# Taylor

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bio website mathtm.blogspot.com location University of California Los Angeles, CA age member for 1 year, 11 months seen 2 days ago profile views 224

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.

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 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). Apr17 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. Apr1 comment Tensor notation and the “Zero-Value Theorem” Does a proof follow the statement of the theorem? Apr1 comment Tensor notation and the “Zero-Value Theorem” $V$ is arbitrary, so presumably he means over every open set $V$. Mar31 answered proving series convergence by definition Feb22 answered An equation, where the solution does not exist, but on solving the equation we got a solution. why this is happening? Feb22 revised The closure of $C^{1}_{0}(\mathbb R)$ in $L^{\infty}$? added 173 characters in body Feb22 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). Feb22 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. Feb22 revised The closure of $C^{1}_{0}(\mathbb R)$ in $L^{\infty}$? added 721 characters in body Feb22 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 $C^{1}_{0}$. Feb22 revised The closure of $C^{1}_{0}(\mathbb R)$ in $L^{\infty}$? added 721 characters in body Feb17 answered How to derive the formula to estimate the stock price probability distribution from call option prices? Feb17 answered The closure of $C^{1}_{0}(\mathbb R)$ in $L^{\infty}$? Feb17 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 $(x,y)\to(0,0)$). Feb17 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. Feb17 comment How to calculate this multivariate limit changing to polar coordiantes? Fixed, thank you! Feb17 answered How to calculate this multivariate limit changing to polar coordiantes? Jan6 revised Uniqueness of harmonic solution (PDE Evans) added 138 characters in body Jan6 answered Uniqueness of harmonic solution (PDE Evans)