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Math Help - 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 g(x) as a series of non-negative powers of x. I was thinking it could be related to g(x) being undefined at 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 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 g(x) as a series of non-negative powers of x. I was thinking it could be related to g(x) being undefined at 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 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 -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 -xg(-x) = xg(x)
    I don't know if you are being sarcastic but if -xg(-x)-xg(x) the function is odd...because if g(x) was even g(-x)=g(x)...so wouldnt -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 -xg(-x)-xg(x) the function is odd...because if g(x) was even g(-x)=g(x)...so wouldnt -xg(-x)=-xg(x) if it was even...sorry if I am being stupid
    I was being sarcastic, but g(x) is odd and xg(x) is even. why ? because g(x) is the product of an odd and even function and 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 g(x) is odd and xg(x) is even. why ? because g(x) is the product of an odd and even function and 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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    <br /> <br />
\sinh \left( x \right) \cdot \cosh \left( {2x} \right) = \frac{{e^{3x}  + e^{ - x}  - e^x  - e^{ - 3x} }}<br />
{4} = \sum\limits_{n = 0}^\infty  {\frac{{3^n  + \left( { - 1} \right)^n  - 1 - \left( { - 3} \right)^n }}<br />
{{4 \cdot n!}}x^n } <br /> <br /> <br />

    Suppose <br />
\frac{1}<br />
{{\sinh \left( x \right) \cdot \cosh \left( {2x} \right)}} = \sum\limits_{n = 0}^\infty  {a_n  \cdot x^n } <br />

    So multiplying we get: <br />
\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 }}<br />
{{4 \cdot n!}}} \right) \cdot x^n } } \right] = 1<br /> <br />

    Collecting terms: <br />
\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} }}<br />
{{4 \cdot \left( {n - k} \right)!}}} \right)} } \right] \cdot x^n }  = 1<br /> <br /> <br />

    Thus we should have <br /> <br />
a_0  \cdot \left( {\tfrac{{3^0  + \left( { - 1} \right)^0  - 1 - \left( { - 3} \right)^0 }}<br />
{{4 \cdot \left( 0 \right)!}}} \right) = 1<br /> <br /> <br />
which is clearly absurd.
    Last edited by PaulRS; April 21st 2008 at 04:56 PM. Reason: I'd thought it was cosh(x) and not cosh(2x)
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