Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Some PreExam problems

  1. Nov 19, 2006 #1
    There are some problems my lecturer gave me which I can't solve ,or I am very unsure about my approach :

    1. Study the nature of this series : [tex] \sum_{n=1}^\infty \frac{a^{n}}{n^{2}} [/tex]

    2. Expand as a series of powers of x the function [tex] f(x)= ln{(1+x)} + \frac {1}{1-x} + e^{2x} [/tex] and determine the convergence radius of the resulting one.


    I know that the last one is done using the Taylor but I ain't sure about my approach, some hints pls.

    Thank you for your time.
     
    Last edited: Nov 19, 2006
  2. jcsd
  3. Nov 19, 2006 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    For the first one, do the terms get larger or smaller as n increases?

    For the second one, what is your approach that you're not sure about?
     
  4. Nov 19, 2006 #3
    Sorry I have an error in the latex code on the first one.

    @Office_Shredder : Depends on the value of a :
    if it is in (-1,1) then the terms become smaller,
    if if is in {-1,1} then they also become smaller,
    else they become larger, as n increases.


    On the secound one, I know some Taylor expansions as I searched for them on wikipedia, and I know the [tex] e^{x} [/tex] expansion, as I have to get e^2x I think I will multiply that expansion by itself...
     
  5. Nov 19, 2006 #4
    Sorry about my Latex problems... I have now finally fixed them. :D
     
  6. Nov 19, 2006 #5
    1. [tex] \sum_{n=1}^\infty \frac{a^{n}}{n^{2}} [/tex]

    [tex] \lim_{n\rightarrow \infty} \frac{a^{n}}{n^{2}}\times n = \frac{a^{n}}{n} = \infty [/tex] (assuming a > 1) So it converges at 1 and diverges for all other x.






    2. [tex] f(x)= \ln{(1+x)} + \frac {1}{1-x} + e^{2x} [/tex]

    [tex] \ln{(1+x)} = \int \frac{1}{1+x} [/tex]
     
    Last edited: Nov 19, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook