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## Main Question or Discussion Point

I have problem understand in one step of deriving the Legendre polymonial formula. We start with:

[tex]P_n (x)=\frac{1}{2^n } \sum ^M_{m=0} (-1)^m \frac{2n-2m)}{m!(n-m)(n-2m)}x^n-2m[/tex]

Where

For 0<=m<=M

[tex]\Rightarrow \frac{d^n}{dx^n}x^2n-2m = \frac{2n-2m)}{m!(n-m)(n-2m)}x^n-2m[/tex]

For M<m<=n

[tex]\Rightarrow \frac{d^n}{dx^n}x^2n-2m = 0[/tex]

[tex]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 [/tex](1)

[tex]\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} [/tex](2)

Notice the [tex]\sum^M_{m=0}[/tex] change to [tex]\sum^n_{m=0}[/tex] from (1) to (2). Can anyone explain this to me?

[tex]P_n (x)=\frac{1}{2^n } \sum ^M_{m=0} (-1)^m \frac{2n-2m)}{m!(n-m)(n-2m)}x^n-2m[/tex]

Where

**M=n/2 for n=even**and**M=(n-1)/2 for n=odd.**For 0<=m<=M

[tex]\Rightarrow \frac{d^n}{dx^n}x^2n-2m = \frac{2n-2m)}{m!(n-m)(n-2m)}x^n-2m[/tex]

For M<m<=n

[tex]\Rightarrow \frac{d^n}{dx^n}x^2n-2m = 0[/tex]

[tex]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 [/tex](1)

[tex]\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} [/tex](2)

Notice the [tex]\sum^M_{m=0}[/tex] change to [tex]\sum^n_{m=0}[/tex] from (1) to (2). Can anyone explain this to me?