# girianshiido

less info
reputation
12
bio website location age member for 8 months seen 32 mins ago profile views 38

# 145 Actions

 May18 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. May18 comment Complex Analysis problem related to T/F statementsNo 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. May11 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. May9 suggested suggested edit on $\epsilon>0$ there is a polynomial $p$ such that $|f(x)-e^{-x}p|<\epsilon\forall x\in[0,\infty)$ May9 comment $\exp(i z^2)$ on $R e^{i \theta}$ with $\theta \in [0, \pi /4]$.Voilà, it works really well. +1 May9 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. May9 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". May9 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 $\mathrm{length}(\gamma_R)\max_{z\in\gamma_R} |e^{iz^2}|$ May9 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? May9 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/… May8 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. May7 comment Calculating the limit $\lim\limits_{x\to0-} (1/x)\cdot e^{1/x}$That's equivalent to $\lim_{t\to +\infty} -te^{-t}$. May7 comment Calculating the limit $\lim\limits_{x\to0-} (1/x)\cdot e^{1/x}$Austin Mohr: The title has been modified : $x\to 0^-$ May7 comment Calculating the limit $\lim\limits_{x\to0-} (1/x)\cdot e^{1/x}$$te^t doesn't give an indeterminate form when t\to +\infty. The other one is well known too. May7 comment Calculating the limit \lim\limits_{x\to0-} (1/x)\cdot e^{1/x}$$t=1/x$ tends to $+\infty$ or $-\infty$ when $x\to 0^+$ or $x\to 0^-$ May7 suggested suggested edit on Calculating the limit $\lim\limits_{x\to0-} (1/x)\cdot e^{1/x}$ May7 revised Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$deleted 7 characters in body May7 comment Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$+1 Good job, Ron! May7 answered Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$ May7 comment Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$Do you want a more detailed answer?