I am struggling to see how P_{k}(x)= [1/(2^{k}(k!))][(d^{k}((x^{2}-1)^{k})/dx^{k}] = [1/(2^{k})][2k choose k][x^{k}] + (terms of lower order)

Help please?

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- Jul 19th 2012, 03:03 AMebbSupposedly simple algebra...
I am struggling to see how P

_{k}(x)= [1/(2^{k}(k!))][(d^{k}((x^{2}-1)^{k})/dx^{k}] = [1/(2^{k})][2k choose k][x^{k}] + (terms of lower order)

Help please? - Jul 19th 2012, 03:16 AMgirdavRe: Supposedly simple algebra...
$\displaystyle (x^2-1)^k$ is a polynomial of degree $\displaystyle 2k$; when you take the $\displaystyle k$-th derivative it's a polynomial of degree $\displaystyle k$. Now, look at the leading term.

- Jul 22nd 2012, 02:56 AMebbRe: Supposedly simple algebra...
Thank you - a great help!