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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 $$ \frac{1} {{n!}}\frac{{d^n }} {{dx^n }}\left( {(x^2 - 1)}^{n} \right) = \sum\limits_{k = 0}^n \left({\frac{{n!}} {{k!\left( {n - k} \right)!}}} \right)^2 \left( {x + 1} \right)^{n - k} \left( {x - 1} \right)^k $$2012-06-11
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    This is just deriving the binomial expansion of the left side $n$ many times.2012-06-11
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    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.2012-06-11
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    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."2012-06-11
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    Yes I know, But I mean only using techniques of sums... I know that use induction , but I'm sure that everyone knows what I mean................2012-06-11
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    Maybe if you tried it for some small $n$ you'd understand how. Let's try $n = 2$. Then $y = (x^2 - 1)^2 = (x+1)^2(x-1)^2$. So $dy/dx = 2(x+1)^2(x-1) + 2(x+1)(x-1)^2$. Then $d^2y/dx^2 = 2(x+1)^2 + 4(x+1)(x-1) + 2(x-1)^2 + 4(x+1)(x-1)$. Understand?2012-06-11
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    That's the price you have to pay for not using induction.2012-06-11
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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?2012-06-11
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    Ok, i'll do it with induction2012-06-11

1 Answers 1

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$(x^2-1)^n = (x-1)^n (x+1)^n = u^n v^n$ 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 $0 \le k \le n$, $$\dfrac{\partial^k}{\partial u^k} u^n = k! {n \choose k} u^{n-k} = \dfrac{n!}{(n-k)!} u^{n-k}$$