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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?

    Sincerely

    Hummingbird25
     
    Last edited: May 8, 2006
  2. jcsd
  3. May 8, 2006 #2

    Curious3141

    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.
     
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