Mathematics - 2023-03-01 (page 1 of 2)

冥王 Hades

01:39

I hate needles. I hate them so much

user 85795

02:34

@copper.hat only if you're a 49er fan

copper.hat

@user85795 i'm a square fan

Ted Shifrin

Be cool, don’t be square?

user 85795

You have to be cruel to be kind.

Ted Shifrin

I do my best.

user 85795

02:52

Too much spoon feeding leaves only the memory of the shape of the spoon.

Ted Shifrin

03:37

Where was spoon feeding in the discussion?

leslie townes

i don't remember. i just remember a round shape. somewhat like a paraboloid. or maybe it was an ellipsoid.

it had lambda-systems and generators

Ted Shifrin

Hyperboloids of one sheet resemble not spoons.

robjohn

@TedShifrin Oh, no... Xander was number 49; that is a square.

leslie townes

i'm assuming the systems have to do with the fact that those things are ruled surfaces?

Ted Shifrin

Doubly ruled, in fact.

robjohn

03:41

@TedShifrin spoons that I don't think would be useful for soup

Ted Shifrin

Maybe a soup with reverse magnetism.

robjohn

Iron enriched

Ted Shifrin

Kale for the win?

Cotton Headed Ninnymuggins

What is happening?

Ted Shifrin

For non-integers, it”s using the gamma function.

Try $(1/2)!$.

Cotton Headed Ninnymuggins

03:48

0.886226925453

Ted Shifrin

Yup. $\sqrt\pi/2$.

robjohn

$\binom{1}{1/2}=\frac4\pi$

I only use integers for the lower coefficient

Ted Shifrin

This was factorial. No binomials.

robjohn

I know, but it can be extended

Ted Shifrin

Yes, of course, but confuzling.

Cotton Headed Ninnymuggins

03:50

interesting

copper.hat

is there one that equals $-{1 \over 12}$?

Ted Shifrin

“How did you like the sweetbreads?” “Interesting.”

Stupid $-1/12$.

copper.hat

never got into innards for food

Ted Shifrin

I can reframe for you, copper.

“How did you like the haggis?” “Yuck.”

copper.hat

yep, would pass on haggis, once is enough

Ted Shifrin

03:55

So much for heritage.

copper.hat

not mine

Ted Shifrin

Close.

robjohn

I thought about trying haggis once when I went to the Tam'O'Shanter. Then I came to my senses.

copper.hat

tripe & drisheen if you want some Irish innards

Cotton Headed Ninnymuggins

@TedShifrin I didn't really have anything to add other than "interesting"

Ted Shifrin

04:02

I’ve had tripe in France, not elsewhere.

OK @Cotton

robjohn

@CottonHeadedNinnymuggins You could have added all the positive integers to get $-\frac1{12}$

copper.hat

generally not a fan of parts, we had a lot of liver & kidney growing up, but i am happy to leave that, along with intestines, tendons, pig's feet & brain behind me

and chicken feet

and frogs

give me potatoes or give me death

robjohn

death au gratin

Ted Shifrin

LOL

Or Death Anna

robjohn

sweet death doesn't convey the potato connection

Koro

04:07

How to show that a holomorphic function can't be radial?

Ted Shifrin

What does that mean?

leslie townes

this is also in ahlfors.

copper.hat

presumably a function of radius only?

robjohn

$f(z)=g(|z|)$

leslie townes

presumably that f(z) depends only on |z|.

Koro

04:08

A radial function F on $R^2$ means that it is constant on every circle centered at 0.

@leslietownes ohh

Ted Shifrin

Look at C-R in polar.

leslie townes

pick your toolkit, really, but C-R definitely jumps out as the low tech way to do this.

Koro

Ah, that's what I thought!

copper.hat

i want to use the open mapping theorem

Cotton Headed Ninnymuggins

@robjohn I saw a Numberphile video on that a while ago, and then Mathologer dunking on them

Koro

04:10

@robjohn and this too. I said that with this f has to be real valued so can't be holomorphic unless it's a constant.

I realize that it is wrong.

copper.hat

why does the $f$ have to be real valued?

Koro

I do need polar here

Ted Shifrin

Or playing with $1$-forms works nicely.

robjohn

@Koro It might be easier to show that if a holomorphic function is radial, then it is constant.

copper.hat

show that $z \mapsto f(e^z)$ is 'independent' of '$y$'.

Ted Shifrin

04:19

That’s good, copper.

robjohn

The mean of a harmonic function over $|z|=r$ is its value at $0$, so if a harmonic function is radial, it is constant.

the real and imaginary parts of a holomorphic function are harmonic

Koro

I used CR to get: $u_r=v_r=0$

where $f(r,\theta)= u+iv= F(r)$, where u=Re F(r), v= Im F(r)

@copper.hat why is it so? $f(e^z)= f(e^x. e^{iy})$

copper.hat

@Koro what is $|e^z|$ in terms of $x,y$ where $z=x+iy$.

$f$ is independent of phase.

Koro

$|e^z|=e^x$

copper.hat

so $f(e^z)$ is 'independent' of $y$. Hence constant.

Koro

04:29

$f(e^z)=|f(e^z) |e^{i \theta}$

copper.hat

the idea is that the range of $f$ is 'one dimensional'

@Koro nooooooo

Koro

ohh

here $\tan \theta= q/p,$ where q= Im f(e^z), p= Re (f(e^z))

do you want to say that $f(e^z)=f(e^x)$
?

I guess not.

copper.hat

If $g(z) = f(e^z)$, what is $g_y$?

leslie townes

as ahlfors would say, 'the argument is harmonic whenever it can be uniquely defined'

copper.hat

have you used that line in court?

Koro

04:35

$g_y(z)= f_y(e^z) e^x (i e^{iy})$

leslie townes

koro this is on the same set of pages as the earlier stuff in ahlfors, if you have that

copper.hat

@Koro no, $g_y = 0$.

Koro

ohh

copper.hat

the idea is that the $e^z$ part 'straightens out' the range of $f$ in some way.

Koro

Using CRE, we get: $u_r=v_r=0$ so $f'(z_0)= 0$ at every $z_0$ and hence f must be constant.

copper.hat

04:37

i am sure there is some manifold mumbo jumbo terminology for that

@Koro yep

Koro

:-)

I still don't understand your approach though. I'll think about it.

copper.hat

what mapping would you use to locally straighten out radial graph paper?

conformal mappings are pure magic

folks used to conformal mappings to solve static flow problems

Koro

ah, I'm yet to study conformal maps 😬

copper.hat

04:53

for me, using this stuff in an engineering setting (radiation patterns, smith charts, etc) helped considerably with developing some intuition

Koro

I see :-).

That's great. I also want to apply my mathematical understanding to engineering.

:-)

Sourav Ghosh

06:07

Please suggest reference to study Quasi cauchy sequence in details.

Not Tfue

07:21

Let
$E ⊂ (0, 1)$
be the set of all real numbers with decimal representation using
only the digits $1$ and $0$:
$E := \{ x ∈ (0, 1) : ∀j ∈ N, ∃d_{−j} ∈ \{1, 0\} \text{ such that } x = 0.d_{−1}d{−2} . . .\}$.
$f : E → ℘(N) \text{ s.t if }
x ∈ E, x = 0.d_{−1}d_{−2} . . .,
f(x) = \{j ∈ N : d_{−j} = 1\}$.

@Koro If it were you how would you prove the bijection of this? Just curious because I feel like my proof doesn't feel that nice.

The one I posted earlier.

Silly Goose

07:33

What does it mean for two algebras defined over the same space to be independent? is it literally just that they share the least amount of elements possible (i.e. only share the same identity...and the non-element)?

Not Tfue

10:19

I'd say the contrapositive statement will be easier to prove.

robjohn

10:38

@NotTfue of what?

oh, your question

Not Tfue

robjohn

@NotTfue what is $f$? (and why not use MathJax?)

Not Tfue

my phone can't render math Jax unfortunately

@robjohn $f(x) = \{j ∈ N : d_{−j} = 1\}$

It produce natural number if d_{-j} is 1 from decimal expansion.

冥王 Hades

Not Tfue

I don't know if this is logical sound but I think if x=x' implies that f(x)=/=f(x') because you can say element of f(x) will not occur in f(x) and vice versa.

robjohn

10:45

@NotTfue you've installed ChatJax? what kind of phone?

Not Tfue

@robjohn I use Samsung browser. Very old phone. Galaxy mini.

robjohn

I can't say for your particular phone, but most phones run ChatJax

Not Tfue

@robjohn have you seen proof of this kind I am very curious how other wrote the proof of bijection between power set and real number in (0,1).

I will try again to install chatjax. It is nightmare.

Well still doesn't work. Works on pc flawlessly.

Anyway I think I can render latex using my imagination.

Not Tfue

11:14

how about subjection? define take any $y\in f(x) x=\sum_{n\in y}1/10^n+1111.... 'so$ f(x)=y $

I wish I was at least gifted with little amount of mathematical intelligence so it won't take me 4 days to just think about particular proof ;_;

@Vicfred best of luck

Koro

11:50

@NotTfue we already discussed this. Didn't we?

mick

12:26

1

Q: $\sin(\frac{\pi}{p}) $ not expressible by positive radicals and $\sin(\frac{\pi}{q_i})$ ??

We have the following identities: $\sin(\frac{\pi}{1})=0$ $\sin(\frac{\pi}{2})=1$ $\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$ $\sin(\frac{\pi}{4})=\frac{1}{2}$ Lets start with a definition. Rules for construction : We start with the set of positive integers and we extend with a finite set $A$ of alg...

Not Tfue

12:49

@Koro Yeah but I thought if there are other way that looks much cleaner.

Thorgott

13:33

@TedShifrin haha, that's neat

I also realized as I've been thinking through it again a couple days ago that the omission of why this "division" works is a more significant gap than I thought

cause that's actually the only part of the argument that doesn't work anymore in higher codimension

Shaun

14:13

in Helpful Commentary, 20 mins ago, by Shaun

Please would someone give some feedback on the following?

-1

A: Union of two affine varieties equals to the product of the varieties

Hint: $k$ is a field and so has no zero divisors.

Koro

14:25

@NotTfue what is pre-image of the empty set, $\emptyset\in P(\mathbb N)$?

Koro

15:34

0

Q: $f: D\to \mathbb C$ is holomorphic such that $diam f(D)=d$, then $|f'(0)|=d/2\implies f$ is linear.

$f: D\to \mathbb C$ is holomorphic such that $\sup_{u,v\in D}|f(v)-f(u)|=d$, then $|f'(0)|=d/2\implies f$ is linear. Here, let's suppose that $D$ is a closed unit disk. By Cauchy's integral formula, since $f(z)$ and $f(-z)$ are both holomorphic, we have for every circle $C_r$ centered at $0$ and ...

complex-analysis cauchy-integral-formula

Mats Granvik

iteration of:
z = z^2 + 1/(1/z + 1/Conjugate[z]) + Zeta[N[z]]

冥王 Hades

Found this on Twitter

I got 45°

PNDas

Is there any papers or articles related to the numbers $n$ satisfying: digit sum of $n^2$= $($digit sum of $n)^2$?

I found a different thing johndcook.com/blog/2018/03/24/squared-digit-sum/….

冥王 Hades

16:03

Is it correct?

Jake

16:19

Hi guys, I'm going a little crazy here.

I'm attempting this questions. The answer comes out to $diag(0.5,0.5,0)$

But, how does that work? P is not made from the eigenvectors of A?

So how is this being diaganolised?

symbolab.com/solver/matrix-diagonalization-calculator/…

e.g. by looking here I can see that P isn't the matrix symbolab says should diagonalise it

yet it happens anyway

Ted Shifrin

@Jake You're just wrong. That's why :) There are lots of $P$'s that work. Lots.

Jake

wait. So you can diagonalise a matrix without its eigenvectors?

I thought the columns of P had to be the eigenvectors of A?

The ones symbolab gives aren't even multiples...

冥王 Hades

Jake

@冥王Hades it's making me go nuts

冥王 Hades

I'm stuck on a matrices problem too

Jake

16:37

@TedShifrin surely, P has to be made of eigenvectors right?

Because you can rearrange to $AP = BP$ and deduce from there

robjohn

@Jake Since two of the eigenvalues are equal, what can we say about the two associated eigenvectors?

Jake

degenerate?

Ted Shifrin

@Jake Yes, that's correct. That's why I said you were wrong earlier.

robjohn

what can you say about any linear combination of the two eigenvectors?

Jake

also an eigenvector

robjohn

16:42

yes, so there are many eigenvectors.

Jake

is that only the case for the degenerate case?

Ted Shifrin

Degenerate is very much the wrong word.

robjohn

It happens for $\begin{bmatrix}\frac12&0&0\\0&\frac12&0\\0&0&1\end{bmatrix}$ also

Ted Shifrin

When you have a repeated eigenvalue, you may or may not have enough corresponding eigenvectors. The words algebraic and geometric multiplicity are relevant.

Jake

I'm a physicist :)

Degenerate is our favourite word

robjohn

16:45

(they all are) ;-)

Jake

@robjohn so could this be done for one without repeated eigenvalues?

robjohn

@Jake If you don't have repeated eigenvalues, the eigenvectors are distinct (up to scalar multiples), they do not form a plane of eigenvectors.

Jake

ahhh I see

metric.ma.ic.ac.uk/metric_public/matrices/…

This article explains it well

This was highly annoying – I'm only tutoring some university student and had my entire math abilities brought to crumbling point

Ted Shifrin

To me the degenerate situation is a repeated eigenvalue with only one (independent) eigenvector.

robjohn

$\begin{bmatrix}\frac12&1&0\\0&\frac12&0\\0&0&1\end{bmatrix}$

Ted Shifrin

16:53

That’ll do nicely, thanks.

robjohn

It was easy to copy/paste and change one value

Prateek Mourya

hello @robjohn

Ted Shifrin

Easy unless on an ipad or phone, yes.

Prateek Mourya

robjohn

@TedShifrin Copy paste is not too difficult on my iPhone. There is a way to select a block of text.

Prateek Mourya

17:06

not sure how it suggests to respective form of Ax Ay and Az

Semiclassical

geometry problem: what's the easiest way to convince someone that $(x(t),y(t))=(\cos t,\cos(t-\phi))$ traces out a diagonal ellipse for $0<\phi<\pi/2$

Prateek Mourya

?

Semiclassical

not sure why you're assuming that's directed at you

one way would be to plot it, either using software or by hand

if i do angle addition formula to expand i get $y= \cos t\cos \phi-\sin t\sin \phi\implies \cos t=x,\sin t =(x \cos\phi-y)/\sin \phi$

hence $x^2+(x\cos \phi -y)^2/\sin^2\phi =1$

which is definitely an ellipse but i dunno if it's obviously diagonal

if i replace $t\mapsto t+\phi/2$ then i guess i get $x,y=\cos t\cos (\phi/2)\pm \sin t\sin (\phi/2)$

so $x+y=2\cos t\cos(\phi/2)$, $x-y=2\sin t\sin(\phi/2)$. at which point i guess it's obvious that it's a diagonal ellipse

@Jake the usual case in physics where you run into degeneracy is in quantum mechanics, in which case you're usually dealing with Hermitian matrices

for which the spectral theorem applies and there's no difference between geometric/algebraic multiplicity of eigenvalues

Ted Shifrin

17:25

@robjohn I agree ... when it works.

What is a diagonal ellipse, Semiclassic?

Semiclassical

axes are $y=\pm x$

Ted Shifrin

Ah.

You make up your own terminology :D

Semiclassical

main reason i'm sticking to diagonal is that it's a 45-degree line segment when $\phi=0$

i.e., a diagonal line

Koro

Let $p$ be an odd prime. $S_p$ has order $p!$. So every Sylow p-subgroup is of order p. Let no. of Sylow p-subgroups be $n_p\equiv 1 \pmod p$. Each of these Sylow subgroups has p-1 elements of order p. So total no. of elements of order p in $S_p$ is $(p-1)n_p\equiv p-1 \pmod p$. But total no. of p-cycles in S_p is (p-1)!. So $n_p(p-1)=(p-1)!\equiv 1 (p-1) \pmod p\implies (p-1)!\equiv -1 \pmod p$. This proves Wilson's theorem. :-)

Ted Shifrin

I'm heading off to get another Covid booster, @Semiclassic, but I'll look at it when I get back.

Semiclassical

17:28

kk

Prateek Mourya

can anyone help

29 mins ago, by Prateek Mourya

not sure how it suggests to respective form of Ax Ay and Az

Ted Shifrin

18:13

@Semiclassical The relevant matrix is $\begin{bmatrix} 1 & -\cos\phi \\ -\cos\phi & 1\end{bmatrix}$, with eigenvalues $1\pm\cos\phi$ and, indeed, eigenvectors $(1,\pm1)$.

Semiclassical

yeah

it's figuring how to make this accessible to students without going into the guts of that

leslie townes

i don't see a way around grappling with the rotation matrix, in some form or another, whether or not you write it down and say 'this is the rotation matrix.'

Semiclassical

well, $t=\phi/2,\phi/2+\pi/2,\phi/2+\pi,\phi/2+3\pi/2$ gives the points on the semi-major/minor axes

and these do lie on the lines $y=\pm x$

i guess it's not exactly obvious that those are the right points tho

Ted Shifrin

Well, high school precalculus used to teach the formula for the rotation of the axes by brute force, no linear algebra.

Semiclassical

i do want to introduce the rotation matrix today if i can, so it's not entirely impossible

Ted Shifrin

18:20

I would suggest that the "easier" alternative is to show symmetry about $y=\pm x$ by showing the equation is preserved by switching $x$ and $y$ up to sign.

Semiclassical

but i don't think i have the luxury of doing the $v^\top A v=1$ form of an ellipse

@TedShifrin true. the way to do that $y=x$ part of that is to take $t\mapsto \phi-t$ so that $(\cos t,\cos(t-\phi))\mapsto (\cos(t-\phi),\cos t)$

actually, once you have one symmetry axis of an ellipse isn't that it? (assuming you know it's an ellipse)

Ted Shifrin

Yes, if you know the axes of symmetry must be orthogonal (also linear algebra).

Semiclassical

true

this is in the context of a very short lab activity, mind

so i have to be judicious about what stuff i include

Ted Shifrin

Yup.

leslie townes

my hs calc teacher assigned an extra credit problem once about finding the volume of a surface of revolution of a graph about the line y = x, and it broke the class. we had none of the prerequisites. i ended up getting a solution, but using stuff that we had not seen in the class.

Semiclassical

18:32

so...volume of a cone

oh

misread

i don't actually know how you'd do that without rotating

leslie townes

Pappus's centroid theorem

In mathematics, Pappus's centroid theorem (also known as the Guldinus theorem, Pappus–Guldinus theorem or Pappus's theorem) is either of two related theorems dealing with the surface areas and volumes of surfaces and solids of revolution. The theorems are attributed to Pappus of Alexandria and Paul Guldin. Pappus's statement of this theorem appears in print for the first time in 1659, but it was known before, by Kepler in 1615 and by Guldin in 1640. == The first theorem == The first theorem states that the surface area A of a surface of revolution generated by rotating a plane curve C about an...

that's what i was able to use (it was not in our textbook)

shintuku

War of the Worlds: Attack of the Centroids

leslie townes

even shintuku must admit that the centroids leaked from a lab in alexandria

shintuku

i'm too much of a ba'ath party fangirl to admit it sorry

Semiclassical

@leslietownes i'd say they should burn it down to prevent any more leaks but i guess they already tried that

leslie townes

18:38

booo semiclassical. too soon!

Semiclassical

one thing i do want to make hay about is the following curve: $(x,y)=(\cos t,\sin t)+ r(\cos (t-\phi),-\sin (t-\phi))$

if you take $(r,\phi)$ as polar coordinates in a $uv$-plane, then you get: $(u,v)=(1,0)$ gives a horizontal segment, $(-1,0)$ gives a vertical segment, $(0,0)$ gives a right-handed circle

points with $r=1$ gives line segments at angle of $\phi/2$ to the $+x$-axis

points with $\phi=0$ give ellipses with semi-axes $1\pm r$. etc

Ted Shifrin

@leslietownes I've assigned this sort of thing as extra credit in Calc II for years. Doing the region between $y=x$ and $y=x^2$. I love the problem. The issue is that you can do it wrong and still get the right answer. Understanding this was a good exercise for me the first time it happened.

Semiclassical

@TedShifrin the joy/horror of howlers

Ted Shifrin

You don't need to rotate. You just need to slice perpendicular to $y=x$ and do a change of variables in the integral.

leslie townes

i forget what the other curve was in my case. i think it was a parabola, but not y = x^2. i think he used one where x was a function of y.

Ted Shifrin

18:45

Probably $x=y^2$ :P

It actually is a great problem. We should give it to everyone here.

PNDas

Can anyone suggest me some books or notes about inverse fourier transform on $L^1$? In the book, they discussed only on $\mathcal S(\mathbb R^n)$.

I forgot write please.

leslie townes

PNDas, i am 10+ years removed from specifics, but are you generally aware of the difficulties in characterizing the range of function spaces under the fourier transform or its inverse? i do not know in this specific case but know in general that knowing exactly what happens under X fourier transform on Y space (where X is empty or inverse) is difficult. and if you had more hypotheses that would definitely help.

ted: anyway, it's what you do if you want to find the one person in 30 people in 1996 who will go on and get a phd in math

Rithaniel

Hola, math chat

PNDas

@leslietownes I know how they define Fourier transform for $L^1,L^2,\mathcal S$ and their properties. I know the properties of inverse fourier transform on $\mathcal S$ and I found a lecture note where they define it for $L^1$. I want to know properties of IFT which holds on $L^1$ but not on $\mathcal S$ etc.

Ted Shifrin

@leslietownes Over the years I had some good students (who did not go particularly far in mathematics) who solved it. There were enough challenging problems when I taught the Spivak course that I did not include this there — just in regular Calc II classes.

Hola @Rithaniel

leslie townes

18:53

at that level of generality i would just direct you to the general textbooks on harmonic analysis. e.g. katznelson who has at least a lot of the forward stuff in his dover-published book.

Semiclassical

@TedShifrin ehh. rotation is just a change of variables itself, so i'm not sure if this is much different

leslie townes

i think ted meant a 1-d change of variable, i.e. at most interchanging x and a function of y.

Koro

$x^2\equiv -1(p)$ has a root iff $p\equiv 1 (4)$. Is this statement true? I don't think so.

leslie townes

if you have multivariable change of variable, game over.

koro can you time travel back to 1801 and tell us?

Ted Shifrin

No, no, totally a different thing, @Semiclassic. You go $u$ units down the line $y=x$ and solve for the $x$-coordinate of the intersection of the perpendicular line with the curve $y=x^2$.

Koro

18:57

($\Rightarrow$) We have two cases: 1) p|(x-1), 2) p|(x+1). In case 1), we get: $x\equiv 1(p)$, i.e., $-1\equiv 1(p)\implies p|2$

Case 2 also gives the same result that $p|2$.

i.e., p=2

Ted Shifrin

Koro: This is a famous result in every algebra book, even mine. One direction is a bit more challenging.

How is $x^2+1$ factoring, exactly? Geez. Pay attention.

Semiclassical

take $p=5$. then $2^2+1=5$

leslie townes

koro it is OK not to approach every result in mathematics like a blank slate

Koro

@TedShifrin Ohhh :(

It's about 12:30 am here.

I don't know how I thought that to be $x^2-1$.

Semiclassical

late night math

Koro

19:00

I don't know why I tend to do that. In my today's exam, I unconsciously wrote $1/(z-1)^2= \frac 1{z^2 (1-1/z^2)}$

Semiclassical

i don't think it's particularly easy to prove that theorem by elementary means

Koro

During my review of my answer script before submission, I noted that and fixed it.

😅

Thorgott

beware the statement is only true for odd primes $p$

Rithaniel

@leslietownes That's a thing I had to learn (but I also think it's a good rule of thumb to re-prove any lemmas/theorems you're using which you don't fully understand)

Koro

@Thorgott no?

x=1 is a solution if p=2.

Thorgott

19:02

last I checked 2 wasn't congruent to 1 modulo 4, though

Semiclassical

$x=1$ doesn't imply $x^2=-1$

Ted Shifrin

One direction is totally elementary. Wilson's Theorem.

The other needs some ring theory.

Semiclassical

what's the simplest non-elementary way to do it, hmm

something something gaussian integers?

Ted Shifrin

Gaussian integers and $\Bbb Z[i]/(p) \cong \Bbb Z_p[x]/(x^2+1)$

Koro

$1^2=1\equiv 1-2 (2)=-1(2)$ @Semiclassical

no?

Semiclassical

19:03

ooo, nice

@Koro sure but how is that relevant

Koro

(,) means $\pmod ,$

Ted Shifrin

I love this proof so much that I even remember the outline :D

Koro

@Semiclassical x=1 is a solution when p=2

*solution of $x^2=-1 (p)$

is it not?

Semiclassical

oh. that's not what thorgott was getting at

Koro

Ohh

he meant the result that I wrote p== 1(4)

Semiclassical

19:05

right

Ted Shifrin

There should be an a priori assumption that $p$ is an odd prime. Geez.

Semiclassical

i guess p=9 is the simplest test

Ted Shifrin

Since $9$ is a well-known prime

Rithaniel

Quick memory check. For a point on a surface, the gradient gives you the direction of steepest descent, right?

Ted Shifrin

What are you talking about, Rithaniel?

Sloppy, sloppy statement/question.

Gradient of what?

shintuku

19:06

it's a hairstyle in vogue i believe

PNDas

Are there infinitely many primes of the form 111....1111? That is $\frac19(10^n-1)$?

Thorgott

the algebraic geometer inside of me (<- a complete fabrication of my deluded mind) says to simply draw Spec(Z[x]) and conclude visually

Ted Shifrin

Conclude what visually?

Koro

Let me look up the solution manual now.

It's good that I have that.

Thorgott

the claim in question, but feel free to conclude anything else you want, too!

Rithaniel

19:08

Sorry. It's been a hot minute since I did multivariable calculus, and I'm trying to remember terminology for the vector which, for a given point in the plane, tells you the direction of steepest descent. I'm just thinking in terms of a real valued function in two real variables (so, a "surface")

Ted Shifrin

@PNDas How many have you tried?

PNDas

Found it en.wikipedia.org/wiki/Repunit

Ted Shifrin

No, you have to specify whether you're graphing a function or looking at a level surface.

If you graph $z=f(x,y)$, then $\nabla f$ gives the direction in the plane along which the surface goes up most quickly, yes. The negative will give you descent.

PNDas

quora.com/Are-there-infinite-no-Of-primes-of-form-1-9-10-n-1

Rithaniel

@TedShifrin Okay, this feels like what I'm remembering

Ted Shifrin

19:10

It follows from understanding directional derivatives.

Semiclassical

@PNDas that's always a multiple of 11 when n is even, so you'd need to restrict to odd n

Ted Shifrin

Divisible by $3$ whenever $n$ is a multiple of $3$.

Semiclassical

ah yeah

Rithaniel

Directional derivatives are probably the thing I should review, in that case

PNDas

@Semiclassical you mean even greater than 2

Obviously

Semiclassical

19:13

yeah

Ted Shifrin

Yes. Or you can watch a few minutes of the appropriate one of my videos :D

Geometry of the gradient is one of my favorite topics in multivariable calculus.

Semiclassical

from mathematica, the first such n after n=2 seems to be n=19

PNDas

If you see the quora link then the answerer mentioned first five such n's

Rithaniel

@TedShifrin Send me a playlist and I'll put them on in the background while I make dinner

PNDas

next n's are 23,317,1031

Actually the main problem is to prove or disprove infinitude of product-smith numbers pndasmathblog.wordpress.com/2023/03/01/product-smith-numbers

Ted Shifrin

19:19

@Rithaniel The link is in my profile. Look at Day 26-27.

Semiclassical

"are there finitely many n such that P(n)" is a format in elementary number theory that can either be easy or entirely intractable

PNDas

This is a weird strategy to promote my blogs.

Semiclassical

@PNDas one reason to not be surprised by the lack of progress is to note that the base-2 version is Mersenne primes

the fact that we don't fully understand the binary version is a pretty good reason to expect that the decimal version is entirely out of reach

PNDas

@Semiclassical I think someone (probably Erdos) said something like : an Idiot can ask a question in number theory which is very hard.

I don't remember

Semiclassical

sounds right

the ratio of question-difficulty to question-complexity can be absurdly high in number theory

PNDas

19:28

I have no relation with number theory. These questions are just out of curiosity.

I wanted to go in number theory but I am bad at abstract algebra.

Ted Shifrin

There are very analytic parts of number theory, but one does need at least a solid first-year graduate knowledge of algebra, regardless.

copper.hat

another unexplained downvote.

Ted Shifrin

Let me go find a copper post to downvote :D

It's probably leslie making trouble, as usual.

Thorgott

Here's a fun proof that took me way too long to work out again:
There's a homomorphism $(\mathbb{Z}/p\mathbb{Z})^{\times}\rightarrow\mathbb{Z}^{\times}$ given by sending $x\mapsto x^{(p-1)/2}$. The domain is a cyclic group of even order, from which it follows that the morphism is non-trivial, hence has kernel of index $2$, and that the set of squares is a subgroup of index $2$. The squares, however, are obviously contained in the kernel, so we have equality. Thus, $-1$ is a square mod $p$ iff $(-1)^{(p-1)/2}=1$ iff $p\equiv1\mod 4$.

this relies on knowing the cyclicity of $(\mathbb{Z}/p\mathbb{Z})^{\times}$, but I'd argue that's still elementary

Ted Shifrin

I'm not sure it's all that much more elementary than the Gaussian integer approach, but it appears so to the outside observer.

I just like the Gaussian integers too much and writing $a^2+b^2 = (a+ib)(a-ib)$ seems a lot less abstract than the group theory result.

Semiclassical

19:43

Gaussian integers feel very concrete to me

Thorgott

yeah, I wouldn't say more/less elementary, it's just two genuinely different proofs, which I think is nice

Semiclassical

i feel like playing with elementary number theory is a decent way to prep one for learning abstract algebra

Ted Shifrin

I think there are other reasons to argue for the Gaussian integer approach. Suppose we want to understand factorization of non-primes.

冥王 Hades

Imagine if voting required passing a calculus 3 test

Ted Shifrin

@Semiclassic Indeed, even though I was never a fan of number theory, for pedagogical reasons I wrote my algebra text incorporating a lot of that from the outset.

@Hades I'd be satisfied if you made that calc 1. It would also be a prerequisite for members of congress, many of whom could never fulfill it.

冥王 Hades

19:47

@TedShifrin I'd be surprised if congress members could even pass basic trig

Ted Shifrin

@Semiclassic I tried to teach my algebra students how to use intelligent exploration of examples to develop intuition and arrive at a proof. But it's not easy to convince a student to do something other than type a bunch of numbers into a calculator :D

@Hades Well, one of the dumbest even went to law school. I don't know what math he had to take for his college degree.

冥王 Hades

@TedShifrin I wouldn't trust them as my lawyers lol

Ted Shifrin

Hell no.

I would happily take the big names from the impeachment proceedings as my lawyers, though. Schiff and Raskin are very impressive.

This question gets an award for worst title and heaviest disguise on a basic linear algebra question!

冥王 Hades

@TedShifrin That reminds me, you should've seen some of the congress hearings when they brought in Mark Suckerberg and even Sundar Pichai. The congress members asked them some of the stupidest questions I've ever heard. I don't like either of them, but I give them a lot of credit for not bursting out in laughter at those questions.

As far as I saw the congress members are tragically ignorant on tech. That shouldn't be the case

Ted Shifrin

Zuckerberg has turned out to be quite the narcissist, too. Pichai I do not know.

冥王 Hades

19:55

@TedShifrin that's why I call him Suckerberg. Pichai is alright I guess. Point is, as mediocre as those two may be, I was feeling bad for them. That's how stupid the congress is

Senator: "How do you sustain a platform where users don't pay for the service?". Suckerberg: "Senator we run ads." Senator: "How many miles do they run?"

Ted Shifrin

There are some very dumb senators. Was this hearing actually in the Senate or in the House? I didn't watch it.

Transcript for

$Mathematics$ Mathematics