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Math Help - A problem!

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    A problem!

    f(x)=x-\frac{3}{2}x^2+\frac{11}{6}x^3-\frac{50}{24}x^4+\frac{274}{120}x^5+...
    This is Taylor series of function f(x)=\frac{ln(1+x)}{1+x}

    My question is:

    f(x)=x-\frac{3}{2}x^2+\frac{11}{6}x^3-\frac{50}{24}x^4+\frac{274}{120}x^5+...=\Sigma^\in  fty_{n=0}(-1)^{n+1}a_n x^n

    What is form of a_n?


    Thank you very much!
    Last edited by Also sprach Zarathustra; June 25th 2010 at 05:50 PM.
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  2. #2
    A Plied Mathematician
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    Well, the denominators look something like a factorial (except for that pesky 3 in the squared term; perhaps you could massage it to look better). Maybe you could set up a recurrence relation to figure out the numerators?
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Ohhh... it's not 2/3 it's 3/2

    Sorry!!!

    (I'm going to change it in my original post)


    Thank you Mr. Keister!
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  4. #4
    A Plied Mathematician
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    Aha. So it is a nice little factorial in the denominator. Well, the recurrence relation idea sounds nice, but I'm not sure how it would work in principle. Why don't you show all the details of the computations for those first few terms? Maybe something'll pop out.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Aha. So it is a nice little factorial in the denominator. Well, the recurrence relation idea sounds nice, but I'm not sure how it would work in principle. Why don't you show all the details of the computations for those first few terms? Maybe something'll pop out.

    You ask for computations ?

    Here it is: Function calculator

    Moreover... look here...

    1 3 11 50 - OEIS Search Results
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  6. #6
    A Plied Mathematician
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    No, that's not what I'm interested in. I'm interested in seeing the step-by-step calculations for computing each coefficient. I'm curious to see if a pattern pops out by examining that process.
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    Unfortunately, I can't do so, it is very complicated, but the thing I can do is to post the original question asked.

    Q.
    Find Taylor polynomial of f(x)=\frac{ln(1+x)}{1+x} around x=0, find formula for f^{(n)}(x).
    Find the radius of convergence of infinite Taylor polynomial(power series).
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  8. #8
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    Hi , it is  H_n =  \frac{1}{1}  +  \frac{1}{2}  + ... +  \frac{1}{n}
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  9. #9
    MHF Contributor Also sprach Zarathustra's Avatar
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    Make sense!

    How you get this result please?
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  10. #10
    MHF Contributor Also sprach Zarathustra's Avatar
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    So, the radius of convergence R of \Sigma^{\infty}_{n=0}(-1)^nH_nx^n is R=0?


    Thanks!
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  11. #11
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    No , the radius is 1 , i obtain the general term by expanding  \ln(1+x) and  \frac{1}{1+x} followed by coefficients' consideration .
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  12. #12
    MHF Contributor chisigma's Avatar
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    The general formula for the product of two series is...

    \displaystyle \sum_{n=0}^{\infty} a_{n}\ x^{n}\ \sum_{k=0}^{\infty} b_{k}\ x^{k} = \sum_{n=0}^{\infty} \sum_{k=0}^{n} a_{k}\ b_{n-k}\ x^{n} (1)

    ... and in Your case is...

    \displaystyle \ln (1+x)= \sum_{n=1}^{\infty} (-1)^{n+1}\ \frac{x^{n}}{n}

    \displaystyle \frac{1}{1+x} = \sum_{k=0}^{\infty} (-1)^{k}\ x^{k} (2)

    Kind regards

    \chi \sigma
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  13. #13
    MHF Contributor Also sprach Zarathustra's Avatar
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    I have difficulties in finding radius of convergence! Why it equals to 1?

    In other words, WHY R=\lim_{n\to\infty}\frac{1}{\sqrt[n]{H_n}} = 1?
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  14. #14
    MHF Contributor Also sprach Zarathustra's Avatar
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    Maybe it from the "sandwich" rule?
    1<H_n<n
    \sqrt[n]{1}<\sqrt[n]H_n<\sqrt[n]{n}

    so when n\to\infty

    H_n \to 1
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  15. #15
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    If \lim_{n \to \infty} \frac{a_{n + 1}}{a_n} exists, then \lim_{n \to \infty} \sqrt[n]{a_n} does as well, and they are equal. Since the first expression exists, you can use it to find the radius of convergence, which will work out to be 1.

    If you think to calculate a_n - a_{n - 1} = \frac{1}{n} from the terms you listed out then it becomes pretty easy to figure out the form of the a_n, though this isn't exactly a rigorous solution.
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