Mathematics - 2022-11-08

Akiva Weinberger

2:24 AM

@geocalc33 So like $\displaystyle\prod_{p\text{ prime}}\frac1{1-p^{-s}}=\sum_{n>0}\frac1{n^s}$ but backwards?

Akiva Weinberger

3:00 AM

I'm sure you can construct artificial examples but I doubt there's anything natural

one potato two potato

3:24 AM

How can I compute the degree of this map $f:S^k\times S^{n-k}\to T^n$ where $n\geq 4$ and $2\leq k\leq n-2$. Here $T^n$ is an $n$-torus

one potato two potato

3:36 AM

Ignore the above question.

Akiva Weinberger

ok

Ajay

@AkivaWeinberger Any interesting math today?

Akiva Weinberger

Showed some people Post's correspondence problem

Most people were apathetic, one was excited

and one person (probably a crank) is trying to debate me about it on Twitter

Say I have the following pairs of black&white sequences:

Ajay

People just won't stop until they prove you wrong.

Akiva Weinberger

Can I arrange these pairs (with repetition allowed) so that the sequence I get from reading across the top is the same as the sequence I get from reading across the bottom?

@Ajay spoiler in ten seconds

So here if you read across either the top or the bottom you get XOXXXXXXO (X=black, O=white).

This is the second pair, then the first pair twice, then the third pair.

@Ajay Now, not all of these sets of pairs have solutions. Some do, some don't for trivial reasons, some don't for less-than-trivial reasons.

The following set of pairs is open (no one has been able to figure out if it has a solution or not):

Ajay

3:52 AM

Top: OX X XXO. Bottom: X XXO O

Akiva Weinberger

Right. The question is, can you find some arrangement of these pairs, allowing repetition, that makes those the same

See the first example

Ajay

Got it

i'm working on it

Akiva Weinberger

For what it's worth, the similar-looking problem $\begin{matrix}O&X&XXO\\X&XXO&O\end{matrix}$ is known to be solvable, but the shortest solution is 75 pairs long

(this differs from the one in the picture in the first pair)

Prithu Biswas

4:08 AM

@AkivaWeinberger So there is no algorithm that , given any sequence of pairs , can correctly decides whether or not a solution exists or not?

Akiva Weinberger

@PrithuBiswas Correct

This problem (Post's correspondence problem, or PCP) is undecidable

It is one of the earliest problems shown to be undecidable, in 1946.

The proof goes by showing that it's equivalent to the halting problem

by simulating Turing machines with these.

Prithu Biswas

@AkivaWeinberger That seems quite non-trivial to me =P

Do you have a reference to a proof I can read?

Akiva Weinberger

Yeah it's super surprising (in my opinion)

I have a print version on a shelf near me...

...but if you Google Post's correspondence problem you should probably be able to find stuff

2

For what it's worth, if PCP[n] means the version of the problem where there are only n pairs, then PCP[2] is known to be decidable, PCP[7] is known to be undecidable, and PCPs[3 through 6] are still open

one potato two potato

4:32 AM

This is a special instance of the fact that the graded commutator of graded derivations is a graded derivation. In particular, the space of graded derivations is a graded Lie algebra with graded commutator as Lie bracket. — Michael Albanese 58 mins ago

@TedShifrin Someone explained the meaning of the formula.

Akiva Weinberger

OK, dumb question

0

Q: If $f$ is analytic defined on $D:=\{z:|z|<1\}$ and $|f(z)|\le 1$, can $f$ be extended to $\overline D$?

If $f$ is analytic defined on $D:=\{z:|z|<1\}$ and $|f(z)|\le 1$, can $f$ always be extended analytically to $\overline D$? Ideally, for $|z_0|=1$, we could simply define $f(z_0)=\lim_{z\to z_0}f(z)$. It seems like a reasonable assumption that this extended version of $f$ (a) exists, (b) is conti...

Bounded analytic functions on open disk can be extended to closed disk, yeah?

Ted Shifrin

I know Michael. I’m not surprised.

Akiva Weinberger

Through limit?

No surprises?

Ted Shifrin

Why must limits exist?

Akiva Weinberger

I can't think of a counterexample

but also I can't show the limits must exist

Ted Shifrin

4:35 AM

What if there is a branch point on the boundary?

Akiva Weinberger

...Ooh.

Like $\sqrt{z-1}$

and that's not analytic at $z=1$

Ted Shifrin

Yup.

Akiva Weinberger

...Ah

Well, backup plan: at least it's continuously extendible

Like, the limits exist

Ted Shifrin

I’m not yet sure.

Akiva Weinberger

1

Q: If $f$ is analytic defined on $D:=\{z:|z|<1\}$ and $|f(z)|\le 1$, can $f$ be extended to $\overline D$?

If $f$ is analytic defined on $D:=\{z:|z|<1\}$ and $|f(z)|\le 1$, can $f$ always be extended continuously to $\overline D$? In other words, can all analytic functions $f:D\to\overline D$ be extended to continuous functions $\tilde f:\overline D\to\overline D$ (such that $\tilde f|_D=f$)? Ideally,...

complex-analysis analytic-functions analyticity

Updated question

Ted Shifrin

4:41 AM

I’m rusty. I’m remembering issues with limits only in certain wedges.

Akiva Weinberger

4:52 AM

Wait is $e^{i\frac{1+x}{1-x}}$ a counterexample

Ted Shifrin

How is that analytic in the disk ?

Akiva Weinberger

Why wouldn't it be? It's defined everywhere but $1$ which isn't in the disk

I guess the $i$ was a mistake

Ted Shifrin

What is $x$

Akiva Weinberger

$z$

1

A: If $f$ is analytic defined on $D:=\{z:|z|<1\}$ and $|f(z)|\le 1$, can $f$ be extended continuously to $\overline D$?

The answer is no. Consider $f(z) = e^{(z+1)/(z-1)}$. The map $z\mapsto \frac{z+1}{z-1}$ takes the unit circle to the imaginary axis and the interior of the unit circle to the left half-plane; therefore $f(z)$ is indeed a map from $D$ to itself. However, $|f(e^{i\theta})|=1$ and thus $f(z)$ tends ...

Ted Shifrin

Why is that bounded?

Akiva Weinberger

4:58 AM

Because $|e^{\rm blah}|$ is $e^{\rm Re(blah)}$ and $(z+1)/(z-1)$ has real part under $0$

Ted Shifrin

Ah, cool. This is probably a good example of that limiting wedge whose name I’m forgetting.

copper.hat

i saw a wandering question there...

Akiva Weinberger

I don't understand why $f(e^{i\theta})$ tends to $1$ as $\theta\to0$. — Akiva Weinberger 4 mins ago

Ted Shifrin

Who says it does?

The point is going to $\infty$ on the imaginary axis. No limit.

leslie townes

5:15 AM

wedge antilles

Cotton Headed Ninnymuggins

To check if a matrix is positive semi-definite, can I simply take the determinant of all upper-left sub-matrices and confirm that they're all greater than or equal to 0?

Ted Shifrin

Ah, Abel was the name I was blocking on. If $\sum a_n$ converges and $f(z)=\sum a_nz^n$ has radius of convergence $1$, then $f(z)\to f(1)$ provided $|1-z|/(1-|z|)$ remains bounded. There's the wedge.

Akiva Weinberger

@TedShifrin You are correct. The answerer messed up.

Ted Shifrin

@CottonHeadedNinnymuggins Yes.

Cotton Headed Ninnymuggins

@TedShifrin perfect, thanks

And I trust you know what you're talking about given you've authored a linear algebra textbook ;)

copper.hat

5:24 AM

Well, I'm not sure about that, I heard that Ted puts ice in his wine.

Ted Shifrin

@copper.hat NEVER

copper.hat

:-) even when the a/c is broken?

Ted Shifrin

@CottonHeadedNinnymuggins But this result is not in my book ;D

copper.hat

i must admit that during a wife beater wearing moment on the front porch once, i did put a cube in my pinot.

Ted Shifrin

@copper.hat what a/c?

copper.hat

5:26 AM

of course, not long after taking a sip, i relocated to another step, dropped my relatively new phone and cracked the screen.

that is wine karma.

Ted Shifrin

More like whine karma.

copper.hat

:-) i'm good at that

my 6'3" son has taken to wearing tank tops when lounging around home, my wife does not approve.

i only recently heard the described as wife beaters. clearly i need to get out more.

Ted Shifrin

Why? It’s not like he’s nekkid.

copper.hat

i think it is just the name association.

Ted Shifrin

Yeah, the colloquial name is repugnant.

copper.hat

5:32 AM

when we grew up, threats like 'i'll kill you" or "i'll skin you to within an inch of your life" were pretty much dinner table stuff, so wife beater just fits right in.

that said, if anyone raised a finger towards any female member of the family, there would be some rapid readjustment.

surprising, given that my sister is the toughest of the siblings.

Cotton Headed Ninnymuggins

@TedShifrin I'm recalling using partial derivatives to find minima of multi-variable functions, it seems easier that way. Is there an intuitive way of understanding why if K is positive definite that it would guarantee a minimum for $p(x,y,z)=x^TKx-2x^Tf+c$?

I don't know how to MathJax in here

copper.hat

if $K>0$ then there is some constant $k$ such that $x^T K x \ge k \|x\|^2$, and from that you can see that $p$ must have a $\min$.

Cotton Headed Ninnymuggins

Is mathjax supposed to work in here? I see a mess

Ted Shifrin

@Cotton Using partial derivatives only locates critical points. It doesn't classify them.

copper.hat

you need to 'install' it in your browser. see the tinyurl links top right

Akiva Weinberger

5:40 AM

The plot is fun

copper.hat

how did you make the plot?

Akiva Weinberger

As $z$ spins around the unit circle, the real and imaginary parts of $e^{(z+1)/(z-1)}$ oscillate infinitely fast between $-1$ and $1$

Ted Shifrin

In the case of the function you just gave, it's a quadratic. The easiest way to do it is to change variables (using the spectral theorem to diagonalize $K$) and then complete the square.

Akiva Weinberger

@copper.hat Plugged the bit labeled 'input' into wolframalpha.com

Ted Shifrin

@AkivaWeinberger You're just looking at $e^{iy}$ for $y\in\Bbb R$, DogAteMy, effectively.

copper.hat

5:41 AM

ahh. i have a vague distaste for wa. can't really explain it.

Akiva Weinberger

Sure, but in this case $y$ is $-\cot(t/2)$

Ted Shifrin

I never use WA, since I paid money to own Mathematica.

Akiva Weinberger

I generally prefer Desmos for plots but it can't handle complex numbers

copper.hat

the complex world produces some nice figures

Ted Shifrin

I was just using the analysis that the LFT $(z+1)/(z-1)$ takes the unit circle to the imaginary axis.

copper.hat

5:43 AM

i used to spend my spare time finding bugs in desmos during a more productive moment of my life

Akiva Weinberger

And indeed $(e^{it}+1)/(e^{it}-1)=-i\cot(t/2)$, validating this.

Ted Shifrin

For @Xander when he arises in the morning: My RAID glued-together USB drives died (one of the drives apparently went defective, even though it was new). I bought a 1 TB SSD today. The backup took a few hours.

Sure, DogAteMy. For my qualitative purposes, it wasn't worth the work.

Akiva Weinberger

@copper.hat Here's one I found a while ago

(The denominator equals 0)

copper.hat

cool. i used to send mine to desmos, i was hoping for a bottle of wine or at least a cup of tea.

Akiva Weinberger

I mean, that's really more of an indictment of floating point, which Desmos uses

$|x|=x$ also fails

but I forgive it because being a graphing calculator is hard

copper.hat

5:45 AM

i used to interact with one of the kings of floating point

Akiva Weinberger

and it expects to draw a curve when it sees $=$, not a region

copper.hat

william (vel) kahan

yeah, most software balks at that sort of thing

Akiva Weinberger

I think the smallest positive number it can register is 2^(-1074)

(64-bit signed floating point, and therefore also Desmos)

which is pretty small and normally isn't an issue, but it also means $(e^{-1/x^2})^{x^2}$ looks weird

copper.hat

not quite desmos, but amusing nonetheless math.stackexchange.com/a/152788/27978

denormalised numbers are interesting.

there is a big fuss about a format called posits. a lot of noise about nothing imo

up there with $\tau$ vs. $\pi$.

Akiva Weinberger

Every time I write $e^{2\pi i\rm stuff}$ I become more of a tauist

especially if I find myself writing eg $e^{2\pi i/10}$ and deliberately avoid simplifying for clarity

copper.hat

5:49 AM

$2 \pi$ or not $2 \pi$

Akiva Weinberger

like if I want to refer to $\zeta_{10}$, the primitive tenth root of unity

copper.hat

$\pi$ is ingrained, just adapt :-)

Akiva Weinberger

I adapt by writing "Let $\tau:=2\pi$" before using it

If I wanted to I could write "Let $c:=\pi^4/90$" or whatever crazy thing and it would be valid

so why not

copper.hat

people like things the way they are, you would get labeled as a tauist

besides, there are no good jokes about $\tau$.

pronounced 'taff' by Greeks

Akiva Weinberger

Indeed

And Euclid is Efklídhis

(dh = voiced th sound as in 'this')

(and also the Modern Greek s is halfway between an English s and an English sh)

(probably the Ancient Greek s was, too, but we can't know that)

I think in Modern Greek Zeus is Zefs?

copper.hat

5:54 AM

Sure we do, I have an old Greek friend, I will ask her

Akiva Weinberger

Yup, just checked

leslie townes

copper will now instruct us as to the difference between a garbanzo bean and a chickpea

copper.hat

thats a load of garbanzo

Akiva Weinberger

The fun thing is that of the three pronunciation systems for the names of the letters of the Greek alphabet (Ancient Greek, Modern Greek, English), all three are different from each other

copper.hat

god, you are worse than me

Akiva Weinberger

5:55 AM

Both languages have changed massively over time

To get from the Modern Greek names to the traditional English names, you have to subtract centuries of Greek evolution and add back centuries of English evolution

copper.hat

i find my own personal language changes over time

Akiva Weinberger

(but don't subtract too many Greek phonological changes! Only go, like, halfway back)

Ancient Greek had no f sound

They had /pʰ/, an aspirated p sound

In Modern Greek, as well as the relevant loanwords into English, this became /f/

copper.hat

i used to know a smattering, unfortunately the words most likely to be remembered describe actions whose description is generally omitted from polite conversation

Akiva Weinberger

hence why we spell it 'ph' in Greek-origin loanwords!

copper.hat

reminds me...

leslie townes

5:58 AM

akiva, you're an aspirated p sound

Akiva Weinberger

pft

copper.hat

there is a restaurant near here called Pho Le, but I cant help but see P Hole

Akiva Weinberger

There's only a handful of words that have both an f and a ph

Francophile spherification filmographies

copper.hat

a bit like another place, a pizza place, called iSlice, but originally their sign was is lice, thankfully i was there to point it out.

i found a website once that had accidentally put something like "this webshite is powered by..."

i pointed it out, they changed it next day, but never replied to me. i felt deprived of a good laugh

juvenile was the term my mother used for me

Cotton Headed Ninnymuggins

6:16 AM

Too funny

Ajay

6:42 AM

Ted, can you lmk if i'm correct?

24 hours ago, 5 minutes total – 3 messages, 1 user, 0 stars

Bookmarked 1 min ago by Ajay

famesyasd

8:37 AM

If I intergrate some function F over a rectangular [a,b] times [c,d] will it give the same value as the integral over (a,b) times (c,d)?

robjohn

As long as it is a function and not a distribution.

some people call distributions "generalized functions"

@AkivaWeinberger photofinish

often spelled as two words though

robjohn

9:12 AM

flophouse

photoflash is usually one word

Jakobian

sandwich is basically a mini-pizza

@famesyasd Lebesgue integrals of functions agreeing over a set of measure zero are the same

here $[a, b]\times [c, d]\setminus (a, b)\times (c, d)$ has measure zero, so the integrals agree

why are the proofs in topology texts always so unintelligible ... like I'm reading this and I'm astonished. I can decipher what the author means but I can never get the thought process behind coming up with something like that, just on the surface level

maybe it's just because people stuff all the tricky stuff in the lemmas before the main statements

Xander Henderson

11:37 AM

@TedShifrin Well, at least you have backups again now?

one potato two potato

11:49 AM

2

A: Give me a hint on Stein complex analysis Chapter 4 Exercise 1

When doing this problem, I really wanted the condition that $\hat{f}(\xi) = 0\; \forall\; x \in \mathbb{R}$. Indeed, by pulling out the textbook, this is the condition given for the problem. First, we get agreement of $A$ and $B$ on the real axis since $\hat{f}(\xi) = 0\; \forall \xi \in \mathb...

In the answer, why $f$ being moderate decrease implies the extension $F$ is bounded?

$$\left|\int_{-\infty}^t f(x)e^{-2\pi iz(x-t)}\ dx\right|\leq\int_{-\infty}^t{A\over 1+x^2}e^{2\pi y}(x-t)\ dx = e^{-2\pi ty}\int_{-\infty}^t{A\over 1+x^2}e^{2\pi yx}\ dx$$

I don't think it follows from MMP. We only know $F$ is bounded on $\Bbb R$.

Oh there's a typo in the middle integral. it should be $e^{2\pi y(x-t)}$.

Here, $y = \operatorname{Im}(z)$.

Jack Ohara

12:45 PM

@robjohn Hello ! can I send you a latex code of my homework so you can check it if it is good?

i have solved couple of problems and it would be very helpful to get some tips and tricks to make latex code look good

very new to this latex stuff

user193319

1:31 PM

Let $\Bbb{F}_p$ be the field with $p$ elements, and let $V_d \subset \Bbb{F}_p[x]$ be the space of polynomials of degree at most $d$ with coefficients in $\Bbb{F}_p$. Consider the evaluation map $\phi_d : V_d \to (\Bbb{F}_p)^p$ sending $f$ to $(f(0),f(1),...,f(p-1))$. I am asked to do two things. First, show that $\phi_d$ is surjective for all $d \ge p-1$, and then for $d=p$ I want to show that $\phi_d$ has a one-dimensional kernel by comparing dimensions.

Goku

Do you ever see a badly written question and attempt to edit and correct, but then halfway you realize "nah, too much work, I'm out"?

No? Just me?

user193319

If I can prove surjectivity, I believe the second part is a trivial application of the rank-nullity theorem. However, I am having trouble showing surjectivity. I was thinking about using Lagrange's interpolation theorem, but that only applies to a tuple of distinct elements.

one potato two potato

I found something strange: Suppose $Q(z)$ is a complex polynomial with distinct $n$-roots on $\Bbb C\setminus\Bbb R$ where $n\geq 2$. Let $\{z_1,...,z_n\}$ be a set of roots of $f$ and assume $\{z_1,...,z_m\}\in\Bbb H^+$ and $\{z_{m+1},...,z_{n}\}\in\Bbb H^-$ then by the Residue theorem,
$$\int_{-\infty}^\infty {1\over Q(x)}\ dx = 2\pi i\sum_{j=1}^m{1\over Q_j(z_j)}$$
where $Q_j = {1\over (z-z_j)}Q(z)$ if we take the upper half circle as a contour and
$$\int_{-\infty}^\infty{1\over Q(x)}\ dx = 2\pi i\sum_{j=m+1}^n{1\over Q_j(z_j)}$$

robjohn

@JackOhara you can post here or make a chat room and test there. Have you installed ChatJax?

Not Tfue

1:50 PM

Why does block multiplication works?

I mean partitioned matrix.

user4539917

Why does anything work in math?

Not Tfue

I don't even have capacity to even solve basic arithmetic problem how will I answer this god like question.

I probably got IQ of smartest dog alive even it will surpass me by solving basic calculus.

user4539917

Calculus was built on the shoulders of Giants.

Not Tfue

Those are geniuses I am a dumb kid who is incapable of even passing basic academic exams.

Calculus was built by lucky people with right gene.

user4539917

2:06 PM

Hard work can take you a long way...

Not Tfue

Dumb people with hard work will never success only smart people can...

if you take iq test of field medalist then all the field medalist will get over 130+ iq points

user4539917

"Smartness" is something built-up over time and a lot of effort.

Without the effort you have nothing.

Not Tfue

this is like initial value problem some people can hardwork and success some will never because they are gifted and people who are not gifted will die doing nothing extraordinary :(

user4539917

Life is a gift.

user193319

Never mind. I solved it using Lagrange interpolation.

Not Tfue

2:15 PM

life is gift for some and suffering for some :(

Peter

@NotTfue I belong to the "suffering"-part.

user4539917

To the Buddhist life is suffering.

Goku

Finally reached 1000

Not Tfue

@Peter welcome to the club

user4539917

Misery loves company.

Xander Henderson

2:20 PM

@user4539917 smbc-comics.com/comic/how-math-works

@NotTfue No. Just... no.

Not Tfue

@XanderHenderson lol

Xander Henderson

There is no calculus gene. Pretty much anyone can learn calculus. But it is hard, and it can take a lot of time and dedicated practice. I mean, I failed calculus the first time I took it.

user4539917

Nicholas Saunderson

Nicholas Saunderson (20 January 1682 – 19 April 1739) was a blind English scientist and mathematician. According to one historian of statistics, he may have been the earliest discoverer of Bayes' theorem. He worked as Lucasian Professor of Mathematics at Cambridge University, a post also held by Isaac Newton, Charles Babbage and Stephen Hawking. == Biography == Saunderson was born at Thurlstone, Yorkshire, in January 1682. His parents were John and Ann Sanderson (or Saunderson), and his father made a living as an excise man. When he was about a year old, he lost his sight through smallpox; but...

Goku

I couldn't even count the number of times I've lost badly to a harder working student even though my IQ score apparently states that I could be "gifted"

user4539917

18 mins ago, by user4539917

Without the effort you have nothing.

Peter

2:29 PM

@NotTfue I wonder who REALLY enjoys life in particular in the terrible time we now have. For "normal" people it is anyway just pain - working hard and having to pay for every "shit". Even rich and prominent people currently have few reasons to be happy.

I disagree however that all fields medalists have IQ 130+ , IQ is not mathematical ability alone and anyway the tests to determine it are more than debatable.

Another common fallacy is that strong chess players must be extraordinary intelligent.

Prithu Biswas

@AkivaWeinberger Hello!

I just watched some videos about the proof of the "Undecidability of posts correspondence problem".

I haven't worked through all the details, but I think I understood it enough to appreciate it =)

Lecture 47/65: Undecidability of the PCP

►

This is the first one from which I understood how the tile matching condition can correspond to valid Computational history. For that we have to choose the correct tiles (or dominos).

But there seems to be a problem. This video proved the undecidability of a simpler version of PCP, where a fixed tile has to appear 1st.

Goku

2:49 PM

@Peter if I recall correctly, Richard Feynman was actually below 130

And sure you may not be the next Ramanujan or whatever but that doesn't mean you can't be great at mathematics

Peter

I am surely greater in mathematics than in chess !

Prithu Biswas

Post Correspondence Problem (PCP) is Undecidable Proof

►

This one modifies the simplified PCP to get to the general PCP. [With timestamp].

geocalc33

3:06 PM

@AkivaWeinberger yes - why don't you think there's anything natural?

Akiva Weinberger

@geocalc33 'cause primes are related to factoring and multiplying

geocalc33

@AkivaWeinberger hmm that makes sense i guess

i was just thinking about additive number theory

SarcasticMathematician

4:06 PM

Surely most mathematician do drugs by looking at what they came up with.

PM 2Ring

4:21 PM

@AkivaWeinberger Here's a Sage version of that complex exponential graph: sagecell.sagemath.org/…

Sage can also do graphs using arbitrary precision arithmetic. It's not necessary for that function, but here's an example anyway.

sagecell.sagemath.org/…

PM 2Ring

4:38 PM

Here's some number theory trivia. 7 is the smallest natural number $k$ such that both $k^2-1$ and $k^2+1$ are divisible by squares. On a related note, $35\sqrt2 - 28\sqrt3$ is very close to $1$.

Not Tfue

is there an article where they show you why block multiplication method is true for all matrices or do you try to find pattern yourself?

Or book.

Akiva Weinberger

I'm sure there are

By bilinearity you only really need to check that this is true for basis matrices, that is, matrices that are 0 everywhere except for a single 1

Not Tfue

4:57 PM

I wish I knew what those words mean.

May be tomorrow I will try to analyze the pattern myself.

I wish time given to learn math was longer so that it will be fun to find something new instead of just preparing for exam and memorizing stuff.

Ted Shifrin

@NotTfue Just draw the diagram with the blocks and use your left finger across rows and right finger down columns.

This is the way to understand visually how matrix multiplication works.

Thorgott

the way to understand where the pattern comes from is learning what matrices have to do with linear maps

Ted Shifrin

@PM2Ring Are there infinitely many such numbers?

Akiva Weinberger

5:18 PM

@NotTfue Ignore what I said, I don't know if it would make it easier anyway

@TedShifrin 107, 207, 307, etc

PM 2Ring

@TedShifrin Yes. But OEIS doesn't have much to say about them. oeis.org/A080666

Ted Shifrin

@AkivaWeinberger I guess I don't even see that. 25 works, but why does 16?

Akiva Weinberger

$(100m+7)^2=10000m^2+1400m+49$

Ted Shifrin

OEIS does give all the $m\times 100 + 7$, though.

Gee, DogAteMy. I never would have known that.

Akiva Weinberger

$4|10000,1400,48$

Ted Shifrin

5:25 PM

Oh, I was using 16, not 4. Dope.

That was all you needed to say.

This is why I would often say to my students, "You're making this too hard."

Akiva Weinberger

You confused me for a minute. I was checking 1400/16 and worrying that I made a mistake.

Ted Shifrin

Well, that was why I was stuck!

One of many reasons I don't dabble in elementary number theory.

PM 2Ring

Number theory can be beautiful, but it's very infuriating. It's so easy to state simple problems that nobody knows how to attack.

Ted Shifrin

I've just never been interested in the sorts of things that most high school math geeks (and plenty older) love with numbers, prime distributions, etc.

Akiva Weinberger

Fun fact: for any recursively enumerable set (this includes the set of primes, set of powers of 2, set of counterexamples to Goldbach) there exists a multivariate polynomial $p(n_1,n_2,\dots,n_k)$ with integer coefficients such that the positive part of its image,$$\{p(n_1,n_2,\dots,n_k):n_1,n_2,\dots,n_k\in\Bbb Z\}\cap\Bbb N_{>0},$$is that set.

PM 2Ring

5:36 PM

I mostly play with semi-practical stuff, like continued fractions. I like finding simple rational approximations for irrational quantities.

leslie townes

"Fun fact" [[citation needed]]

Akiva Weinberger

Leslie you're in a math chat

Ted Shifrin

Oh dear. Leslie and Ted agree again.

leslie townes

yes, in this we are unanimous.

PM 2Ring

@AkivaWeinberger Eg, en.wikipedia.org/wiki/… has the prime number poly.

I was initially amazed when I first learned about that poly. But then I realised it's totally useless for actually generating primes. :)

Q: In the Bernoulli trials, how many Bernoullis were found guilty? A: All of them.

leslie townes

6:07 PM

PM i haven't clicked but is deciding positivity the obstacle?

that's my internal vibe, wondering if it's true

PM 2Ring

@leslietownes Kind of. You have to randomly plug sets of numbers in to the poly (in 26 variables) and see if the result is positive.

You don't really have any control over which primes it spits out.

Goku

6:38 PM

Why can't I get myself to actually make edits to correct otherwise interesting questions?

Every time, half way between an edit, I just go "nah this is too much work, I'm out" and just leave it at that resulting in the question getting closed.

Ted Shifrin

7:22 PM

I find that a lot of the edits that people are doing on old posts are not worth much.

PseudoLooped

7:35 PM

transcribing images in old posts is good, because eventually imgur will dump old photo links, and when that happens those old posts are going to become invalidated, and also it makes it easier for people to read who use screen readers (like one of my favorite prof who likes to share posts from this site to his students)

(and the passive learning and hidden gems you’ll find is worth it tbh)

also, bringing in image links into the body is good, because it negates the premise of a risky link click to see an image (clicking links in general when you don’t know where it goes is very risky tbh), which is something you’ll see in old posts, so avoiding that all together is very good

Ted Shifrin

I don't dispute what you said. But a number of 5+ yr old posts all of a sudden come to the top of my feed because of minor grammatical edits. And the posts weren't interesting to start with.

PseudoLooped

That fair, i’m so used to doing what i mentioned (and adding MathJax to very old posts, as Anon was telling me there was a time Mathjax wasn’t on the site that people would just post raw latex and you had to mentally render it yourself) that i haven’t seen those ones

Xander Henderson

7:54 PM

@Peter I mean, we don't even have a very good definition of what "intelligent" means, and IQ tests were never meant to measure individual intelligence, anyway (they were meant as a statistical measure of performance in French schools, so that resources could be better allocated to those schools which were in need).

Starting a sentence with "My IQ is..." is asinine.

copper.hat

edits as a route to jump suit, there's a line of action for me

PseudoLooped

I’m not familiar with the term “jump suit”, like the thing skydivers where?

Thorgott

I think all edits that improve a post, no matter how minor, are good in principle. The issue isn't that people make minor edits, but the design that makes it so that any minor edit bumps posts to the homepage.

it would be a massive benefit to have a way of editing posts without bumping them

PseudoLooped

I strongly wish suggested edits where anonymous, bc it would prevent people from rejecting edits based on individuals and not on the content

copper.hat

i don't like tweaking edits.

leslie townes

8:03 PM

whateve

copper.hat

8:15 PM

replaced by scrunching up one's face as if angostura bitter had been consumed

Akiva Weinberger

@leslietownes Basically the output is positive if one of the variables is $p$, another encodes the sequence of all pairs between $1$ and $p$ exclusive (eg encodes $((2,2),(2,3),(3,2),\dots,(p-1,p-1))$), and if none of those multiply to $p$. In that case the output is $p\cdot(1-blah)$ where $blah$ is $0$ iff none of those multiply to $p$

Well, that's one way to do it. That particular one might use Wilson's theorem

But you could build a polynomial the first way, using some nonobvious machinery to check that condition using the tools available in a Diophantine setting

In essence, you could build a polynomial whose outputs are primes from just shoving the definition of primes into a polynomial-building algorithm

PseudoLooped

Final thing before peace: transcribing images also makes it easier to find past problems on MSE, which makes it easier for problems to be discovered or striking existing questions as dupes

Ted Shifrin

8:34 PM

@Thorgott Concur.

Cotton Headed Ninnymuggins

9:06 PM

I want to find the closest point in a subspace spanned by 3 vectors $\in \mathbb{R}^4$ to a vector $b\in \mathbb{R}^4$. Only 2 of those 3 vectors are linearly independent so my "answer" depends on "z". Would it make sense that there are an infinite number of closest points to b or did I do something wrong?

Ted Shifrin

You did something wrong. For any subspace whatsoever there is a unique closest point.

Cotton Headed Ninnymuggins

That was my suspicion

Akiva Weinberger

9:29 PM

desmos.com/calculator/mdeuiyxmnh

Cross section of $e^{(z+1)/(z-1)}$ over the unit circle

The x-axis is the real part of the input; the slider is the imaginary part. Blue and purple are the real and imaginary parts of the output

Akiva Weinberger

11:22 PM