Help regarding Legendre Rodrigue's formula problem.

[yungman]

Homework Statement

Question:
Use Rodrigues' formula and integral by parts to show:

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

(As a convention f^{(0)}(x)=f(x)

Homework Equations

Rodrigues' Formula: P_n(x)=\frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n

Bonnet's recurrence relation:

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

P'_{n+1}(x)=P'_{n-1}(x)+(2n+1)P_n(x)

The Attempt at a Solution

I tried
1)\int^1 _{-1}f(x)P_n (x)dx=\frac{1}{2^n n!}\int^1_{-1}f(x)\frac{d^n}{dx^n}(x^2 -1)^n dx

Then use U=f(x), dV=\frac{d^n}{dx^n}(x^2 -1)^n dx

That did not go far.

2) P_n(x)=\frac{1}{2n+1}[P'_{n+1}(x)-P'_{n-1}(x)]

substitude into \int^1 _{-1}f(x)P_n (x)dx

That went no where.

Please give me some advice to proof the relation.

What is f^{(n)}(x)? is it \frac{d^n}{dx^n}f(x)?

 

[Pengwuino]

What happened when you did integration by parts as in your first attempt? That should immediately pop out the answer with n-integration by parts and you just need to deal with the term you receive from integrating by parts with the limits.

And yes, f^{(n)}(x)=\frac{d^n}{dx^n}f(x)

 

[yungman]

What happened when you did integration by parts as in your first attempt? That should immediately pop out the answer with n-integration by parts and you just need to deal with the term you receive from integrating by parts with the limits.

And yes, f^{(n)}(x)=\frac{d^n}{dx^n}f(x)

I did not attempt to perform integrate by parts after the first time because for one, n can be any number and the answer I got after the first time don't seems to be any simplier than the original equation. Also something tell me that there is a trickier way to proof than to integrate it out.

Do you mean there is no short cut but to really do it n times? I would go through it if that is the only way.
Thanks.

 

[Pengwuino]

Yes, n can be any number, but if you do it n-times, you know all that happens is the f(x) acquires n-derivatives by the fact that the Legendre polynomials explicitly have \frac{d}{dx} in them. For example, if you simply had

\int f(x)\frac{d^n}{dx^n}g(x)dx

doing integration by parts n-times is simple and it is exactly what you're doing here. You're literally pulling n-derivatives off g(x) and putting them onto f(x). Now the actual work here is that m-applications of integration by parts creates m-terms of

[f^{(m)} (x)\frac{{dg^{n - m} (x)}}{{dx^{n - m} }}]_{x = - 1}^{x = 1}

that need to be evaluated. At this point this is where the recursion relationships should come in handy. Do one integration by parts and look at the constant term and see if it goes away or not.

And actually, after giving it some thought, you may not even need the recursion relations.

 

[yungman]

Yes, n can be any number, but if you do it n-times, you know all that happens is the f(x) acquires n-derivatives by the fact that the Legendre polynomials explicitly have \frac{d}{dx} in them. For example, if you simply had

\int f(x)\frac{d^n}{dx^n}g(x)dx

doing integration by parts n-times is simple and it is exactly what you're doing here. You're literally pulling n-derivatives off g(x) and putting them onto f(x). Now the actual work here is that m-applications of integration by parts creates m-terms of

[f^{(m)} (x)\frac{{dg^{n - m} (x)}}{{dx^{n - m} }}]_{x = - 1}^{x = 1}

that need to be evaluated. At this point this is where the recursion relationships should come in handy. Do one integration by parts and look at the constant term and see if it goes away or not.

And actually, after giving it some thought, you may not even need the recursion relations.

Thanks, I'll try it tomorrow. My grandson is running a fever and he is being difficult!

 

[yungman]

I want to verify:

\int \frac{d^n}{dx^n}(x^2-1)^n dx=\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^{n-1} +C_1

\int \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^{n-1}dx=\frac{d^{n-2}}{dx^{n-2}}(x^2-1)^{n-2}+C_2

 

[Pengwuino]

No, those statements are not true. What do you know about \int \frac{d}{dx}g(x) dx?

 

[yungman]

No, those statements are not true. What do you know about \int \frac{d}{dx}g(x) dx?

It equal to g(x)+C. I did not write the C out. But that don't apply here.

I worked it out, this is my answer:

1) \int^1 _{-1}f(x)P_n (x)dx=\frac{1}{2^n n!}\int^1_{-1}f(x)\frac{d^n}{dx^n}(x^2 -1)^n dx

Then use U=f(x), dV=\frac{d^n}{dx^n}(x^2 -1)^n dx

\int^1 _{-1}f(x)P_n (x)dx = \frac{1}{2^n n!}[f(x)\frac{d^{n-1}}{dx^{n-1}}(x^2 -1)^n]_{x=-1}^{x=1} - \frac{1}{2^n n!}\int^1_{-1} f^{(1)}(x) \frac{d^{n-1}}{dx^{n-1}}(x^2 -1)^n dx


[f(x)\frac{d^{n-1}}{dx^{n-1}}(x^2 -1)^n]_{x=-1}^{x=1} =[f(x)2x\frac{d^{n-2}}{dx^{n-2}}(x^2 -1)^{n-1}]_{x=-1}^{x=1} =0

because of the x inside and equal zero when substitude x=1 and subtract x=-1. Therefore:



\int^1 _{-1}f(x)P_n (x)dx = - \frac{1}{2^n n!}\int^1_{-1} f^{(1)}(x) \frac{d^{n-1}}{dx^{n-1}}(x^2 -1)^n dx

= -\frac{1}{2^n n!}[f^{(1)}(x)\frac{d^{n-2}}{dx^{n-2}}(x^2 -1)^n]_{x=-1}^{x=1} + \frac{1}{2^n n!}\int^1_{-1} f^{(2)}(x) \frac{d^{n-2}}{dx^{n-2}}(x^2 -1)^n dx


= \frac{1}{2^n n!}\int^1_{-1} f^{(2)}(x) \frac{d^{n-2}}{dx^{n-2}}(x^2 -1)^n dx

= \frac{1}{2^n n!}[f^{(2)}(x)\frac{d^{n-3}}{dx^{n-3}}(x^2 -1)^n]_{x=-1}^{x=1} - \frac{1}{2^n n!}\int^1_{-1} f^{(3)}(x) \frac{d^{n-3}}{dx^{n-3}}(x^2 -1)^n dx

= - \frac{1}{2^n n!}\int^1_{-1} f^{(3)}(x) \frac{d^{n-3}}{dx^{n-3}}(x^2 -1)^n dx

The step repeat until:

= (-1)^{n-1}\frac{1}{2^n n!}[f^{(n)}(x)(x^2 -1)^n]_{x=-1}^{x=1} + (-1)^n \frac{1}{2^n n!}\int^1_{-1}f^{(n)}(x)(x^2 -1)^n dx

In this case [(x^2 -1)^n]_{x=-1}^{x=1}=0 Therefore

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


But it does not work if f(x) is a lower order polynomial than n!

 

[Pengwuino]

There are no integration constants when you are using definite integrals. Also, in the terms that go like

\frac{1}{2^n n!}[f(x)\frac{d^{n-1}}{dx^{n-1}}(x^2 -1)^n]_{x=-1}^{x=1}

You can't simply plug in x=-1,1. You have to determine how that derivative acts first. Now that might be a problem... I think this is where the recurrence relationships will come in handy.

 

[yungman]

There are no integration constants when you are using definite integrals. Also, in the terms that go like

\frac{1}{2^n n!}[f(x)\frac{d^{n-1}}{dx^{n-1}}(x^2 -1)^n]_{x=-1}^{x=1}

You can't simply plug in x=-1,1. You have to determine how that derivative acts first. Now that might be a problem... I think this is where the recurrence relationships will come in handy.

Actually I thought about the constant and I change the post already as you wrote this! We crossed.

I'll look at what you say and work it out in a little bit.

Thanks

 

[yungman]

Yes I see what happen:

[f(x)\frac{d^{n-1}}{dx^{n-1}}(x^2 -1)^n]_{x=-1}^{x=1} =[f(x)2x\frac{d^{n-2}}{dx^{n-2}}(x^2 -1)^{n-1}]_{x=-1}^{x=1} =0

because of the x inside function equal zero when substitude x=1 and subtract the same function with x=-1.

I updated the last post. Can you take a look whether I got it or not. I still have question on the last line about if F(x) is a polynomial with power less than n. That still would not work.

Again thanks for your time. Please let me know.

Sincerely
Alan

 

[yungman]

Anyone can confirm my work?