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Math Help - Taylor & MacLaurin series

  1. #1
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    Taylor & MacLaurin series

    Hi!

    I understand the concept of these series but sometimes I don't quite understand the problems.

    For example
    f(x) = 1 + x + x^2
    Represent this as a Taylor series.
    But the series isn't infinite, it stops, right?

    Also,
    Find the Maclaurin's series for:
    f(x) = 1/(1+x)^2
    f(x) = 1/(1+x)^3

    Thanks
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Volcanicrain View Post
    Hi!

    I understand the concept of these series but sometimes I don't quite understand the problems.

    For example
    f(x) = 1 + x + x^2
    Represent this as a Taylor series.
    But the series isn't infinite, it stops, right?

    Also,
    Find the Maclaurin's series for:
    f(x) = 1/(1+x)^2
    f(x) = 1/(1+x)^3

    Thanks
    For the first one use the fact that a taylor series is given by

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

    For the other two notice this \frac{D[\frac{1}{1+x}]}{dx}=\frac{-1}{(1+x)^2}

    Now using the fact that \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n

    We see that \frac{-1}{(1+x)^2}=\sum_{n=1}^{\infty}n(-1)^{n}x^{n-1}

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

    For the second one use the fact that \frac{1}{(1+x)^3}=\frac{1}{2}\frac{d^2[\frac{1}{1+x}]}{dx^2} and use the same trick
    Last edited by Mathstud28; June 1st 2008 at 09:14 AM.
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  3. #3
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    For the first one:
    The function only derives up to n = 3, so... That means that it's not really infinite, right, as all other terms would be zero?

    And for the second problem: Can we really derive series like that?? And why is it a negative series? ( I don't see where that comes from)
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Volcanicrain View Post
    For the first one:
    The function only derives up to n = 3, so... That means that it's not really infinite, right, as all other terms would be zero?

    And for the second problem: Can we really derive series like that?? And why is it a negative series? ( I don't see where that comes from)
    You are right so therefore since f^{(n>3)}(x)=0

    We have f(c)=\sum_{n=0}^{3}\frac{f^{(n)}(c)(x-c)^{n}}{n!}

    And yes...this is the most common form of series finding....for example

    find the series for \ln(1-x)


    Since \ln(1-x)=-\int\frac{dx}{1-x}=-\int\sum_{n=0}^{\infty}x^{n}=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}


    If you are unsure take a value in the interval of convergence and compare it to the actual function

    \ln(1-.5)=-.693=-\sum_{n=0}^{\infty}\frac{(.5)^{n+1}}{n+1}
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  5. #5
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    Interesting.
    Okay, hm. So any function could be represented as a Taylor series, but that series isn't infinite; it only goes as far as the nth derivative goes (that is, isn't equal to 0)

    Thanks Mathstud
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  6. #6
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    Quote Originally Posted by Volcanicrain View Post
    So any function could be represented as a Taylor series
    No. Only functions which are infinitely differenciable. And even if the function is infinitely differenciable the Taylor series does not necessarily converge to the function.
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  7. #7
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    So the function 1 + x + x^2 is not a Taylor Series even when you put into that form?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Volcanicrain View Post
    So the function 1 + x + x^2 is not a Taylor Series even when you put into that form?
    It is a taylor series

    It would be \sum_{n=0}^{\infty}\frac{f^{(n)}(c)(x-c)^{n}}{n!}

    which in this case would yield f(c)+f'(c)(x-c)+\frac{f''(c)(x-c)^2}{2!}+\frac{f'''(c)(x-c)^3}{3!}+0+0+0+0+0...
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  9. #9
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    Oh I understand. Mental slip, the function is infinitely differentiable, but the fact that it yields zeros for n greater than 3 makes it easier to calculate.

    Okay, thanks! You've helped me conceptualize it better.

    Now why would it not always necessarily converge to the function? (If you don't mind. I find this stuff interesting, but my teacher is so vague when teaching this stuff)
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Volcanicrain View Post
    Oh I understand. Mental slip, the function is infinitely differentiable, but the fact that it yields zeros for n greater than 3 makes it easier to calculate.

    Okay, thanks! You've helped me conceptualize it better.

    Now why would it not always necessarily converge to the function? (If you don't mind. I find this stuff interesting, but my teacher is so vague when teaching this stuff)
    Because numbers outside of the interval of convergence would make the series diverge

    For example \ln(1-x)=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1} \only on [-1,1)

    because if you put in any other values it would make the series diverge


    for example since 1 ins excluded I will show why

    -\sum_{n=0}^{\infty}\frac{1^{n+1}}{n+1}=-\sum_{n=0}^{\infty}\frac{1}{n+1}

    Which diverges by the integral test \int_0^{\infty}\frac{1}{x+1}=\ln(x+1)\bigg|_0^{\in  fty}=\infty

    Thus since the series diverges for x=1 it can't be used...the same goes for all other values not included in the interval of convergence
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  11. #11
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    Oh! You just blew my mind O_O

    So is it right to assume that if the Taylor series of a function converges, it converges to the function?
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  12. #12
    MHF Contributor arbolis's Avatar
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    And even if the function is infinitely differenciable the Taylor series does not necessarily converge to the function.
    Can you give me an example please? I know it's true, but don't know any example.
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  13. #13
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    Quote Originally Posted by arbolis View Post
    Can you give me an example please? I know it's true, but don't know any example.
    Define, f(x) = \left\{ \begin{array}{c}e^{-1/x^2} \mbox{ if }x\not = 0 \\ 0 \mbox{ if }x=0 \end{array} \right..

    Show that f^{(n)}(0) exists and is equal to 0.
    This would imply that f has a zero Taylor series.

    Note its Taylor series does not converge to the function f on any interval (-r,r) ( r>0).
    Thus, f cannot be expressed as an infinite series at 0.
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Volcanicrain View Post
    Oh my God you just blew my mind O_O

    So is it right to assume that if the Taylor series of a function converges, it converges to the function?
    Yes...I am not quite sure what relevance that has but yes....and just so you know

    theyre is a taylor series for any function about c...this is due to the fact that even if the series for something diverges basically everywhere it converges at its center

    For example if found a function that could would diverge at any value it would still diverge at its center c...this is due to the definition of the taylor series

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

    I dont care if it diverges for any values basically it will always converge at c...this is because if you imput c you get

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

    Which as I just showed is convergent...thus all functions have convergent power series at their centers
    Last edited by Mathstud28; May 11th 2008 at 08:17 PM.
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  15. #15
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    Interesting. Well thank you very much Mathstud, you've been a great help tonight ^^
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