# Taylor

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bio website mathtm.blogspot.com location University of California Los Angeles, CA age member for 1 year, 10 months seen Mar 2 at 20:54 profile views 216

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.

 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 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 $C^{1}_{0}$. 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! Dec13 comment Solutions to Dirichlet problem on the half space with $L^{\infty}$ boundary data. I assume it is just the restrition of $u$ to the boundary with the definition given as above; i.e. just act like $u$ is a function on the boundary and pair it with a test function defined there. Dec12 comment Probability of $\alpha\log n$ consecutive successes in a Bernoulli process for $\alpha$ small It is finite when defining the $A_{n}$, and yes about the i.o. Dec10 comment What direction does the n vector (normal to the surface) have to be when doing Stokes' theorem? Orientation; to get a real (rigorous) answer, however, you'll have to wait until you take differential geometry/topology. A simple answer is what you've already indicated: the curve is oriented clockwise, and therefore the surface normal is outward (a physicist would tell you to apply the right-hand rule). At the end of the day however, unless you're applying your result to other computations, it really doesn't matter since your answer will only differ by a sign. Dec6 comment A convergence result for functions in L^2 Cauchy-Schwarz and the pointwise bound $|f_{n}(x)||g(x)|\leq|x|^{-1/3}|g(x)|$ may be useful. Dec6 comment (Obvious?) Half-Space Poisson Kernel Estimate I added the page where the theorem/proof appears. See (1) Dec6 comment (Obvious?) Half-Space Poisson Kernel Estimate Sorry, I'm looking closer at the passage in the text and it looks like $y$ is actually held fixed. The reason I was tempted to say it is not is that in my linked question the equivalent of $y$ there (i.e. $t$) is allowed to vary, but that is why (I think) there is the additional constant involved in the estimate, namely $\beta$. So anyway, yes, $y$ is fixed. Dec6 comment (Obvious?) Half-Space Poisson Kernel Estimate I believe $y>0$ is allowed to vary. $\alpha$ is the only parameter fixed, and $A$ is only allowed to depend on $\alpha$. Dec4 comment Constructing a Distributional Solution to the Inhomogeneous C.R. Equations @Post - See the update to my question. I'd like to avoid any reference to complex analysis if it's possible. Nov20 comment Analytic functions of absolute value 1 on the boundary of the unit disc (Or the strong form of the maximum modulus principle.) Nov18 comment Mean-value like theorem for holomorphic functions. It implies the existence of some neighborhood $V$ of $z$ such that $g$ fails to be one to one and this implies the existence of distinct $s,t$ such that $g(s)=g(t)$, which is the desired conclusion. I'm thoroughly amused by the fact that all three questions I posted had solutions which depended on some suitably chosen auxiliary function and an application of some major theorem. Nov18 comment For $f\in H(U)$, find a bound on $|f(0)|$ given separate bounds of $|f|$ on $\partial U^{+}$ and $\partial U_{-}$. How does $|g|$ being a constant imply $g$ is? What about $re^{it}$? Nov18 comment For $f\in H(U)$, find a bound on $|f(0)|$ given separate bounds of $|f|$ on $\partial U^{+}$ and $\partial U_{-}$. Hmm...in regards to the second part of your comment, I see now that even if $|g|\geq\frac{1}{4}$ too, it would not imply $g$ was constant; however, it does imply $|g|$ is, no? Nov18 comment For $f\in H(U)$, find a bound on $|f(0)|$ given separate bounds of $|f|$ on $\partial U^{+}$ and $\partial U_{-}$. I made a correction to my comment as you were writing yours (though I didn't have time to finish rewriting the conclusion as $|f(0)|\leq\frac{1}{2}$ before the comment edit expiration. I was trying to write out the details of the answer in my head and accidently also reasoned that the minimum of $|g(z)|$ on the boundary was $\frac{1}{4}$ as well, which is not the case. Nov18 comment For $f\in H(U)$, find a bound on $|f(0)|$ given separate bounds of $|f|$ on $\partial U^{+}$ and $\partial U_{-}$. Okay, so then we have $||g||_{T}=||f||_{T^{+}}||f||_{T^{-}}.$ By symmetry and the conditions on $f$, we conclude $||g||_{T}\leq\frac{1}{4}.$ Since $g$ is holomorphic and continuous on $\bar{U}$, we have by the maximum modulus principle $|g(z)|\leq\frac{1}{4}.$ Therefore, $f(z)=\frac{1}{4f(-z)}$ and so $|f(0)|\leq\frac{1}{4}|\frac{1}{f(0)}|$, i.e. $|f(0)|=\frac{1}{2}.$