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Math Help - Taylor serie

  1. #1
    Newbie Klaus's Avatar
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    Lightbulb Taylor serie

    Hello,
    How to show that the Taylor serie of \frac{1}{1-x} converges ?
    I found that the serie was \sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}.

    Should I use Lagrange's remainder ?

    Thank you
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Klaus View Post
    Hello,
    How to show that the Taylor serie of \frac{1}{1-x} converges ?
    I found that the serie was \sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}.

    Should I use Lagrange's remainder ?

    Thank you
    \frac{1}{a-x}=\frac{1}{a}\frac{1}{1-\frac{x}{a}}=\frac{1}{a}\sum_{n=0}^{\infty}\left(\  frac{x}{a}\right)^n\quad\forall{x,a}\backepsilon\t  ext{ }\left|\frac{x}{a}\right|<1

    This is just an extension of the geometric series.


    Is this what you were asking?
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  3. #3
    Newbie Klaus's Avatar
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    In fact I would like to show that
    \sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}
    converges...
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Klaus View Post
    In fact I would like to show that
    \sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}
    converges...
    Ok, except it doesnt always now does it?

    If a=0 and x=10 for example

    But utilizing the Root test we can see that

    \sum_{n=1}^{\infty}\frac{(x-a)^{n-1}}{(1-a)^n} converges

    iff \bigg|\frac{x-a}{1-a}\bigg|<1
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  5. #5
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by Klaus View Post
    In fact I would like to show that
    \sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}
    converges...
    \sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}

    simplifies to

    \sum_{n=1}^{\infty} \frac{(x-a)^{n}}{(x-a)(1-a)^n}

    \frac{1}{x-a} \sum_{n=1}^{\infty} \left(\frac{(x-a)}{(1-a)}\right)^n

    It converges as a geometric sum if |x - a| < |1 - a| which means it will definitely converge for values of x < 1
    Last edited by colby2152; June 23rd 2008 at 09:21 AM.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by colby2152 View Post
    \sum_{n=1}^{\infty} \frac{(x-a)^{n-1}}{(1-a)^n}

    simplifies to

    \sum_{n=1}^{\infty} \frac{(x-a)^{n}}{(x-a)(1-a)^n}

    \frac{1}{x-a} \sum_{n=1}^{\infty} \left(\frac{(x-a)}{(1-a)}\right)^n

    It converges as a geometric sum if x - a < 1 - a which means it will definitely converge for values of x < 1
    Didn't you neglect an absolute value there?
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  7. #7
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Didn't you neglect an absolute value there?
    Good point or else there could be an infinite large negative sum.

    Also, note that a \ne 1 & x \ne a
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