255 reputation
18
bio website ondrejcertik.com
location Los Alamos, NM
age 31
visits member for 2 years, 4 months
seen Jul 28 at 4:07

Nov
18
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.
Jan
16
comment Is odd continuous function differentiable at $x=0$?
@N.S. Please do, I didn't want to put your solution into answers myself.
Jan
16
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.
Jan
16
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.
Jan
16
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.)
Jan
16
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!
Sep
21
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.
Sep
21
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.
Sep
21
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!
May
9
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?
May
9
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...