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Series expansion help

  1. Jan 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that the first non-zero coefficient in the expansion of
    ##e^{-x}-\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}##
    in ascending powers of x is that of x^5

    2. Relevant equations
    Series expansion, logarithmic series

    3. The attempt at a solution
    Let ##y=e^{-x}-\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}##

    ##{\ln y={\ln \left[\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}\right]\ }\ }##

    ##{\ln y\ }={\ln \left(1-x\right)\ }-\frac{1}{2}{\ln \left(1-x^2\right)-\frac{1}{3}{\ln \left(1-x^3\right)\ }\ }##

    ##{\ln y\ }=\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}\right)-\frac{1}{2}\left(-x^2-\frac{x^4}{2}\right)-\frac{1}{3}\left(-x^3\right)##

    ##{\ln y\ }=-x-\frac{x^5}{5}##

    ##y=e^{-x}\cdot e^{-\frac{x^5}{5}}##

    ##ye^x=e^{-\frac{x^5}{5}}##

    ##\left[\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}\right]e^x=e^{-\frac{x^5}{5}}##



    I know I'm incredibly close to the answer, however it seems like I can't solve it.
     
  2. jcsd
  3. Jan 20, 2015 #2

    RUber

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    You seem to have left off the e^(-x) term in your first step with the log.
     
  4. Jan 21, 2015 #3
    I'm aware of this, however the reason why I do this is to find the series expansion of the (1-x)/((1-x^2)^(1/2) (1-x^3)^(1/3)) first, then use it to subtract from the series expansion of e^-x.
     
  5. Jan 21, 2015 #4

    RUber

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    Then perhaps you simply mislabeled your y.
    If ## y = \frac{ 1-x }{ (1-x^2)^{1/2}(1-x^3)^{1/3}} ## then I agree that you might write
    ## y = e^{-x - x^5/5}##
    Your expression then is ##e^{-x} - e^{-x - x^5/5} ##
    Take a look at the series representations of those two exponentials. The result should be clear.
     
  6. Jan 22, 2015 #5
    How do you get e^-x -e^(-x -x^5 /5) ?
     
  7. Jan 22, 2015 #6

    RUber

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    The way you expanded ln(y) was for the fraction part without the ##e^{-x}##. So ##e^{-x} - e^{ln y}## should be your expression.
    As you said in post #3, you wanted to subtract the fractional part from the exponential part.
     
  8. Jan 22, 2015 #7
    Wasn't if I use e^(ln y), the expression would change to e^(-x -x^5 /5) ?
     
  9. Jan 22, 2015 #8

    RUber

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    Back to my original point that in your first post you left ##e^{-x}## off when you took the log of y. So using the definition of y which I gave in post 4, you have the expression ##e^{-x}-y##. You did all the work to express ## y = e^{-x -x^5/5}##. Now you just have to subtract that from ##e^{-x}##.
     
  10. Jan 22, 2015 #9
    Now I have e^-x -e^(-x-x^5/5) Should I expand both as a series?
     
  11. Jan 23, 2015 #10

    RUber

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    If you do, your answer will be evident.
     
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