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    The question is: Find the first four non-vanishing terms of:

    iii.  \frac{x}{e^x -1}

    So I've tried two methods, one method I substituted  y=e^x but that got complicated quickly and I wasn't sure what to do with it once I subbed back in for  e^x next I tried the old fashion differentiate it each time method and that didn't work because I kept ending up with 3 products to differentiate (and we haven't learnt a product rule for 3 things multiplied together).

    Attached is the worked solution which I don't even understand, come someone please help me with this question as I am well and truly stuck!

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    (Original post by physicsboy1)
    The question is: Find the first four non-vanishing terms of:

    iii.  \frac{x}{e^x -1}

    So I've tried two methods, one method I substituted  y=e^x but that got complicated quickly and I wasn't sure what to do with it once I subbed back in for  e^x next I tried the old fashion differentiate it each time method and that didn't work because I kept ending up with 3 products to differentiate (and we haven't learnt a product rule for 3 things multiplied together).

    Attached is the worked solution which I don't even understand, come someone please help me with this question as I am well and truly stuck!

    write the exponential series at the bottom
    cancel the 1
    then cancel an x all they through
    then ...
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    (Original post by TeeEm)
    write the exponential series at the bottom
    cancel the 1
    then cancel an x all they through
    then ...
    It's still \dfrac{1}{(...)}, so we need some way to convert the (...) to the numerator.

    OP, the product rule for 3 is (fgh)' = f'gh + fg'h + fgh', you can prove this by doing (fgh)' = ((fg)h)' = (fg)'h + (fg)h' = (f'g + fg')h + fgh' = f'gh + fg'h + fgh'.

    Now for the answer you posted, the writer just substituted in Y = \dfrac{e^x - x - 1}{x}, because the taylor series of Y = \dfrac{x}{2} + \dfrac{x^2}{6} + ..., so at the Y^4 boundary, it is bound to stop (you won't need to consider any higher Y powers).

    I tried integrating \displaystyle \dfrac{x}{e^x - 1}, and then get the taylor series of the integral and then differentiate, but the integral is not nice at all, so that fails. I think the writer of the solution just wanted to force a substitution. So you can attempt various substitution via trial and error, for which \dfrac{x}{e^x - 1} in terms of the substitution Y has a nice taylor series, and you can sub your values back in and do that hog-mess of algebra.
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    (Original post by 0x2a)
    It's still \dfrac{1}{(...)}, so we need some way to convert the (...) to the numerator.

    OP, the product rule for 3 is (fgh)' = f'gh + fg'h + fgh', you can prove this by doing (fgh)' = ((fg)h)' = (fg)'h + (fg)h' = (f'g + fg')h + fgh' = f'gh + fg'h + fgh'.

    Now for the answer you posted, the writer just substituted in Y = \dfrac{e^x - x - 1}{x}, because the taylor series of Y = \dfrac{x}{2} + \dfrac{x^2}{6} + ..., so at the Y^4 boundary, it is bound to stop (you won't need to consider any higher Y powers).

    I tried integrating \displaystyle \dfrac{x}{e^x - 1}, and then get the taylor series of the integral and then differentiate, but the integral is not nice at all, so that fails. I think the writer of the solution just wanted to force a substitution. So you can attempt various substitution via trial and error, for which \dfrac{x}{e^x - 1} in terms of the substitution Y has a nice taylor series, and you can sub your values back in and do that hog-mess of algebra.
    do you know the binomial expansion?
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    (Original post by TeeEm)
    do you know the binomial expansion?
    Ohhhh of course
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    (Original post by 0x2a)
    Ohhhh of course
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    in fact I am saying silly things too

    factorize minus out
    -x/(1- ex) then binomial
 
 
 
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