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Taylor series for xln(x)

  1. May 13, 2010 #1
    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:

    [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.
     
  2. jcsd
  3. May 13, 2010 #2
    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]?
     
  4. May 13, 2010 #3
    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
  5. May 13, 2010 #4
    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.
     
  6. May 13, 2010 #5
    Sorry, it was supposed to be t in there, anyway I'll see if I can figure out what you meant.
     
  7. May 13, 2010 #6
    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?
     
  8. May 13, 2010 #7
    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.
     
  9. May 13, 2010 #8
    I made a typo. I was supposed to write f'(1) = 0. But yeah, I was wrong anyway.
     
  10. May 13, 2010 #9
    I'm still not having much luck figuring this out, any chance you could be a bit more specific about what you meant?
     
  11. May 13, 2010 #10
    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?
     
  12. May 13, 2010 #11
    [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: May 13, 2010
  13. May 14, 2010 #12
    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.
     
  14. May 14, 2010 #13
    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)
     
  15. May 14, 2010 #14
    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.
     
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