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bio website mathtm.blogspot.com
location University of California Los Angeles, CA
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visits member for 1 year, 11 months
seen 2 days ago

I am a recent graduate in applied mathematics at UCLA and currently trying to break into the quantitative investment/trading industry while also continuing to pursue advanced graduate-level mathematics as both a hobby and career necessity.


2d
comment Integral of $\sin^{-1}(1/2 - \sin x) dx$
A quick check on Wolfram reveals that the answer is likely no (by solution, I assume you mean a closed formula for the indefinite integral).
Apr
17
answered Let $M$ be a bounded subset of the space $C_{[a,b]}$. Prove that the set of all functions $F(x)=\int^{x}_{a}f(t)dt$ with $f\in{M}$ compact.
Apr
1
comment Tensor notation and the “Zero-Value Theorem”
Does a proof follow the statement of the theorem?
Apr
1
comment Tensor notation and the “Zero-Value Theorem”
$V$ is arbitrary, so presumably he means over every open set $V$.
Mar
31
answered proving series convergence by definition
Feb
22
answered An equation, where the solution does not exist, but on solving the equation we got a solution. why this is happening?
Feb
22
revised The closure of $C^{1}_{0}(\mathbb R)$ in $L^{\infty}$?
added 173 characters in body
Feb
22
comment How to derive the formula to estimate the stock price probability distribution from call option prices?
@Raphael. The part where I mention "...as long as $\phi$ and $S\phi$ are integrable we're fine" is the justification for differentiation under the integral sign (see the link for Leibniz's rule).
Feb
22
comment How to derive the formula to estimate the stock price probability distribution from call option prices?
Sorry, wrote the answer too hastily and made a sign mistake at the end. Thanks Daniel Fischer for the edit.
Feb
22
revised The closure of $C^{1}_{0}(\mathbb R)$ in $L^{\infty}$?
added 721 characters in body
Feb
22
comment The closure of $C^{1}_{0}(\mathbb R)$ in $L^{\infty}$?
I made an edit to my answer above which deals with the closure of $C^{1}_{0}$.
Feb
22
revised The closure of $C^{1}_{0}(\mathbb R)$ in $L^{\infty}$?
added 721 characters in body
Feb
17
answered How to derive the formula to estimate the stock price probability distribution from call option prices?
Feb
17
answered The closure of $C^{1}_{0}(\mathbb R)$ in $L^{\infty}$?
Feb
17
comment How to calculate this multivariate limit changing to polar coordiantes?
Simple examples: (1) $x^{2}+y^{2}=r^{2}\to0$ as $r\to0$ no matter what $\theta$ is; (2) $y(x^{2}+y^{2})=r^{3}\sin\theta\to0$ as $r\to0$ no matter what $\theta$ is; (3) $\frac{x^{2}y^{2}}{x^{4}+3y^{4}}=\frac{r^{4}\cos^{2}\theta\sin^{2}\theta}{r^{4}(‌​\cos^{4}\theta+3\sin^{4}\theta}=\frac{\sin^{2}(2\theta)}{4(\cos^{4}\theta+3\sin^{‌​4}\theta)}$ which depends on $\theta$ (and indeed, the limit does not exist as $(x,y)\to(0,0)$).
Feb
17
comment How to calculate this multivariate limit changing to polar coordiantes?
$(x,y)\to(0,0)$ is equivalent to $r\to0$. That's why we replace the limit by $r\to0$ when working polar coordinates. However, as you no doubt have seen, $(x,y)\to(0,0)$ can be accomplished in several different ways (paths), which can sometimes lead to different limiting values. When working in polar coordinates, $r\to0$ is approach along a ray to the origin. If $\theta$ is not involved, or if its presence in the limiting expression is dominated by $r\to0$, then there is no problem. If $\theta$ is involved, then the approach can vary depending on $\theta$, and this must be considered.
Feb
17
comment How to calculate this multivariate limit changing to polar coordiantes?
Fixed, thank you!
Feb
17
answered How to calculate this multivariate limit changing to polar coordiantes?
Jan
6
revised Uniqueness of harmonic solution (PDE Evans)
added 138 characters in body
Jan
6
answered Uniqueness of harmonic solution (PDE Evans)