Mathematics - 2023-09-01

nuwe

00:00

One thing I know: field should not be null or empty. This would be an unexpected mistake made by the user. Anyways
1) user could have forgotten a whole column
2) bunch of nulls could've been inserted in "chunks" e.g., 20% of nulls are at the first 20% rows, then, 40% at the end, etc.
3) I could use random sampling to estimate this efficently. Cases from 2) could be false positives if read in order.

Shaun

Hi :) I thought I'd advertise a bounty:

5

Q: Show, using presentations, that $\Bbb Z_m\times\Bbb Z_n\cong\Bbb Z_{{\rm lcm}(m,n)}\times \Bbb Z_{\gcd(m,n)} .$

Note: This is an alternative-proof question and so is not a duplicate. The Question: Show, using group presentations, that $$\Bbb Z_m\times\Bbb Z_n\cong\Bbb Z_{{\rm lcm}(m,n)}\times \Bbb Z_{\gcd(m,n)}.$$ Motivation: I was trying to answer this question in particular . . . $\Bbb Z_m \times \Bbb ...

That is all.

GratefulDisciple

00:42

@XanderHenderson Good argument. Reminds me of a language where you soon run out of numbers when counting and use "many". This paper references key researches such as Brian Butterworth's 2008 paper Numerical thought with and without words.

Xander Henderson

@GratefulDisciple Huh... interesting that neither paper cites any work by Lakoff or Nuñez, which is where my understanding of the cognition of numbers comes from.

But their work was done... uh... 30 years ago? So maybe it is just that I am very out of touch with modern linguistics / anthropology.

Anywho... time to teach.

John Zimmerman

01:25

@XanderHenderson hope you have a great time teaching :P

Dannyu NDos

I've sometimes wondered, why don't we titlecase "abelian" when we titlecase "Cauchy" and "Hausdorff"?

John Zimmerman

cauchian, hausdorffian

Dannyu NDos

So the -ian suffix is the culprit?

John Zimmerman

Not sure just a guess

Semiclassical

01:47

27

Q: Mathematical concepts named after mathematicians that have become acceptable to spell in lowercase form (e.g. abelian)?

I would like to collect a list of mathematical concepts that have been named after mathematicians, which are now used in lowercase form (such as "abelian"). This question is partly motivated by my former supervisor, who joked (something like): You know you've made it as a mathematician when ...

big-list terminology mathematicians

"You know you've made it as a mathematician when they start using your name in lowercase."

John Zimmerman

nice

is that true for Riemann?

Semiclassical

hmm, not really

it does feel sorta random

there's some speculation in the comments for that one whether it's due to influence from other languages

see similarly math.stackexchange.com/a/1299426/137524

John Zimmerman

@Semiclassical that is interesting.

I like math history

Semiclassical

same

the one which i've wondered about is the $\int f(x)\,dx$ ordering

in math that seems pretty hardened into convention, whereas in physics you still see $\int dx f(x)$ occasionally

John Zimmerman

@Semiclassical have you heard of John Harrison? incredible story

Semiclassical

02:01

nope

John Zimmerman

John Harrison

John Harrison (3 April [O.S. 24 March] 1693 – 24 March 1776) was an English carpenter and clockmaker who invented the marine chronometer, a long-sought-after device for solving the problem of calculating longitude while at sea. Harrison's solution revolutionized navigation and greatly increased the safety of long-distance sea travel. The problem he solved had been considered so important following the Scilly naval disaster of 1707 that the British Parliament was offering financial rewards of up to £20,000 (equivalent to £3.35 million in 2023) under the 1714 Longitude Act, though Harrison was never...

Semiclassical

ahh

John Zimmerman

underdog story

Harrison set out to solve the problem directly, by producing a reliable clock that could keep the time of the reference place. His difficulty was in producing a clock that was not affected by variations in temperature, pressure or humidity, remained accurate over long time intervals, resisted corrosion in salt air, and was able to function on board a constantly-moving ship. Many scientists, including Isaac Newton and Christiaan Huygens, doubted that such a clock could ever be built

Thomas Finley

02:38

0

Q: Need some help in understanding the orientations of fibres.

I was studying about "Tensor Decompositions and it's Applications" from a small handout, i.e Tensor Decompositions and Applications https://www.kolda.net/publication/TensorReview.pdf. This topic is relatively new to me. So, I need to start from basic stuffs. In addition, I am self studying this t...

definition tensors

Anyone interested in this?

Ted Shifrin

02:59

The first figure is mislabeled. It should be like the standard way we draw $x,y,z$ axes, so $i$ should be where $j$ is, etc.

The vocabulary is probably CS vocabulary, definitely not math.

one potato two potato

03:10

@ThomasFinley not to me, but why don't you ask to person who asked you to read?

leslie townes

a useful tip for self-study is, if you encounter something that looks like it might be mislabeled, or a transposition away from what you expect it to be, to delay worrying about it until you see whether (a) it matters at all and (b) if it wasn't just a typo or typo equivalent.

the algorithm "read until something doesn't make sense, then stop every time" is far from optimal for self study.

2

instead of stopping, maybe just write in a note for yourself. or comment on the PDF if you aren't on paper.

7

Q: Closed form for $\sum\limits_{n=1}^\infty\frac{W(n^2)}{n^2}$

Apologies if this has been asked before. I am wondering if the following series has a closed form: $$\alpha=\sum_{n=1}^\infty\frac{W(n^2)}{n^2}\tag{1}$$ where $W(x)$ is the Lambert W function. I am interested in this series as an extension of the series: $$\sum_{n=1}^\infty\frac{\ln{n^2}}{...

sequences-and-series closed-form lambert-w

it's not every day someone responds to a years-old post with something new.

one potato two potato

if something is weird, stop reading and google it, look for another source. If there's no other source then good luck.

Ted Shifrin

@leslietownes I just did that because the idiot bot modified an old post and brought it to the front page.

Boy, typing on the iPad without seeing what I type is a recipe for mess.

Soumik Mukherjee

03:41

@ThomasFinley you may take a look at this

This can give you some intuitions about "higher order analogue of matrix rows and columns"

John Zimmerman

04:44

$$\varphi^2(x)<P_k\le\varphi(x)$$

I'm trying to show this inequality is tight

$$P_k=\bigg(1-\frac{\pi(10^k)}{10^k},1-\big(1-10^{-k}\big)\bigg)$$

where $k \in \Bbb N$

$\pi(\cdot)$ is the prime counting function.

when I manipulate the inequality I believe I'm doing legal moves - but since $P_k$ is of that form I hesitate to be sure.

where $\varphi(x)=e^{\frac{1}{\log x}}$ and $x\in(0,1).$

I do the usual moves to reduce the inequality. I start by taking the logarithm the outer sides and hit $P_k$'s $y-$coordinate withe log as well.

$\frac{2}{\log x}<\bigg( 1-\frac{\pi(10^k}{10^k}, \log(1-(1-10^{-k})\bigg) \le \frac{1}{\log x}$

divide by two

$$\frac{1}{\log x}<\bigg( \frac{1}{2}(1-\frac{\pi(10^k}{10^k}), \frac{1}{2} \log(1-(1-10^{-k})\bigg) \le \frac{1}{2\log x}$$

multiply by $\log x$ and reverse inequality signs

$$1>\bigg( \frac{1}{2}(1-\frac{\pi(10^k}{10^k}), (\log x)\frac{1}{2} \log(1-(1-10^{-k})\bigg) \ge \frac{1}{2}$$

Let $k \to \infty$ and let $x \to 1^{-}.$

I made a mistake: the $\frac{1}{2}$ in the $x$ coordinate should be deleted.

with those limits taken we obtain:

$$ \frac{1}{2} \le \big( 1,0 \big) \le 1$$

which checks out

the idea is to "squeeze" the $P_k$

Thomas Finley

05:11

0

Q: Write the function $\log(1+x), x\in(-1,1]$ by expanding it into an infinite series by application of Taylor's Theorem.

Write the function $\log(1+x), x\in(-1,1]$ by expanding it into an infinite series by application of Taylor's Theorem. I tried solving this as follows: We know that, (Taylor's Theorem in Cauchy's form of remainder) If $f(x)$ is a function such that, $f^{n-1}(x)$ is continuous on $[a,a+h]$ $f^n(...

real-analysis logarithms taylor-expansion

I am gonna need a little help with this

@leslietownes Thanks! This is what I needed.

I loved your suggestions and I found them really useful, @leslietownes!

Bye for now! GTG

Thomas Finley

05:42

1

Q: Show that the vectors $(1, 1, 0), (1, 0, 1),$ and $(0, 1, 1)$ generate $F^3.$

Show that the vectors $(1, 1, 0), (1, 0, 1),$ and $(0, 1, 1)$ generate $F^3.$ I tried solving the problem as follows: Let $a,b,c\in F$ such that $a(1,1,0)+b(1,0,1)+c(0,1,1)=(x,y,z)$ for some $(x,y,z)\in F^3.$ This means, $(a,a,0)+(b,0,b)+(0,c,c)=(x,y,z).$ So, we obtain the following system of e...

solution-verification vector-spaces

Everyone in the comment says, nothing is given about F and the result is false if characteristic of F is 2. But, when I say, should I skip this? Someone replies that I shouldn't

He says my calculation is valid, I am confused more.This is becoz I did all the calculation as if, $F=\Bbb R.$

Everything seems too mysterious.

Can someone explain, what's going on?

Coz I don't seem to have the slightest idea bout it...

Ok, but to solve this, if we assume say, $2\neq 0$ in $F_2$, what should we do to solve this problem. My calculations are clearly wrong, as I assumed, $F=\Bbb R$ .

Isn't it?

John Zimmerman

0

Q: Inequality proof verification

Define a function that counts primes, by the set of points: $$P_k=\bigg(1-\frac{\pi(10^k)}{10^k},1-\big(1-10^{-k}\big)\bigg)$$ for $k\in \Bbb N.$ Here $\pi(10^k)$ is the prime counting function. Then consider the following inequality we'd like to prove: $$ \varphi^2(x)<P_k\le\varphi(x) $$ where $...

inequality solution-verification prime-numbers analytic-number-theory upper-lower-bounds

copper.hat

i don't get this dumping questions here after asking them on the main site.

Thomas Finley

@copper.hat Ugh...you need to understand despair first :)

copper.hat

you think i have not?

Thomas Finley

@copper.hat Nah

:P

copper.hat

05:56

despair is hardly a 4 hour wait.

Thomas Finley

But jokes apart, can you help me with my last comment?

I waited 48 hours. Is it enough? :?)

leslie townes

john: it would improve the legibility of the post if you either (1) explained somewhere in the post what you mean by an ordered pair satisfying an inequality, or (2) rewrote as to involve inequalities involving only things that are clearly numbers and not ordered pairs of numbers. option (2) would maybe allow you to get rid of the unusual symbol from neptune, king of the sea about what is going on with the y coordinate.

123

Hello Everyone...

Thomas Finley

@123 hallo

John Zimmerman

@leslietownes Thanks haha. $P_k$ is a (decreasing) function though so I think it's kosher. Does that make sense?

I just don't have an explicit closed form for $P_k$

Soumik Mukherjee

06:11

@ThomasFinley You can't assume $2 \neq 0$ in $F_2$, do you mean $2 \neq 0$ in $F$?

leslie townes

john: sorry, what does (a,b) <= c mean? in terms of a, b, and c?

and same with d <= (a,b). i don't think it's clear at all. i can guess, but then i'm conscious of guessing.

to be clear, i don't have any ideas for your problem, but i think you might be restricting the audience for your post if you assume that a reader will invest the time to figure that stuff out when it isn't 100% standard.

John Zimmerman

I will include a picture

Thomas Finley

@SoumikMukherjee oh, yeah sorry

John Zimmerman

essentially I'm trying to prove upper and lower bounds on the discrete set $P_k$ which acts sort of like a prime counting function

the same way you can achieve upper and lower bounds on the prime counting function

Lemon

06:38

https://math.stackexchange.com/questions/164472/proving-that-sequentially-compact-spaces-are-compact/4201658#4201658 I don't know if I am reading this answer wrong, but "Firstly there is a 𝛿>0
such that any ball 𝐵(𝑥,𝛿)
with 𝑥∈𝐴
is contained in some 𝑈𝑖". Since the cover is open, and any point x must belong in one of these open sets, can't we just use the openness to show the exists of $\delta$?
.

leslie townes

07:08

lemon: unrolling just the definition of openness of the cover, you get that given x, there is a delta (generally depending on x) with the property that B(x,delta) is contained in one of the elements of the cover. the assertion there appears to be stronger than that - that you can find a single positive number delta that will "work" in this sense no matter what x is.

i.e. you can find a delta that depends only on the set A and the cover {U_i}, and not on the choice of x in A.

i haven't read the rest of the argument but there's definitely more being asserted there than just that {U_i} is an open cover of A. the existence of such a delta will use the sequential compactness of A.

such a number delta is sometimes called a 'lebesgue number' of the covering (implicitly it also depends on A, but usually when discussing these things A is fixed and only perhaps the cover is varying). en.wikipedia.org/wiki/Lebesgue%27s_number_lemma

Koro

08:08

where have they used the fact that F is a relative homeomorphism?

p is the quotient map from X disjoint union D^n to X U_f D^n.

Also, why is p invertible?

[if my confusions could create electricity, there would be free electricity in the whole world.]

😅

Soumik Mukherjee

09:01

Suppose we have $X= [0,1] \times [0,1]$, We can get a Möbius strip or a Klein bottle by quotienting out X by some suitable equivalent relation $\sim$. The least $n$ for which these objects can be embedded in $\Bbb{R}^n$ is $3, 4$ respectively. My question is how far can we get? i.e. what is the highest value of $n$ such that we get an object $X/\sim$ that can be embedded in $\Bbb{R}^n$ but not in any $\Bbb{R}^m$ for $m<n$?

Jakobian

@Koro shouldn't be in general, why

You can write things like $(F\sqcup i)\circ p^{-1}$ even if $p$ doesn't have an inverse

it usually means that the function thus obtained is well-defined

Koro

@Jakobian Ohh, doesn't that mean a map?

Jakobian

No, you can treat it more as composition of relations with $p^{-1}$ the inverse relation to $p$

Koro

It's even more confusing now.

Mad

@Semiclassical i understood everything you said until this point

Jakobian

09:10

So $((F\sqcup i)\circ p^{-1})(x)$ is being defined as $(F\sqcup i)(y)$ where $p(y) = x$

it's just that it will be constant on the fibers of $p$

($F\sqcup i)\circ p^{-1}$ will be a map, but $p^{-1}$ doesn't have to be

this notation is useful since if $q$ is a quotient map, you often want to use that $f$ is continuous iff $f\circ q^{-1}$ is, assuming $f$ is constant on fibers of $q$

you have a quick compact way of writing what you mean

Koro

@Jakobian that makes sense

So where have they used relative homeomorphicity of F?

Jakobian

well, I'd have to dive into the details of this argument

that's why I didn't answer this point

Koro

The existence of g is by pushout universal property.

Jakobian

I'm guessing it's to claim that $g$ is bijective

but it's better for you to write out all the details and check where the property is used

that way you'll have a clear view of the argument, without any hidden details

sunny

A function that is of "constant sign"...does this mean it is non-zero?

Koro

09:22

@Jakobian yup

leslie townes

sunny: probably, but more context would help. it's not common to refer to the "sign" of something that can be zero without expressly providing information about how to handle that case, or context that makes clear how to handle that case

sunny

@leslietownes ok, I'll provide some more context

Jakobian

@sunny No. For example, in Sturm's theorem, the sign switching relies on it switching from positive to negative. So $0$ is treated as neutral

Not always at least

sunny

ok

here's the context, and I'm quoting from a book:

Jakobian

I'd say a function of constant sign is one that doesn't switch from positive to negative, or vice versa

sunny

09:25

> A differential equation of first order is said to have separable variables if it of the form $$x'=g(x)h(t).$$ We assume that $g$ and $h$ are continuous, real-valued, and that $g$ is of constant sign. Then the equation can be rewritten $$\frac1{g(x)}x'=h(t)$$ (the variables $x$ and $t$ have been separated).

I don't understand how we can obtain the second without assuming $g(x)\neq 0$.

Soumik Mukherjee

@Jakobian can you take a look at the above topology question? I wonder if the answer is something trivial

Mad

Sunny, you do assume it is none zero,

It doesnt need to be globally none zero, you look for the points where it is not

Jakobian

@SoumikMukherjee any Peano continuum is a quotient of $X$

so we can obtain $[0, 1]^n$ this way for all $n$

sunny

@Mad ok

Jakobian

and even the Hilbert cube $[0, 1]^\omega$

Soumik Mukherjee

09:27

ohh

Jakobian

the question would be more interesting if the quotient map was special somehow, say, a covering map?

those would be "more natural" quotient maps

sunny

I was confused about the fact that my book was only explicitly writing that $g$ is continuous and of constant sign. I don't know if this implies that it is non-zero.

Soumik Mukherjee

Let's add the condition of covering map then

Jakobian

@sunny It doesn't really imply that, as leslie said it's context-dependent

here it's obvious that it implies that since we divide by $g(x)$

sunny

I see.

Jakobian

09:38

@SoumikMukherjee Well, the quotient of [0, 1]^2 onto a Mobious strip is not a covering map. But maybe we can get away with the weaker property of having finite fibers

I have results about what you can get away with when fibers are bounded in size

I just need to find them

Well, just under the assumption of finite fibers I can tell you that a quotient of $[0, 1]^2$ with finite fibers can't be embedded into $\mathbb{R}^n$ for $n\leq 1$

If we assume the quotient map is additionally open, I can tell you more

If $f:[0, 1]^2\to X$ is an open quotient map with countable fibers, then $X$ must have dimension $2$. Consequently, $X$ can be embedded into at most $\mathbb{R}^5$

If $X$ is additionally a smooth manifold, like the Mobious strip or Klein bottle, then $X$ can be embedded into $\mathbb{R}^4$

@SoumikMukherjee

@Jakobian finite fibers*

oh btw I'm assuiming $X$ is a metric space, and the quotients considered here are metric spaces

@Jakobian this also won't be open though

冥王 Hades

10:20

Maybe I should become Wyvern Rhadamanthys

Jam

I m stuck to the following thing: Consider the galois group of $X^7-1$ which is the gallois group of the cyclotomic polynomial $x^6+x^5+x^4..+1$ now the field extention is $Q(e^\frac{2πi}{7})$ and the Galois group must be a cyclic of order 6. Consider the Q-aytomorphism which send $f(\frac{e^2πi}{7})=e^\frac{4πi}{7}$ where $f \in G$ tghe galois group. This has order 4? which is impossible in a group of order 6 what am i doing wrong.??

$f(e^\frac{2πι}{7})=e^\frac{4πi}{7}$

leslie townes

can you say more about why f has order 4? if z = exp(2 pi i/7) and f is a homomorphism with f(z) = z^2 then f(f(f(z))) = z^8 = z because z^7 = 1?

Jam

10:42

how u got f^3=z^8?

because of wahat you said exactle f^4(z)=z the id so of order 4

nvm i was adding the powers $f^4(z)=(z^2)^4$ which is wrong apparently it is $z^(2^4)$

leslie townes

11:16

well, not "apparently," so much as, directly, in the sense that it comes out of the homomorphism property

if f is a homomorphism and f(z) = zz then f(f(f(z))) = f(f(zz)) = f(f(z)f(z)) = f(zzzz) = f(z)f(z)f(z)f(z) = zzzzzzzz

applying f to a string of z's will double the number of z's. with hopefully a clear generalization to what happens if f(z) = z^k for some other positive integer k

Alessandro Codenotti

@Jakobian regarding the 2nd countability of X given that of X/G, I cannot find counterexamples even to far weaker questions. Like what is an example with $f\colon X\to Y$ is continuous with finite fibers, $Y$ second countable and $X$ not second countable?

Soumik Mukherjee

11:58

@Jakobian Thanks for the response, I need some time to get a full grasp of everything that you have written.

John Zimmerman

12:36

carpe diem

@leslietownes I found a way to make everything work in my question we discussed.

I just set up the inequality and don't manipulate it!

then no need for those tridents lol

Jakobian

12:52

@AlessandroCodenotti I have no idea

porridgemathematics

13:08

is the following true? If $x,y,z,w$ are positive numbers, satisfying $x+y \leq z+w$ and $x \leq z,w \leq y$, then $\frac{1}{x} + \frac{1}{y} \geq \frac{1}{w} + \frac{1}{z}$

Alessandro Codenotti

@Jakobian I'm not sure whether that can happen

porridgemathematics

this book im reading is making me ask really fundamental questions, because all the high level steps figure, and then there is a step that is equivalent to what I just asked, which I find harder to figure, lol

well, not fundamental, but basic sounding

porridgemathematics

13:33

nvm, just figured it out

rogerroger

14:18

what do people mean when they say a function has linear growth

say $f:\mathbb{R}^n\to \mathbb{R}$...

Jakobian

@AlessandroCodenotti there's just not a lot of sources discussing those kind of things

and if there are, it's usually about $X$ having good properties while $Y$ does not, not the other way around

anyway I'd focus more on what we can prove rather than on what we can't

If you're wondering about your question you can always ask it on the site, there's people smarter than us here

Thomas Finley

14:34

How did they assert that if $\phi$ is continuous at $c\in A_1\subseteq A$ then $\phi$ is continuous as well on the restricted domain $A_1$ as well?

Koro

H= quaternions group.

$HP^n$ is constructed from $HP^{n-1}$ by attaching a 4n cell.

Thomas Finley

If $c$ was an isolated point of A, then we have no problem. But if not, then how can we gurantee that $c$ is a cluster point in $A_1$ besides being a limit point of A.

Koro

Say, I take the quotient map $q: S^{4n+3}\to HP^n$. Noting that $S^{4n+3}/S^1= CP^{n+1}$, q induces a map from CP^{n+1} to HP^n. How do I calculate the homology induced by this map?

Hi @Ted!!

Ted Shifrin

Hi @Koro.

Koro

I'd asked this question few days ago with no clue. I ask again with some clue. The first part of it was to show that $HP^n$ is formed from $HP^{n-1}$ by attaching a 4n cell. This is complete.

Ted Shifrin

14:43

You’re meaning $2n+1$.

Koro

@Koro For 5), a map could be constructed like this. But not sure how inducing homology using this.

yes, I meant 2n+1 there.

Ted Shifrin

I honestly have never thought about quaternionic projective space. Have you thought through the analogous procedure with $\Bbb R$ and $\Bbb C$?

Koro

@Koro the 4) follows from this by constructing an appropriate F.

Alessandro Codenotti

@Jakobian I'll think about it a bit and then maybe ask

Koro

the good thing about the lemma I referred to (replied to) is general and can be used for any field $K\in \{\mathbb H, \mathbb R,\mathbb C\}$.

Ted Shifrin

14:46

I guess there is no analogue.

Jakobian

Also, I realized that my "locally connected separable metric spaces have countable basis consisting of connected sets" contains an error.

Koro

Suppose that d= dim K over R as vector space, then $HK^n= HK^{n-1}\sqcup_f D^{dn}$

Jakobian

I should be going by sets of diameter $< \frac{1}{n}$ for each $n$ and obtain a covering for each one. But it's not clear that it's a covering

ah. Sorry, just use Lindelof property

Koro

i.e., we attach a dn cell.

for H, d=4; for C, it's 2.

Ted Shifrin

What is the map when $n=1$?

Koro

14:51

12 mins ago, by Koro

Say, I take the quotient map $q: S^{4n+3}\to HP^n$. Noting that $S^{4n+3}/S^1= CP^{n+1}$, q induces a map from CP^{n+1} to HP^n. How do I calculate the homology induced by this map?

Instead of CP^{n+1}, it is CP^{2n+1}

$HP^n= S^{4n+3}/S^3$

John Zimmerman

15:36

How do you show an inequality $f(x)\le h(s) \le g(x)$ where $h$ is a parametric function?

I think it'd be easier if $h(s)$ could be expressed as a function of $x$.

particularly $h(s)=\bigg( 1-\frac{\pi(s)}{s}, 1-(1-\frac{1}{s}) \bigg)$

I'm pretty sure that can't be a common function if $\pi(s)$ is involved.

Jakobian

15:53

I didn't even need metrizability!

Soumik Mukherjee

Let $f : [0,1] \to [0,1]^2$ be an Osgood curve. Is the upper bound of Lebesgue measure of $f([0,1])$ known?

Semiclassical

17:09

o/

Ted Shifrin

17:48

@JohnZimmerman You continue to write stuff that makes no sense. First, inequalities with a vector bounded by scalars. Then this.

John Zimmerman

@TedShifrin I developed it further and posted on mathoverflow where it was answered successfully

0

A: "Squeezing" the primes?

As demonstrated by your link, you're letting $$ x=1 - \frac{\operatorname{li}(10^s)}{10^s} \approx 1 - \frac{1}{s\log 10} $$ for $1\le s\le8$ and then noting that $$ \exp\left(\frac{1}{\log x}\right) <10^{-s}< \exp\left(\frac{2}{\log x}\right) $$ or in other words $$ \frac{1}{\log x} <-s\log 10< ...

Semiclassical

@Mad There's two sentences there, so I'm not sure which part confused you. But 1) if $B^2 v = (\lambda^2-a^2)v$, then $(B^2-\lambda^2+a^2)v=0$. Hence $B^2-\lambda^2+a^2$ has $v$ as eigenvector with eigenvalue $0$.

that's what i was getting at with it "sufficing" to look at $B^2-\lambda^2+a^2$ rather than $B^2$ itself.

Mad

Okay so you need to multiply matrix I next to labda and a

Semiclassical

yeah, i'm being sloppy in the same way that Schwabel was

Mad

I understand now the steps mathematically, but how do i now understand why we are doing this

I am not having a general picture

John Zimmerman

18:00

I also came to the conclusion using my own method: $$\frac{1}{2} \le \big( 1,0 \big) \le 1$$

very similar result to Charles' result

Semiclassical

well, you're trying to find solutions to the Dirac solutions

John Zimmerman

I used $\pi(x)$ there Charles used $\mathrm{Li}(x)$

Semiclassical

if you find the eigenstates, you can build up other solutions by linear combinations

Ted Shifrin

It makes no sense as you write it. Even the guy that answered you said that. Make the effort to write basic mathematics so that it makes sense.

2

Semiclassical

so we're trying to find said eigenstates

Ted Shifrin

18:03

It’s super frustrating when you talk about super sophisticated stuff in terms that make no sense without a translation.

Semiclassical

the reason I proceeded as I did was to go from a 2-by-2 matrix whose elements are themselves 2-by-2 matrices, to just a 2-by-2 eigenvalue problem

with the cost being that we get that funky-looking $B^2-(\lambda^2-a^2)I$ combination

however, you can sorta bracket that by letting $M=B^2,\mu=\lambda^2-a^2$

at which point your problem is $Mv=\mu v\leftrightarrow (M-\mu I)v=0$

so we're down to just an eigenvalue problem

Mad

@Semiclassical Yea, but i mean in respect to the matrix B squared

Semiclassical

i don't understand what you're specifically confused about then

are you confused about why $B^2$ is relevant, or about how to solve the $B^2v=(\lambda^2-a^2)v$ problem?

John Zimmerman

@TedShifrin I completely understand your frustration - it's frustrating for me too sometimes. I like to race to the end and then once I find a solution that feels right I fill in the details. But since I focus too much on racing to the end I have become weak over time in computational skills. That's why I am working on committing to becoming more self sufficient and align myself to mainstream methods

Ted Shifrin

18:19

I think you’re thinking that a (parametric) curve lies between the graphs of two functions, but you cannot say that correctly as you’re trying to.

John Zimmerman

Yes - and I will for sure study Charles answer meticulously

Ted Shifrin

@Koro so you get an exact sequence of homotopy groups, but for homology you have to figure it out on the level of cells, I guess. The usual Hopf map is boring.

Koro

18:38

one fact is: lot of such induced maps are the 0 maps; while the non-trivial ones Z-->Z are not known.

not sure how to find them.

top row is CW chain complex of CP^{5} and the bottom one is for HP^2.

? ain't known

(i.e. n=2)

Ted Shifrin

Have you started with $n=1$? I expect the lower $n$ get used in higher ones.

user 726941

19:42

@JohnZimmerman +1, have you slayed Maxwell's demon yet?

robjohn

22:37

@JohnZimmerman Did you ever say what $(0,1)$ means as a real number (or how you are comparing it with real numbers)?

@JohnZimmerman There is a difference between $\operatorname{Li}(x)$ and $\operatorname{li}(x)$. I think you intended the latter.

leslie townes

rob this has maybe been worked out elsewhere but the apparent meaning of john's "f(x) <= (a,b) <= g(x)" is that f(a) <= b <= g(a), i.e. that the point (a,b) lies above the graph y = f(x) and below the graph of y = g(x). there might be more to it when there are other symbols involved. there have been various nudges made toward encouraging more conventional notation

Ted Shifrin

LOL @nudges

leslie townes

:)

user 726941

(:

robjohn

$(\cdot$

Happy cyclops

@leslietownes I remember seeing Ted nudge last night, but I never saw the meaning. Thanks for providing one.

Gotta take the pups for part 1 of a 2 part play date. BBL

user 726941

22:54

cya

EM4

quick question

the limit of arctan(tan(x)) x approaching infinity is it pi/2.

Or I am just wrong, if so I will go back to drawing board.

Ted Shifrin

@EM4 Tell me why it has a limit.

leslie townes

what ted said. also, the style police would have something to say about the use of juxtaposition in "the limit of f(x) x approaching"

Ted Shifrin

That was hardly a well-formed sentence, so Ted skipped his grammar police role.

EM4

never mind I know where I messed up.

how are you guys doing?

Ted Shifrin

23:09

Where?

EM4

outcome of this limit is indeterminate.

leslie townes

the style police would say "the limit does not exist"

you're right, it doesn't. :)

Ted Shifrin

Can a periodic function ever have a limit at infinity?

leslie townes

that's the most informative generalization of this example. i guess there's the minor nit of handling the static in the domain of this particular f.

Ted Shifrin

Right. @leslie, for you?

leslie townes

23:25

although i think even a frenchman would agree that this function of tan x has a limit at infinity. or is it we who would agree with the frenchman that it does.

ooh, interesting one. they might want some kind of convexity there too. it feels like a place where convexity might come up. i'll think about it.

Ted Shifrin

Huh?

leslie townes

to simplify the argument, not because it's necessary.

oh i see, they're looking for a necessary condition. i dunno about that. it might be telling if rudin only says "find other sets for which this is true" and not "characterize those sets for which this is true," i dunno.

Jakobian

@leslietownes why are French people wrong?

Ted Shifrin

23:42

Apparently they define limits only when the function is continuous.

leslie townes

they aren't wrong, jakobian. i was just joking about that continental tendency to define, or to refuse to define, limits in certain cases. i couldn't even remember which

but having a cluster of points running off to infinity on which our function wasn't defined was enough to make me wonder what the academie francaise would think

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$Mathematics$ Mathematics