- #1
ognik
- 626
- 2
This looked straight forward, but I've gone wrong somewhere ...
Using Rodriquez' formula, find: $ \int_{-1}^{1}{x}^{n}P_n(x) \,dx $ Rodriquez: $ P_n(x)=\frac{1}{2n}\frac{1}{n!}\d{^{n}{({x}^{2}-1)^n}}{{x}^{n}} $
So find $ \int_{-1}^{1}{x}^{n} \frac{1}{2n}\frac{1}{n!}\d{^{n}{({x}^{2}-1)^n}}{{x}^{n}} \,dx $. I'll keep $ \frac{1}{2n}\frac{1}{n!} $ aside for now, and let $ (x^2-1) = f $, and not keep writing in the limits ...
Then, by parts, $ \int {x}^{n} \d{^{n}{f}^n}{{x}^{n}} dx = $ $ {x}^{n} \d{^{n-1}{f}^n}{{x}^{n-1}}|^1_{-1} - \int \d{^{n-1}{f}^n}{{x}^{n-1}} n{x}^{n-1} $ - and repeating by parts n times
I would appreciate knowing how to state better the following arguments around simplifying the problem ...
1st term: After doing 3 Successive differentiations of $ \d{^{n-1}{f}^n}{{x}^{n-1}} $ ALL have a $ ({x}^{2}-1) $ term in them $=(x+1)(x-1)$, therefore I claim all terms with $ \d{^{i}{f}^n}{{x}^{i}} $ in [-1,1] vanish.
I also claim that repeated integration by parts will always have a 1st term which includes $ \d{^{i}{f}^n}{{x}^{i}} $, and to which the above claim applies. So the problem is simplified to the sequence of the integral part only.
Inside the repeated integral will always have 2 parts:
The $x^n$ part: $ x^n $ is differentiated with each integration (always making it the 'u' term for parts). Therefore after n integrations it becomes $n!x^{0}=n! $
The $ \d{^{n}{f^n}}{{x}^{n}} $ part: Because this is always 'v' by parts, each successive integral has a reduced derivative, until finally we are left with just $f^n $
So finally I can find $ \frac{1}{2n}\frac{1}{n!} n! \int_{-1}^{1}f^n \,dx = \frac{1}{2n} \frac{{(x^2-1)}^{n}}{n(n+1)}|^1_{-1}$
I can't see what I have done wrong above?
Using Rodriquez' formula, find: $ \int_{-1}^{1}{x}^{n}P_n(x) \,dx $ Rodriquez: $ P_n(x)=\frac{1}{2n}\frac{1}{n!}\d{^{n}{({x}^{2}-1)^n}}{{x}^{n}} $
So find $ \int_{-1}^{1}{x}^{n} \frac{1}{2n}\frac{1}{n!}\d{^{n}{({x}^{2}-1)^n}}{{x}^{n}} \,dx $. I'll keep $ \frac{1}{2n}\frac{1}{n!} $ aside for now, and let $ (x^2-1) = f $, and not keep writing in the limits ...
Then, by parts, $ \int {x}^{n} \d{^{n}{f}^n}{{x}^{n}} dx = $ $ {x}^{n} \d{^{n-1}{f}^n}{{x}^{n-1}}|^1_{-1} - \int \d{^{n-1}{f}^n}{{x}^{n-1}} n{x}^{n-1} $ - and repeating by parts n times
I would appreciate knowing how to state better the following arguments around simplifying the problem ...
1st term: After doing 3 Successive differentiations of $ \d{^{n-1}{f}^n}{{x}^{n-1}} $ ALL have a $ ({x}^{2}-1) $ term in them $=(x+1)(x-1)$, therefore I claim all terms with $ \d{^{i}{f}^n}{{x}^{i}} $ in [-1,1] vanish.
I also claim that repeated integration by parts will always have a 1st term which includes $ \d{^{i}{f}^n}{{x}^{i}} $, and to which the above claim applies. So the problem is simplified to the sequence of the integral part only.
Inside the repeated integral will always have 2 parts:
The $x^n$ part: $ x^n $ is differentiated with each integration (always making it the 'u' term for parts). Therefore after n integrations it becomes $n!x^{0}=n! $
The $ \d{^{n}{f^n}}{{x}^{n}} $ part: Because this is always 'v' by parts, each successive integral has a reduced derivative, until finally we are left with just $f^n $
So finally I can find $ \frac{1}{2n}\frac{1}{n!} n! \int_{-1}^{1}f^n \,dx = \frac{1}{2n} \frac{{(x^2-1)}^{n}}{n(n+1)}|^1_{-1}$
I can't see what I have done wrong above?
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