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Math Help - power series

  1. #1
    Junior Member
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    power series

    Determine the power series expansion of f(z) at z_0 without computing derivatives
    (i) f(z) = sinz , z_0 = i;

    (ii) f(z )= \frac{1}{z^2-3z+2} , z_0 = 0;

    (iii) f(z) = Log z, z_0 = 1;

    So how the heck would I do this? If someone could show me step by step how to do one of them I'm pretty sure I could do the rest, but I don't know how to do it without derivatives as I would have just taken the taylor series
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  2. #2
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    You basically have to know what the power series are for \sin x, \frac{1}{x}, \ln x. Then you can just use substitution.
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  3. #3
    MHF Contributor
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    To elaborate, for number 2, you would use the power series centered at x = 2 for f(x) =\frac{1}{x}, which is:

    f(2) + \frac{(x - 2)f'(2)}{1!} + \frac{(x-2)^2f''(2)}{2!} + ...

     = \frac{1}{2} - \frac{(x-2)}{2^2} + \frac{(x-2)^2}{2^3} - \frac{(x-2)^3}{2^4} + ... + \frac{(x-2)^n(-1)^n}{2^{n+1}} + ...

    Then you substitute x = z^2 - 3z + 2.
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  4. #4
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    Here's what I did for(1):

    \sin(z-i)=\sum_{n=0}^{\infty}\frac{(-1)^n(z-i)^{2n+1}}{(2n+1)!}

    Now, how do we express this in terms of sin(z)? Well:

    \sin(z-i)=\sin(z)\cos(i)-\sin(i)\cos(z)
    \cos(z-i)=\sin(z)\sin(i)+\cos(z)\cos(i)

    Now, I'll leave it to the reader to show:

    \sin(z)=\cosh(1)\sum_{n=0}^{\infty}\frac{(-1)^n(z-i)^{2n+1}}{(2n+1)!}+i\sinh(1)\sum_{n=0}^{\infty}\f  rac{(-1)^n(z-i)^{2n}}{(2n)!}

    Guess that's what you meant Iceman? Or no?
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