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 $\frac{A}{x(x-a)^m}=\frac{A_1}{x}+\frac{A_2}{(x-a)^m}$?

A screenshot from Paul's Online Math Notes pertaining to said statement.

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?

  • $\begingroup$ Why is it not just something like Simply put, because it just doesn't work out. Try to determine $A_1, A_2$ so that your "simple" form holds identically, and you'll soon find that it's not possible. $\endgroup$ – dxiv Jan 15 '18 at 5:24
  • $\begingroup$ So in order to prove this statement I have to systematically prove by contradiction every other instance where the number of terms is < m that they don't work? $\endgroup$ – Chung Ren Khoo Jan 15 '18 at 5:25
  • $\begingroup$ No, it's not a matter of trial and error. But it's not entirely clear whether you want to use or you want to prove the form of partial fraction decomposition. Either way, see this link for example. $\endgroup$ – dxiv Jan 15 '18 at 5:29

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.


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