1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Series expansion of xln((x+1)/x)

  1. Oct 9, 2011 #1
    basically i have to check if
    [tex] xln\frac{(x+1)}{x} [/tex]→ 1 as x→∞


    the first term is 0 as x→∞

    in the answers they say they used maclaurin series and got

    [tex]x(\frac{1}{x} + O\frac{1}{1^{2}}) [/tex]

    but dont show how they did it.

    would the first term in the series be

    [tex]a(ln(\frac{a+1}{a}))[/tex]

    there a = 0, but then the ln function is defined?

    im getting a little confused about have a=0 but we want to figure out where the function goes to as x→∞
     
  2. jcsd
  3. Oct 9, 2011 #2
    Looks to me it should be:

    [tex]1/x(x+O(k))[/tex]

    if instead of taking the limit at infinity, we take the limit at zero of the expression:

    [tex]1/x\ln\left(\frac{1+1/x}{1/x}\right)[/tex]

    Now we can express the log expression as a MacLaurin series.
     
  4. Oct 12, 2011 #3
    sorry i dont understand what you mean.

    are you rewriting our original expression?

    how can i write the ln(1+(1/x)) expression so that its defined for x=0
     
  5. Oct 12, 2011 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Why is jackmell's suggestion useful?

    [itex]\displaystyle \frac{1+1/x}{1/x}=x+1[/itex]
     
  6. Oct 12, 2011 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You should not care what happens near x = 0; after all, the question is asking about what happens as x --> infinity. So, if x is large, y = 1/x is small, and surely you know how ln(1+y) behaves near y = 0.

    RGV
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Series expansion of xln((x+1)/x)
  1. Taylor series for xln(x) (Replies: 13)

Loading...