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

1. May 26, 2013

### Alv95

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

$f(x) = \frac{x}{1+e^{\frac{1}{x}}}$

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

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

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? Thanks

2. May 26, 2013

### Office_Shredder

Staff Emeritus
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
$$\frac{1}{2} \left(1-\frac{e^u-1}{2} +\left(\frac{e^u-1}{2} \right)^2+.... \right)$$

3. May 26, 2013

### Alv95

Thank you very much Office_Shredder

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

Thank you again

4. May 26, 2013

### Alv95

... 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 I should replace $\left( \frac{e^u - 1}{2}\right)$ with his expansion, right?

EDIT: Done! Thanks

Last edited: May 26, 2013
5. May 26, 2013

### Ray Vickson

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
6. May 26, 2013

### Alv95

I don't know yet what Bernoulli numbers are (we don't do them in my school) Thanks anyways for your help
...
From what I have understood from Wikipedia () 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!

7. May 26, 2013

### Ray Vickson

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.

8. May 26, 2013

### vela

Staff Emeritus
This function doesn't have a Taylor expansion at infinity because it diverges in that limit. Did you perhaps mean a Laurent expansion instead?

9. May 26, 2013

### Alv95

ok

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. May 26, 2013

### Ray Vickson

OK, so take it from there!

11. May 26, 2013

### Alv95

@ Vela

I have found an oblique asymptote:

$y = \frac{x}{2} - \frac{1}{4}$