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Math Help - Simple taylor expansion? (Complex)

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    Simple taylor expansion? (Complex)

    Using

    \frac{1}{1+z} = \frac{1}{3+(z-2)}

    expand \frac{1}{1+z} is a Taylor series about z=2.

    ---
    Do I just find all the derivatives or something? Perhaps I need to use a geometric series here? Partial fractions? How do I start?

    EDIT: It also mentions I need the first four non-zero terms of the series.
    Last edited by scorpion007; May 14th 2010 at 03:11 AM.
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    Quote Originally Posted by scorpion007 View Post
    Using

    \frac{1}{1+z} = \frac{1}{3+(z-2)}

    expand \frac{1}{1+z} is a Taylor series about z=2.

    ---
    Do I just find all the derivatives or something? Perhaps I need to use a geometric series here? Partial fractions? How do I start?

    \frac{1}{1+z}=1-z+z^2-z^3+\ldots=\sum^\infty_{n=0}(-1)^{n+1}z^n\,,\,|z|<1\Longrightarrow \frac{1}{1+z}=\frac{1}{3}\,\frac{1}{1+\frac{z-2}{3}}= \frac{1}{3}\left(1-\frac{z-2}{3}+\frac{(z-2)^2}{3^2}-\ldots\right)= \sum^\infty_{n=0}\frac{(z-2)^n}{3^{n+1}} , \left|\frac{z-2}{3}\right|<1\iff |z-2|<3 .

    Tonio
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  4. #4
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    Quote Originally Posted by scorpion007 View Post
    Using

    \frac{1}{1+z} = \frac{1}{3+(z-2)}

    expand \frac{1}{1+z} is a Taylor series about z=2.

    ---
    Do I just find all the derivatives or something? Perhaps I need to use a geometric series here? Partial fractions? How do I start?

    Assume you can write \frac{1}{1 + z} as a Taylor polynomial.


    So \frac{1}{3 + (z - 2)} = c_0 + c_1(z - 2) + c_2(z - 2)^2 + c_3(z - 2)^3 + \dots.

    By letting z = 2 we can see that c_0 = \frac{1}{3}.


    Differentiate both sides

    -\frac{1}{[3 + (z - 2)]^2} = c_1 + 2c_2(z - 2) + 3c_3(z - 2)^2 + 4c_4(z - 2)^3 + \dots.

    By letting z = 2 we can see that c_1 = -\frac{1}{3^2}.


    Differentiate both sides

    \frac{2}{[3 + (z - 2)]^3} = 2c_2 + 3\cdot 2c_3(z - 2) + 4\cdot 3c_4(z - 2)^2 + 5\cdot 4c_5(z - 2)^3 + \dots.

    By letting z = 2 we can see that c_2 = \frac{1}{3^3}.


    Differentiate both sides

    -\frac{3\cdot 2}{[3 + (z - 2)]^4} = 3\cdot 2c_3 + 4\cdot 3 \cdot 2c_4(z - 2) + 5\cdot 4 \cdot 3 c_5(z - 2)^2 + 6\cdot 5 \cdot 4 c_6(z - 2)^3 + \dots.

    By letting z = 2 we can see that c_3 = -\frac{1}{3^4}.


    Differentiate both sides

    \frac{4\cdot 3 \cdot 2}{[3 + (z - 2)]^5} = 4 \cdot 3 \cdot 2c_4 + 5 \cdot 4 \cdot 3 \cdot 2c_5(z - 2) + 6\cdot 5 \cdot 4 \cdot 3 c_6 (z - 2)^2 + 7\cdot 6 \cdot 5 \cdot 4 c_7(z - 2)^3 + \dots.

    By letting z = 2 we can see that c_4 = \frac{1}{3^5}.



    I think you can see that the series is taking the form

    \frac{1}{1 + z} = \frac{1}{3} - \frac{1}{3^2}(z - 2) + \frac{1}{3^3}(z - 2)^2 - \frac{1}{3^4}(z - 2)^3 + \frac{1}{3^5}(z - 2)^4 + \dots - \dots

    \frac{1}{1 + z} = \sum_{k = 0}^{\infty}\frac{(-1)^k}{3^{k + 1}}(z - 2)^k.


    Now you need to check the values for which this series will converge.

    Since this is a geometric series with common ratio r = -\frac{1}{3}(z - 2), it only converges for

    \left|-\frac{1}{3}(z - 2)\right| < 1

    -1 < -\frac{1}{3}(z - 2) < 1

    3 > z - 2 > -3

    -3 < z - 2 < 3

    -1 < z < 5.
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    Thanks a lot guys! Tonio, did you mean
    \sum^\infty_{n=0}(-1)^{n}z^n?

    and
    <br />
\sum^\infty_{n=0}\frac{(-1)^n(z-2)^n}{3^{n+1}}<br />
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    Quote Originally Posted by scorpion007 View Post
    Thanks a lot guys! Tonio, did you mean
    \sum^\infty_{n=0}(-1)^{n}z^n?

    and
    <br />
\sum^\infty_{n=0}\frac{(-1)^n(z-2)^n}{3^{n+1}}<br />

    Yes. I just forgot the sum begin at zero and not at 1...

    Tonio
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    Re: Simple taylor expansion? (Complex)

    multiply top and bottom by 1/3 and re-express the function as 1/3*(-1/(2-z)/3) now substitute (2-z)/3 into the geometric expansion and the rest is algebra! God I wish I could express myself better symbolically. Much less messy than ProveIt's approach!
    Last edited by dave0147; November 14th 2012 at 01:37 PM.
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