# Ondřej Čertík

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bio website ondrejcertik.com location Los Alamos, NM age 30 member for 1 year, 11 months seen Apr 12 at 18:48 profile views 39

 Nov18 comment Derivative of big O symbol $\sin(x^2)$ is infinitely differentiable at $x=0$ and the series is just a polynomial, so for it $O'(1) = O(1)$. You might be expanding around $x=\infty$, while I am asking about $x=0$. I clarified the question about this. Jan16 comment Is odd continuous function differentiable at $x=0$? @N.S. Please do, I didn't want to put your solution into answers myself. Jan16 comment Solution of functional equation $f(x/f(x)) = 1/f(x)$? Very nice! So if the function $f(x)$ is analytic, then you prove that $f(x)=1$. Assuming only that $f(x)$ is continuous, then the only possible other solutions are not analytic. That helps a lot. Jan16 comment Solution of functional equation $f(x/f(x)) = 1/f(x)$? Right. So the domain of $g(x)$ must be an interval (e.g. not a union of two disjoint intervals). However, what if we only solve this on an interval [1, 10], let's say? I guess we would run into some contradictions with domains and ranges in the functional equation. Jan16 comment Solution of functional equation $f(x/f(x)) = 1/f(x)$? Because $g(x)=1/x$ is also a solution of $g(g(x))$ and somehow it got eliminated. So something is not right. (Of course, it would get eliminated later anyway due to $g'(0)=1$, but that's not the point.) Jan16 comment Is odd continuous function differentiable at $x=0$? N.S., you are right! $\frac{f(x)}{x} = \sin\frac{1}{x^2}$ which oscillates between -1 and 1 and so the limit does not exist. So this function is not differentiable at $x=0$. Thanks! Sep21 comment How to prove asymptotic limit of an incomplete Gamma function I verified your steps, I think it's all correct. I am accepting your answer as it gives a simple proof. Sep21 comment How to prove asymptotic limit of an incomplete Gamma function Thanks! As a matter of fact, I actually wanted to prove that $\gamma(z, x)\over\Gamma(z)$ goes to zero, but thought it'd be easier to do it with the upper incomplete gamma function. Sep21 comment How to prove asymptotic limit of an incomplete Gamma function So you proved that $\gamma(z, x) < {x\over z-1-x}\Gamma(z)$ for $z-1 > x > 0$ and from that the limit follows immediately. Nice! May9 comment How to prove that Legendre polynomials form a complete basis using functional analysis Ah -- I forgot about the boundary condition -- so requiring the solutions to satisfy some conditions (for example being non-singular, or I can even require something stronger, like being $\pm 1$ at the endpoints), then the only allowed $\lambda$ are of the form $n(n+1)$, thus proving that the spectrum is discrete. If we allow Legendre functions, then the spectrum is continuous --- because I think that the Legendre functions can be normalized to (Dirac) delta function (using physics terminology). What about 2? May9 comment How to prove that Legendre polynomials form a complete basis using functional analysis @J.M., for non-integer $n$ (i.e. a general $\lambda$), the solutions are Legendre functions (en.wikipedia.org/wiki/Legendre_function). So it would seem that all $\lambda$ are actually allowed, thus the spectrum being continuous...