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Math Help - Taylor series

  1. #1
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    Taylor series

    f(x) =ln (1-x)

    a) Compute f'(x), f''(x), f'''(x). Spot the pattern and give an expression for f ^(n) (x) [the n-th derivative of f(x)]

    b) Compute the MacLaurin series of f(x) (i.e. the Taylor series of f(x) around x=0)

    c) Compute the radius of convergence and determine the interval of convergence of the series in b).

    d) Determine the Taylor series of f'(x) around x=0. Can you do so without using b)?

    e) How would you have computed part b) if you had first done part d)?

    I did part a) (hope its right)

    f'(x) = -1/(1-x)
    f''(x) = -1/(1-x)^2
    f'''(x) = -2/(1-x)^3

    f^(n) (x) = -((n-1)!)/(1-x)^n
    for n = 1,2,3,...

    I am not too sure for the other ones, thanks for any help.
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  2. #2
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    Quote Originally Posted by @_@ View Post
    f(x) =ln (1-x)

    a) Compute f'(x), f''(x), f'''(x). Spot the pattern and give an expression for f ^(n) (x) [the n-th derivative of f(x)]
    f(x)=\ln (1-x)
    f(0)=\ln (1-0)=0
    Thus,
    a_0 =\frac{f(0)}{0!}=0

    f'(x)=-\frac{1}{1-x}
    f'(0)=-\frac{1}{1-0}=-1
    Thus,
    a_1=\frac{f'(0)}{1!}=-1

    f''(x)=\frac{1}{(1-x)^2}
    f''(0)=\frac{1}{(1-0)^2}=1
    Thus,
    a_2=\frac{f''(0)}{2!}=\frac{1}{2}

    f'''(x)=-\frac{2}{(1-x)^3}
    f'''(0)=-\frac{2}{(1-0)^3}=-1
    Thus,
    a_3=\frac{f'''(0)}{3!}=-\frac{1}{3}

    It seems the pattern is,

    \ln (1-x)=-\frac{x}{1}+\frac{x^2}{2}-\frac{x^3}{3}+...
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  3. #3
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    but its asking for the n-th derivative of f(x), would my answer then be correct?
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    Quote Originally Posted by @_@ View Post
    but its asking for the n-th derivative of f(x), would my answer then be correct?
    f^{(n)}(0)=\frac{(-1)^n}{n}
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  5. #5
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    oh. i see, then how do you computer radius of convergence and its interval?
    thanks.
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    Quote Originally Posted by @_@ View Post
    oh. i see, then how do you computer radius of convergence and its interval?
    thanks.
    The power series that we get is,*)
    \sum_{k=1}^{\infty}\frac{(-1)^k}{k}x^k

    In order to find the radius of convergence we will use the generalized ratio test for power series:
    \lim_{k\to \infty} \left| \frac{(-1)^{k+1}}{k+1}\cdot \frac{k}{(-1)^k} \right|=1
    Thus the reciprocal of that is also 1.
    Meaning the radius of convergence is 1.
    Thus,
    |x|<1 you can also write -1<x<1
    The series converges absolutely.


    *)Which is in fact not the power series for \ln (1-x).
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  7. #7
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    why is it not the power series? Then where did you get this equation?
    thanks.
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  8. #8
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    Quote Originally Posted by @_@ View Post
    why is it not the power series? Then where did you get this equation?
    thanks.
    Because if a power series does exist. Then this has to be it. It does not mean that this is it. In fact this power series is for \ln |1-x| for |x|<1.
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  9. #9
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    i am kind of lost right now, sorry about that.

    so far, we have found part a) which is the n-th derivative of f(x),
    so what about MacLaurin series of f(x), is it (-x/1) +(x^2/2) - (x^3/3) +...?

    and also for f''(x), i don't know why i kept getting f''(x) = -1/(1-x)^2, which you don't have a negative.

    thanks.
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  10. #10
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    Quote Originally Posted by @_@ View Post
    i am kind of lost right now, sorry about that.

    so far, we have found part a) which is the n-th derivative of f(x),
    so what about MacLaurin series of f(x), is it (-x/1) +(x^2/2) - (x^3/3) +...?

    and also for f''(x), i don't know why i kept getting f''(x) = -1/(1-x)^2, which you don't have a negative.

    thanks.
    Because you have,
    -\frac{1}{(1-x)^2}
    When you take the derivative you need to use the chain rule, the inner function is,
    1-x and you need to multiply by its derivative thus, -1. But you already have a minus sign, so they die and you have a positive sign. Basic differenciation.
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  11. #11
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    Lightbulb

    The general formula for a Taylor Series of f(x) around the point x=a is

    f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n where f^{(n)}(x) is the nth derivative.

    The MacLaurin Series is just a Taylor Series centered around a=0.
    Last edited by putnam120; November 27th 2006 at 08:42 AM.
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  12. #12
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    Quote Originally Posted by putnam120 View Post
    The general formula for a Taylor Series of f(x) around the point x=a is

    f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x)}{n!}(x-a)^n where f^{(n)}(x) is the nth derivative.

    The MacLaurin Series is just a Taylor Series centered around a=0.
    To add, "If a function is infinitely differenciable at x=a then the Taylor series is definied as a power series...."

    But, the Taylor series of a function does not always represent the function. You need to show that the remainder term sequence is convergent to zero. Like here.
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  13. #13
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    the taylor series would just be like this? f(x)= (-x/1) +(x^2/2) - (x^3/3) +...

    how would you put it in a form like the one putnam120 provide?
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  14. #14
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    Quote Originally Posted by ThePerfectHacker View Post
    To add, "If a function is infinitely differenciable at x=a then the Taylor series is definied as a power series...."

    But, the Taylor series of a function does not always represent the function. You need to show that the remainder term sequence is convergent to zero. Like here.
    o yes. its just that most of the functions that i work w/ the remainder tends to zero. but its not too hard to prove.
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  15. #15
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    Quote Originally Posted by @_@ View Post
    the taylor series would just be like this? f(x)= (-x/1) +(x^2/2) - (x^3/3) +...

    how would you put it in a form like the one putnam120 provide?
    f(x)=\sum_{k=1}^{\infty} \frac{(-1)^k}{k}x^k
    That is the summation form.

    (Again, f(x)\not = \ln (1-x))
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