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  1. #1
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    taylor series question

    Hi everybody, I'm having trouble with this problem, if you could help it would be greatly appreciated...
    find the Taylor series summation for f(x) = x/((2x+1)(3x-2)) based at 1. Then give an interval where the Taylor series converges.

    Thanks in Advanced
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  2. #2
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    Quote Originally Posted by juntheking@yahoo.com View Post
    Hi everybody, I'm having trouble with this problem, if you could help it would be greatly appreciated...
    find the Taylor series summation for f(x) = x/((2x+1)(3x-2)) based at 1. Then give an interval where the Taylor series converges.

    Thanks in Advanced
    by parital fractions we get

    f(x)= \frac{1}{7} \left(\frac{2}{3x-2}+\frac{1}{2x+1} \right)

    f'(x)=\frac{1}{7} \left(\frac{2\cdot 3(-1)}{(3x-2)^2}+\frac{1 \cdot 2 (-1)}{(2x+1)^2} \right)

    f''(x)=\frac{1}{7} \left(\frac{2\cdot 3^2(-1)(-2)}{(3x-2)^3}+\frac{1 \cdot 2^2 (-1)(-2)}{(2x+1)^3} \right)

    f^{n}(x)=\frac{1}{7} \left(\frac{2\cdot 3^n(-1)^nn!)}{(3x-2)^{n+1}}+\frac{1 \cdot 2^n (-1)^nn!}{(2x+1)^{n+1}} \right)

    you can use this to generate the coeffeints for x=1 to get

    \frac{1}{3}-\frac{8}{9}(x-1)+\frac{70}{27}(x-1)^2-\frac{626}{81}(x-1)^3+...
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  3. #3
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    I understand why you did that, but how do I describe this in terms of a summation comparing this to the known summation of 1/1-x based at 0?

    thanks again
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  4. #4
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    Ok

    Quote Originally Posted by TheEmptySet View Post
    by parital fractions we get

    f(x)= \frac{1}{7} \left(\frac{2}{3x-2}+\frac{1}{2x+1} \right)

    f'(x)=\frac{1}{7} \left(\frac{2\cdot 3(-1)}{(3x-2)^2}+\frac{1 \cdot 2 (-1)}{(2x+1)^2} \right)

    f''(x)=\frac{1}{7} \left(\frac{2\cdot 3^2(-1)(-2)}{(3x-2)^3}+\frac{1 \cdot 2^2 (-1)(-2)}{(2x+1)^3} \right)

    f^{n}(x)=\frac{1}{7} \left(\frac{2\cdot 3^n(-1)^nn!)}{(3x-2)^{n+1}}+\frac{1 \cdot 2^n (-1)^nn!}{(2x+1)^{n+1}} \right)

    you can use this to generate the coeffeints for x=1 to get

    \frac{1}{3}-\frac{8}{9}(x-1)+\frac{70}{27}(x-1)^2-\frac{626}{81}(x-1)^3+...
    Assuming the faux emptyset is correct what you should do is once you have split the partial fractions divide the bottom to get it into a form similar to \frac{1}{1-x} and go from there..
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  5. #5
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    Quote Originally Posted by juntheking@yahoo.com View Post
    I understand why you did that, but how do I describe this in terms of a summation comparing this to the known summation of 1/1-x based at 0?

    thanks again
    You do not need to use the geometric series for this problem.

    The form of the taylor series centered at x=1 is

    \frac{f(1)}{0!}(x-1)^0+ \frac{f'(1)}{1!}(x-1)^1 + \frac{f''(1)}{2!}(x-1)^2+... =\sum_{n=0}^{\infty}\frac{f^{n}(1)}{n!}(x-1)^n

    evaluating a few gives us

    f(1)=\frac{1}{3}

    f'(1)=-\frac{8}{9}

    Take note that if you want to do something cool evaluate the nth derivative at x=1 and we get...

    f^{n}(1)=\frac{1}{7}\left[ \frac{2 \cdot 3^n}{1^{n+1}}+\frac{2^n}{3^{n+1}} \right](n!)(-1)^n= \frac{(-1)^n n!}{7}\left[ \frac{2 \cdot 3^{n+1}+2^n}{3^{n+1}}\right]

    Plugging this into the sum formula for the Taylor Series gives

    \sum_{n=0}^{\infty}\frac{\frac{(-1)^n n!}{7}\left[ \frac{2 \cdot 3^{n+1}+2^n}{3^{n+1}}\right]}{n!}(x-1)^n= \sum_{n=0}^{\infty}\frac{(-1)^n}{7}\left[ \frac{2 \cdot 3^{n+1}+2^n}{3^{n+1}}\right](x-1)^n

    I hope this helps
    Last edited by TheEmptySet; April 13th 2008 at 05:55 PM. Reason: left out an "n"
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  6. #6
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    I guess I need to work on my algebra, but how did you do the partial fraction and extract a (1/7) term?

    thanks
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  7. #7
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    Quote Originally Posted by juntheking@yahoo.com View Post
    I guess I need to work on my algebra, but how did you do the partial fraction and extract a (1/7) term?

    thanks
    since they are both linear factors we get

    \frac{x}{(2x+1)(3x-2)}=\frac{A}{(2x+1)}+\frac{B}{(3x-2)}

    Clearing the fractions we get

    x=A(3x-2)+B(2x+1)
    if we evalue this at x=\frac{2}{3},x=-\frac{1}{2} we get

    \frac{2}{3}=A(2-2)+B(\frac{4}{3}+1) \iff \frac{2}{3}=\frac{7}{3}B \iff B=\frac{2}{7}

    Now the otherone

    -\frac{1}{2}=A(\frac{-3}{2}-2)+B(-1+1) \iff -\frac{1}{2}=-\frac{7}{2}A \iff A=\frac{1}{7}

    \frac{\frac{1}{7}}{(2x+1)}+\frac{\frac{2}{7}}{(3x-2)}=\frac{1}{7} \left[ \frac{1}{2x+1}+\frac{2}{3x-2}\right]

    Good luck.
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