Mathematics

4:00 AM

Remember to make sure that your integral has du or dx at the end, and remember not to interchange the two.

So, I have a point on the complex plane, call it $z_0$ such that it generates an electric field at every point on the plane. The intensity of the field varies as the distance between $z_0$ and the point and is directed along the line from $z_0$ to the point. I have to show that $F(z) = \frac{1}{ \bar z- \bar z_0}$.

The Great Duck

@Semiclassical it wasnt a question. it was a statement. I know it is equivalent. I just wanted to make you werent referring to a different meaning of "heat equation" such as an equation for determing the heat output by a thermodynamic engine with a certain efficiency.

Semiclassical

I see.

The Great Duck

I was saying to consider 3D modelling related uses to deal with the heat equation.

orbit-stabilizer

I can do it if it varies as teh square of the distance, but otherwise cannot

WDUK

4:05 AM

ok so i get 1/8*(ln(4x^2 + 1) + C

orbit-stabilizer

Take the derivative, does it give you back what you started with?

WDUK

no

arctan(2x)/2 does how ever

so either im doing u substitution wrong still

Tanner Swett

@WDUK You're closer, but it looks like you still made a mistake.

WDUK

or im not getting it

well i have 1/u du where u is 4x^2 + 1 derivative being 8x so i apply 1/8

Tanner Swett

Specifically, it looks like you mistakenly substituted 1/8*du for dx.

Semiclassical

4:08 AM

@TannerSwett Though, oof. Finding the probabilities for that doesn't seem great.

The Great Duck

2

Q: Verifying an integration method for step function integrals.

I should note that this is an integration algorithm and therefore intermediate steps DO appear to be unjustified. The purpose of this question is to justify or reject this algorithm as always giving correct solutions. Suppose we have a function $f(x)$ that has no vertical asymptotes and can be w...

500 rep bounty to whoever proves this

please help guys

WDUK

oh im missing an x

Tanner Swett

Right.

WDUK

but wouldnt that give 1/8 coefficient for 1/u x du that doesn't seem right

Tanner Swett

This much is, in fact, correct: if u = 4x^2 + 1, then the integral of 1/(4x^2 + 1) dx equals the integral of 1/8 * 1/u * x du.

... assuming I haven't made any mistakes there.

WDUK

4:15 AM

yeh thats what i get

Tanner Swett

So, what we've found out here is correct, but it's not useful, since it still has an x in there.

WDUK

so that means the substitution doesn't work/help ?

unless it can still be solved that way

Tanner Swett

I think that's right. It's been a few years since I did integrals, so I don't remember this very well.

WDUK

so there must be some way to identify when u substitution won't be a good option here without having wasted 15 minutes :P

orbit-stabilizer

Right, but using $u = 4x^2 + 1$, we have that $x = \pm\sqrt{\frac{u-1}{4}}$.

Semiclassical

4:18 AM

ah crap, I think i infinite-looped my mathematica

orbit-stabilizer

So you could substitute that in, but you'd see you end up with the same problem again

Practice a lot. You'll learn what works and what doesn't work

WDUK

is there no known rule to know what will and won't ?

Semiclassical

Experience helps

@TannerSwett I should also point out that mathematically you're doing a 1D random walk: Start at x=50, walk 10 right with probability p and walk 10 left with probability 1-p, subject to the expected change in position after one step being -0.5. The walk halts when you reach x=0.

orbit-stabilizer

finding antiderivatives is not like finding derivatives. there's no algorithm

Semiclassical

There are patterns you'll come across, of course.

orbit-stabilizer

4:23 AM

i learned that $f(z) = \frac{1}{z}$ corresponds to rotating the riemann sphere! so cool!

Semiclassical

Moebius Transformations Revealed

►

WDUK

this is all i can really find on when u sub is viable:
A u-sub can be done whenever you have something containing a function (we'll call this g), and that something is multiplied by the derivative of g

Semiclassical

$\int f(x)\,dx=\int f(g(u))g'(u)\,du$

Which of course is just the substitution rule.

The thing with a u-sub is that you're hoping that $f(x)$ is something simple enough to integrate by hand

WDUK

ah so there is a rule thats what i was looking for

Semiclassical

yeah. but that's just using $x=g(u)\implies dx=g'(u)\,du$

You can swap the roles of x and u in there, of course

orbit-stabilizer

4:27 AM

Wow. That is genuinely beautiful.

Semiclassical

ikr

orbit-stabilizer

@Semiclassical thanks for sharing that

Semiclassical

np

Secret

@Semiclassical aka inverse chain rule

orbit-stabilizer

Riemann was a smart guy.

Honestly one of the most incredible things I've seen.

Semiclassical

4:28 AM

Mobius transformations are wicked cool eh

orbit-stabilizer

Would this extend to higher dimensions? Doesn't seem like it depends on any particular properties about $\mathbb{C}$.

So, we have a 4 dimensional sphere 'rotating' and that corresponds to some transformation of 3-space

Semiclassical

It's relying on stereographic projection

so there's definitely higher analogues

orbit-stabilizer

Right. We have a 1-1 map between the sphere and the plane

Secret

Hmm... Is moebius transformation make sense on quarternions?

orbit-stabilizer

mostly one to one

Semiclassical

4:29 AM

I know there's some crazy stuff involving the Hopf fibration which I suspect is relevant here

Hopf fibration -- fibers and base

►

WDUK

i hope to learn quarternions at some point i use them a lot in programming but have no real understanding of the numbers

orbit-stabilizer

Why do you use them in your programming? Seems like a niche thing to use

Semiclassical

Rotations.

WDUK

you get gimbol lock issues in computer games

orbit-stabilizer

Ah, for gaming. got it

Semiclassical

4:31 AM

Right. Though that's something I more know of than actually understand

Secret

quaternions are one of the easiest way to code rotations in gaming

Semiclassical

I feel like the music does a good job of differentiating the two videos.

WDUK

none of the programmers i know actually understand them lol just take the functions for granted for the most part

Eric Silva

conformal maps are not as interesting in dim > 2

orbit-stabilizer

oh, the riemann sphere also provides a proof why the cardinality of hte unit disc is the same as the rest of the plane excluding the unit disc

Semiclassical

4:33 AM

First one is simple and elegant, second one is rather more baroque

Eric Silva

so the higher analogues of mobius transforms are way less fun

orbit-stabilizer

dat 1 - 1 map

Semiclassical

@EricSilva Sad!

Hmm

Eric Silva

they're still cool but they're less rich because you loos the complex analysis

Semiclassical

Is there no version of stereographic projection for the 3-sphere to R^3?

Eric Silva

4:34 AM

there is

Semiclassical

Hmm.

Eric Silva

but you still dont get all the nice conformal maps because there's no riemann mapping theorem

Semiclassical

Then I think the original idea would still work, though not as nicely as in the Riemann sphere case

ahh

orbit-stabilizer

yeah, i dont get the hopf bundle one.

Semiclassical

So there'd be a correspondence, but the transformations you get aren't going to be nice

Eric Silva

4:36 AM

in higher dim there's only inversive guys, homotheties and isometries

orbit-stabilizer

gotta go back to my complex hmwk

Secret

Stereographic Projection of a 3-Sphere

►

Eric Silva

well they're all "nice" but they're certainly less rich

Semiclassical

hmm, okay

Eric Silva

cf. one of the many theorems named after liouville

Semiclassical

4:37 AM

lol

Tanner Swett

@Semiclassical Right, it's the expected time to reach a particular point in a biased random walk. I haven't been able to find a formula for that on Google.

Semiclassical

I think you've got the right result, but yeah

Tanner Swett

Liouville's theorem is nice, but it's not as nice as the Liouville theorem, nor the theorem of Liouville. :D

Eric Silva

conformal maps are still the isometries of $\mathbb{H}^{n}$ though so that's something cool about them

Semiclassical

@TannerSwett mathoverflow.net/questions/124242/…

Narcissus jewel

4:39 AM

@LeakyNun I meant that the proof lifts verbatim, saving you that one line.

Tanner Swett

@Semiclassical Huh, that's the answer to a harder version of my question. :D

Semiclassical

Yeah, it's the entire distribution

Narcissus jewel

@LeakyNun I like to math-golf :P.

Secret

Fun fact: It is easy to work out who is ________ because there will be a very sharp discontinuity in the graph of math chat

Even though that functionality is designed for espionage purposes, they did not aware there is a fatal flaw that will reveal their presences

We are experts at sniffing out the hiders who are in service of The One True Enemy of our kind, for we have at least 15 years of experience on that kind of socially evasive behavior that had so far escaped from persecution

Narcissus jewel

What is this meant to mean @Secret?

Secret

4:48 AM

It's encrypted. Try to decode it, but it is not very important. It makes a nice anime plot though as balarka noted

Narcissus jewel

Wait what does encrypted mean here?

Tanner Swett

@Semiclassical Well, I'm a bit too sleepy at the moment to extract a formula from that...

Narcissus jewel

I see no code?

Secret

actually, I am not sure if riddle count as an encryption type...

Semiclassical

same.

WDUK

4:50 AM

i'll guess, ronald mcdonald is the answer

Tanner Swett

If N is the number of bets made before losing all your money, B is your initial bankroll, and L is your expected loss on each bet, is the formula just E(N) = B/L, as expected?

Semiclassical

@TannerSwett That's my impulse as well.

But there's a way to settle it: math.stackexchange.com/q/2400386/137524

Take the answer there and let $a\to\infty$

Narcissus jewel

Semiclassical

In this case: $z=5$, $p=0.475$, $q=0.525$, $a=\infty$

Secret

turns out riddle and encryption are two separate concepts, o well, guess I need to update my terminology

so, it is a riddle, basically

Semiclassical

4:55 AM

looks like their answer reduces to $D_z=z/(q-p)$ since $p<q$ and $a\to\infty$

but the expected change after one event is $(1)p+(-1)q=-(q-p)$, so this amounts to the answer you anticipated @TannerSwett

Tanner Swett

Nice.

This question is inspired by a true story, by the way.

I came in with $50. My first two bets were $15, and all my subsequent bets were $10.

I won the first bet and lost the following six bets.

Semiclassical

lol

Tanner Swett

What are the chances of that? The answer, of course, depends on what one means by "that".

Secret

Narcissus: My riddles are designed in such a way that after 3 years, even I cannot solve them

They are so heavily dependent on memory at a certain moment in time and place

But as far I knew, they are very effective at doing what it is intended to do

Tanner Swett

What are the chances of getting that exact sequence of wins and losses in 6 bets? 1.01%.

I mean 7 bets, of course.

The probability of getting that exact sequence of wins and losses in 6 bets is 0. :)

Semiclassical

5:11 AM

A joking response to that: "The chances were 100%. It happened, after all."

Icche Guri

hi

Semiclassical

(there's probably a deep philosophical statement inside of that, but i'm not going to venture it)

Secret

Pr(Stuff happens) = 100%
Pr(When it happens) < 100%

That is, we knew it happened, but we cannot predict in advance when or how frequently it will happen given any sequence

Semiclassical

Pr(X|X)=1

Secret

yeah, because given X had happened

Semiclassical

5:17 AM

right-o

Tanner Swett

... if X is a non-empty event.

Semiclassical

true

if Pr(B)=0, then the usual formula that Pr(A|B)=Pr(A and B)/Pr(B) doesn't make a heck of a lot of sense.

Tanner Swett

Yeah, I don't even know how Pr(A|B) is defined when Pr(B) is 0.

Semiclassical

the above is the Kolmogorov definition

Secret

I guess we can go one step backwards, cause bayes theorem is actually Pr(A|B)Pr(B)=Pr(A and B). Thus if Pr(B) is zero, then Pr(A and B) must be zero thus the value of Pr(A|B) is irrelevant

The probability of A given B is some value, but B never happened, thus the premise is invalid, I think...

Semiclassical

5:23 AM

From the wiki page on conditional probability: "Some authors, such as de Finetti, prefer to introduce conditional probability as an axiom of probability: $P(A\cap B)=P(A|B)P(B)$. Although mathematically equivalent, this may be preferred philosophically; under major probability interpretations such as the subjective theory, conditional probability is considered a primitive entity."

Secret

I suspect since B never happened, "given B happened" is vacuously true...? I need to revise my logic...

Semiclassical

A little later: "If P(B) = 0, then according to the simple definition, P(A|B) is undefined. However, it is possible to define a conditional probability with respect to a σ-algebra of such events (such as those arising from a continuous random variable)."

But it also goes on to indicate that such a definition runs into problems when B is a set of zero measure...

which apparently gives this:

Borel–Kolmogorov paradox

In probability theory, the Borel–Kolmogorov paradox (sometimes known as Borel's paradox) is a paradox relating to conditional probability with respect to an event of probability zero (also known as a null set). It is named after Émile Borel and Andrey Kolmogorov. == A great circle puzzle == Suppose that a random variable has a uniform distribution on a unit sphere. What is its conditional distribution on a great circle? Because of the symmetry of the sphere, one might expect that the distribution is uniform and independent of the choice of coordinates. However, two analyses give contradic...

Tanner Swett

@Secret The "given" here isn't a logical connective (so it's not the logical connective of material implication).

Semiclassical

"The concept of a conditional probability with regard to an isolated hypothesis whose probability equals 0 is inadmissible." -Kolmogorov

So the basic rule of how to define P(A|B) when P(B)=0 is: Don't.

Tanner Swett

Right... but can it be defined if we equip the probability space with appropriate additional information?

Semiclassical

5:31 AM

shrug

Ted Shifrin

6:42 AM

@Akiva: A simple comparison test isn't going to cut it for infinite products, unless you go via the conversion to sums we were talking about.

Secret

Tanner, Semi: Yeah, it seems particularly for null sets, the measure is important as well the pdf, because the different ways in approaching the null set via the limiting case of a family of events will give different conditional pdfs

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$Mathematics$ Mathematics