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

Alv95
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I have some problems finding Taylor's expansion at infinity of

<br /> f(x) = \frac{x}{1+e^{\frac{1}{x}}} <br />

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

<br /> g(u) = \frac{1}{u} \cdot \frac{1}{1+e^u} \hspace{10 mm} \mbox{ where } \hspace{10 mm} u = 1/x <br />

in order to then use the known expansion of \frac{1}{1+t} but the problem is that I can not do it because :
\lim_{ u \to 0 } e^{u} = 1 \hspace{10 mm} \mbox{ and not } 0Any ideas on how to do it? :smile: Thanks :smile:
 
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You should express that denominator uas
\frac{1}{2+(e^{u}-1)}
\frac{1}{2} \frac{1}{1+ (e^{u}-1)/2}
Now if u is close to 0 (eu-1)/2 is close to zero and we can expand inthis
\frac{1}{2} \left(1-\frac{e^u-1}{2} +\left(\frac{e^u-1}{2} \right)^2+... \right)
 
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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:
 
... The problem is now that I get a different expansion depending on how far I go in the approximation ... those

\left( \frac{\cdots \, \, - 1}{2}\right)^n

change the first term of the expansion ...

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

EDIT: Done! Thanks :smile:
 
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Alv95 said:
I have some problems finding Taylor's expansion at infinity of

<br /> f(x) = \frac{x}{1+e^{\frac{1}{x}}} <br />

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

<br /> g(u) = \frac{1}{u} \cdot \frac{1}{1+e^u} \hspace{10 mm} \mbox{ where } \hspace{10 mm} u = 1/x <br />

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


Any ideas on how to do it? :smile: Thanks :smile:

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
 
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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: \frac{1}{1+e^u} as a "generating function" so that it will then generate the required expansion ?
...
I get \frac{1}{1+e^u} = \frac{e^u - 1}{e^{2u}-1} but it is a little bit different from the standard forms of the generating functions ...

Any insight on Bernoulli numbers is welcome! :biggrin:
 
Alv95 said:
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: \frac{1}{1+e^u} as a "generating function" so that it will then generate the required expansion ?
...
I get \frac{1}{1+e^u} = \frac{e^u - 1}{e^{2u}-1} but it is a little bit different from the standard forms of the generating functions ...

Any insight on Bernoulli numbers is welcome! :biggrin:

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.
 
Alv95 said:
I have some problems finding Taylor's expansion at infinity of

<br /> f(x) = \frac{x}{1+e^{\frac{1}{x}}} <br />
This function doesn't have a Taylor expansion at infinity because it diverges in that limit. Did you perhaps mean a Laurent expansion instead?
 
ok :wink:

I have got:

\frac{1}{y^2-1} = \frac{1}{2(y-1)}-\frac{1}{2(y+1)}

therefore:

\frac{1}{y+1} = \frac{1}{y-1}-\frac{2}{y^2-1}

My function is then:

\frac{1}{e^u+1} = \frac{1}{e^u-1}-\frac{2}{e^{2u}-1}
 
  • #10
Alv95 said:
ok :wink:

I have got:

\frac{1}{y^2-1} = \frac{1}{2(y-1)}-\frac{1}{2(y+1)}

therefore:

\frac{1}{y+1} = \frac{1}{y-1}-\frac{2}{y^2-1}

My function is then:

\frac{1}{e^u+1} = \frac{1}{e^u-1}-\frac{2}{e^{2u}-1}

OK, so take it from there!
 
  • #11
@ Vela

I have found an oblique asymptote:

y = \frac{x}{2} - \frac{1}{4}
 
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