12:15 AM
Is there a simplification for $ln(x) \times ln(x)$?
if x = 1, yes
in general, no :)
12:45 AM
$(ln x)^2$.
yeah, that
1:02 AM
or $\log(x^{\log(x)})$ if you're a lunatic
1:13 AM
You call that lunatic simplification?
it is technically simpler
in the sense of being composed of just two operations instead of three
good morning
How is composition of log and squaring three?
1:30 AM
composition is a free operation
I meant the very first expression
which involves two applications of log and one multiplication, i.e. three operations
Oh. Irrelephant.
i'm gonna go back to my original answer of 'no'
Well, if you break powers into individual multiplications … Your exponentiation is more complex, for sure.
LOL @ leslie
@Leslie Next time I make a trek to Costa Mesa for Hi Time, we should do lunch at Din Tai Fung.
Probably misspelled.
agreed. and you got it. i sometimes write tae instead of tai but it's tai.
around the office some of the younger people would suggest a lunch trip there by asking "hey.. DTF?" the millennials find this funny. i don't know why.
1:43 AM
Oh oh … that’s another agreement!
Need some confirmation
The bottom of the ladder is falling down at a speed of 3m/s. The length of the ladder is 17m. What is the speed of the top of the ladder falling down when the bottom is 8m away from the wall?
should the bottom of the ladder be skidding away from the wall at 3 m/s?
@leslietownes yes
2:06 AM
i get 45/8 m/s, but i did a lot of it in my head.
it's a little faster than 3m/s but not too much faster, so at least that feels right
@Leslie, it's $dy/dt=-3$, no?
i interpreted dx/dt as 3. i probably messed up the numbers.
maybe the ladder is in freefall and both the bottom and the top are falling down. we would need to do a surface integral to figure out the net rotational force on the ladder.
what is the density of the air?
@TedShifrin thanks, sorry for yet another inane Q. I realised how dumb that was 20 minutes later
A whole 20? ;D
I love the interesting physics problem about when the ladder pulls away from the wall, but that is too sophisticated.
2:23 AM
So what answers did you all get?
Lol just a simple related rates@leslietownes. No need to think much:)
I assume that b²+h²=289 and db/dt=3, is that right?
Thus getting dh/dt=-1.6, or 1.6 if we ignore the negative sign.
how can the bottom of the ladder be falling down if it is on the ground?
@copper.hat what do you mean? The bottom is sliding away from the wall. The top is falling down.
2:38 AM
@xxxx036 the question says "bottom of the ladder is falling down..."
Apologies for some misunderstanding
The top is falling down XD.
${dh \over dt} = - {x \over h} {d x \over dt}$ with $x=8, {d x \over dt} = 3, l = 17$.
and $h=\sqrt{17^2-8^2} = 15$.
2:56 AM
@TedShifrin A whole 20, I estimate with 60% confidence. It was certainly more than 0 and less than 120.

5 hours later…
8:20 AM
Hi, @Balarka
Hi, @Balarka Sayan here. Can I talk to you?

3 hours later…
11:36 AM
@Thorgott I'm afraid I didn't see why it is separable (am really out of it), but it was quite easy using some equivalent characterisations of purely inseparable field extensions, so I'm content with that
12:13 PM
Hi everyone
I wanted tp find singular solutions to $z=px+qy+4\sqrt{1+p^2+q^2}$
this is clairut's equation, so the general solution is
$z=ax+by+4\sqrt{1+a^2+b^2}$
1:04 PM
How does discriminant help in finding nature of roots ?
Like I know till that when finding turning point for a Q.E
We get on y axis - D/4a
As I see it,D/4a helps in finding nature of roots
1:31 PM
Anyone have a neat idea of how I could to rewrite the PDE

$$\partial_t u(t,x)=\text{div}\Big(\nabla f(x)\Big)+ \text{div}\Big( A \nabla u(t,x) \Big)~~~(1)$$

in the form :

$$\partial_t u(t,x)=\text{div}\Big( A \big( \nabla u(t,x) +\nabla g(x)\big) \Big)~~~~~~~~~~(2)$$
2:01 PM
Is there a Cech cohomology way to see why flasque sheaves should be acyclic with respect to the global sections functor?
@SayanChattopadhyay idk, maybe try lifting the difference
2:22 PM
Hello, I have a question.
Apollonius showed that a circle can be defined as the set of points in a plane that have a specified ratio of distances to two fixed points. https://en.wikipedia.org/wiki/Circles_of_Apollonius
Is there a 3d equivalent for the sphere? A set of points in 3d that have a specified ratio of distances to three fixed points? I tried googling but couldn't find much.
is there a way to create tags on this website?
3:20 PM
@TedShifrin proved.
In the context of elliptic curves, what would be a local field complete w.r.t a discrete valuation? I can't seem to find a clear definition of a local field. My guess now is: we have a discrete valuation on our field which induces a metric (and therefore a topology), and w.r.t. to this metric the space is complete. If this is correct: is there a reason why they add that the field has to be local (i.e., would there be examples of non-local fields (whatever that me may) that arise this way?)
This is probably trivial, but what does $P_n(t)$ mean in this post? (Note: t stands for log(x) in that post)
3:49 PM
Can anyone help me with this? Thanks!
0

Suppose $Z\sim N(\mu,1)$ and $V$ is independent of $Z$ with distribution $\chi^2_m$. Then $T=\frac{Z}{(V/m)^{1/2}}$ is said to have a noncentral $t$ distribution with noncentrality $\mu$ and $m$ degrees of freedom. I want to show that $$P(T\leq t)=2m\int_0^\infty \Phi(tw-\mu)f_m(mw^2)wdw$$ where ...

I think what I said is correct (considering some more notation in the book), and I'll assume that the assumption that the space is locally compact is necessary

1 hour later…
4:56 PM
1

Prove that $a^2x^2+ (b^2 + a^2 - c^2)x + b^2 = 0$ does not have real roots if $a+b>c$ and $|a-b|<c$. $a,b,c \in \mathbb{R}$. Solution I found online: The discriminant is given as: $$D = (b^2 + a^2 - c^2)x - 4a^2b^2$$ $$(a^2 + b^2)^2 - 2c^2(a^2 + b^2) + c^4 - 4a^2b^2$$ Let $c^2= t$. You get : ...

5:35 PM
@Rover Great.
6:00 PM
@Monty Yes, but it would be best to discuss it on meta before to see if was a tag and has been removed for some reason.
it my equities continue their slide i will be moving to my under freeway location.
the sp500 seems to have stopped sliding anyway. i'm sometimes glad that my job basically prevents me from owning individual stocks.
Can anyone help me with this? Thanks!
0

Suppose $Z\sim N(\mu,1)$ and $V$ is independent of $Z$ with distribution $\chi^2_m$. Then $T=\frac{Z}{(V/m)^{1/2}}$ is said to have a noncentral $t$ distribution with noncentrality $\mu$ and $m$ degrees of freedom. We deduced that $$P(T\leq t)=2m\int_0^\infty \Phi(tw-\mu)f_m(mw^2)wdw$$ where $fm(... the stomach churning helps with my digestion :-) On Page 31 here, while proving the one-dimensional Brunn-Minkowski inequality, they assume that A and B are compact. I read that this is due to the inner regularity of the Lebesgue measure on R^n. Could someone explain please? http://library.msri.org/books/Book31/files/ball.pdf 6:16 PM You can always find a compact subsets that approximates the measure of your (finite measure) set up to any$\epsilon$Hmm, agreed, that is inner regularity! I can approximate A and B by compact subsets from within, yes but why is it allowed to assume A and B to be compact here? like, how do we make sure it doesn't mess with the inequality We gotta have a formal proof right @AlessandroCodenotti day 3 of no exercise. contemplating adding a question about positive numbers summing to$-{1 \over 12}$. Would it make sense to make a post for this? @epsilon-emperor its a long running joke/annoyance i have. 6:32 PM tempted to ask math.SE to find$dy/dx$given that$e^y + e^x = \sinh y$and see if anybody gets the joke. busy trying to get the joke... there's no such y. if you write sinh in terms of exponentials its clear that sinh y - e^y is always negative. i mean i guess you're finding slopes of level curves of something, but that equation doesn't define y as a function of x. there were a number of problems like that in the 'calculus for the business and social sciences' textbook i had to teach out of. not with sinh but same technical problem. the author probably didn't know they were joking. The empty joke. my favorite kind it would help if i could keep my minus signs in order :-) continuing my search for the even primes 6:46 PM Maybe you need exorcise more than exercise, i'm going to work on the triple prime conjecture, which asserts that there are infinitely many primes p for which p, p+2, p+4 are all prime. the exorcism should have started decades ago... i should keep a file of these jokes. the empty joke database. maybe it could be a wiki. call it the extreme point wiki, maybe you could attract some tea party funding? i have a generic question, why is there so much craze for differential equations but not integral equations 6:54 PM @copper.hat Ahahah I was asking if it made sense to make a post for my question related to inner regularity earth: i dunno, historical accident? many equations just more naturally present themselves in differential form? once you get beyond the 'standard' stuff that everybody who is math-adjacent is forced to take, there is far less of a division between the two. for many simple families of ODE, the mathematically simplest way of proving existence is by rewriting as an integral equation. but this is simple in terms of theorems and proofs, maybe not in terms of expressions you write down. the fact that if a function has a nice expression, its derivative likely does too, although its integral might not, may reinforce the bias, at least in introductory classes. @EarthIsNotFlat have wondered that myself. integral equations avoid many fine points that differential equations present. there's this idea that if you don't have an elementary formula for something, it's "worse," creating an impression that differentiation is 'easier' than integration because it is less likely to mess up clean formulas. by most technical standards, in terms of proving theorems, differentiation is "worse" d/dx e^(kx) = k e^(kx) is a nice formula. it also shows you that if you put any norm on any function space including all functions e^(kx), d/dx is going to be an unbounded operator, which kinda sucks also d/dx tends to kick out stuff that isn't necessarily in the domain of d/dx anymore, which also sucks integration would never$^{\star}$do this to you$^{\star}$Narrator: integration will sometimes do this to you it bothered me in my early engineering days to look at a forced ode where the input had noise, because surely this is not differentiable. those were nice simple days. 7:09 PM most of the differential equations ive looked at were motivated or came directly from physics where they were reductions of some system based on$F=ma$, but there is something that feels more physical about integrals imo. @epsilon-emperor You want$\mu(A+B)^{1/n}\geq\mu(A)^{1/n}+\mu(B)^{1/n}$, fix a sequence of compact sets$K_m\subseteq A$with$\mu(K_m)\to\mu(A)$and$C_m\subseteq B$with$\mu(C_m)\to\mu(B)$. Then you prove Brunn-Minkowski for those, getting, for all$m$,$\mu(K_m+C_m)^{1/n}\geq\mu(K_m)+\mu(C_m)$and you take the limit as$m\to\infty$is there any parallel analogue of differential geometry (maybe integral geometry) (and you always have$\mu(A+B)^{1/n}\geq\mu(K_m+C_n)^{1/n}$, this is clear) integral geometry is a field that exists @AlessandroCodenotti I forgot a few exponents there, anzway the point is that you get to$\mu(A+B)^{1/n}\geq\mu(K_m+C_m)^{1/n}\geq\mu(K_m)^{1/n}+\mu(C_m)^{1/n}$and take$m\to\infty$on the rhs 7:25 PM I am building a statistical model. From the theortical properties of the entity being modelled, two features which when interchanged must lead to an exactly negated prediction. The model however need not meet this requirement but ideally should. I have decided to call this "anti-symmetry consistency" or "symmetric consistency". I want to know if this is an abuse of existing mathematical terms. If the input is (x, y) and the prediction is z, the prediction must "ideally" be -z for the input (y, x). I am not sure if there is an existing term for this and whether the names I have come up with for this property are reasonable. @AlessandroCodenotti Understood, thanks a lot! it doesn't seem like an abuse of terms to me. i might call it just "symmetry" (with the precise sense spelled out as you have done here). i say this only because in the abstract, "consistency" sounds like it's something that it is very bad not to have, and if you're willing to consider models that don't have the property, "this does not have the symmetry property" sounds more neutral than that it lacks a form of "consistency." but this is in the realm of what sounds nice to my ear, not what contravenes established conventions. i don't think there are any. @leslietownes It's bad to not to have that property. The model needs to respect it to be scientifically valid. However, they don't have to be perfect consistency; some slight deviations are expected but it should be more or less consistent. 7:50 PM okay, you do you. i don't know what is or isn't scientifically valid. i have the same terminological quibble with 'bias,' in the sense of 'biased estimator.' people treat biased estimators like hot coals. when in many instances they can be just fine. but that terminology is definitely here to stay 8:07 PM @leslietownes so you're saying that they're biased against biased estimators. @leslietownes prime beef is beef that is not the product of two other beefs? Does that make it immaculate beef? :-) @Thorgott Indeed, see my thesis and a subsequent paper. How's @copper's exorcism going? the equity & exercise daemons are winning... They too will pass. 8:24 PM ah, I didn't know you did some integral geometry too though I guess I shouldn't be surprised given that the field partially originated in Chern's work 8:38 PM Well, yes, the kinematic formula he did for submanifolds of$\Bbb R^n$. My thesis was for complex submanifolds of$\Bbb CP^n$. oh wow, one of my professors does integral geometry and I just checked one of his papers and your thesis is referenced in there the world is a small place 8:58 PM Very cool. Who? Andreas Bernig 9:13 PM @robjohn it makes the beef unitarian. 1 hour later… 10:26 PM Hey all is there an algorithm that factors a number into a square and non-square part? Specifically$e= c^2 d$find$c$and$d$given$e\$?

1 hour later…
11:27 PM
i don't know of one. i'm not sure that it's even known whether there is a complexity gap between determining primality and fully factoring an integer as a product of primes. i am also not familiar with the best known methods of each.