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Thread: Series expansion question.

  1. #1
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    Series expansion question.

    The first part i had no trouble with.

    Its just the explanation for part 2 for why it is not possible to express $\displaystyle g(x)$ as a series of non-negative powers of x. I was thinking it could be related to $\displaystyle g(x)$ being undefined at $\displaystyle x=0 $ but then it can still be expressed as a series of non-negative power of (x-a) though, and why does it work for $\displaystyle xg(x)$? i am struggling to give a decent explanation for this part

    Thanks

    Bobak
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by bobak View Post
    The first part i had no trouble with.

    Its just the explanation for part 2 for why it is not possible to express $\displaystyle g(x)$ as a series of non-negative powers of x. I was thinking it could be related to $\displaystyle g(x)$ being undefined at $\displaystyle x=0 $ but then it can still be expressed as a series of non-negative power of (x-a) though, and why does it work for $\displaystyle xg(x)$? i am struggling to give a decent explanation for this part

    Thanks

    Bobak
    Well...I know the reason that the the powers are even is when you add the two powers of x on the denominator then subtract one you get an even number
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    Well...I know the reason that the the powers are even is when you add the two powers of x on the denominator then subtract one you get an even number
    I also know the reason the powers are even, thanks for answering a question I didn't ask. the function is even $\displaystyle -xg(-x) = xg(x)$
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by bobak View Post
    I also know the reason the powers are even, thanks for answering a question I didn't ask. the function is even $\displaystyle -xg(-x) = xg(x)$
    I don't know if you are being sarcastic but if $\displaystyle -xg(-x)-xg(x)$ the function is odd...because if g(x) was even $\displaystyle g(-x)=g(x)$...so wouldnt $\displaystyle -xg(-x)=-xg(x)$ if it was even...sorry if I am being stupid
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    I don't know if you are being sarcastic but if $\displaystyle -xg(-x)-xg(x)$ the function is odd...because if g(x) was even $\displaystyle g(-x)=g(x)$...so wouldnt $\displaystyle -xg(-x)=-xg(x)$ if it was even...sorry if I am being stupid
    I was being sarcastic, but $\displaystyle g(x)$ is odd and $\displaystyle xg(x)$ is even. why ? because $\displaystyle g(x)$ is the product of an odd and even function and $\displaystyle xg(x)$ is the product of an odd and odd function.

    Bobak
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by bobak View Post
    I was being sarcastic, but $\displaystyle g(x)$ is odd and $\displaystyle xg(x)$ is even. why ? because $\displaystyle g(x)$ is the product of an odd and even function and $\displaystyle xg(x)$ is the product of an odd and odd function.

    Bobak
    I am sorry...I misunderstood your question...and you need not be so hostile
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  7. #7
    Super Member PaulRS's Avatar
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    $\displaystyle

    \sinh \left( x \right) \cdot \cosh \left( {2x} \right) = \frac{{e^{3x} + e^{ - x} - e^x - e^{ - 3x} }}
    {4} = \sum\limits_{n = 0}^\infty {\frac{{3^n + \left( { - 1} \right)^n - 1 - \left( { - 3} \right)^n }}
    {{4 \cdot n!}}x^n }


    $

    Suppose $\displaystyle
    \frac{1}
    {{\sinh \left( x \right) \cdot \cosh \left( {2x} \right)}} = \sum\limits_{n = 0}^\infty {a_n \cdot x^n }
    $

    So multiplying we get: $\displaystyle
    \left( {\sum\limits_{n = 0}^\infty {a_n \cdot x^n } } \right) \cdot \left[ {\sum\limits_{n = 0}^\infty {\left( {\tfrac{{3^n + \left( { - 1} \right)^n - 1 - \left( { - 3} \right)^n }}
    {{4 \cdot n!}}} \right) \cdot x^n } } \right] = 1

    $

    Collecting terms: $\displaystyle
    \sum\limits_{n = 0}^\infty {\left[ {\sum\limits_{k = 0}^n {a_k \cdot \left( {\tfrac{{3^{n - k} + \left( { - 1} \right)^{n - k} - 1 - \left( { - 3} \right)^{n - k} }}
    {{4 \cdot \left( {n - k} \right)!}}} \right)} } \right] \cdot x^n } = 1


    $

    Thus we should have $\displaystyle

    a_0 \cdot \left( {\tfrac{{3^0 + \left( { - 1} \right)^0 - 1 - \left( { - 3} \right)^0 }}
    {{4 \cdot \left( 0 \right)!}}} \right) = 1


    $ which is clearly absurd.
    Last edited by PaulRS; Apr 21st 2008 at 04:56 PM. Reason: I'd thought it was cosh(x) and not cosh(2x)
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