# Taylor series for xln(x)

1. May 13, 2010

### Refraction

1. The problem statement, all variables and given/known data

For f(x) = xln(x), find the taylor series expansion of f(x) about x = 1, and write the infinite series in compact form.

2. The attempt at a solution

I can find the expansion itself fine, these are the first few terms:

$$0 + (x-1) + \frac{(x-1)^{2}}{2!} - \frac{(x-1)^{3}}{3!} + \frac{2(x-1)^{4}}{4!} - \frac{6(x-1)^{5}}{5!} + \frac{24(x-1)^{6}}{6!}$$

but I'm having trouble finding a pattern for the nth term (and I can't write it in compact form without that). f(1) and the first 7 derivatives of f(x) at x = 1 end up being 0, 1, 1, -1, 2, -6, 24, and it's mainly that part that seems to make finding it a problem.

2. May 13, 2010

### Dickfore

Do you know the Taylor series expansion for $\ln(1 + t)$? What is the relation between $x$ and $t$?

3. May 13, 2010

### Refraction

I guess it would be t - (t^2)/2 + (t^3)/3 - (t^4)/4 + (t^5)/5 etc. if it was about t = 0? Not sure what you mean by the relationship between x and t though.

Last edited: May 13, 2010
4. May 13, 2010

### Dickfore

Why do you use $x$ in your expansion of $\ln(1 + t)$? And, if you do not know what I mean by relation between $x$ and $t$, all I can say is think harder how this helps with regards to your problem.

5. May 13, 2010

### Refraction

Sorry, it was supposed to be t in there, anyway I'll see if I can figure out what you meant.

6. May 13, 2010

### DJ Hobo

I don't understand why it isn't:

$$0 + 0 + \frac{(x-1)^{2}}{2!} - \frac{(x-1)^{3}}{3!} + \frac{2(x-1)^{4}}{4!} - \frac{6(x-1)^{5}}{5!} + \frac{24(x-1)^{6}}{6!}$$

Doesn't f'(0) = 0?

7. May 13, 2010

### Refraction

f'(x) is ln(x) + 1, and in this question it's at x = 1 not x = 0 (which wouldn't be defined for f'(x) anyway)
so it's f'(1) = ln(1) + 1 = 1 in that case.

8. May 13, 2010

### DJ Hobo

I made a typo. I was supposed to write f'(1) = 0. But yeah, I was wrong anyway.

9. May 13, 2010

### Refraction

I'm still not having much luck figuring this out, any chance you could be a bit more specific about what you meant?

10. May 13, 2010

### Dickfore

If you need the Taylor expansion of a function $f(x)$, but you know the Taylor expansion of $f(1 + t)$ around $t = 0$, then how should you express this $t$ in terms of the original $x$? Also, to what value of $x$ would $t = 0$ correspond to?

11. May 13, 2010

### Refraction

$t=x-1$ and when $t=0, x=1$ ? I just realised that for a function like $xln(1+x)$ you can find the series for $ln(1+x)$ and multiply the compact form by $x$ to give the expansion for $xln(1+x)$.

It looks like I need to do something similar here, but I'm not sure what that last step would be now. I thought I might be able to multiply the taylor series of $ln(x)$ by $x$ to give the series for $xln(x)$, but that didn't work out the same way.

Last edited: May 13, 2010
12. May 14, 2010

### Dickfore

The coefficient in front of $ln(1+t)$ is not $t$ anymore after this substitution had been made. You get two terms. Then you need to add the series term by term and simplify everything.

13. May 14, 2010

### Refraction

By the coefficient $t$ for $ln(1+t)$, do you mean when it would be $t ln(1+t)$? (making it $(x-1)ln(x)$ after the substitution)

14. May 14, 2010

### Dickfore

No. Will you please provide a step by step procedure for your solution, so that we can tell you where you are wrong. We are not allowed to give detailed solutions.