1)^2}+\frac{1}{4(v+1)^2}-\frac{1}{4(v-1)}+\frac{1}{4(v+1)} \end{align} $$

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  • $\begingroup$ Yep, but you made mistakes anyway. $\endgroup$ – Yves Daoust Jul 14 '18 at 14:11
  • $\begingroup$ @YvesDaoust Thank you. $\endgroup$ – saulspatz Jul 14 '18 at 14:18

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