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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.

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    Ok, i'll do it with induction2012-06-11

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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 $0 \le k \le n$, $\dfrac{\partial^k}{\partial u^k} u^n = k! {n \choose k} u^{n-k} = \dfrac{n!}{(n-k)!} u^{n-k}$