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Math Help - Finding a Taylor series expansion.

  1. #1
    C.E
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    Finding a Taylor series expansion.

    Hi, can somebody please show me how to find the following Taylor series expansion? (3+2x)^\frac{1}{3} (about x=0).

    I get that f(0)= 3^\frac{1}{3}
    f'(0)= \frac{1}{3} 3^\frac{-2}{3}
    f''(0)= \frac{-2}{9} 3^\frac{-5}{3}
    etc...

    and I know that f(x)=f(0)+f'(0)x+ f''(0)x^2/2! + ...

    but I am having a lot of trouble writing it as a sum. Any ideas?
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  2. #2
    Super Member Gamma's Avatar
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    Quote Originally Posted by C.E View Post
    Hi, can somebody please show me how to find the following Taylor series expansion? (3+2x)^\frac{1}{3} (about x=0).

    I get that f(0)= 3^\frac{1}{3}
    f'(0)= \frac{1}{3} 3^\frac{-2}{3}
    f''(0)= \frac{-2}{9} 3^\frac{-5}{3}
    etc...

    and I know that f(x)=f(0)+f'(0)x+ f''(0)x^2/2! + ...

    but I am having a lot of trouble writing it as a sum. Any ideas?
    You sure you are doing your derivatives right in the MacLaurin Polynomial?
    f(x)=(3+2x)^{\frac{1}{3}}
    f'(x)=\frac{1}{3}(3+2x)^{\frac{-2}{3}}2=\frac{2}{3}(3+2x)^{\frac{-2}{3}}
    f''(x)=\frac{-2^2}{3^2}(3+2x)^{\frac{-5}{3}}2=\frac{-2^3}{3^2}(3+2x)^{\frac{-5}{3}}
    f'''(x)=\frac{2^35}{3^3}(3+2x)^{\frac{-8}{3}}2=\frac{2^45}{3^3}(3+2x)^{\frac{-8}{3}}


    I think you forgot a few things like chain rule and I am not sure what else went wrong when you did it.

    So i get like
    f(x) = 3^{\frac{1}{3}} + \frac{2}{3}3^{\frac{-2}{3}}x+\sum_{n=2}^{\infty}\frac{(-1)^{n+1}2^n2*5*8*11*...*3n-4}{3^nn!}3^{\frac{-(3n-1)}{3}}x^n

    I did it kinda fast, so you should look over it to be sure, but it looks like this works out. I think the pattern really gets going here starting at the second term of the expansion. There might be a slicker way of writing it, but I think this will do.
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  3. #3
    C.E
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    How could you find the radius of convergence of a series like this one? I was expecting it to come out a lot simpler so, the ratio test could be easily applied.
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  4. #4
    Super Member Gamma's Avatar
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    ratio test, and carefully, most of the nasty stuff will cancel out with the ratio test and you will just have kind of an ugly coefficient out front. I would guess you could also probably make use of the alternating series test, that (-1) out there probably makes a big difference. It has been quite some time since I have worked with these things, you are probably more fresh on it than me to be honest.
    Last edited by Gamma; May 20th 2009 at 01:28 PM. Reason: Expounding
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    Quote Originally Posted by C.E View Post
    Hi, can somebody please show me how to find the following Taylor series expansion? (3+2x)^\frac{1}{3} (about x=0).

    I get that f(0)= 3^\frac{1}{3}
    f'(0)= \frac{1}{3} 3^\frac{-2}{3}
    f''(0)= \frac{-2}{9} 3^\frac{-5}{3}
    etc...

    and I know that f(x)=f(0)+f'(0)x+ f''(0)x^2/2! + ...

    but I am having a lot of trouble writing it as a sum. Any ideas?
    Try expanding (3+2x)^\frac{1}{3}by the Binomial Theorem. This should give you the series in a simpler form.
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