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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!

  • 1
    First you can do $x^n-1 = (x-1)*$something, then substitute the right thing in for $x$ to get your case.2011-11-08

3 Answers 3

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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.

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    Oh I see Thanks guys!2011-11-09
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Hint: $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}.b+\cdots+b^{n-1})=(a-b)\sum_{i+j=n-1}a^ib^j$

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    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.2011-11-08
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use the identity $b^n - a^n = (b-a)\sum^{n-1}_i b^{n-i} a^i$