I'm having difficulty understanding why / how to go about proving the fact that a partial fraction decomposition of $\frac{A}{x(x-a)^m}=\frac{A_1}{x}+\frac{A_2}{(x-a)^1}+\frac{A_3}{(x-a)^2}+ … +\frac{A_m}{(x-a)^m}$ .
Why is it not just something like $\frac{A}{x(x-a)^m}=\frac{A_1}{x}+\frac{A_2}{(x-a)^m}$?
I've read an answer on MSE : Partial Fraction Decomposition? but I don't understand it. Is there a way of proving this statement without resorting to vector spaces and basis thingamajigs?
Best Answer
Take the simplest case $\dfrac1{x(x-a)^2} $.
If $\dfrac1{x(x-a)^2} =\dfrac{u}{x}+\dfrac{v}{(x-a)^2} $ then $\dfrac{u}{x}+\dfrac{v}{(x-a)^2} =\dfrac{u(x-a)^2+vx}{x(x-a)^2} =\dfrac{ux^2+(v-2ua)x+ua^2}{x(x-a)^2} $ and you need to have $u = 0, v-2ua=0, ua^2 = 1$.
This is three equations in two unknowns and, in this case, has no solution.
You need three parameters, which are gotten by writing $\dfrac1{x(x-a)^2} =\dfrac{u}{x}+\dfrac{v}{x-a}+\dfrac{w}{(x-a)^2} $.
This becomes $1 =u(x-a)^2+vx(x-a)+wx $ and this can be solved for $u, v, w$.
The general case is $\dfrac1{x(x-a)^m} =\dfrac{c}{x}+\sum_{j=1}^m \dfrac{b_j}{(x-a)^j} $ and this can be solved for $c$ and the $b_j$.
If we clear fractions, this becomes $1 = c(x-a)^m+\sum_{j=1}^m b_jx(x-a)^{m-j} = c(x-a)^m+x\sum_{j=1}^m b_j(x-a)^{m-j} $.
Setting $x=0$ gives $1 = c(-a)^m$ so $c = \dfrac1{(-a)^m} $ and $1 = (x-a)^m(-a)^{-m}+x\sum_{j=1}^m b_j(x-a)^{m-j} $.
Putting $x=a$ gives $1 = ab_m $ so $b_m = \dfrac1{a} $.
The successive $b_j$ can be similarly determined.