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Math Help - Taylor series, complex

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    Taylor series, complex

    Hi there. I have this exercise, which asks me to develop the Taylor series for the given function, and to determine the convergence ratio r, and the ratio for which the series converges to the function R.

    The function is:
    f(z)=\frac{1}{z-1} around z_0=2

    So the series looks like a geometric series to me.
    f(z)=\frac{1}{z-1}=\frac{-1}{1-z}

    Now the thing is that its centered at z_0=2, so I'm not sure if this would be right:
    f(z)=\frac{1}{z-1}=\frac{-1}{1-z}=-\sum_0^{\infty}(z-2)^n

    I put (z-2) as the sequence ratio because it was centered at z_0=2, but I'm not sure if this is right.
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    Re: Taylor series, complex

    Quote Originally Posted by Ulysses View Post
    Hi there. I have this exercise, which asks me to develop the Taylor series for the given function, and to determine the convergence ratio r, and the ratio for which the series converges to the function R.

    The function is:
    f(z)=\frac{1}{z-1} around z_0=2

    So the series looks like a geometric series to me.
    f(z)=\frac{1}{z-1}=\frac{-1}{1-z}

    Now the thing is that its centered at z_0=2, so I'm not sure if this would be right:
    f(z)=\frac{1}{z-1}=\frac{-1}{1-z}=-\sum_0^{\infty}(z-2)^n

    I put (z-2) as the sequence ratio because it was centered at z_0=2, but I'm not sure if this is right.
    You have a good idea, but this may help clean up the details.

    First let w=z-2 \implies z=w+2

    This gives the function

    f(w)=\frac{1}{w+2-1}=\frac{1}{1+w}=\frac{1}{1-(-w)}=\sum_{n=0}^{\infty}(-1)^nw^n=
    \sum_{n=0}^{\infty}(-1)^n(z-2)^n
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