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For the same series $$\sum_{n=0}^{\infty}\binom{3n}{2n} x^n$$

I am trying to calculate te sum by using residue theory.

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At the last line, I need to find the roots of $z^2-(z+1)^3x=0$ and one of the roots is insinde the unit circle. I cannot find this root. Next, I need to calculate $$res(\frac{(1+z)^3}{z^2}, the root)$$

But I cannot all of them. Please can you show me? Thanks

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  • $\begingroup$ I changed the title to make it a little more descriptive. $\endgroup$ – Antonio Vargas Oct 11 '13 at 3:05
  • $\begingroup$ For small $x$, $z^2 - (z+1)^3 x$ has two roots near $z = 0$ and one root with large $|z|$. It will be simpler to deform the contour and pick up the "negative" of the residue from the root at large $|z|$ instead. $\endgroup$ – achille hui Oct 11 '13 at 4:32
  • $\begingroup$ Dear @achillehui I already have thought all of them. But really I cannot do and calculate. Please can you help ne thanks alot:) $\endgroup$ – B11b Oct 11 '13 at 6:51
  • $\begingroup$ Dear @achillehui I even used wolframalpha to calculate these, but I cannot. :( $\endgroup$ – B11b Oct 11 '13 at 6:59
  • $\begingroup$ The key is don't find the root. Instead, use it to constraint what your integral could be. It turns out the integral is a root of a cubic polynomial and the cubic polynomial is very "similar" to a Chebyshev polynomial. The end result is we can express the integral using trigonometric functions instead of radicals. $\endgroup$ – achille hui Oct 11 '13 at 7:33
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Let $\;\;\displaystyle f(z;x) = z - \frac{(z+1)^3 x}{z}\;\;$, the integral $\mathscr{I}$ we want to calculate is

$$\mathscr{I} = \frac{1}{2\pi i}\int_{(0+)}\frac{dz}{f(z;x)} = \frac{1}{2\pi i}\int_{(0+)}\frac{z\,dz}{z^2-(z+1)^3 x}$$

where $(0+)$ stands for a circular contour surrounding $0$ in ccw direction.

If one pick a small but non-zero real $x$ and make a plot of the denominator $z^2 - (z+1)^3 x$, one will notice it has two roots $\lambda_1, \lambda_2$ near $z = 0$ and a root $\lambda_3$ with $|\lambda_3| > 1 > |\lambda_1|, |\lambda_2|$. Instead of summing the residues from $\lambda_1, \lambda_2$, it will be simpler if one compute the contribution from $\lambda_3$.

To do this, we deform the contour to a ccw circle at $\infty$. In the process, we will leave a small cw circlular loop around $\lambda_3$. i.e.

$$\mathscr{I} = \frac{1}{2\pi i}\left( \int_{(\infty+)} + \int_{(\lambda_3-)}\right) \frac{dz}{f(z;x)} $$ Since $|f(z; x)| \sim O(|z|^2)$ for large $z$, the integral around $\infty$ vanishes and

$$\mathscr{I} = \frac{1}{2\pi i}\int_{(\lambda_3-)}\frac{dz}{f(z;x)} = -\text{Res}\left(\frac{1}{f(z;x)}; \lambda_3\right) = -\frac{1}{f'(\lambda_3;x)} $$ Notice $$\begin{align} & f'(z;x) = 1 - \frac{(z+1)^3 x}{z} \left(\frac{3}{z+1}-\frac{1}{z}\right)\\ \implies & f'(\lambda_3;x ) = 1 - \lambda_3\left(\frac{3}{\lambda_3+1}-\frac{1}{\lambda_3} \right) = \frac{2 - \lambda_3}{\lambda_3+1} \end{align}$$ We get $$\mathscr{I} = \frac{\lambda_3 + 1}{\lambda_3 - 2}\quad\iff\quad\lambda_3 = \frac{2 \mathscr{I} + 1}{\mathscr{I} - 1}$$ Substitute this back into $f(\lambda_3;x) = 0$, we obtain following constraint of $\mathscr{I}$: $$ f(\frac{2 \mathscr{I} + 1}{\mathscr{I} - 1}; x ) = \frac{(4 - 27x)\mathscr{I}^3 - 3\mathscr{I} - 1}{(\mathscr{I}-1)^2(2\mathscr{I}+1)} = 0 $$ Let $\alpha = \sqrt{1 - \frac{27}{4}x}$, this reduce to $$4\alpha^2\mathscr{I}^3 - 3\mathscr{I} - 1 = 0 \quad\iff\quad T_3(\alpha\mathscr{I}) = 4 (\alpha\mathscr{I})^3 - 3 \alpha\mathscr{I} = \alpha $$ where $T_3(t)$ is the Chebyshev polynomial of first kind with degree 3:

$$T_3(t) = 4 t^3 - 3 t = \cos(3\cos^{-1}(t)) = \cosh(3\cosh^{-1}(t))$$

Since $\mathscr{I}$ is the root of $T_3(\alpha\mathscr{I}) = \alpha$ which $\to 1$ as $x \to 0$, we can conclude $$ \mathscr{I} = \frac{1}{\alpha}\cos\left(\frac13\cos^{-1}\alpha\right) = \frac{1}{\sqrt{1 - \frac{27}{4}x}}\cos\left(\frac13\cos^{-1}\sqrt{1 - \frac{27}{4}x}\right) $$

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  • $\begingroup$ That ıs, what you found is the root. Right? Well, how can I calculate its residue? $\endgroup$ – B11b Oct 11 '13 at 9:33
  • $\begingroup$ By the way thank you so much:) $\endgroup$ – B11b Oct 11 '13 at 9:34
  • $\begingroup$ No, $\mathscr{I} = $ the "integral" is a root of a different polynomial. The root we need from the polynomial $z^2 - (z+1)^3 x$ is $\lambda_3$. The residue at $\lambda_3$ is $\frac{1}{f'(\lambda_3; x)}$ because $\lambda_3$ is a simple root. $\endgroup$ – achille hui Oct 11 '13 at 9:37

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