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How can I prove the following equality?

$$ \frac{1} {{n!}}\frac{{d^n }} {{dx^n }}\left( {\left( {x^2 - 1} \right)^n } \right) = \sum\limits_{k = 0}^n {\left( {\frac{{n!}} {{k!\left( {n - k} \right)!}}} \right)} ^2 \left( {x + 1} \right)^{n - k} \left( {x - 1} \right)^k $$

And without the use of induction. Only with knowledge of derivatives and sums.

EDITED

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What you have is incorrect, the RHS is $(2x)^n$ while the LHS is not. You probably meant $$ \frac{1} {{n!}}\frac{{d^n }} {{dx^n }}\left( {(x^2 - 1)}^{n} \right) = \sum\limits_{k = 0}^n \left({\frac{{n!}} {{k!\left( {n - k} \right)!}}} \right)^2 \left( {x + 1} \right)^{n - k} \left( {x - 1} \right)^k $$ –  user17762 Jun 11 '12 at 1:53
    
This is just deriving the binomial expansion of the left side $n$ many times. –  Eugene Jun 11 '12 at 1:54
    
But I don't know how to derivate "n-times" if there exist some expresion for it, I don't know. Obviously If I try with induction will work , but I don't want it. –  Matias Jun 11 '12 at 1:59
    
Any statement in proving something for all of natural numbers must invoke induction at some point. Quoting @BillDubuque, "A proof that a statement is true for all integers must - at some point or another - employ mathematical induction. The use of induction may not be obvious - it may be hidden (far) down the inference chain in some other theorem or lemma invoked." –  user17762 Jun 11 '12 at 2:06
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I'm curious: why won't you use induction? After all, this is not basic H.S. calculus, so induction must be well known by now (in fact, over here induction is H.S. stuff), so why not? –  DonAntonio Jun 11 '12 at 2:36
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1 Answer 1

up vote 2 down vote accepted

$(x^2-1)^n = (x-1)^n (x+1)^n = u^n v^n$ where $u=x-1$ and $v=x+1$. Now since $\dfrac{d}{dx} f(u,v) = \dfrac{du}{dx} \dfrac{\partial f}{\partial u} + \dfrac{dv}{dx} \dfrac{\partial f}{\partial v} = \dfrac{\partial f}{\partial u} + \dfrac{\partial f}{\partial v}$, which can be written as $\dfrac{d}{dx} = \dfrac{\partial}{\partial u} + \dfrac{\partial}{\partial v}$, we have, by the Binomial Theorem, $$ \dfrac{d^n}{dx^n} u^n v^n= \left(\dfrac{\partial}{\partial u} + \dfrac{\partial}{\partial v}\right)^n u^n v^n= \sum_{k=0}^n {n \choose k} \dfrac{\partial^k}{\partial u^k} \dfrac{\partial^{n-k}}{\partial v^{n-k}} u^n v^n$$ Now note that $\dfrac{\partial^k}{\partial u^k} u^n = \dfrac{n!}{(n-k)!} u^{n-k}$ and similarly $\dfrac{\partial^{n-k}}{\partial v^{n-k}} v^n = \dfrac{n!}{k!} v^{k}$

Well, that might reasonably be done by induction, or you could use a Taylor series and the binomial theorem: if $g(t) = (t+u)^n$, then $$g(t) = \sum_{k=0}^\infty \dfrac{t^k}{k!} g^{(k)}(0) = \sum_{k=0}^\infty \dfrac{t^k}{k!} \dfrac{\partial^k u^n}{\partial u^k}$$ but also $$ g(t) = (t+u)^n = \sum_{k=0}^n {n \choose k} t^k u^{n-k} $$ and comparing the terms in $t^k$ shows that for $0 \le k \le n$, $$\dfrac{\partial^k}{\partial u^k} u^n = k! {n \choose k} u^{n-k} = \dfrac{n!}{(n-k)!} u^{n-k}$$

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