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Binomial Series

  1. Nov 20, 2004 #1
    (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=0} ^{\infty} \frac{n^2}{2^n} [/tex]

    [tex] \hline[/tex]

    Here is what I've got:

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

    (b)

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

    Thank you
     
    Last edited: Nov 20, 2004
  2. jcsd
  3. Nov 20, 2004 #2
    Did you write the question correctly? I don't see how you can use partial fractions in the last equation you give to get the question again. I don't think you can either because for x = 2 the equivalency isn't there.
     
    Last edited by a moderator: Nov 20, 2004
  4. Nov 20, 2004 #3

    arildno

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    That last equality is incorrect!
     
  5. Nov 20, 2004 #4
    I see. You guys are right about that last equality.

    If I have

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

    I may need to rewrite

    [tex] \frac{n^2}{2^n} [/tex]

    and compute it with the aid of the f(x) I found in part (a). What I can do is:

    [tex] \frac{n^2}{2^n} = n^2 \left( \frac{1}{2} \right) ^n [/tex]

    but the connection with f(x) still is not clear.
     
  6. Nov 20, 2004 #5
    I've just made some corrections to my 1st post.... did I get it right this time?

    Thanks.
     
  7. Nov 20, 2004 #6
    vsage,

    I'm sorry. I wrote it incorrectly last time. It isn't:

    [tex] \frac {x^2+x}{1-x^3} [/tex]

    I've just edited my 1st post to fix it. Here it is:

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

    Sorry again, and thanks for your help. :smile:
     
  8. Nov 20, 2004 #7
    Consider that [tex]\frac {1}{(1-x)^3} = \frac {d^2\frac{1}{2} \frac{1}{1-x}}{dx^2}[/tex]
     
  9. Nov 20, 2004 #8
    Well, I see that:

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

    This method is, in fact, simpler. The directions of this problem in part (a) are to expand f(x) as a power series, so it's ok to do that. But, I've got this exercise from the "Binomial Series" section in my calc book, so I tried to stick to it. That's why I used it in my 1st post.

    Maybe your tip was about the connection between f(x) and the series in part (b). To be honest, I don't see how it may be related to it.
     
  10. Nov 21, 2004 #9
    Folks, let me rephrase what I have. I was able to develop the solution quite a bit farther, but there still are some unclear areas.

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

    (b)

    [tex] \sum _{n=1} ^{\infty} \frac{n^2}{2^n} = 6 = f\left( \frac{1}{2} \right) \qquad \fbox{UNCLEAR 2} [/tex]

    Comments:

    [tex] \fbox{UNCLEAR 1}: [/tex] I don't understand the transition from the series above to this result.
    [tex] \fbox{UNCLEAR 2}: [/tex] I've found the sum of the series ("6") with the aid of my calculator. Later, I evaluated a guess ("1/2"), which gave the correct answer. However, it still isn't explicit the connection between

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

    Thank you very much.
     
    Last edited: Nov 21, 2004
  11. Nov 21, 2004 #10
    Guys, I may have found the rest of the 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.
     
    Last edited: Nov 21, 2004
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