Legendre 證明題

The Legendre polynomials {P_n(x)}[n=0,1,2...] form an orthogonal set of functions in the following sense: = 定積分[x=-1 to 1] {P_n(x)*P_m(x)} dx =0 , if n 不等於m.
On the other hand, P_m(x)'s are polynomials of exactly degree m, so x^m can always be represented as a linear combination of P_0, P_1, ...P_m. i.e. x^m= summation [i=0 to m] c_i*P_i(x). Thus 定積分[x=-1 to 1] {x^m*P_n(x)} dx = summation [i=0 to m] c_i* 定積分[x=-1 to 1] {P_i(x)*P_n(x)} dx =0 , for m 不等於n. So the statement is true, even when m>n.

2010-01-16 22:07:26 補充：
as m>n, it is not always true., because c_m may not be zero. Thanks to 天助 reminding. The correct generalization is this: when m>n, it will be true for n*m is odd, since c_n=0 for sure, while m*n is even , the statement is not true.

2010-01-17 21:40:38 補充：
綜合以上, 定積分[x=-1 to 1] {x^m*P_n(x)} dx =0 , for all m小於等於n; 當m大於等於n時, 敘述仍然成立若m,n是一奇一偶, or [m+n 是奇]{但非[m*n是奇],因n有可能是0,再次感謝天助斧正}; 至於當m大於等於n時且m,n是同奇或同偶時,定積分[x=-1 to 1] {x^m*P_n(x)} dx =0一定不成立.證明部分全繫於c_n係數為0或否.

2010-01-19 10:51:50 補充：
定積分[x=-1 to 1] {x^m*P_n(x)} dx =0 , for all m小於等於n --->定積分[x=-1 to 1] {x^m*P_n(x)} dx =0 , for all m小於n

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To:
001老師:m >= n時不一定成立!

2010-01-17 02:08:11 補充：
m*n為奇數時,也不對!
應該是m+n為奇數時必成立(與m,n大小無關)