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Math Help - Taylor development

  1. #1
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    Taylor development

    Hi there. I have this exercise which I'm trying to solve now. It says:

    Using that \displaystyle\sum_{n=0}^{\infty}x^n=(1-x)^{-1} find one Taylor development for the function f(x)=\ln(1-x)

    So, I've made some derivatives:
    f^1(x)=\displaystyle\frac{-1}{(1-x)},f^2(x)=\displaystyle\frac{-1}{(1-x)^2},f^3(x)=\displaystyle\frac{-2}{(1-x)^3},f^4(x)=\displaystyle\frac{-6}{(1-x)^4},f^5(x)=\displaystyle\frac{-24}{(1-x)^5}

    And then:

    \displaystyle\sum_{n=0}^{\infty}\displaystyle\frac  {f^b(x_0)(x-x_0)^n}{n!}=-\displaystyle\frac{(x-x_0)}{(1-x_0)}-\displaystyle\frac{(x-x_0)^2}{2(1-x_0)^2}-\displaystyle\frac{2(x-x_0)^3}{6(1-x_0)^3}-\displaystyle\frac{6(x-x_0)^4}{24(1-x_0)^4}-\displaystyle\frac{24(x-x_0)^5}{120(1-x_0)^5}+\ldots+-\displaystyle\frac{(x-x_0)^n}{n(1-x_0)^n}

    I have two problems with this. In the first place, the general expression that I've found (which is probably wrong) doesn't work for n=0, it does for the others values of n. I thought of starting the summation at 1, but I'm not sure if this is valid. In the second place I don't know how to use the relation the problem gives at the beginning. I can see that I have (1-x_0) for every term, but I couldn't make it fit inside the summation.

    So this is what I got: \displaystyle\sum_{n=1}^{\infty}-\displaystyle\frac{(x-x_0)^n}{n(1-x_0)^n}

    Bye there, thanks for your help and suggestions.
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  2. #2
    Behold, the power of SARDINES!
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    You didn't use the hint

    Notice that

    f(x)=\ln(1-x) \implies f'(x)=\frac{-1}{1-x}

    But now by the hint we know this is a geometric series and gives

    f'(x)=\frac{-1}{1-x}=-\sum_{n=0}^{\infty}x^n

    But we want the power series for f not its derivative so lets integrate to get

    \int f'(x)dx =-\int \sum_{n=0}^{\infty}x^ndx=-\sum_{n=0}^{\infty}\int x^ndx

    This gives

    f(x)=C-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}

    and since f(0)=0 \implies C=0

    f(x)=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}
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  3. #3
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    Very nice. Thank you. I would never realized of it :P

    BTW, the problem is implicitly telling that the function is taken on an interval where its uniformly convergent, right?
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  4. #4
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    Oops, I thought wrong; sorry. :)
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  5. #5
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    I think not but now I don't need 'em anyway :P
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  6. #6
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    Quote Originally Posted by Ulysses View Post
    I think not but now I don't need 'em anyway :P
    Yeah, I missed that you accounted for the sign alternations in the denominators! :P
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by Ulysses View Post
    Very nice. Thank you. I would never realized of it :P

    BTW, the problem is implicitly telling that the function is taken on an interval where its uniformly convergent, right?
    Yes the geometric series is uniformly convergent on |x|<1
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  8. #8
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    Thank to all of you. Now I'm trying to solve the same exercise, but with the function f(x)=(1+4x^2)^{-1} I've tried to make some algebraic work to get the expression I'm looking for, (1-x)^{-1} but I didn't get it.

    I've made the first derivative, and a few more, but I think its just necessary the first, as before. I get:
    f'(x)=\diplaystyle\frac{-8x}{(1+4x^2)^2}

    I thought of making f(x)=\diplaystyle\frac{1}{(1+4x^2)}= \diplaystyle\frac{1-4x^2}{(1+4x^2)(1-4x^2)}=\diplaystyle\frac{1-4x^2}{(1-16x^2)}
    So, when I take the derivative in this last form I get:

    f'(x)=\diplaystyle\frac{-256x^3}{(1-16x^4)^2}

    I'm not sure how to proceed.
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  9. #9
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    Quote Originally Posted by Ulysses View Post
    Thank to all of you. Now I'm trying to solve the same exercise, but with the function f(x)=(1+4x^2)^{-1} I've tried to make some algebraic work to get the expression I'm looking for, (1-x)^{-1} but I didn't get it.

    I've made the first derivative, and a few more, but I think its just necessary the first, as before. I get:
    f'(x)=\diplaystyle\frac{-8x}{(1+4x^2)^2}

    I thought of making f(x)=\diplaystyle\frac{1}{(1+4x^2)}= \diplaystyle\frac{1-4x^2}{(1+4x^2)(1-4x^2)}=\diplaystyle\frac{1-4x^2}{(1-16x^2)}
    So, when I take the derivative in this last form I get:

    f'(x)=\diplaystyle\frac{-256x^3}{(1-16x^4)^2}

    I'm not sure how to proceed.
    It is just function composition.


    Let y=-4x^2

    Then we know that

    f(x)=\frac{1}{1+4x^2}=\frac{1}{1-y}=f(y)

    But this is just a geometric series so we get

    f(y)=\frac{1}{1-y}=\sum_{n=0}^{\infty}y^n

    Now just plug back in y=-4x^2

    f(x)=\sum_{n=0}^{\infty}(-4x^2)^n=\sum_{n=0}^{\infty}(-1)^n4^nx^n
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