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Thread: power series expansion

  1. #1
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    power series expansion

    g(x) = 2 / (1-x)^3 center c=0
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by harveyo View Post
    g(x) = 2 / (1-x)^3 center c=0
    notice that $\displaystyle \frac 2{(1 - x)^3}$ is the second derivative of $\displaystyle \frac 1{1 - x}$, and we know the power series for that. simply derive the power series for $\displaystyle \frac 1{1 - x}$ twice and that's your answer.
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    notice that $\displaystyle \frac 2{(1 - x)^3}$ is the second derivative of $\displaystyle \frac 1{1 - x}$, and we know the power series for that. simply derive the power series for $\displaystyle \frac 1{1 - x}$ twice and that's your answer.

    $\displaystyle 2/ (1-x)^3 = 2 * 1/(1-x)^3 $


    $\displaystyle = 2 * -1/2 d/dx [1/(1-x)^2] $

    $\displaystyle = -1 d/dx [1/(1-x)^2] $

    $\displaystyle = -1 d/dx [-1/(1-x)] $

    ?????????
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  4. #4
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    You haven't done anything that jhevon suggested!

    He suggested that you write out the power series for $\displaystyle \frac{1}{1- x}$ and differentiate that twice!

    What is the power series for $\displaystyle \frac{1}{1- x}$?

    (Hint: the sum of the geometric series $\displaystyle \sum_{n=0}^\infty ar^n$ is $\displaystyle \frac{a}{1- r}$.)
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  5. #5
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    power series

    1/ 1-x = 1 + x +x^2 + x^3 ......

    therefore

    1/ (1-x)^2 = 1+ 2x +3x^2 +4x^3 ......

    now

    1/(1-x)^3 = 2 + 6x + 12x^2 .....

    ?
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    Quote Originally Posted by harveyo View Post
    g(x) = 2 / (1-x)^3 center c=0
    Another approach:

    Write $\displaystyle g(x) = 2 (1-x)^{-3}$

    and apply the Binomial Theorem (for negative exponents).
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by harveyo View Post
    power series

    1/ 1-x = 1 + x +x^2 + x^3 ......

    therefore

    1/ (1-x)^2 = 1+ 2x +3x^2 +4x^3 ......

    now

    1/(1-x)^3 = 2 + 6x + 12x^2 .....

    ?
    the left side of your third equation is incorrect. and when i said differentiate the power series for 1/(1 - x) i was talking about differentiating the expression behind the summation sign in $\displaystyle \sum_{n = 0}^\infty x^n$, for $\displaystyle |x|<1$, twice

    or you can follow awkwards suggestion if you learned the negative binomial theorem
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