meta user
network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 8 months |
| seen | 32 mins ago | |
| stats | profile views | 38 |
|
May 18 |
comment |
Proving an inequality: $|1-e^{i\theta}|\le|\theta|$ Draw the unit circle and the two points $1$ and $e^{i\theta}$. Draw the line segment between these points. Compare the length of this segment with the arc joining the two points. |
|
May 18 |
comment |
Complex Analysis problem related to T/F statements No you're not. Every entire function is bounded on every compact subset of $\Bbb C$. Liouville's theorem states that every entire function bounded on $\Bbb C$ is constant. |
|
May 11 |
comment |
Why does $ \frac{1}{2\pi i} \int_{\gamma} \frac{f'(z)}{f(z)} dz = \eta( f \circ \gamma, 0) $ ($ \gamma $ is a regular closed curve) You're right, the $\frac 1{2i\pi}$ is missing, since it wasn't "absorbed" by the change of variable. You found the error yourself and corrected it! Good job. |
|
May 9 |
suggested | suggested edit on $\epsilon>0$ there is a polynomial $p$ such that $|f(x)-e^{-x}p|<\epsilon\forall x\in[0,\infty)$ |
|
May 9 |
comment |
$\exp(i z^2)$ on $R e^{i \theta}$ with $\theta \in [0, \pi /4]$. Voilà, it works really well. +1 |
|
May 9 |
comment |
$\exp(i z^2)$ on $R e^{i \theta}$ with $\theta \in [0, \pi /4]$. Hint : use $\sin(x)\geq \frac{2}{\pi}x$ for all $x\in[0,\frac \pi 2]$. I wouldn't use the Jordan Lemma, it's useless. |
|
May 9 |
comment |
$\exp(i z^2)$ on $R e^{i \theta}$ with $\theta \in [0, \pi /4]$. I see, I was talking about the so-called "estimation lemma". |
|
May 9 |
comment |
$\exp(i z^2)$ on $R e^{i \theta}$ with $\theta \in [0, \pi /4]$. By the way, are you sure you're really using Jordan's lemma ? I think it gives the upper bound $\mathrm{length}(\gamma_R)\max_{z\in\gamma_R} |e^{iz^2}|$ |
|
May 9 |
comment |
$\exp(i z^2)$ on $R e^{i \theta}$ with $\theta \in [0, \pi /4]$. Since $|e^u|=e^{\mathrm{Re}(u)}$ you can study the real part of $iz^2-iz$ when $z=Re^{i\theta}$ with $\theta\in[0,\pi/4]$ and find its maximum. It's not too hard. Maybe there's a shorter way? |
|
May 9 |
comment |
Closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$. A bit tough but it can be done with residue theorem. Check math.stackexchange.com/questions/384780/… |
|
May 8 |
comment |
Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$ @RonGordon : Nice document & Bonus. Indeed the algebra involved here is quite terrifying. The $\cos(b e^it)$ formula is helpful. |
|
May 7 |
comment |
Calculating the limit $\lim\limits_{x\to0-} (1/x)\cdot e^{1/x}$ That's equivalent to $\lim_{t\to +\infty} -te^{-t}$. |
|
May 7 |
comment |
Calculating the limit $\lim\limits_{x\to0-} (1/x)\cdot e^{1/x}$ Austin Mohr: The title has been modified : $x\to 0^-$ |
|
May 7 |
comment |
Calculating the limit $\lim\limits_{x\to0-} (1/x)\cdot e^{1/x}$ $te^t$ doesn't give an indeterminate form when $t\to +\infty$. The other one is well known too. |
|
May 7 |
comment |
Calculating the limit $\lim\limits_{x\to0-} (1/x)\cdot e^{1/x}$ $t=1/x$ tends to $+\infty$ or $-\infty$ when $x\to 0^+$ or $x\to 0^-$ |
|
May 7 |
suggested | suggested edit on Calculating the limit $\lim\limits_{x\to0-} (1/x)\cdot e^{1/x}$ |
|
May 7 |
revised |
Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$ deleted 7 characters in body |
|
May 7 |
comment |
Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$ +1 Good job, Ron! |
|
May 7 |
answered | Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$ |
|
May 7 |
comment |
Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$ Do you want a more detailed answer? |
