Could it be that in some very simple cases the "inverse proportionality rule" does not work?

What rule is that?

urgh nevermind I think I found it

it has ntohing to do with proportions...

@TedShifrin this one

the thing people learn when they are 10

and still drawing with space crayons

ah, the rule that everything is linear.

or i guess not everything. but within some problem, there's something that's going to be linear.

what a great principle

bob and mary can bake a cake at 350F in 30 minutes. how long will it take them if alice and james help, and they bake it at 700F?

15 minutes, duh.

Duh.

oh, i got 7.5 minutes. the temp cuts it by half and then alice and james cut it by another half.

it's the proportionality principle

Alice and James are a red herring.

The kitchen isn't big enough for them to help.

How many minutes is the expectation if there is 50% chance that alice doesn't come

james also employs a mixed strategy. with probability .5 he helps with the cake and with probability .5 he attempts to impede its completion.

I d rather compute the deception instead of the expectation. Much easier: infinity

And Alice drops the eggs on the floor. Cake kaput.

Alice eats the cake, and grows to be 40 feet tall, thereby destroying the kitchen utterly.

Contradictory models.

Has there ever been a demonstration of ‘centripetal force due to gravitational force’ in a lab?

But the main question is, what should they cook to reach the nash equilibrium

Brownies, not cakes.

I mean for example: given two particles of mass m and M which are distance r apart, the gravitational force between them is given by Newton’s law of gravitation as: $\frac{GmM}{r^2}$.

apparently it's group optimal if they all rat each other out to the police.

What does this have to do with centripetal acceleration?

That’s associated with circular motion.

We consider a star of mass M and a planet of mass m. If the planet revolves around the start then assuming circular orbit, the planet needs a centripetal force.

they have definitely measured the gravitational attraction between spheres.

What gives that centripetal force? Gravitational force.

So the planet revolves around the star.

Assuming the right initial conditions.

Ordinarily elliptical, not circular.

are you treating the star as fixed or also moving

Now the question is: Has this ever been demonstrated on a lab scale? Because on lab scale, masses will be extremely small and hence feeble gravitational force. So I think that it can’t be demonstrated.

what i'd expect to be demonstrated in a lab is the equivalent with positive/negative charges

Ted: yes, assuming the circular orbit.

@Semiclassical I mean imaginings a fixed particle in centre and another particle revolving around it due to gravitational force in a lab.

Good thing our resident physicist is here.

depends on the size of the lab. a lot of astronomy is a demonstration of this stuff

Yes, I was going to say the earth in the solar system lab.

(not that all of astronomy is applied gravitational mechanics: for instance, using spectral lines to deduce the composition of stars. but i'd still say that's the majority of it)

Probably not much is known about gravity yet. Newton’s universal law of gravitation does say that the force has so and so value but does not explain why the force exists.

Einstein's theory of general relativity would like a word with you

Science doesn't answer the question "why". That is for philosophers.

a lot of early criticism of newton emphasized that. i'm not sure that it's criticism.

the only orbiting you see at lab scale in this house is people orbiting around the delicious cake that bob and mary made as they wait for it to cool.

You missed the dropped eggs.

the cat took care of that problem.

No cake without eggs.

there have been questions about whether gravity gets modified at small distances. people who wanted extra dimensions to be a thing were hopeful about that, but...nah

So 26-D string theory isn’t?

see for instance arstechnica.com/science/2020/04/…

well, i should say that what's (so far) been ruled out is extra dimensions being observable

is 50 micrometers small?

fully ruling out the concept is essentially impossible b/c you can just say "oh, it's there but we haven't reached the right scale"

@Semiclassical I read that formula for force in Newton’s law of gravitation has now been revised now.

Salut @Astyx

Salut

Je passe en coup de vent

@Koro only modifications i know of are for general relativity

A bientôt

which is not what i'm talking about

it seems like it would be hard to get one thing orbiting around another thing in an actual lab. the forces involved would be so tiny and any friction would dampen the festivities.

@leslietownes yes!!

astyx: i don't know the origin of the question. to me it is small but not that small.

its the width of a human hair, roughly.

I'm just grasping at straws to justify string theory

oh. i was grasping at human hair.

@Astyx how about dividing volume of a rain drop by volume of the earth?

ok. why?

that would yield a number, not a distance!

Oh, you wanted distance.

i will say: the fact that Newtonian gravitation works well (but not perfectly!) up to the planetary level, and apparently works down to the width of human hair

is pretty impressive

If we change coordinates to polar and we pick x+1 = r cos(phi) and y+1 = r sin(phi), why is the determinant of the Jacobian matrix still r

i'm surprised they can measure it on that small of a scale.

From Cartesian to polar

(a quick question not regarding gravity=

@VLC if you write the determinant and do the calculation, you’ll see that you get r.

x= r cos(phi)-1, y= r sin (phi) -1

@Koro Yes but does that work only for k + x = r sin(phi), where k is a real number. Or in what case does it work?

Is there a theorem regarding this ?

What is the derivative of a constant?

I see it's zero...

So it works whenever we add or subtract a constant

@Astyx I grasp at strings to justify straw theory.

one way to see this is the chain rule + the fact that the jacobian of a translation is the identity

I want to ask how this makes sense: finding largest positive a such that: $\int_0^3 f + \int _0^7 f^{-1}\ge a$ for every continuous and surjective $f:[0,\infty)\to [0,\infty)$.

finding largest a? What does that mean?

Suppose it is 21, then I can find some f such that LHS is 25.

it is also given that f is increasing in the strict sense.

I think it should be smallest a.

Young’s inequality

seems like you're trying to find the minimum value of the LHS over that set of f. the greatest lower bound. so yeah, largest positive a seems right.

That’s a good interpretation. Thanks.

maybe it's not a priori clear that there is such a positive a. if ted were here he'd say, try drawing a picture. but i would never say that.

It’s clear from a picture.

Sometimes I use so much oil when cooking I am affraid the US army might invade my kitchen.

I'd be more concerned about the Russian Army.

I hear that they are always in a hurry.

@Koro I'd say $a=6$

@Koro that is not true for every $f$

@robjohn Is there a way to explain why changing the order of integration in this answer appears to be valid? I could integrate by parts before changing the order of integration, but then things just get messy.

@XanderHenderson sometimes I do absolutely nothing yet I am still affraid they might suddenly take over my garden

They have been standing in front of my window for days now. Stating they just want peace.

@RandomVariable The only thing I can see would be to integrate by parts, change order, and hope that integrate by parts will simplify things after. That looks like a touchy integral.

on an unrelated note I have a random guy stating he is willing to "do maths" for me if I pay him

@RandomVariable That will work with the inner integration, I don't know if that can be done with the outer integral.

isn't there some extremely difficult math equation I could give him to have a bit of fun? :p

@LandonZeKepitelOfGreytBritn is he truly random, or just pseudo-random?

nah some guy on reddit

I ll tel him that if he can solve this equation <insert difficult theorem/equation here> I m open to talk

I ll give him 10 minutes or so :p

no need to make him loose too much time

He minored in math apparently

and finds calculus 3 ridiculously easy

Is it true that if we go from cartesian to polar coordianates and say that x = r sin(phi) and that y = r cos(phi), everything else is still the same ?

Can we do that

found something: integral(1/(x^5+2))

So that sin and cosine have different rolles

landon: it's kinda fun to see how high n can get before wolfram alpha or your favorite software no longer evaluates int 1/(x^n + 1) dx (or its variants, e.g. with your +2) exactly.

last i checked it didn't occur to WA to try series as an alternative.

Yhea haha :)

But I am far from being an expert myself. I just need calculus on occasions. This one seemed quite funny because for the uninitiated the result mihgt look very surprising

I just threw that into WA

@VLC What do you mean “everything else is the same”?

I mean that the angle is still the same, it is not reversed or anything like that @TedShifrin

@VLC The angle starts at noon and goes clockwise, not 3 and counterclockwise

Of course the angle is different if you switch sin and cos.

@robjohn @TedShifrin Would the trigonometric identities still hold ?

by that I mean addition of angles etc

The trig functions would be the same.

@VLC Consider the following transformation of the plane: $(x, y)\mapsto (y, x)$. Now introduce the usual polar coordinates after this tranformation. This is basically what you are doing.

You've just changed coordinates. matrix is $\begin{bmatrix}0&1\\1&0\end{bmatrix}$

@Koro It is curious how division by zero is allowed in this exercise

Not to mention $f^{-1}$ denoting $1/f$

@Jakobian You’re wrong on both. It’s the inverse function.

The inverse multi-function?

f was given to be strictly increasing. koro stated that separately a line or two below the other stuff.

Oh, okay, this makes sense now

sometimes problems like this have cute generalizations if the inverse multifunction is not too badly behaved. i dunno about this one, though. my mental picture really needs it to be the way it is.

i used to work with an economist who had a book full of multifunction-based inequalities. economists need them for some reason. i guess because what goes up must come down.

unless it's property values. those only go up, right?

i wish i knew what that book was called. it was a type of book i hadn't seen before. kinda a long list of formulas and theorems and identities, with a lot of emphasis on explicit equations for everything. but some proofs and pretty broad coverage. game theory one page, schauder's fixed point theorem on the next.

very eclectic.

@robjohn It's a shame because it works out so nicely, and it seems to be valid in other similar cases.

@RandomVariable It might be, but it certainly requires some justification.

Perhaps you could try $\frac{\sin(x)}{x^a}$ as $a\to1^+$

There is no dominating function so we can't use DCT

This problem is known to be one of the hardest problem in KSAT.

@Alex I ended up getting 4/6 points even though I didn't complete the proof, so I guess the support was enough.

What is KSAT?

@robjohn I basically asked about that a few months ago here. I don't really understand the self-contained proof in the accepted answer, but the "quick proof" is what I was looking for.

under: that's a good strategy. i used to do that, at least if i had a half-formed strategy with an identifiable gap in it that was not equivalent to the original problem.

i liked seeing that when grading, too. it conveys more information than a skipped problem.

Agreed, as long as it’s not writing BS and all sorts of irrelevancies.

@RandomVariable There has to be a nice simple solution to that. Have you looked at $\arcsin(\sin(x))$? I am tempted to answer that question.

I used to dock people for writing false things in an effort to get points by throwing spaghetti on the wall.

yeah, i wouldn't give partial credit for stuff like that.

it never occurred to me to dock people. that's next level.

Yeah, I try not to do that. I did the proof up until where I didn't where to go. Then I described what I thought the next step might be, but explained I wasn't certain.

That’s fine.

It makes me cringe when I turn in stuff that might be completely false.

i think sometimes the better the student, the more they feel like they have to be perfect or skip the problem. struggling students are more than happy to fill a page.

@TedShifrin (south) Korea SAT.