Question on orthogonal Legendre series expansion.

[yungman]

This start out as homework but my question is not about helping me solving the problem but instead I get conflicting answers depend on what way I approach the problem and no way to resolve. I know the answer. I am not going to even present the original question, instead just the part that I have issue with. The difference is by using:

A) \int_{-1}^1 f(x)P_n(x)dx = \frac{1}{2^n n!}\int_{-1}^{1} f^n(x)(x^2-1)^n dx

And

B) (n+1)P_{n+1}(x) + nP_{n-1}(x) = (2n+1)xP_n(x)



The function f(\theta) = cos(\theta) and I want to compute:

\int_0^{\pi} f(\theta) sin(\theta) P_n(cos(\theta)d\theta

= \int_0^{\pi} cos(\theta) sin(\theta) P_n(cos(\theta)d\theta \hbox { we let }x=cos(\theta) \Rightarrow dx=-sin(\theta) d \theta

\Rightarrow \int_0^{\pi} cos(\theta) sin(\theta) P_n(cos(\theta)d\theta = \int_{-1}^{1} x P_n(x)dx (1)

Using A) This imply f(x) is only first degree and \int_{-1}^{1} x P_n(cos(x)dx = 0 for n larger than 1. The answer is 0 for n=0 and 2/3 for n=1 and the rest is 0 for n>1.

But using B)

\int_{-1}^{1} x P_n(x)dx = \frac{n+1}{2n+1}\int_{-1}^{1} P_{n+1}(x)dx + \frac{n}{2n+1}\int_{-1}^{1} P_{n-1}(x)dx

Recall \int_{-1}^{1} P_{n}(x)dx = 0 \;\Rightarrow\; \int_{-1}^{1} P_{n+1}(x)dx = \int_{-1}^{1} P_{n-1}(x)dx = 0

\Rightarrow \int_{-1}^{1} x P_n(x)dx = 0

So you see you get two different answers using A) and B). I double and triple check this and I cannot see where I did wrong.

 

[Mute]

Recall \int_{-1}^{1} P_{n}(x)dx = 0.

This is not true for all n. P_n(-x) = (-1)^nP_n(x), from which you can show that your identity is only valid if n is odd.

 

[yungman]

I was wrong on this:

\int_{-1}^{1} P_{n}(x)dx = 0 \hbox { only if n not equal to 0, equal to 2 if n=0 }

I just post the question to confirm n cannot be negative. I am still waiting for the answer. If n cannot be negative, then one cannot use B) because \int_{-1}^{1} P_{n-1}(x)dx is not legal for n=0. This will clarify my question that there is only one way I can get the answer.

And I was wrong on n=1. It is not zero. for n=1, I do get 2/3 as the answer, so it agree in both ways of calculation.

 

[Mute]

I was wrong on this:

\int_{-1}^{1} P_{n}(x)dx = 0 \hbox { only if n not equal to 0, equal to 2 if n=0 }

That's still not correct. I'm not sure if you are just stating that it was wrong and you realize it now or if you think this is the corrected statement, so I will clarify: the correct statement is

\int_{-1}^{1} P_{n}(x)dx = \left\{\begin{array}{c c} 0~\mbox{if n is odd}\\2~\mbox{if n is even}\end{array}\right.

As for your other question, n cannot be negative but you case still use n = 0 in the identity you labeled B): the P_{n-1}(x) term vanishes in this case because it is multiplied by n, and indeed, you find that the identity tells you P_1(x) = xP_0(x), which is true.

 

[yungman]

That's still not correct. I'm not sure if you are just stating that it was wrong and you realize it now or if you think this is the corrected statement, so I will clarify: the correct statement is

\int_{-1}^{1} P_{n}(x)dx = \left\{\begin{array}{c c} 0~\mbox{if n is odd}\\2~\mbox{if n is even}\end{array}\right.

As for your other question, n cannot be negative but you case still use n = 0 in the identity you labeled B): the P_{n-1}(x) term vanishes in this case because it is multiplied by n, and indeed, you find that the identity tells you P_1(x) = xP_0(x), which is true.

Thanks for your help. If n=0,1,2...and P_{n-1}(x) term disappeared for n=0, then that answer all my question.

Regarding the formula, it is correct and written in the book. To proof it is correct, we use the identity:

P'_{n+1}(x) = P'_{n-1}(x) +(2n+1)P_n(x) \hbox { and } P_{n+1}(1) = P_{n-1}(1) = 1 \hbox { and } P_{n+1}(-1) = P_{n-1}(-1) = (-1)^{n+1}P_{n-1}(1)

\Rightarrow \int_{-1}^{1} P_{n}(x)dx = \frac{1}{2n+1} [P_{n+1}(x) - P_{n-1}(x)]_{-1}^1 = \frac{1}{2n+1}[P_{n+1}(1) - P_{n-1}(1) - P_{n+1}(-1) + P_{n-1}(-1)] = 0

This is actually the bases of orthogonal relation of Legendre polynomial.

Thanks for your help.

Alan