@Gyromagnetic gross

but mathematica gives an answer in terms of Bessel functions

first thing: let $x=au$ so that the integral becomes $a\int_0^\infty e^{-(ab)u}\sqrt{u(1+u)}\,du$. So now the integral is a function of $s=ab$ alone

according to mathematica, the integral is $e^{s/2}K_1(s/2)/(2s)$

Wikipedia seems to be down. Maybe I should have made a local backup on my computer. :-)

it was very in and out for me. some pages loaded, some didn't

at least some chunk of what serves wikipedia is out

It seems that it works again - at least for me.

It's working for me

@robjohn I am not sure whether you wanted to keep this room alive or to let it freeze. It is getting close to the 14-day limit.

Of course, I understand that for a mod this is basically a moot point - a moderator can unfreeze a room at any time and even post messages into a room which is frozen.

@hyper-neutrino There is a difference between the (boolean) algebra of formulae and its Lindenbaum algebra quotient.

Note to self: elementary number theory is hard.

is there any known result for finding maximum and minimum value of nowhere differentiable function defined on an interval?

I don't understand this question: Suppose that V is finite dimensional vector space. $T\in L(V,W)$ is given to be surjective. Show that there exists a subspace U of V such that $T|_U$ is an isomorphism of U onto W.

I think that since V is FDVS, we have by fundamental theorem of algebra that null T={0}. Ah no, I confused T: V$\to W$ with a linear operator.

The question is clear to me now.

@MartinSleziak Yeah, I use it when needed, but sometimes it is not needed for a while and so I unfreeze it when needed. Sometimes I just go in to polish the brass, or straighten the books.

Is there a standard way to do a best fit curve (say a polynomial) for a bunch of points in the plane but with an added condition that it's a lower (or upper) bound for all the points? Like a naive approach would be to take the best approximation and then lower it until it's a bound, but I feel like there's a better way, like "take a convex hull and then best-fit that and then lower it" or something.

mark that's an interesting question. could be tough, as you note, just looking at the convex hull is an obvious solution but without any regularity.

i asked a question once on math.se kind of like this. if you look at finitely many points on the graph of a convex function, pairs of points satisfy difference quotient inequalities reflecting the fact that the second derivative is positive (but not referring to the second derivative).

i asked, roughly, if you have a set of points in the plane satisfying these inequalities, or perhaps a larger set reflecting the positivity of higher-order derivatives too, is there a "best-fit" schema that produces an approximant of sufficient regularity that it actually has these derivatives and they are always positive (and not e.g. only positive at various points).

nobody answered it.

math.SE doesn't seem like the natural home for it, but i couldn't think of a better one.

Thanks for your response. At least it's good to know someone else was thinking about this sort of stuff

@MarkS. how are you measuring "best" fit?

Not picky about it, but I was imagining, say, the sum of the square vertical errors

i dunno about mark's q but when i asked it, that was part of the question. i was more concerned about 'unique ways of producing convex approximants from convex data'. the motivation was an applied problem where for various applied reasons the curve fitting the data had to be convex to a certain degree.

In the specific case that spurred this question, I'm looking at about 400 points that follow a nice shape with a bit of noise and want a good lower bound sextic, say

mathoverflow.net/questions/181559/… seems related, but all the answer says about my question is basically "pick a high-enough degree and then shift it vertically"

@MarkS. If there are two points at the same $x$, do you want the function to be less that the least of the values at that $x$ or less than the greatest of the values at that $x$?

@robjohn I've made sure there are not any such pairs, but if I made a mistake I'd want less than the least of the values.

It seems this might be related to "convex optimization", so I'm trying to read up on that

Yeah, just telling Mathematica "use your quadratic optimization techniques to minimize the sum of the square errors subject to the inequalities at each of the 400 x-coordinates" worked fine. No theoretical understanding required.

Given an ideal $I\vartriangleleft A$ of a commutative ring, when is $1+I$ its set theoretic complement in $A$?

do you have an example where $I$ is not the unique maximal ideal of a local ring?

this holds iff $A/I\cong\mathbb{Z}/2\mathbb{Z}$

tautologically so, but I don't think there's any reasonable alternative description

@AlessandroCodenotti $I=2\mathbb{Z}$ in $A=\mathbb{Z}$. this also usually doesn't hold for the maximal ideal in a local ring. maybe you misread

@Thorgott If $R$ is local with maximal ideal $M$ isn't $1+m$ a unit for all $m\in M$ and $R\setminus M$ the set of units?

consider the local ring $k[[x]]$, $k$ a field. the units are cosets of the maximal ideal naturally indexed by the units in $k$ and $1+(X)$ is only one of those (unless $k=\mathbb{F}_2$)

ah of course I had a brainfart

Ah ok I see your point now. Saying $R\setminus M=1+M$ is just a convoluted way of saying that $M$ has one coset in $R$, which in turn is a convoluted way of saying $R/M\cong\Bbb Z/(2)$

yeah

@Thorgott, I see. I should have asked my question for the "more saturated" multiplicative set $A^\times +I$.

@Thorgott since this monoid need not itself be saturated, perhaps most natural is to ask the question for the saturation $(1+I)_\mathrm{sat}$. That is, when are $I,(1+I)_\mathrm{sat}$ set-theoretic complements?

that happens iff $I$ is maximal

if $\pi\colon A\rightarrow A/I$ is the natural quotient map, $I=\pi^{-1}(0)$ and $(1+I)_{\mathrm{sat}}=\pi^{-1}((A/I)^{\times})$, so $A$ is the disjoint union of $I$ and $(1+I)_{\mathrm{sat}}$ iff $A/I$ is the disjoint union of $0$ and $(A/I)^{\times}$, which is the case iff $A/I$ is a field

Anyone else here that uses Lang's Algebra for studying?

@Thorgott jolly good. Thanks!

smoken, apparently not :) i used it a little during an algebra class.

Room is pretty quiescent, I would wait for a reply when more people are active.