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Math Help - Taylor Series

  1. #1
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    Taylor Series

    Hello,

    I have this problem, that requires us to derive the taylor series based at x = -1 for f(x) from scratch, then find an interval of converge for the series, including endpoint behavior, and finally approximating:
    ln((-1 + 2*Pi)/(Pi)) accurate to within 10^-25.

    f(x) = ln(x + 2)

    Okay, so this is what I have done.

    I first found the first 6 derivatives to find a pattern. So they are as follows:

    f'(x) = 1/(x+2)
    f''(x) = -1/(x+2)^2
    f^3(x) = 2/(x+2)^3
    f^4(x) = -6(x+2)^4
    f^5(x) = 24/(x+2)^5
    f^6(x) = -120/(x+2)^6

    And so forth.

    So then I expanded it to:

    1*(x+2)^(-1)
    1*(-1)*(x+2)^(-2)
    1*(-1)*(-2)*(x+2)^(-3)
    1*(-1)*(-2)*(-3)*(x+3)^(-4)

    And so forth.

    Okay, so the coefficients are factorials. The coefficient for the nth derivative is just:

    (+ or -)(n-1)!

    Now, the nth derivative can be written as follows:

    (+ or -)(n-1)!(x+2)^(-n)

    The signs alternate, so I need a factor of (-1)^n or (-1)^(n+1), whichever matches up correctly

    For n = 1,2,3,..., the signs are +,-,+,..., therefore I need to use the (-1)^(n+1)

    So I have (-1)^(n+1)(n-1)!(x+2)^(-n)

    What I really wanted was the nth derivative evaluated at x=-1, so substituted -1 in for x to get:

    (-1)^(n+1)(n-1)!(1)^(-n)= (-1)^(n+1)(n-1)!

    Now I have to use the Taylor series expansion formula:

    S(x) = sum(a_n(x-x_0)^n, where a_n = [S^n(x_0)]/n!

    So now I have sum((A_n/n!)(x+1)^n, x, 0, infinity).

    A_n is the nth deriv. evaluated at x = -1.

    Now what. How do I evaluate the other parts and further expand this if its not done so already. How do I get an approximation accurate to within such a small error bound.

    Book has the following result: Sum from n=1 to infinity of [(-1)^(n+1) / n] * (x+1)^n ... I was close just don't really see it. How would I do the remaining parts for accuracy and approximating the function above?

    Thanks for your help.
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  2. #2
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    Definition:
    A function which is infinitely diffrenciable its "Taylor Series centered at c" is defined as the power series:
    g(x)=\sum^{\infty}_{k=0}\frac{f^{(k)}(c)(x-c)^k}{k!}

    You have,
    y=\ln (x+2)
    It is centered at c=-1
    Then, (infinitely diffrenciable for x=2)
    y'(x)=\frac{1}{x+2}
    y''(x)=-\frac{1}{(x+2)^2}
    y'''(x)=\frac{2}{(x+2)^3}
    y^{(4)}(x)=-\frac{6}{(x+2)^4}
    In general, n\geq 1
    y^{(n)}(x)=(-1)^{n-1}\frac{(n-1)!}{(x+2)^n}
    When evaluated at x=-1 we have,
    y^{(n)}(-1)=(-1)^{n-1}\frac{(n-1)!}{(-1+2)^n}=(-1)^{n-1}(n-1)!

    Thus, its Taylor Series is,
    g(x)=f(0)+\sum^{\infty}_{k=1}\frac{(-1)^{k-1}(k-1)!(x+1)^k}{k!}
    Which simplifies to,
    g(x)=f(0)+\sum^{\infty}_{k=1}\frac{(-1)^k(x+1)^k}{k}
    Thus,
    g(x)=\ln(2)-\frac{x+1}{1}+\frac{(x+1)^2}{2}-\frac{(x+1)^3}{3}+...

    Further we can prove that the Lagrange remainder is convergent to 0, thus this power series is a representation for \ln(x+2)
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    Definition:

    Thus,
    g(x)=\ln(2)-\frac{x+1}{1}+\frac{(x+1)^2}{2}-\frac{(x+1)^3}{3}+...

    Further we can prove that the Lagrange remainder is convergent to 0, thus this power series is a representation for \ln(x+2)
    When |x+1|<1

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack
    When |x+1|<1

    RonL
    If that did that in front of Cauchy I would be dead.
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  5. #5
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    I continued to derive the Taylor's Series for it using my way...when I last was on here, I was at this point:

    So now I have sum((A_n/n!)(x+1)^n, x, 0, infinity).

    I substituted what A_n was and simplified.

    I ended up with summation(([(-1)^(n+1)]/n)*(x+1)^n, n = 1...infinity) since n = 0 is zero, therefore does not appear.

    I think that's the right derivation (or hope, since everything will be messed up).

    For part b, it asks to find the interval of convergence for this series, including endpoint behavior.

    I decided to use the RATIO TEST to determine this.

    So: [[(x+1)^(n+1)]/(n+1)]/[(x+1)^n]/n...which simplifies to:

    [n*(x+1)]/(n+1)...now the ratio test says to take the limit of that as
    n -> infinity. If the limit is < 1, it converges. If it is > 1, it diverges. If it = 1, it is inconclusive.

    I found the limit is x + 1. Oh, I forgot to mention that its the absolute value of that. So, if |x+1| < 1, it converges. If |x+1| > 1, it diverges.

    So: will not converge when |-x - 1| > 1 or |x + 1| > 1 or (x < -2 OR x > 0)

    Therefore, x needs to be between -2 and 0 for it to converge.

    For endpoints, to check convergencejust plug those two points into the original taylor series? And then what?

    Now, part c, I have no idea how to approach this part:

    "Use your results, along with other ideas, to approximate
    ln((-1 + 2*Pi)/(Pi)) accurate to within 10^(-25).

    THANKS!
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