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Thread: Laurent series of a square root function example

  1. #1
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    Laurent series of a square root function example

    Hello,

    I'm trying to figure out this example:

    \begin{equation*}\begin{split} 1 - 1/z^2 &= x \\ 1 - x &= 1/z^2 \\ 1/z^2 &\geq 1 \:\: z \in [-1,1] \\ \implies g(z) &= z\sqrt{1 - 1/z^2} \:\: z \not\in [-1,1] \\ f(z)^2 &= z^2(1-1/z^2)=z^2-1. \end{split}\end{equation*} So $f(z)=\text{SQRT}{(z^2-1)}$ and $f$ has a square root on $\mathbb{C} \backslash [-1,1]$.

    Using the binomial theorem, for $(1/z)^k<1$, we find $$z\sqrt{1-1/z^2} = z \sum_{n=0}^\infty {{1/2}\choose{k}} (-1)^k \Big(\frac{1}{z^{2k}}\Big)^k = \sum_{n=0}^\infty {{1/2}\choose{k}} (-1)^k \frac{1}{z^{2k-1}}. $$

    If $r>1$ then $$\frac{1}{2 \pi i} \int_{|z|=r} \text{SQRT}(z^2-1)dz = \frac{1}{2 \pi i} \int_{|z|=r} \sum_{n=0}^\infty {{1/2}\choose{k}} (-1)^k \frac{1}{z^{2k-1}} dz \overset{?}{=} \frac{1}{2 \pi i} \int_{|z|=r} {{1/2}\choose{1}} (-1) \frac{1}{z^{2-1}} dz = -1/2 $$

    The problem I'm having is justifying the equality with the question mark over it.

    The example also includes the information $\lim_{z \to \infty}\frac{f(z)}{z}=\sqrt{1}=1 \neq 0$. I haven't seen how this can be related/applies (if it does at all.)

    Any ideas?
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  2. #2
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    Re: Laurent series of a square root function example

    The only pole with a non-zero residue at $0$ occurs when $k=1$

    That's why they can go from the summation to the single value of $k=1$
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  3. #3
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    Re: Laurent series of a square root function example

    OK I think I understand.

    $$\frac{1}{2 \pi i} \int_{|z|=r} \text{SQRT}(z^2-1)dz = \frac{1}{2 \pi i} \int_{|z|=r} \sum_{n=0}^\infty {{1/2}\choose{k}} (-1)^k \frac{1}{z^{2k-1}} dz = \frac{1}{2 \pi i} \sum_{n=0}^\infty \int_{|z|=r} {{1/2}\choose{k}} (-1)^k \frac{1}{z^{2k-1}} dz. $$

    Now I think you use the fact that $\text{Res}_{z_0} f = \frac{1}{2 \pi i} \int_\gamma f(z) dz = 0$ when $\gamma$ is a simple counter clockwise curve around $z_0$ and $z_0$ is a removable singularity of $f$.

    So like for this example,

    https://en.wikipedia.org/wiki/Residu...Series_methods

    The only term of the series expansion which has a non-zero residue occurs at $\frac{\sin{1}}{z-1}$. Since that's the only term that has a non-removable singularity.

    And for this example

    $$\sum_{n=0}^\infty {{1/2}\choose{k}} (-1)^k \Big(\frac{1}{z}\Big)^k = \sum_{n=0}^\infty {{1/2}\choose{k}} (-1)^k \frac{1}{z^{k-1}} $$

    the only non-removable singularity would occurs when $k=2$.
    Last edited by bkbowser; Apr 15th 2018 at 04:11 PM.
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