Taylor series expansion for xln(x) with x = 1

[Refraction]

Homework Statement

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.

 

[Dickfore]

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

 

[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.

 

[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.

 

[Refraction]

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

 

[DJ Hobo]

Homework Statement

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!}

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?

 

[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.

 

[DJ Hobo]

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.

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

 

[Refraction]

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.

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

 

[Dickfore]

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

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?

 

[Refraction]

t=x-1 and when t=0, x=1 ? I just realized 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.

 

[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.

 

[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)

 

[Dickfore]

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)

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.