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Math Help - Analysis

  1. #1
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    Post Analysis

    For this problem I need to apply Taylor's Formula
    the function is f(x)= lnx
    where x0=1

    So the Formula would be:
    lnx= (Sigma) k=0 to n (ak (x-1)^k + M(x-1)^((n+1)))

    first how do I get the general formula for ak and M.
    __________________________________________________ _____________

    2.
    Let f(x)= x^2 and epsilon=.01 . Try a value delta for which
    |x-y| < .delta implies |x^2 - y^2| < .01

    for all x,y in [0,117].

    For this problem, lets say I choose delta to be 1
    then what do we do??
    Thank You
    Last edited by Sally_Math; December 5th 2009 at 06:56 AM.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Sally_Math View Post
    For this problem I need to apply Taylor's Formula
    the function is f(x)= lnx
    where x0=1

    So the Formula would be:
    lnx= (Sigma) k=0 to n (ak (x-1)^k + M(x-1)^((n+1)))

    first how do I get the general formula for ak and M.

    Thank You
    First lets calculate some derivatives.

    f(x)=\ln(x)
    f'(x)=\frac{1}{x}
    f''(x)=\frac{(-1)}{x^2}
    f'''(x)=\frac{(-1)(-2)}{x^3}
    f^{(4)}(x)=\frac{(-1)(-2)(-3)}{x^4}
    Notice the pattern so we get the nth derivative is
    f^{(n)}(x)=\frac{(-1)(-2)\cdots(-n+1)}{x^{n}}=\frac{(-1)^{n-1}(n-1)!}{x^n}

    Since f(1)=\ln(1)=0

    Now by the defintion of Taylor series we get

    \ln(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}}{n!}(x-1)^n=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n=

    \ln(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^n

    Notice that the series is alternating and I hope this helps.
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    First lets calculate some derivatives.

    f(x)=\ln(x)
    f'(x)=\frac{1}{x}
    f''(x)=\frac{(-1)}{x^2}
    f'''(x)=\frac{(-1)(-2)}{x^3}
    f^{(4)}(x)=\frac{(-1)(-2)(-3)}{x^4}
    Notice the pattern so we get the nth derivative is
    f^{(n)}(x)=\frac{(-1)(-2)\cdots(-n+1)}{x^{n}}=\frac{(-1)^{n-1}(n-1)!}{x^n}

    Since f(1)=\ln(1)=0

    Now by the defintion of Taylor series we get

    \ln(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}}{n!}(x-1)^n=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n=

    \ln(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^n

    Notice that the series is alternating and I hope this helps.
    so can we say that the general formula for ak is
    \ln(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}}{n!}(x-1)^n
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by Sally_Math View Post
    so can we say that the general formula for ak is
    \ln(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}}{n!}(x-1)^n
    Using the defintion of the Taylor series expansion at the point a

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

    So in your case n \ge 1

    a_n=\frac{f^{n}(1)}{n!}=\frac{(-1)^{n-1}}{n}
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  5. #5
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    Thanks
    I don't know why I am having a hard time with this
    so
    lets say if x=.75 and n=3, how do I find c:____

    and for M is it ((-1)^n (n-1)!)/n!

    I'm kind of confused, :S
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