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!

Homework Help: HELP: A complex valued power series

  1. May 8, 2006 #1
    Hi All,

    I have this here power series which is complex valued

    [tex]\frac{2n+1}{2^n} i^n = \frac{4}{25} + \frac{22}{25}i[/tex]

    My task is to prove that this power series has the above mentioned sum.

    To do this I seperate the sum into two sums

    [tex]S_0 + S_1 = (i/2)^n + 2* (i/2)*n*(i/2)^{(n-1)}[/tex]

    the first is easy since can use

    [tex]S_0 = \frac{1}{1-(i/2)} = 4/5+2/5i[/tex]

    The second is quite hard to compute. I'm to told that I need to differentiate, by using the sum

    [tex]\sum _{n=0} ^{\infty} x^n = \frac{i}{2}[/tex]

    then by differentiating the sum I get

    [tex]f'(x) = \sum_{n=0} ^{\infty} n x^{n-1}[/tex]

    In my textbook it says that x^n converge to 1/(1-x)

    Then since |i/2| < 1, I differentiate both sides of the equation.

    [tex]\frac{1}{(1-(i/2))^2} = \sum_{n=0} ^{\infty} n (i/2)^{n-1}[/tex]

    I multiply by i/2 on both sides of the equation to get.

    [tex]\frac{i/2 * 1}{(1-(i/2))^2} =\sum_{n=0} ^{\infty}(i/2) n (i/2)^{n-1} [/tex]

    then since |i/2| < 1, then I multiply by a factor of two on both sides of the eqation

    [tex]2\frac{i/2 * 1}{(1-(i/2))^2} = \sum_{n=0} ^{\infty}2 (i/2) n (i/2)^{n-1}[/tex]

    [tex]-16/25 + 12/25i = \sum_{n=0} ^{\infty}2 (i/2) n (i/2)^{n-1}[/tex]

    Finally [tex]\frac{4}{5}+ \frac{2}{5i}+ (\frac{-16}{25} + \frac{12}{25i}) = \frac{4}{25} + \frac{22}{25}i[/tex]

    I have the feeling that I'm doing something wrong?


    Last edited: May 8, 2006
  2. jcsd
  3. May 8, 2006 #2


    User Avatar
    Homework Helper

    Sorry, I don't really have the time to really scrutinise the working you gave. There does seem to be a little typo in the LaTex, but it seems generally OK.

    There's a much easier way than differentiation to work out sums of the form

    [tex]S = \sum{nx^{-n}}[/tex] or variants on the theme.

    Just multiply by x to get

    [tex]Sx = \sum{nx^{1-n}}[/tex]

    and compare like powers of Sx to S term by term.

    You will find that [tex]Sx - S = S(1-x)[/tex] will amount to a very easy geometric progression and you can just divide that by (1-x) to find S.

    Much easier than differentiation, IMHO.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook