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Math Help - Taylor Polynomials and Series..

  1. #1
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    Taylor Polynomials and Series..

    Hello!

    I need help figuring out the Taylor series..I keep getting undetermined form in these functions..Please help!

    Q1) Suppose f(x) = cos(2x) is replaced by its Taylor Polynomial P2(X) about the point X(0) = pi/2 on the interval 0<= x <= pi. Estimate the error?

    Q2) Derive the Taylor Series for f(x) = ln(1+x) with X(0) = 0 and determine for which value of X the series converges.

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

    I need help figuring out the Taylor series..I keep getting undetermined form in these functions..Please help!

    Q1) Suppose f(x) = cos(2x) is replaced by its Taylor Polynomial P2(X) about the point X(0) = pi/2 on the interval 0<= x <= pi. Estimate the error?

    Q2) Derive the Taylor Series for f(x) = ln(1+x) with X(0) = 0 and determine for which value of X the series converges.

    Thanks!
    For the first one this is an alternating series, so use the appropriate error test.

    For the second one

    Consider that

    \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n\quad\forall{x}\in(-1,1)

    and that

    \ln(1+x)=\int\frac{dx}{1+x}
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  3. #3
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    Edit: sorry, ignore this, didn't see mathstud's post

    Q1) Suppose f(x) = cos(2x) is replaced by its Taylor Polynomial P2(X) about the point X(0) = pi/2 on the interval 0<= x <= pi. Estimate the error?
    The formula you are probably using is
    error \leq M\frac{r^{n+1}}{(n+1)!}
    where M is the maximum value of f^{(n+1)} on the interval and r is the maximum distance from the point around which the taylor series is expanded and a point in the interval.

    Q2) Derive the Taylor Series for f(x) = ln(1+x) with X(0) = 0 and determine for which value of X the series converges.
    If this doesn't make sense I don't know how you are supposed to do this, but I am sure there is a way.

    \log(1+x) = \int\frac{1}{x+1}dx
    =\int\frac{1}{1--x}
    using geometric series, with |-x|<1
    =\int \sum_{n=0}^{\infty}(-x)^n
    =\sum_{n=0}^{\infty}\int (-x)^n
    =\sum_{n=0}^{\infty}\frac{1}{n+1}(-x)^{n+1}
    =\sum_{n=1}^{\infty}\frac{1}{n}(-x)^{n}
    =\sum_{n=1}^{\infty}\frac{(-1)^n}{n}x^{n}

    The series converges whenever |x|<1 because 1 is the distance to the nearest value of x that produces a singularity in C.
    Last edited by badgerigar; June 25th 2008 at 05:58 PM. Reason: ninjaed
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by badgerigar View Post
    If this doesn't make sense I don't know how you are supposed to do this, but I am sure there is a way.

    \log(1+x) = \int\frac{1}{x+1}dx
    =\int\frac{1}{1--x}
    using geometric series, with |-x|<1
    =\int \sum_{n=0}^{\infty}(-x)^n
    =\sum_{n=0}^{\infty}\int (-x)^n
    =\sum_{n=0}^{\infty}\frac{1}{n+1}(-x)^{n+1}
    =\sum_{n=1}^{\infty}\frac{1}{n}(-x)^{n}
    =\sum_{n=1}^{\infty}\frac{(-1)^n}{n}x^{n}

    The series converges whenever |x|<1 because 1 is the distance to the nearest value of x that produces a singularity in C.
    Uhm, the other way would be using the definition of the Maclaurin series

    f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)x^n}{n!} and notcing the pattern

    And the interval of convergence is (-1,1)

    By the root test.
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  5. #5
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    Uhm, the other way would be using the definition of the Maclaurin series

    and notcing the pattern
    oh. good point .

    I just looked at the wikipedia entry for the root test and it looked useful too
    Thanks
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by badgerigar View Post
    oh. good point .

    I just looked at the wikipedia entry for the root test and it looked useful too
    Thanks
    How is it that you are working with power series but are ignorant of the Root test (not at all meant to be hostile, I am just curious)
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    How is it that you are working with power series but are ignorant of the Root test (not at all meant to be hostile, I am just curious)
    I dunno. I'm a second year student at Melbourne uni and I don't recall having seen it. The most advanced stuff I have done in power series was in a semester of complex analysis, although Taylor series have been a major part of every single subject I have done since starting uni except my first year programming subjects. I also learn stuff as I read on this forum, which gives me a more patchy education on certain topics, but that doesn't really apply to power series because I have been doing that as part of my course.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by badgerigar View Post
    I dunno. I'm a second year student at Melbourne uni and I don't recall having seen it. The most advanced stuff I have done in power series was in a semester of complex analysis, although Taylor series have been a major part of every single subject I have done since starting uni except my first year programming subjects. I also learn stuff as I read on this forum, which gives me a more patchy education on certain topics, but that doesn't really apply to power series because I have been doing that as part of my course.
    Hmm, that is interesting. You have taken complex analysis, with all its Laurent series and such I am sure then that you are well aquainted with poewr series, but usually the Root test which is an integral part of series convergence/divergence you skipped. It's not a bad thing I suppose, just peculiar . You should probably learn it
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