- #1
yungman
- 5,741
- 294
I have problem understand in one step of deriving the Legendre polymonial formula. We start with:
P_n (x)=\frac{1}{2^n } \sum ^M_{m=0} (-1)^m \frac{2n-2m)}{m!(n-m)(n-2m)}x^n-2m
Where M=n/2 for n=even and M=(n-1)/2 for n=odd.
For 0<=m<=M
\Rightarrow \frac{d^n}{dx^n}x^2n-2m = \frac{2n-2m)}{m!(n-m)(n-2m)}x^n-2m
For M<m<=n
\Rightarrow \frac{d^n}{dx^n}x^2n-2m = 0
P_n (x)=\frac{1}{2^n n!} \sum ^M_{m=0} (-1)^m \frac{n!)}{m!(n-m)}\frac{d^n}{dx^n}x^2n-2m(1)
\Rightarrow P_n (x)=\frac{1}{2^n n!}\frac{d^n}{dx^n} \sum ^n_{m=0} (-1)^m \frac{n!)}{m!(n-m)}(x^2)^{n-m}(2)
Notice the \sum^M_{m=0} change to \sum^n_{m=0} from (1) to (2). Can anyone explain this to me?
P_n (x)=\frac{1}{2^n } \sum ^M_{m=0} (-1)^m \frac{2n-2m)}{m!(n-m)(n-2m)}x^n-2m
Where M=n/2 for n=even and M=(n-1)/2 for n=odd.
For 0<=m<=M
\Rightarrow \frac{d^n}{dx^n}x^2n-2m = \frac{2n-2m)}{m!(n-m)(n-2m)}x^n-2m
For M<m<=n
\Rightarrow \frac{d^n}{dx^n}x^2n-2m = 0
P_n (x)=\frac{1}{2^n n!} \sum ^M_{m=0} (-1)^m \frac{n!)}{m!(n-m)}\frac{d^n}{dx^n}x^2n-2m(1)
\Rightarrow P_n (x)=\frac{1}{2^n n!}\frac{d^n}{dx^n} \sum ^n_{m=0} (-1)^m \frac{n!)}{m!(n-m)}(x^2)^{n-m}(2)
Notice the \sum^M_{m=0} change to \sum^n_{m=0} from (1) to (2). Can anyone explain this to me?
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