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Math Help - Taylor series

  1. #1
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    Taylor series

    Find the Taylor series of the given function at a = 0

    <br />
f(x)=\frac{1}{(1-x)^2}<br />


    Your help is very appreciated!
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  2. #2
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    I got to the point where I found that

    <br />
f^{(n)}(a) = (n + 1)!<br />

    But I am not sure how to proceed from here...
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  3. #3
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    Quote Originally Posted by hasanbalkan View Post
    Find the Taylor series of the given function at a = 0

    <br />
f(x)=\frac{1}{(1-x)^2}<br />


    Your help is very appreciated!
    It's been a while since I've done these, but let's give it a try.

    f(x)=\frac{1}{(1-x)^2}

    So now we need to start taking some derivatives, and see what kind of pattern shows up.

    f'(x) = \frac{2}{(1-x)^3}

    I can see a pattern now just from this derivative that a few things are going to happen. The sign is going to alternate each term, the power in the denominator will increase by 1, and the numerator will have a coefficient of (n+1)!, where n is the nth derivative.

    So generally, f^{n}(x) = \frac{(-1)^{n+1}(n+1)!}{(1-x)^{n+2}}

    I think that looks good, but I apologize if I messed up. Also what's nice is since you are doing this for a=0, the denominator will always be 1.

    As for the rest of the Taylor formula...

    f(x) = \sum_{n=0}^{\infty} \frac{f^{n}(a)}{n!}(x-a)^n

    Can you take it from here?
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  4. #4
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    Quote Originally Posted by hasanbalkan View Post
    I got to the point where I found that

    <br />
f^{(n)}(a) = (n + 1)!<br />

    But I am not sure how to proceed from here...
    I think the sign will alternate for the derivatives. When you plug this in to Taylor's formula, the (n+1)! and n! can reduce nicely.

    Once you do that, write out the first few terms and see what pattern you can see.
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  5. #5
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    Quote Originally Posted by hasanbalkan View Post
    Find the Taylor series of the given function at a = 0

    <br />
f(x)=\frac{1}{(1-x)^2}<br />


    Your help is very appreciated!
    Do you how to find a Taylor series?

    You need to find derivatives of the function for the following formula.

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  6. #6
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    Thank you for the quick response! Now I know what I need to do.
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by hasanbalkan View Post
    Find the Taylor series of the given function at a = 0

    <br />
f(x)=\frac{1}{(1-x)^2}<br />


    Your help is very appreciated!
    Use the fact that the maclaurin series for \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n

    Now notice that \frac{D[\frac{1}{1-x}]}{dx}=\frac{1}{(1-x)^2}

    Now that we have shown this relationship between \frac{1}{1-x} and \frac{1}{(1-x)^2} we can make the logical next step

    Since the derivative of \frac{1}{1-x} is \frac{1}{(1-x)^2}

    and we know that \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n

    THen it is logical to assume that the power series for \frac{1}{(1-x)^2} is the derivative of \sum_{n=0}^{\infty}x^n

    Which is \sum_{n=1}^{\infty}nx^{n-1}

    But now that we have doen this we must also recall that taking the derivative of a series can change its endpoint behavior...so therefore we know this series converges on (-1,1) because that is the radisu of convergence of \sum_{n=0}^{\infty}x^{n}

    But now we must check -1 and 1

    To do this we imput the two values...if we imput 1 we get

    \sum_{n=1}^{\infty}n1^{n-1}=\sum_{n=1}^{\infty}n

    and using the n-th term test we see that \lim_{n\to\infty}n=\infty
    so this new series diverges at x=1

    Now we must check when x=-1
    Maknig the series \sum_{n=1}^{\infty}n(-1)^{n-1}

    But we dont even have to check the Alternating series test because as we showed \lim_{n\to\infty}a_n=\lim_{n\to\infty}n=\infty
    So the second prerequisite of the alternating series test has already failed so this series is divergent at x=-1

    So we no know that the series is \sum_{n=1}^{\infty}nx^{n-1} and is convergent on (-1,1)

    Just for fun and to check \frac{1}{(1-.5)^2}=4

    and \sum_{n=1}^{\infty}n(.5)^{n-1}=4

    haha I know that was much more than was needed but I hoped it would be helpful in learning how to find these!

    Mathstud
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