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Thread: Taylor Series

  1. #1
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    Taylor Series

    I'm suppose to use the Taylor series at x = 0 for
    $\displaystyle
    \frac{1}{1-x} which, I believe, is 1 + x + x^2 + x^3 + ...,
    $

    for the following function: $\displaystyle \frac{x}{1 + x^3}$

    I have no idea how I 'apply' the first to the second. What does that mean?

    What does $\displaystyle
    \frac{1}{1-x} = 1 + x + x^2 + x^3 + ...,
    $

    have to do with the second function and how do I 'apply' it?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Why don't write in the expression $\displaystyle \frac{1}{1-x}$ instead of $\displaystyle x$ [for example...] $\displaystyle -x^{3}$ ?...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    I don't understand, what do I replace with what, and where, and why?

    (I'm still confused).
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  4. #4
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    you know that $\displaystyle \frac1{1-x}=\sum x^n$ which is the Taylor series at $\displaystyle x=0$ (well known as McLaurin series), then put $\displaystyle x\mapsto -x^3,$ thus $\displaystyle \frac1{1+x^3}=\sum (-1)^nx^{3n}$ so multiply both sides by $\displaystyle x$ and you're done.
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  5. #5
    MHF Contributor chisigma's Avatar
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    For $\displaystyle |x|<1$ is...

    $\displaystyle \frac{1}{1-x} = 1 + x + x^{2} + \dots$ (1)

    If we replace in (1) $\displaystyle x$ with $\displaystyle -x^{3}$ we obtain...

    $\displaystyle \frac{1}{1+x^{3}} = 1 - x^{3} + x^{6} - \dots$ (2)

    ... and now we multiply both terms of (2) by $\displaystyle x$ we obtain...

    $\displaystyle \frac{x}{1+x^{3}} = x - x^{4} + x^{7} - \dots$ (3)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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