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Just need a quick explanation (Power Series)

  1. Nov 21, 2004 #1
    Problem:

    (a) Expand

    [tex] f(x)=\frac{x+x^2}{\left( 1-x\right) ^3} [/tex]

    as a power series.

    (b) Use part (a) to find the sum of the series

    [tex] \sum _{n=1} ^{\infty} \frac{n^2}{2^n} [/tex]

    [tex] \hline[/tex]

    Solution:

    (a)

    [tex] f(x)=\frac{x+x^2}{\left( 1-x\right) ^3} = \left( x + x^2 \right) \left[ 1 + (-x) \right] ^{-3} = \left( x + x^2 \right) \sum _{n=0} ^{\infty} \binom{-3}{n} (-x) ^n = \left( x + x^2 \right) \sum _{n=0} ^{\infty} (-1) ^n \binom{-3}{n} x ^n [/tex]

    [tex] f(x) = \left( x + x^2 \right) \left[ \binom{-3}{0} - \binom{-3}{1}x + \binom{-3}{2} x^2 - \binom{-3}{3} x^3 + \binom{-3}{4} x^4 - \cdots \right] [/tex]

    [tex] f(x) = \left( x + x^2 \right) \left[ 1 - \frac{(-3)}{1!}x + \frac{(-3)(-4)}{2!}x^2 - \frac{(-3)(-4)(-5)}{3!}x^3 + \frac{(-3)(-4)(-5)(-6)}{4!}x^4 -\cdots \right] [/tex]

    [tex] f(x) = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} \frac{\left( n+2 \right)!}{n!}x^n \right] = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} (n+2)(n+1)x^n \right] [/tex]

    [tex] f(x) = x + x^2 + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+1} \right] + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+2} \right] [/tex]

    [tex] f(x) = x + x^2 - \frac{x\left( 3x-3x^2+x^3\right)}{\left( x-1 \right) ^3} - \frac{x^2\left( 3x-3x^2+x^3\right)}{\left( x-1 \right) ^3} \qquad \fbox{UNCLEAR} [/tex]

    [tex] f(x) = x + x^2 - \left[ x +\frac{1}{\left( x-1 \right) ^2} +\frac{1}{\left( x-1 \right) ^3} \right] - \left[ x^2 + \frac{1}{ x-1} + \frac{2}{\left( x-1 \right) ^2} + \frac{1}{\left( x-1 \right) ^3} \right] [/tex]

    [tex] f(x) = -\frac{1}{ x-1} -\frac{3}{\left( x-1 \right) ^2} - \frac{2}{\left( x-1 \right) ^3} = \frac{x\left( x+1 \right)}{\left( x-1 \right) ^3} = x \frac{d}{dx} x \frac{d}{dx} \left[ \frac{1}{1-x} \right] = x \frac{d}{dx} x \frac{d}{dx} \left[ \sum _{n=0} ^{\infty} x^n \right] [/tex]

    [tex] f(x) = \sum _{n=1} ^{\infty} n^2 x^n [/tex]

    (b)

    [tex] \sum _{n=1} ^{\infty} \frac{n^2}{2^n} = \sum _{n=1} ^{\infty} n^2 \left( \frac{1}{2} \right) ^n = f\left( \frac{1}{2} \right) = \frac{\frac{1}{2}+\left( \frac{1}{2} \right)^2}{\left( 1-\frac{1}{2}\right) ^3} = 6 [/tex]

    Comments:

    [tex] \fbox{UNCLEAR}: [/tex] I don't understand the transition from the series above to this result.

    Thank you :smile:
     
  2. jcsd
  3. Nov 21, 2004 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    You know

    [tex]g(x) = \sum _{n = 0} ^{\infty}\frac{g^{(n)}(a)}{n!}(x-a)^n[/tex]

    Can you find some g such that you can express:

    [tex] \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+1} \right][/tex]

    in the form above?
     
  4. Nov 22, 2004 #3
    Thanks for the tip!

    (a)

    [tex] f(x)=\frac{x+x^2}{\left( 1-x\right) ^3} = \left( x + x^2 \right) \left[ 1 + (-x) \right] ^{-3} = \left( x + x^2 \right) \sum _{n=0} ^{\infty} \binom{-3}{n} (-x) ^n = \left( x + x^2 \right) \sum _{n=0} ^{\infty} (-1) ^n \binom{-3}{n} x ^n [/tex]

    [tex] f(x) = \left( x + x^2 \right) \left[ \binom{-3}{0} - \binom{-3}{1}x + \binom{-3}{2} x^2 - \binom{-3}{3} x^3 + \binom{-3}{4} x^4 - \cdots \right] [/tex]

    [tex] f(x) = \left( x + x^2 \right) \left[ 1 - \frac{(-3)}{1!}x + \frac{(-3)(-4)}{2!}x^2 - \frac{(-3)(-4)(-5)}{3!}x^3 + \frac{(-3)(-4)(-5)(-6)}{4!}x^4 -\cdots \right] [/tex]

    [tex] f(x) = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} \frac{\left( n+2 \right)!}{n!}x^n \right] = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} (n+2)(n+1)x^n \right] [/tex]

    [tex] f(x) = x + x^2 + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+1} \right] + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+2} \right] [/tex]

    [tex] f(x) = x + x^2 + \left\{ \frac{x}{2} \frac{d^2}{dx^2} \left[ x^2 \sum _{n=0} ^{\infty} x^n \right] - x \right\} + \left\{ \frac{x^2}{2} \frac{d^2}{dx^2} \left[ x^2 \sum _{n=0} ^{\infty} x^n \right] - x^2 \right\} [/tex]

    [tex] f(x) = x + x^2 + \left\{ \frac{x}{2} \frac{d^2}{dx^2} \left[ x^2 \left( \frac{1}{1-x} \right) \right] - x \right\} + \left\{ \frac{x^2}{2} \frac{d^2}{dx^2} \left[ x^2 \left( \frac{1}{1-x} \right) \right] - x^2 \right\} [/tex]

    [tex] f(x) = x + x^2 - \frac{x\left( 3x-3x^2+x^3\right)}{\left( x-1 \right) ^3} - \frac{x^2\left( 3x-3x^2+x^3\right)}{\left( x-1 \right) ^3} [/tex]

    [tex] f(x) = x + x^2 - \left[ x +\frac{1}{\left( x-1 \right) ^2} +\frac{1}{\left( x-1 \right) ^3} \right] - \left[ x^2 + \frac{1}{ x-1} + \frac{2}{\left( x-1 \right) ^2} + \frac{1}{\left( x-1 \right) ^3} \right] [/tex]

    [tex] f(x) = -\frac{1}{ x-1} -\frac{3}{\left( x-1 \right) ^2} - \frac{2}{\left( x-1 \right) ^3} = \frac{x\left( x+1 \right)}{\left( x-1 \right) ^3} = x \frac{d}{dx} x \frac{d}{dx} \left[ \frac{1}{1-x} \right] = x \frac{d}{dx} x \frac{d}{dx} \left[ \sum _{n=0} ^{\infty} x^n \right] [/tex]

    [tex] f(x) = \sum _{n=1} ^{\infty} n^2 x^n [/tex]

    (b)

    [tex] \sum _{n=1} ^{\infty} \frac{n^2}{2^n} = \sum _{n=1} ^{\infty} n^2 \left( \frac{1}{2} \right) ^n = f\left( \frac{1}{2} \right) = \frac{\frac{1}{2}+\left( \frac{1}{2} \right)^2}{\left( 1-\frac{1}{2}\right) ^3} = 6 [/tex]
     
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