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Let $P_n(x) = \dfrac{1}{2^n n!}\dfrac{d^n}{dx^n} [(x^2-1)^n]$, we know that $\sum\limits_{n=0}^{\infty} P_n(x)t^n=(1-2tx+t^2)^{-1/2}$

How can we proved

$P_{2n}(0)=\dfrac{(-1)^n(2n)!}{4^n(n!)^2}$ and $P_{2n+1}(0)=0$

I tried to set $x=0$, then we get $f(t)=\sum\limits_{n=0}^{\infty} P_n(0)t^n=(1+t^2)^{-1/2}$, we then have $f^{(n)}(0)=n! P_n(0)$. But I do not know how to get the required answer by differentiating $(1+t^2)^{-1/2}$ $n$ times. Any ideas? (any answers using Rodrigues's formula would also be appreciated, though I do not see any obvious way of doing that)

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You might want to show that $P_{2n+1}(x)$ is odd; that should be sufficient to show that it is zero for zero argument. The series you got after setting $x=0$ in the generating function shows that you're almost there. –  J. M. Apr 7 '13 at 16:30
    
@J.M. I cannot see how that is easier. (or at least, using the Rodrigue's forula or the generating function), but thanks, I did not know that –  Lost1 Apr 7 '13 at 16:33
1  
Really? Did you not notice that $\frac1{\sqrt{1+t^2}}$ is even? –  J. M. Apr 7 '13 at 16:34
    
@J.M. touche..., but wait, how does that prove $P_{2n+1}(x)$ is odd? –  Lost1 Apr 7 '13 at 16:36

1 Answer 1

up vote 1 down vote accepted

By the binomial theorem: $$ f(t) = (1 + t^2)^{-1/2} = \sum_{n \ge 0} \binom{-1/2}{n} t^{2 n} $$ The respective binomial coefficient is: $$ \binom{-1/2}{n} = \frac{(-1)^n}{2^{2n}} \binom{2n}{n} $$ so you have your result.

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gosh, that was easy ... 0_0. Thank you! I feel stupid now. –  Lost1 Apr 7 '13 at 16:34
    
@Lost1, you should see some long-winded, convoluted computations I've done, just to get pointed out (in very public class) that the result really is obvious by what I just had explained before... [No, I won't publish them here, not even anonymously ;-] –  vonbrand Apr 7 '13 at 16:54

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