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Taylor expansion at infinity of x/1+e^(1/x)

  1. May 26, 2013 #1
    I have some problems finding Taylor's expansion at infinity of

    [itex]
    f(x) = \frac{x}{1+e^{\frac{1}{x}}}
    [/itex]

    I tried to find Taylor's expansion at 0 of :

    [itex]
    g(u) = \frac{1}{u} \cdot \frac{1}{1+e^u} \hspace{10 mm} \mbox{ where } \hspace{10 mm} u = 1/x
    [/itex]

    in order to then use the known expansion of [itex] \frac{1}{1+t} [/itex] but the problem is that I can not do it because :
    [itex] \lim_{ u \to 0 } e^{u} = 1 \hspace{10 mm} \mbox{ and not } 0 [/itex]


    Any ideas on how to do it? :smile: Thanks :smile:
     
  2. jcsd
  3. May 26, 2013 #2

    Office_Shredder

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    You should express that denominator uas
    [tex] \frac{1}{2+(e^{u}-1)} [/tex]
    [tex] \frac{1}{2} \frac{1}{1+ (e^{u}-1)/2} [/tex]
    Now if u is close to 0 (eu-1)/2 is close to zero and we can expand inthis
    [tex] \frac{1}{2} \left(1-\frac{e^u-1}{2} +\left(\frac{e^u-1}{2} \right)^2+.... \right) [/tex]
     
  4. May 26, 2013 #3
    Thank you very much Office_Shredder :smile:

    I had thought of adding/subtracting 1 but not of factorising the 2 :redface:

    Thank you again :wink:
     
  5. May 26, 2013 #4
    ... The problem is now that I get a different expansion depending on how far I go in the approximation ... those

    [itex]\left( \frac{\cdots \, \, - 1}{2}\right)^n[/itex]

    change the first term of the expansion ...

    EDIT: Sorry :smile: I should replace [itex]\left( \frac{e^u - 1}{2}\right)[/itex] with his expansion, right? :smile:

    EDIT: Done! Thanks :smile:
     
    Last edited: May 26, 2013
  6. May 26, 2013 #5

    Ray Vickson

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    Do it in terms of Bernoulli numbers. Bernoulli numbers appear in the series for ##1/(e^x - 1)##. You need to get the series for ##1/(e^x+1)##, which you can do by expressing ##1/(e^x+1)## in terms of ##1/(e^x - 1)## and ##1/(e^{2x}-1)##; in other words, you need to express ##1/(y+1)## in terms of ##1/(y-1)## and ##1/(y^2 - 1)##.

    For more on Bernoulli numbers, see, eg., http://mathworld.wolfram.com/BernoulliNumber.html
     
    Last edited: May 26, 2013
  7. May 26, 2013 #6
    I don't know yet what Bernoulli numbers are (we don't do them in my school) :smile: Thanks anyways for your help :smile:
    ...
    From what I have understood from Wikipedia (:rolleyes:) I should rewrite my function: [itex]\frac{1}{1+e^u}[/itex] as a "generating function" so that it will then generate the required expansion ?
    ...
    I get [itex]\frac{1}{1+e^u} = \frac{e^u - 1}{e^{2u}-1}[/itex] but it is a little bit different from the standard forms of the generating functions ....

    Any insight on Bernoulli numbers is welcome! :biggrin:
     
  8. May 26, 2013 #7

    Ray Vickson

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    Google is your friend---look them up!

    I would not write ##1/(y+1) = (y-1)/(y^2-1)##; that just makes the problem much worse!

    Suppose I asked you to do the reverse, which would be to write ##1/(y^2-1)## in as simple a way as possible in terms of ##1/(y-1)## and ##1/(y+1)##. What would you do? Have you ever encountered partial fractions?

    I don't think it matters whether or not you have seen Bernoulli numbers in your school; all that matters is that you are allowed (I suppose) to go to the library and look things up in a book, or (nowadays) go on-line to find information. Just *cite* your sources, whether they are books or web pages.
     
  9. May 26, 2013 #8

    vela

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    This function doesn't have a Taylor expansion at infinity because it diverges in that limit. Did you perhaps mean a Laurent expansion instead?
     
  10. May 26, 2013 #9
    ok :wink:

    I have got:

    [itex]\frac{1}{y^2-1} = \frac{1}{2(y-1)}-\frac{1}{2(y+1)}[/itex]

    therefore:

    [itex]\frac{1}{y+1} = \frac{1}{y-1}-\frac{2}{y^2-1}[/itex]

    My function is then:

    [itex]\frac{1}{e^u+1} = \frac{1}{e^u-1}-\frac{2}{e^{2u}-1}[/itex]
     
  11. May 26, 2013 #10

    Ray Vickson

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    OK, so take it from there!
     
  12. May 26, 2013 #11
    @ Vela

    I have found an oblique asymptote:

    [itex]y = \frac{x}{2} - \frac{1}{4}[/itex]
     
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