Problem with a partial fraction decomposition

Question

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.

Answer 1

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

Answer 2

$$\frac{1}{(v-1)^2(v+1)^2}=\frac{A}{(v-1)}+\frac{B}{(v-1)^2}+\frac{C}{(v+1)}+\frac{D}{(v+1)^2}$$ 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 $A,B,C,D$.

Answer 3

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

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

Answer 4

An easy way to do this particular one is $$\begin{align} \frac{1}{(v-1)(v+1)}&=\frac12\left(\frac{1}{v-1}-\frac{1}{v+1}\right)\\ \frac{1}{(v-1)^2(v+1)^2}&=\frac14\left(\frac{1}{(v-1)^2}+\frac{1}{(v+1)^2}-\frac{2}{(v-1)(v+1)}\right)\\ \frac{1}{(v-1)^2(v+1)^2}&=\frac{1}{4(v-1)^2}+\frac{1}{4(v+1)^2}-\frac{1}{4(v-1)}+\frac{1}{4(v+1)} \end{align} $$

Answer 5

$$\frac{1}{(v-1)^2(v+1)^2} = \frac{A_1}{v-1} + \frac{A_2}{(v-1)^2} + \frac{A_3}{v+1} + \frac{A_4}{(v+1)^2}$$

This isn't the short way, but I enjoy doing it this way because I enjoy watching it get simpler as I go along.

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

Let $v = -1$, and you get

$$1 = 4A_4 \implies A_4 = \frac 14$$

$$4 = 4(v-1)(v+1)^2A_1 + 4(v+1)^2A_2 + 4(v-1)^2(v+1)A_3 + (v-1)^2$$

$$-(v^2-2v-3) = 4(v-1)(v+1)^2A_1 + 4(v+1)^2A_2 + 4(v-1)^2(v+1)A_3$$

$$-v+3 = 4(v-1)(v+1)A_1 + 4(v+1)A_2 + 4(v-1)^2A_3$$

Let $v=-1$, again, and you get

$$4 = 16A_3 \implies A_3 = \frac 14$$

$$-v+3 = 4(v-1)(v+1)A_1 + 4(v+1)A_2 + (v-1)^2A_3$$

$$-(v^2-v-2) = 4(v-1)(v+1)A_1 + 4(v+1)A_2$$

$$-(v-2) = 4(v-1)A_1 + 4A_2$$

Let $v=1$, and you get

$$1 = 4A_2 \implies A_2 = \frac 14$$

$$-v+2 = 4(v-1)A_1 +1$$

$$-v+1 = 4(v-1)A_1 \implies A_1 = -\frac 14$$