Taylor series for xln(x)

  • Thread starter Refraction
  • Start date
  • #1
21
0

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:

[tex]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!}[/tex]

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.
 

Answers and Replies

  • #2
2,967
5
Do you know the Taylor series expansion for [itex]\ln(1 + t)[/itex]? What is the relation between [itex]x[/itex] and [itex]t[/itex]?
 
  • #3
21
0
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:
  • #4
2,967
5
Why do you use [itex]x[/itex] in your expansion of [itex]\ln(1 + t)[/itex]? And, if you do not know what I mean by relation between [itex]x[/itex] and [itex]t[/itex], all I can say is think harder how this helps with regards to your problem.
 
  • #5
21
0
Sorry, it was supposed to be t in there, anyway I'll see if I can figure out what you meant.
 
  • #6
2
0

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:

[tex]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!}[/tex]
I don't understand why it isn't:

[tex]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!}[/tex]

Doesn't f'(0) = 0?
 
  • #7
21
0
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
2
0
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.
 
  • #9
21
0
if you do not know what I mean by relation between [itex]x[/itex] and [itex]t[/itex], 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?
 
  • #10
2,967
5
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 [itex]f(x)[/itex], but you know the Taylor expansion of [itex]f(1 + t)[/itex] around [itex]t = 0[/itex], then how should you express this [itex]t[/itex] in terms of the original [itex]x[/itex]? Also, to what value of [itex]x[/itex] would [itex]t = 0[/itex] correspond to?
 
  • #11
21
0
[itex]t=x-1[/itex] and when [itex]t=0, x=1[/itex] ? I just realised that for a function like [itex]xln(1+x)[/itex] you can find the series for [itex]ln(1+x)[/itex] and multiply the compact form by [itex]x[/itex] to give the expansion for [itex]xln(1+x)[/itex].

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 [itex]ln(x)[/itex] by [itex]x[/itex] to give the series for [itex]xln(x)[/itex], but that didn't work out the same way.
 
Last edited:
  • #12
2,967
5
The coefficient in front of [itex]ln(1+t)[/itex] is not [itex]t[/itex] 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
21
0
By the coefficient [itex]t[/itex] for [itex]ln(1+t)[/itex], do you mean when it would be [itex]t ln(1+t)[/itex]? (making it [itex](x-1)ln(x)[/itex] after the substitution)
 
  • #14
2,967
5
By the coefficient [itex]t[/itex] for [itex]ln(1+t)[/itex], do you mean when it would be [itex]t ln(1+t)[/itex]? (making it [itex](x-1)ln(x)[/itex] 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.
 

Related Threads on Taylor series for xln(x)

  • Last Post
Replies
4
Views
1K
Replies
3
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
16
Views
14K
  • Last Post
Replies
2
Views
696
  • Last Post
Replies
1
Views
1K
Replies
1
Views
5K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
4
Views
51K
Replies
4
Views
22K
Top