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Math Help - maclaurin series

  1. #1
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    maclaurin series

    differentiate the maclaurin series for 1/(1-x) twice to find the maclaurin series of 1/(1-x)^3. Thanks
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  2. #2
    Super Member Deadstar's Avatar
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    \frac{1}{1-x} = \sum^{\infty}_{n=0} x^n \text{ for }|x| < 1

    If you differentiate \frac{1}{1-x} twice you get \frac{2}{(1-x)^3}

    You just differentiate each term in the sum above and divide by 2. After a few terms you should see a pattern then just turn it into an infinite sum.
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  3. #3
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    Thanks, but what would the maclaurin series be? Is it something like
    sum_(n=0)^infinity1/2 n (n-1) x^(n-2). How do you get that?
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  4. #4
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Brotha View Post
    Thanks, but what would the maclaurin series be? Is it something like
    sum_(n=0)^infinity1/2 n (n-1) x^(n-2). How do you get that?
    \frac{1}{(1-x)} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + etc...

    So differentiate both sides...

    -\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + 5 x^4 + 6x^5 + etc...

    differentiate again...

    \frac{2}{(1-x)^3} = 2 + 6x + 12x^2 + 20x^3 + 30x^4 + etc...

    This is a series of the form...

    \sum_{n=0}^{\infty} (n+1)(n+2)x^n.

    How did I get this? Just differentiate x^n twice. You don't even need to all that stuff I did above. Just it's good to see it done so you know what's going on.

    Differentiating x^n twice gives you n(n-1)x^{n-2}. I.e. (n+2)(n+1)x^n

    So summarizing...

    \frac{2}{(1-x)^3} = \sum_{n=0}^{\infty} (n+2)(n+1)x^n

    => \frac{}{(1-x)^3} = \frac{1}{2} \sum_{n=0}^{\infty} (n+2)(n+1)x^n
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