Given that $$\frac{\mathrm{d}^{2n}}{\mathrm{d}x^{2n}}(x^{2}-1)^{n} = (2n)!$$
How can we find $$\left[\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{2}-1)^{n}\right]^{2}\quad ?$$
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1Did you try anything so far? – user88595 May 09 '14 at 13:21
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I don't think so, it is not true that $(n!)^2=(n^2)!$ – 7raiden7 May 09 '14 at 13:27
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1Try with $n=1$ what happen? – rlartiga May 09 '14 at 13:28
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@Chinny84 how does that relate to this question? – user2850514 May 09 '14 at 13:36
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@user2850514 absolutely right! – Chinny84 May 09 '14 at 13:44
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1Compare with Rodrigues formula for Legendre polynomials. – Start wearing purple May 09 '14 at 13:52
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@O.L. could you please explain us your idea? – rlartiga May 09 '14 at 15:24
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@rlartiga If the asker wants an explicit sum expression, then the solution would be to expand using binomial formula, differentiate and simplify as below. But in case he would like to identify this sum with something known, I note that it is proportional to the squared $n$th degree Legendre polynomial. – Start wearing purple May 09 '14 at 15:39
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$\left(\dfrac{d^n}{dx^n}(x^2-1)^n\right)^2$
$=\biggl(\dfrac{d^n}{dx^n}\sum\limits_{k=0}^nC_k^n(-1)^kx^{2n-2k}\biggr)^2$
$=\biggl(\sum\limits_{k=0}^n\dfrac{(-1)^kn!(2n-2k)!x^{n-2k}}{k!(n-k)!(n-2k)!}\biggr)^2$
$=\biggl(\sum\limits_{k=0}^\left[\frac{n}{2}\right]\dfrac{(-1)^kn!(2n-2k)!x^{n-2k}}{k!(n-k)!(n-2k)!}\biggr)^2$
Harry Peter
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