Vtyjkyuk

I am trying to prove that $frac1n!space2^nfracd^ndx^n(x^2-1)^n=P_n(x)$, where $P_n(x)$ is the Legendre Polynomial of order n.

I've been told that the proof uses complex analysis, of which I know nothing, isn't there a proof with elementary methods? (If there isn't, I'm still interested in the other one).

1. Check that the left side indeed defines an $n$th order polynomial.




2. Check that $displaystyle int_-1^1P_n(x)P_m(x)dx$ vanishes for $mneq n$ (integration by parts).




3. Check the normalization condition $P_n(1)=1$ (Leibniz rule).

Added: As you almost correctly write in the comment below, the result of integration by parts (assuming that $m<n$ and transferring the derivatives from $P_n$ to $P_m$) can be written as
$$int_-1^1P_m(x)P_n(x)dx=sum_k=1^nc_mnkleft[fracd^m+k-1 (x^2-1)^mdx^m+k-1fracd^n-k(x^2-1)^ndx^n-kright]_-1^1,$$
where $c_mnk$ is some irrelevant constant. Consider the second factor in the square brackets. There you have a polynomial having $n$th order zeros at $x=pm 1$ which we differentiate $n-k$ times. The result will therefore have $k$th order zeros at these points, which implies vanishing of the integral.

If $n$ is integer
begineqnarray*
fracd^nd x^n (x^2-1)^n &=& fracd^nd x^n left [
sum_k=0^n (-1)^k fracn!k!(n-k)! x^2n-2k right ] \
&=& sum_k=0^n (-1)^k fracn!k! (n-k)! frac(2n-2k)!(n-2k)! x^n-2k.
endeqnarray*
The sum above does not go up to $k=n$, since after $k=[n/2]$, the derivatives
are 0 then we write

begineqnarray*
fracd^nd x^n (x^2-1)^n &=&
sum_k=0^[n/2] (-1)^k fracn!k! (n-k)! frac(2n-2k)!(n-2k)! x^n-2k.
endeqnarray*
It follows from the infinite series truncated to the Legendre polynmial that

begineqnarray*
P_n(x) = frac12^n n! fracd^ndx^n (x^2-1)^n.
endeqnarray*

The approach followed here is in reverse order. We started with Rodriguez's formula
and showed that it corresponds to a Legendre polynomial. A more intuitive approach
is to start at the polynomials

begineqnarray*
y(x)= (1-x^2)^n.
endeqnarray*
and take derivates, and verifty that the derivatives taken $n$ times will get you
to the Legendre differential equation. That is, we have that

begineqnarray*
y' = -2 n x (1-x^2)^n-1
endeqnarray*
which we can write as

begineqnarray
(1-x^2) y' + 2n x y = 0.
labeltole
endeqnarray
and starts looking a bit like a Legendre differential equation.

We want to differentiate this equation $k$ times and use the Leibniz rule.
That is, if we call $u=1-x^2$,

begineqnarray*
frac d^kdx^k [u y'] = sum_j=0^k binomkj
u^(j) y^(k-j+1)
endeqnarray*
Given that $u$ is a second order polynomial only three terms of this
sum will survive. That is

begineqnarray*
frac d^kdx^k [u y'] &=& u y^(k+1) + k u' y^(k) + k(k-1) u^(2) y^(k-1) \
&=& (1-x^2)y^(k+1) - 2 k x y^(k) -2 frack(k-1)2 y^(k-1) = 0
endeqnarray*
Likewise we use the Leibniz rule for the product $2nxy$ where only two terms will
survive. That is

begineqnarray*
frac d^kdx^k [2 n x y] &=& 2 n x y^(k) + 2 n k y^(k-1),
endeqnarray*
we combine the two results above to find

begineqnarray*
(1-x^2)y^(k+1) - 2 k x y^(k) - k(k-1) y^(k-1)
+ 2 n x y^(k) + 2 n k y^(k-1) = 0
endeqnarray*
At this point we observe that if $k=n+1$, we find

begineqnarray*
(1-x^2)y^(n+2) - 2(n+1) x y^(n+1) - n(n+1) y^(n)
+ 2 n x y^(n+1) + 2 n (n+1) y^(n) = 0
endeqnarray*
which simplifies to

begineqnarray*
(1-x^2) y^(n+2) - 2 x y^(n+1) + n(n+1) y^(n)=0.
endeqnarray*

and this is the Lagrange differential equation with $y^n=P_n$.
We then showed that

begineqnarray*
fracd^ndx^n(1-x^2)^n
endeqnarray*
satisfies the Lagrange differential equation. The factor $1/(2^n n!)$ is
included to make $P(1)=1$.

(The general formula of Legendre Polynomials is given by following equation:
$$
P_k(x)=sum_m=0^frack2frac(-1)^m(2k-2m)!2^km!(k-m)!frac1(k-2m)! x^k-2m
$$

The Rodrigues' formula is:

$$ frac12^kk!fracd^kdx^k[(x^2-1)^k] $$

The Binomial theorem is as follow:
$$(x+y)^k=sum_i=0^kfrack!i!(k-i)!x^k-iy^i$$

Then $$(x^2-1)^k=sum_i=0^kfrack!i!(n-i)!(x^2)^k-i(-1)^i$$

$$frac12^kk!fracd^kdx^k[(x^2-1)^k]=frac12^kk!fracd^kdx^kbig[sum_i=0^kfrack!i!(n-i)!(x^2)^k-i(-1)^ibig]$$

So
$$ =frac12^kk!sum_i=0^kfrack!i!(n-i)!fracd^kdx^k(x)^2k-2i(-1)^i .....(1)$$

$$fracd^kdx^k(x)^2k-2i=frac(2k-2i)!k-2ix^k-2i......(2)$$

By compensate (2) into (1), we get:

$$
P_k(x)=sum_m=0^frack2frac(-1)^i(2k-2i)!2^ki!(k-i)!frac1(k-2i)! x^k-2i
$$

Change dummy variable (i) to (m), we get the same general formula of Legendre Polynomials :

$$
P_k(x)=sum_m=0^frack2frac(-1)^m(2k-2m)!2^km!(k-m)!frac1(k-2m)! x^k-2m
$$