# Why does $\frac{1}{(w-z)^n}-\frac{1}{(w-a)^n}=(\frac{1}{(w-z)}-\frac{1}{(w-a)}) \sum_{i+j=n-1}\frac{1}{(w-z)^i}.\frac{1}{(w-a)^j}$?

If $n\in \mathbb{N}$ can anyone explain why

$$\frac{1}{(w-z)^n}-\frac{1}{(w-a)^n}=\left(\frac{1}{(w-z)}-\frac{1}{(w-a)}\right) \sum_{i+j=n-1}\frac{1}{(w-z)^i}.\frac{1}{(w-a)^j}$$

It's part of a proof of any complex function that can be differentiated once can be differentiated infinitely many times. I would very much appreciate understanding why this identity holds!

Thanks!

-
For the same reason that $a^n - b^n = (a - b) * \sum a^i b^j$ holds; develop the right hand side and telescope what comes out. – Gunnar Þór Magnússon Nov 8 '11 at 19:20
First you can do $x^n-1 = (x-1)*$something, then substitute the right thing in for $x$ to get your case. – GEdgar Nov 8 '11 at 20:13

This follows easily from the identity $$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \dots + y^{n-2} x + y^{n-1})$$ where $x = 1/(w-z)$ and $y=1/(w-a)$. I am writing this so that you know this very useful identity, which works in any commutative ring. To prove it, you can simply expand the factor on the right.

EDIT : I am putting it here explicitly because of J.M.'s comment and because I never actually noticed it could be shown in this way, but when you look at the identity $$\frac{1-z^n}{1-z} = z^{n-1} + \dots + z + 1$$ , you can let $z = y/x$ and see that $$1-\left( \frac yx \right)^n = \left( 1 - \frac yx \right) \left( \left( \frac yx \right)^{n-1} + \dots + \frac yx + 1 \right).$$ Multiplying by $x^n$ on the left hand side and by $x\cdot x^{n-1}$ on the right hand side, we also get the identity.

-
A little typo: $x$ and $y$ should be without $n$. – Tapu Nov 8 '11 at 19:36
Ok, right, so how does this identity work? – Freeman Nov 8 '11 at 21:49
@Swapan : Thanks. – Patrick Da Silva Nov 8 '11 at 22:18
@LHS : What do you mean by "how does it work?" Do you want me to prove it holds? – Patrick Da Silva Nov 8 '11 at 22:19
@LHS: $x^n-y^n=x^n\left(1-\left(\frac{y}{x}\right)^n\right)$. You are familiar with summing a geometric series, no? – J. M. Nov 9 '11 at 0:25

Hint: $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}.b+\cdots+b^{n-1})=(a-b)\sum_{i+j=n-1}a^ib^j$

-
I am probably missing something obvious here, but how does this identity work? – Freeman Nov 8 '11 at 21:49
You mean "how this formula was derived"? or, how to apply this formula to your problem? Well, in the first case, note that $a^n-b^n$ becomes zero when $a=b$; so $(a-b)$ is a factor,the other factor can be obtained by long division. In the second case, please note that $a=\frac{1}{w-z}$ and $b=\frac{1}{w-a}$ gives your desired relation. – Tapu Nov 8 '11 at 22:02

use the identity $b^n - a^n = (b-a)\sum^{n-1}_i b^{n-i} a^i$

-