Mathematics - 2021-08-20

Thorgott

00:00

@TedShifrin perhaps, the only thing that wasn't a waste about that experience if so

Ted Shifrin

Yeah, measure theory is quite formal.

Thorgott

@TedShifrin perhaps, the only thing that wasn't a waste about that experience if so

leslie townes

formalizing the reals is indeed one of the more empty and unrewarding tasks in mathematics. they can be approached axiomatically without a construction.

rapasite

I am merely trying to answer my own question, I have no pûrpose

Thorgott

thankfully, I also managed to pick up a degree of geometric thinking along the way

oops, double send, my internet just stuttered

Ted Shifrin

00:01

Got it, Thor.

Did you learn any linear algebra, @rapasite?

rapasite

yeah 101 but not soo mutch more

Ted Shifrin

Since you have physics interest, maybe some intro diff geo might be more fun for you.

rapasite

I did vector space eigenvalue and other basic stuff

leslie townes

thorgott i've been having that a lot too. not just here but on office stuff.

Ted Shifrin

OK, cool. Plenty.

Thorgott

00:03

I believe learning a construction of the reals is something one should've seen at some point in their lives, but without any sense of urgency

Ted Shifrin

Download my free diff geo notes, @rapasite. Link in profile.

rapasite

goemetry diferantial?

Ted Shifrin

@Thorgott I disagree, for the typical math student.

Thorgott

I'm just not a fan of working with objects whose existence you've never proven in the long run (emphasis on "in the long run")

leslie townes

ha, my wife just went looking for some new clothes and the cat was in a hamper of folded laundry and attacked her when she looked inside of it. feel free to ask me more about the benefits of cat ownership.

Ted Shifrin

00:13

My cat is growing yuger and mostly bites me.

leslie townes

livvy jumped out of the hamper, grabbed my wife's leg, bit it, and then ran away. there will be a bruise.

cats are very strange, in evolutional terms. they're horrible but somehow we have them domestically.

Ted Shifrin

They can be affectionate, but the hunter instinct is there. My cats in GA actually did hunt outside and then were mostly sweet inside. Sadly, the coyotes were also hunting them, outside.

rapasite

ted thanks for you course but what is the link between the exemple

leslie townes

growing up we had a cat who would leave us pieces of squirrels on the doorstep. we don't let livvy out for this reason.

she likes watching wildlife from the window, and killing spiders and moths as she finds them in the house. that's enough.

god, it was digusting. i'd get ready for school, open the door, and it's four squirrel limbs and a tail on the doormat.

cats are predators.

Ted Shifrin

I don’t understand your question, @rapasite. Nothing to do with constructing real numbers. Just more interesting mathematics.

That is less abstract.

rapasite

00:24

never mind it start to make sense at page

Ted Shifrin

Yes, I had mice and the occasional bird, too, leslie. Ugh.

rapasite

6

Ted Shifrin

First pages are lots of examples of parametric curves. Most are important for later.

rapasite

00:57

ted I can see the tangent vector T and the curvature as the rayon of the tangent circle, what is the intuition for Torsion? the angle between them?

they should be perpendicular

Ted Shifrin

01:10

No, torsion is twisting of the osculating plane.

rapasite

osculating?

ok so it's kind of a curvature of this plane

Semiclassical

i feel like i should know this, but

obviously, there are rational points on a circle. but the only one i know (assuming 0<x<y<1 for definiteness) which has a finite decimal expansion is (0.6,0.8)

Are there any others? I don't think so but it seems unlikely

rapasite

01:26

rational points?

Ted Shifrin

Infinitely many … see exercises at end of section 1.1 of my diff geo notes. A recipe for all Pythagorean triples.

Semiclassical

yeah. coming up with rational points is easy

but the examples i know would have repeating decimal representations

Thorgott

(0.28,0.96) is another one

Ted Shifrin

You’re asking for finite decimals. I see.

Semiclassical

ahh, nice

Ted Shifrin

01:30

So you need factors of 2 and 5 only for the hypotenuse.

Semiclassical

right

and that's not too bad, come to think of it

Ted Shifrin

The rational parametrization of the circle should generate these.

Semiclassical

since the hypotenuse in a pythagorean triple needs to itself be a sum of squares

Thorgott

my example is just the square of your example, I realized

I don't know whether powers of this example are the only examples

Avra

any heard about total order please?

Ted Shifrin

01:33

Plugging rationals into the rational parametrization gives all points. Just plug in appropriate rationals.

shintuku

so, I'm trying to figure out if there exists functions that fill up intervals of the real line in the following way: (of course, they don't ever finish filling it up)

a
			f(3)
		f(1)
			f(4)
	f(0)
			f(5)
		f(2)
			f(6)
b

etc, where [a,b] is an interval on the real line. anyone has a clue? above, f(0) is halfway between a and b, f(1) is halfway between f(0) and a, f(4) is halfway between f(0) and f(1), etc.

Semiclassical

eh. that moves the problem back to finding t such that (1-t^2)/(1+t^2) and 2t/(1+t^2) have finite decimal expansions

which doesn't seem much easier

@shintuku this feels Cantor-ish

so basically looking for a function which would index Cantor's set

Thorgott

shin, you're describing an enumeration of the dyadic rationals if a=0,b=1

Ted Shifrin

You need $t$ so that $1+t^2$ is divisible only by 2 and 5?

shintuku

bless you both for these search terms

Semiclassical

01:36

given that 0<t<1, that's definitely not the correct way to understand that

unless divisibility of rationals means something different

Ted Shifrin

No, $t$ can be an integer, silly.

Semiclassical

um

f?

oh. true (assuming you meant t)

Ted Shifrin

You parametrize the circle by $\Bbb R$.

I hate typing on phone.

Semiclassical

fair

yeah, that works

had to convince myself that the minus signs in the usual parametrization were irrelevant

Ted Shifrin

I have no idea what you’re talking about.

Semiclassical

01:42

for t>1, x = (1-t^2)/(1+t^2) would be negative

but that's just b/c of how i wrote the parametrization. could've just as well done (t^2-1)/(1+t^2)

Ted Shifrin

Oh. Then you use $1/t$.

Semiclassical

sure, but then you're back to having 0<t<1

Ted Shifrin

But you found it with integers.

Semiclassical

right

Ted Shifrin

@rapasite Read section 1.2. Planes don’t have curvature. Rate of turning of its normal, yes.

Semiclassical

01:57

there's probably a more efficient way for me to do this, but doing a brute-force check of whether $2^j 5^k=1+t^2$ for rational $t$ has only given me $t=2,7$ so far (checking j,k up to 200)

t=2 is the usual (0.6,0.8) example, t=7 is the (0.28,0.96) example

so maybe there's no other examples? (not that checking this much is anywhere near definitive)

Ted Shifrin

I haven’t tried. I would start with the modular arithmetic and Chinese remainder.

Or just let the computer run. I don’t be even know why this is a good question.

rapasite

since it depend on our decimal writing i am not sure it's fundamental

Ted Shifrin

Yup, excellent comment.

Semiclassical

because it's interesting if there's only a finite number of examples for a given base

Semiclassical

03:38

For something entirely different from that

i just stumbled across an example of two polytopes which are combinatorially equivalent (same number of vertices, edges, ..., facets) but not affinely equivalent

so that's neat

i mean, it's obvious that it can occur just judging from the 2D case of rectangles and squares. rather rmore startling to have it show up in a 8D case

(i also have no idea why they're combinatorially equivalent, i'm just going based on Polymake)

Ted Shifrin

03:55

But rectangles and squares are affinely equivalent.

I have no knowledge of this stuff …

Semiclassical

...yes, that was dumb

squares and trapezoids

Ted Shifrin

They are too.

Semiclassical

really? take the unit square [0,1]^2 and move the (1,1) vertex to (1,2). that's a trapezoid, but there's no linear transformation that would send (0,0)->(0,0), (0,1)->(1,0), (0,1)->(0,1), and (1,1)->(1,2)

Ted Shifrin

Affine transformation, not linear.

Semiclassical

i'm still confused. affine transformations should preserve parallelism, shouldn't they?

Ted Shifrin

04:04

Yes. I need to check my algebra book to see what I’m vaguely remembering.

Semiclassical

also if (0,0) -> (0,0) then x-> Ax+b requires b=0, so linear. (of course i'm assuming (0,0) -> (0,0) which is a bit of a cheat)

Ted Shifrin

Oh, all trapezoids are affinely equivalent to an isoceles traoezoid.

Semiclassical

hmm. yeah, i can see that

Ted Shifrin

Parallelepiped affinely equivalent to cube.

Semiclassical

I'd share the underlying example, but 1) I think it might be too boring, and 2) I'm writing it up as a question right now

Ted Shifrin

04:09

No problem.

PM 2Ring

04:46

@Semiclassical There are infinite examples. The primitives triples (a, b, c) with c a power of 5. The next one is (44, 117, 125). See here

Semiclassical

yeah, i figured that out. next one should be (336,527,625)

PM 2Ring

You can composite sums of squares using en.m.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity Also see en.m.wikipedia.org/wiki/Brahmagupta%27s_identity

Semiclassical

ooo, but that is a snazzy way to do it

yeah, i devolved to having mathematica do it for me

and it does validate the idea that doing the (3,4,5) case suffices to generate the rest

just higher and higher powers of 3-4i

PM 2Ring

Triads can be organised into a tree. en.wikipedia.org/wiki/Tree_of_primitive_Pythagorean_triples That article mentions a couple of solutions.

Semiclassical

huh, nice, i haven't seen that before

PM 2Ring

04:51

We just need to show there's no triad with $c=2^n$. That's not too hard.

Semiclassical

i thought as much but didn't immediately see how to deduce it

(though i didn't really think about it)

PM 2Ring

Pythagorean triads are a fun application of Gaussian integers.

Semiclassical

i'm guessing the point is that, while you can write $c=a^2+b^2$ in that case, you'll never have coprime $a,b$

so it always reduces down to a smaller case

PM 2Ring

$(u+iv)^=(u^2-v^2) + 2uvi$ and $|(u^2-v^2) + 2uvi| = u^2+v^2$, which leads us to the standard way to generate triads.

Semiclassical

nice

actually, it looks like the case of $2^n$ divides into two

PM 2Ring

04:59

The "problem" is that the Gaussian factorization of 2 is $(1+i)(1-i)$. So when we take even powers of either factor, we don't get mixed complex Gaussians (which would give us a triad), just pure reals or pure imaginaries. I.e., one of the legs of our triad is always zero.

Semiclassical

if $n$ is even, then $c=2^{2k}=(2^k)^2$ and it doesn't seem as if $c$ can be written as a sum of two squares

if $n$ is odd, then $c=2^{2k}=(2^k)^2+(2^k)^2$ which doesn't generate a valid triple

so I think $2^n$ can never be written as a sum of distinct squares

PM 2Ring

Right

Semiclassical

I take it that's the analogue what you just said, without any details to back it up :P

PM 2Ring

:)

Semiclassical

if we take $c=2\cdot 5^k$, is it obvious that that reduces to the case $c=5^k$?

the case $c=4\cdot 5^k$ definitely does

PM 2Ring

05:05

The Gaussian integers are a unique factorization domain. And only (real) primes not of the form 4n+3 can be factorized in the Gaussians. And of course, that factorization gives us a sum of 2 squares.

2 is the oddball case. It's not 4n+3, so it can be factored. But it's not 4n+1, and it's (obviously) the only one where the real & imag parts of the factor are equal.

Semiclassical

right

PM 2Ring

I think Conway said: "All primes apart from 2 are odd, and 2 is the oddest prime of all".

Semiclassical

so this is a problem that's doable but awkward

b/c of the 2's

PM 2Ring

Well, once we show that 2 gives us a degenerate triad for $0^2 + 2^2=2^2$, we're basically done, because none of the higher even powers can work either.

Otherwise, there'd be an alternative Gaussian factorization of that power of 2, which is impossible.

Semiclassical

ah, neat

PM 2Ring

05:24

I'm getting lots of timeout errors here. My last 2 posts just disappeared. :( Oh well.

Semiclassical

same

PM 2Ring

Multiplying two complex numbers on the unit circle corresponds to rotation. And for these rational points like $(.6+.8i)$, each $(.6+.8i)^n$ is unique, so we know that $\tan^{-1}(.6/.8)$ can't be a rational multiple of $2\pi$.

Semiclassical

neat

PM 2Ring

05:45

A silly way to plot the complex unit circle is to choose any point on it, take the square root, randomly choose to multiply by -1, plot & repeat. You quickly cover a high proportion of the points. This is basically the backwards iteration method of plotting the Julia set of $z^2 - 1 = 0$

Sorry. That should be $z = z^2$. Oops

It's been a while since I played with this stuff...

in Python on Stack Overflow Chat, Oct 11 '18 at 14:58, by PM 2Ring

Here's a silly circle drawing program I put together a couple of hours ago. It's based on the old formula for generating Pythagorean triads: a=u²-v², b=2uv, c=u²+v². Also note that for complex z=u+iv, z² =a+ib, so if (u,v) is a point on the unit circle with angle θ from the +ve X axis, then (a,b) is also on the unit circle with angle 2θ.

That program is doing the forward iteration, starting from $(1+2i)/\sqrt5$

PM 2Ring

06:21

Here's a variation in Sage. It uses translucent points so you can (sort of) see which regions get hit the most.

S.M.T

06:43

If we have : $ 5^3 X 5^2 ÷7^2 $

Then , can we write$ 5^{3+2} $ ÷ $7^2$

robjohn

07:49

@schn By itself, the definition of little-o says $\lim\limits_{hu\to0}\frac{f(h,u)}{(hu)^m}=0$. However, in my answer, there is the context of the integral in which the estimate is being used. The difficulty is that this concept cannot be abstracted into a calculus in the way that you seem to want to do. Context tells us whether $h$ and $u$ are each bounded away from $0$ or not.

Peter John

08:58

0

Q: If $P$ is a Sylow $2$-subgroup of $D_{2n}$ then $N_{D_{2n}}(P)=P$.

Let $2n= 2^ak$ where $k$ is odd. Prove that the number of Sylow 2-subgroups of $D_{2n}$ is $k$. [Prove that if $P\in Syl_2(D_{2n})$ then $N_{D_{2n}}(P)=P$.] This question was already asked here. But none of the answers proved $N_{D_{2n}}(P) = P$. I want to prove the statement by proving $N_{D_{...

Question related to Sylow theorem. The question was asked before but I found none of the answers proved the statement by proving the suggested hint.

Koro

10:17

1

Q: Getting strict inequality in $|f(x)|\le \frac 1x$, $x\gt 0$, where $f(x)=\int_x^{x+1} \sin t^2 \, dt$

Given that $f(x)=\int_x^{x+1} \sin t^2 \, dt$ Substituting $u=t^2$ gives: $f(x)=\int_{x^2}^{(x+1)^2}\frac{\sin u}{2\sqrt u}\,du=\frac{\cos x^2}{2x}-\frac{\cos (x+1)^2}{2(x+1)}-\int_{x^2}^{(x+1)^2}\frac{\cos u}{4u^{3/2}}\, du$ It follows that $|f(x)|\le\frac 1{2x}+\frac 1{2(x+1)}+\frac 1{2x}-\frac...

real-analysis riemann-integration

How do I prove the strict inequality?

Semiclassical

A slightly more compact way to put it: show that $\Big|\int_x^{x+1} \sin(t^2)\,dt\Big|<1/x$ for $x>0$

more compact titles are generally better

Koro

any suggestion on how to solve the inequality problem @Semiclassical

6 mins ago, by Koro

1

Q: Getting strict inequality in $|f(x)|\le \frac 1x$, $x\gt 0$, where $f(x)=\int_x^{x+1} \sin t^2 \, dt$

Given that $f(x)=\int_x^{x+1} \sin t^2 \, dt$ Substituting $u=t^2$ gives: $f(x)=\int_{x^2}^{(x+1)^2}\frac{\sin u}{2\sqrt u}\,du=\frac{\cos x^2}{2x}-\frac{\cos (x+1)^2}{2(x+1)}-\int_{x^2}^{(x+1)^2}\frac{\cos u}{4u^{3/2}}\, du$ It follows that $|f(x)|\le\frac 1{2x}+\frac 1{2(x+1)}+\frac 1{2x}-\frac...

real-analysis riemann-integration

the strict inequality looks correct desmos.com/calculator/1xx7qgrvam

Semiclassical

one idea is to rewrite the problem as showing $-1<g(x)<1$ where $g(x)=x f(x)$

with the advantage that now you can find critical points

hmm

actually, i say that but

the condition becomes $g'(x)=f(x)+x f'(x)=0$

and good luck solving that

Koro

it complicates things a lot. doesn't it?

Semiclassical

actually, i take that back somewhat. If $g'(x)=0$ then the value at this $x$ is $g(x)=-xf'(x)=x \sin(x^2)-x\sin((x+1)^2)$

no, this is nonsense

Koro

10:31

:(

Semiclassical

i mean, it's not wrong but it's not useful

Sayan Chattopadhyay

10:47

How hard is it to read mathematical french?

Is there something I should look out for? I have read this paper by Deligne

shintuku

might be easier to read than mathematical german, less declensions

Semiclassical

best i can see (which is interesting but not exactly useful) is that, if you linearize $t^2$ about $t=x+1/2$, then the integral comes out to $f_1(x)=(x+1/2)^{-1} \sin(x+1/2)\sin((x+1/2)^2)$

which is quite a good approximation of the original integral, getting better as $x$ increases

Semiclassical

11:07

one other somewhat interesting point, derived from the above and some numerical experimentation: $|f(x)|$ seems to almost satisfies the stronger inequality $|f(x)|<1/(x+1/2)$

Semiclassical

11:22

it only fails at about $x=4.33$, and even there only by about $4\times 10^{-4}$. but it does fail there. (by contrast, $f_1(x)$ does satisfy this bound. both satisfy the weaker $<1/x$ bound)

Markos Andres

12:07

Can scan someone help me with this series?

((x^2n)/sqrt(n))log(17x²/sqrt(n))

Semiclassical

help you do what with it?

Markos Andres

how can i proceed? I calculated the limit of an and it tends to 0, so they converge. i don't know how to calculate for which x to converge

Semiclassical

so evidently you're asking for the interval of convergence of that series. (if you don't say that in the first place we have no way of knowing)

that said...yeah, that looks yikes

the big red flag being that x^2 inside the log

i'm also not seeing how you're so confident of the limit. plotting with wolfram, it looks like the terms diverge as n->infinty when x>1:

link

the growth is slower if you reduce x from 2 to 1.1, but it still inexorably diverges

Markos Andres

12:37

lim of that logarithm came to me 0 :(

huzaifa abedeen

Polar form of -2i and -4

What is Polar form of -2i and -4??

I wrote it as 2(cos pi/2 + i sin pi/2) and 4(cos 3pi/2 + i sin 3pi/2)

But my answer is incorrect

Semiclassical

what are sin(pi/2) and sin(3pi/2)? (and cos(3pi/2), for that matter)

@MarkosAndres the log by itself would, sure. but you're multiplying it by x^(2n)

take x=2, for instance. then you've got 4^n/sqrt(n) * log(68/sqrt(n))

actually, ignore what i said about "the log by itself"

there's two points to consider here. first, it is of course true that log(68/sqrt(n)) is initially decreasing: you're making 68/sqrt(n) smaller. but once sqrt(n)>68, i.e., n>68^2, then log(68/sqrt(n)) will become increasingly more and more negative. so log(68/sqrt(n)) will eventually start getting bigger in magnitude

second is 4^n/sqrt(n). the sqrt(n) will make things decrease, sure, but 4^n grows much much faster

so you've got one function which slowly gets small and then slowly gets bigger, and one function which is always getting bigger and bigger

together, that means the function blows up

(assuming x>1. if x<1 then x^(2n) rapidly shrinks, so goes to zero. if x=1 you've got 1/sqrt(n) * log(17/sqrt(n)), and the 1/sqrt(n) out front will ensure it the function converges to zero. but it'll converge much more slowly)

geocalc33

14:24

Happy Friday

user193319

Let $f$ be an essentially bounded function on a measure space $(X, \mu)$ (we can assume $\mu(X) < \infty$ if necessary). I know the simple functions form a dense set in $L^{\infty}(X,\mu)$, so there exists a sequence of simple functions $\phi_n$ converging to $f$ in $||\cdot||_{\infty}$. May I conclude from this that $\int_{X} f d \mu = \lim_{n \to \infty} \int_{X} \phi_n d \mu$?

leslie townes

15:14

yeah. |int f - int phi_n| <= int |f - phi_n| <= mu(X) ||f - phi_n|| which by hypothesis goes to zero.

unless i'm missing something.

sorry for not texing that.

schn