Timeline for An integral and a Jensen-type formula
Current License: CC BY-SA 4.0
9 events
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| Nov 30, 2023 at 10:34 | history | edited | 12321 | CC BY-SA 4.0 |
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| Nov 30, 2023 at 10:33 | comment | added | 12321 | @Conrad thank you for the comment! | |
| Nov 30, 2023 at 3:03 | comment | added | Conrad | with $w=e^{i\theta}$ use that $\Im(z_0 \bar w)=\frac{z_0/w-\bar z_0 w}{2i}$ and $\log |w|=0$ to transform the log integrand to $\log |z_0w-Miz_0+Mi\bar z_0 w^2-Mw^2|$ and the inside function $f(w)=z_0w-Miz_0+Mi\bar z_0 w^2-Mw^2$ is a quadratic so clearly analytic so you can apply the result above with $R=1$ and need find the roots of the quadratic inside the unit disc... | |
| Nov 29, 2023 at 17:32 | history | edited | 12321 | CC BY-SA 4.0 |
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| Nov 29, 2023 at 17:28 | history | edited | 12321 | CC BY-SA 4.0 |
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| Nov 29, 2023 at 17:25 | comment | added | 12321 | \begin{align*} \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-i\theta}\log(|f(Re^{i\theta})|)d\theta=\frac{1}{2}R\frac{f'(0)}{f(0)}+\frac{1}{2}\sum_{\rho}\left(\frac{R}{\rho}-\frac{\overline{\rho}}{R} \right) \end{align*} where $\rho$ are the roots in $D(0,R)$ I edit the post | |
| Nov 29, 2023 at 17:19 | comment | added | Sidharth Ghoshal | What is the “well known formula” in the earlier case? | |
| S Nov 29, 2023 at 17:12 | review | First questions | |||
| Nov 29, 2023 at 18:12 | |||||
| S Nov 29, 2023 at 17:12 | history | asked | 12321 | CC BY-SA 4.0 |