Mathematics - 2018-07-21

Fargle

02:34

9

To complete Student's answer : $\mathbb Z_p \subset \mathbb Q_p \subset \mathbb C_p$, and since $\mathbb C_p$, the algebraic closure of $\mathbb Q_p$, is isomorphic to $\mathbb C$, the field of the complex numbers, we know that $$ |\mathbb R| = |\mathbb Z_p| \le |\mathbb Q_p| \le |\mathbb C_p| =...

In short, yes.

Balarka Sen

I am here to chew ass and kick bubblegum.

2

EnjoysMath

03:13

0

Q: Would this suffice in a visual type theory to define a list?

See the image. I got that from: wikipedia article. I want to make a visual type theory so that we aren't stuck comprehending pure text for eternity. Also, is the abstract type List a product of some sort? The diagram doesn't indicate this since it's based on a product diagram of $E \times L$...

Do you think that searching for isomorphic subgraphs of a category diagram is a bad idea?

Handmade diagrams don't get super huge

The user can then a the click of a button apply a rule defined by a diagrma

Failed to be a Mathematician

04:29

math.stackexchange.com/users/464147/… @secret Am I correct?

Secret

which?

Failed to be a Mathematician

proof

math.stackexchange.com/users/464147/…

Alpha Romeo

05:27

Hey, can someone please verify/give reference to this: If partial derivates exist at every point in a neighborhood of a $R^2$ -> $R$ function, then the partial derivatives are continuous as well.

Secret

05:51

in The Symposium, 16 mins ago, by Secret

> Come now, I will tell you ... the only routes of inquiry that are for thinking: the one, that it is and that it is not possible for it not to be, is the path of Persuasion (for it attends upon Truth), the other, that it is not and that it is right that it not be, this indeed I declare to you to be a path entirely unable to be investigated: For neither can you know what is not (for it is not to be accomplished) nor can you declare it. [2=B2]
For the same thing is for thinking and for being. [3=B3]

Erik M

06:20

What’s the most surprising math result you know of?

Secret

06:38

@FailedtobeaMathematician there are tons of proof questions, which?

Chandler Watson

@Jagerber48 hello!

Jagerber48

Hello! Thanks for having a look at my question!

Chandler Watson

Of course! Just took an intro course to diff geo and the question looked really well thought out :)

Jagerber48

Thanks :)

It sounds like you're getting the idea of what I'm asking

Chandler Watson

No, totally -- it's strange to me too

The reason stuff like the Hodge star and musical isomorphisms and what have you come up is because to properly represent things in a coordinate-invariant manner, we use what are called differential forms

Have you heard the term before?

Jagerber48

06:40

Yes I've heard of them but my understanding is not great.

I've read about them and self-teach myself but I've never taken a proper differential geometry course.

Chandler Watson

No problemo at all, honestly my own understanding is not too deep yet

Jagerber48

I understand decently well what something like dx is as a covector

and it seems like differential forms are anti-symmetric tensor products of covectors like dx

Chandler Watson

Nice, okay, so you're familiar with the idea of the cotangent space and everything?

Jagerber48

but I can't get my head around why that would be interesting or the properties of the objects

Chandler Watson

Okay, that's actually a great way of putting it

Why it's interesting (afaik, anyone feel free to correct me at any point) is because we have two parallel ways of seeing things

We have the standard calculus way, where we work with fields, and have div, grad, and curl

And then we have working with differential forms

We can kinda switch back and forth between the two using the "musical isomorphism" -- does that sound familiar?

Jagerber48

06:46

I'm in physics so I'm familiar with raising and lowering operators on tensors/vectors using the metric tensor. it seems like the musical isomorphism is capturing exactly this sort of thing. Thinking of it as converting between a "standard calculus" way of working with fields etc. and the way of working with differential forms sounds interesting to me.

Chandler Watson

That's exactly it. So you're pretty much free using that isomorphism to flip back and forth between the tangent space and the cotangent space (i.e. vector field to differential form and back again)

Jagerber48

ok

Are you able to explain how grad and div come into the discussion of forms?

Chandler Watson

Yup, just piecing together the details (took the class in winter, trying to remember most of it! :P)

Jagerber48

ok great! thank you!

Chandler Watson

np!

So, the way of taking the gradient is to take your function from a 0-vector to a 0-covector (pretty much the same thing), take the exterior derivative to get a 1-covector, then convert it back from a 1-covector to a 1-vector. And that's it :)

The gradient of a vector field is just a nicely-wrapped exterior derivative

When you try curl, things get a tiny bit tricker, but not a whole lot worse.

Jagerber48

06:59

ok, I can kind of follow for grad but I'm curious to hear about div. I've struggled with that one and definitely don't understand the Hodge star, other than understanding that like, 3-2 = 1 and 3-1 = 2 has something to do with it...

Chandler Watson

Yeah, exactly. At first, the Hodge star seems a little arbitrary and strange.

I'll do curl first, because it also uses the Hodge star, but it takes us "internally" from 1-covectors to 2-covectors

Jagerber48

ok

Chandler Watson

So curl, we start with a 1-vector and convert it to a 1-covector

We then take the exterior derivative as usual, but then we're stuck with a 2-covector

And we need some way of getting out a 1-vector when we're all done

Jagerber48

How do you think about the exterior derivative here?

Chandler Watson

So we use the Hodge star operator. Since we're working in 3D, the Hodge star takes us from a k-covector to a (3 - k) covector

The exterior derivative here gets a little tricky, since we have to worry about all those wedge products now

So intuition understandably gets a little trickier.

Wikipedia's way of thinking about it talks about thinking about the differential forms as measuring flux through small parallelpipeds, and taking the exterior derivative just adds another dimension to that parallelpiped.

I've heard good things about the approach taken in this document tho, and it discusses intuition on Hodge star as well (tbh I might read that later tonight :)

In any case, we use the Hodge star to get us back from a 2-covector to a 1-covector, and then we can just go back to a 1-vector as normal

Jagerber48

07:07

ok

Chandler Watson

Intuitively, the Hodge star swaps out everything that's "there" for everything that's "not there", i.e. the hodge star of dx^dy is dz

Jagerber48

hmm ok

Chandler Watson

That's up to sign though, cyclic permutations preserve the sign but doing something like dy^dx yields -dz

Jagerber48

ok

Chandler Watson

So you can see that it clearly takes you from 2-covectors to 1-covectors, tho of course it does seem a little funny at first.

For div (the grand finale) you go from 1-vector to 1-covector, take a hodge star first this time to get to a 2-covector (we'd already done 0-covectors and 1-covectors, so this is the intuitive next step), then take the exterior derivative to get a 3-covector.

We want out a 1-vector tho, so we hodge star again to get a 1-covector and then just send it back.

So grad is pretty much a fancy exterior derivative on 0-forms, curl is a fancy exterior derivative on 1-forms, and div is a fancy exterior derivative on 2-forms, but they're all pushed and hodged around so that they play nice with 3D vector fields

What's fun is that you can actually do out those computations on a generic vector field in Euclidean space, and get out the good old expressions for grad, div, and curl :)

Jagerber48

07:14

Yeah I think that is what I would like to try to do. That would probably help with my question

Your explanation of Hodge star seems pretty straightforward and helpful

The thing I wouldn't know how to calculate is the exterior derivative of a 1-form or 2-form

Chandler Watson

Oh, gotcha -- the rules are pretty nice, so I can just give a quick example

so if you have something like x dx + y dz, taking the exterior derivative just means that you wedge things together, i.e. you get dx^dx + dy^dz = dy^dz, because anything wedge itself is zero by antisymmetry

Jagerber48

ok

So I guess it is important that there is a the presence of both a function and a covector

because you need to take the differential of the function to get the second covector in the product

yeah that's not too bad

Chandler Watson

it is indeed :) and the wiki page has some nice extra details if you need them

like if you have xy dz, it's the sum of all the partial derivatives, i.e. y dx^dz + x dy^dz

also, d(df) = 0 for any function f is useful

Jagerber48

right

cool thanks! Yeah I think I'll be able to unravel these definitions a bit now.

I guess my original question still stands about the $\nabla$ operator standing on its own

Chandler Watson

of course, I've starred your question because I'm pretty curious about that too

Jagerber48

07:23

It seems like the answer must be wrapped in this differential geometry stuff somewhere!

But yeah otherwise it just seems like the operator can be used and manipulated successfully in too many ways to be more than "just" a convenience.

Iza_lazet

Just an aside: the "blood-related" generalization of vector calculus is geometric calculus (Clifford algebra) and working with multi-vectors (where grad, div, curl are defined on multivectors). Though for diff geo purposes, working with multi-covectors (differential forms) is sometimes more convenient thanks to pullbacks (we can't pullback a vector field, but we can pullback a differential form). The correspondence between multi-vectors and multi-covectors is not always canonical, unfortunately.

Chandler Watson

Right? It's true that because we have coordinate free definitions of each "nabla operator" your observations must hold, but I'd love to see if there's some clean intuitive way to see why that does.

@Iza_lazet whoa, huh -- any source recommendations for geometric calculus?

Iza_lazet

What was the question about nabla again?

Jagerber48

4

Q: Coordinate free definition of $\nabla$ operator

There are a number of posts on this site asking similar questions and some of them have been answered (to my taste) at least partially but none give a complete answer that I am satisfied with. See links at the bottom of this question for a small selection of posts asking related (or even the same...

differential-geometry notation vectors

Thanks for the input on geometric calculus/Clifford algebra. I've distantly heard about that in the context of spinors but maybe I'll have to look into more detail

Iza_lazet

that is easy. In fact, it is exactly analogous to how the exterior derivative is defined: by first agreeing on the 0-dimensional case, then by listing the algebraic properties one wants the operator to have, so that we get a unique operator. Although obviously, students are usually taught just the formula.

I learned geometric calculus from Alan Macdonald's books and notes. It is a very immature field, so there is some disagreement on certain notations (such as contraction) and what properties people want the operators to have.

his books, unfortunately, do not contain the BAC-CAB family of formulas, but somehow his online notes do.

Chandler Watson

07:35

Interesting--thanks so much!

I'm going to nod off for the night here, take care you two

Jagerber48

Have a good night! Thank you for all of the information and taking the time to discuss with me!

Chandler Watson

Ofc! It was interesting and really informative for me too :))

Jagerber48

@Iza_lazet are you able to elaborate on what you mean by the 0-dimensional case we have to agree on and what algebraic properties we would want the nabla operator to have? Or perhaps you could provide an answer to the linked question whenever you have the time?

Iza_lazet

I think either Herstenes or Alan provided the derivation in their books. Basically you want things like associativity, Leibniz's rule, change in grading, etc. The 0-dimensional just means for a function f (0-dimensional), $\nabla f$ is the usual gradient of the function.

also multilinearity of course

Failed to be a Mathematician

0

Q: If $A$ is an ordered set has the Least upper bound property iff it has the greatest lower bound property.

Mentioned as an easy exercise in 'Topology', James R. Munkres, 2e, Pearson[page No:25] If $A$ is an ordered set has the Least upper bound property iff it has the greatest lower bound property. An ordered set $A$ is said to have the least upper bound property if every nonempty subset $A_0$...

proof-verification order-theory supremum-and-infimum

I got the help :)

Secret

07:46

I see

That theorem is very important, so study it carefully.

Iza_lazet

an honest curious question, when do people use this theorem outside the exercise?

Secret

If I recall, you need that to prove nested intervals theorem

and that will allow you to prove Bolzano–Weierstrass theorem, which is needed to show every sequence in a closed real interval has a convergent subsequence, which is used in proving things related to limits

Iza_lazet

08:01

Oh ok. So that is yet another strategy, among many equivalent ones, to approach the fundamental property of real numbers.

Secret

yeah, and it is because the reals obey lub that you can say so much about continuous functions on the reals. But again I am not an expert, as I mostly learn all my real analysis from this chat as I never get a chance to fit a real analysis course in my undergrad

Iza_lazet

personally, I have always preferred the Cauchy-completeness approach, which can be generalized to general metric spaces. Well, most professors like to teach LUB and Dedekind cuts though.

Secret

I also like the dedekind cuts approach more, because it generalises easily to the surreals

also I really don't like to see more than 3 inequality signs in a single line, which is common for the cauchy approach

though I have some understanding of both approaches and Iam actually planning to restudy those constructions and proofs later because I need them to understand dense linear orderings in general

Jasmine

13:17

Consider $dx/dy -dy/dx + d^2x/dy^2=7$ what will be order and degree

Silent

@TedShifrin, you mention homework problems in your video lectures several times. Can I have access to them?

Jasmine

Can anyone please help me on this?

12 mins ago, by Jasmine

Consider $dx/dy -dy/dx + d^2x/dy^2=7$ what will be order and degree

I think that order should be 2 and degree should be undefined

rschwieb

13:48

@Fargle Ugh of course :) The construction of the $p$-adic numbers really distracted me from where they live! Thanks

Ted Shifrin

14:47

@Silent Well, you'd need the textbook for those ...

Secret

16:06

hmm...

Which definition and assumption so I blow up in order to cause $\text{lub} \to \text{glb}$ to fail?

Just for curiosity to see how failure of least upper bounded sets look like

Source: math.stackexchange.com/a/2858264/180683

Ah I see, so by "de-lub" the set $A$, you can have $\ell_0 \not\in L(B)$ or $\ell_0 \not\in B$

The rationals bounded by irrational endpoint open intervals $(-r,r) \cap \Bbb{Q}$ is the easiest counterexamples as the would be lub will become $r$ thus falling outside the set, hence breaking the proof as required

But is there more interesting way to create a set which does not satisfy lub?

In particular, can we have $\ell_0 \in L(B)$ and yet lub fails?

Meanwhile, the failure of the proof due to empty L(B) will correspond to the case where B has no lower bound, thus it cascade down forever regardless of which filter you pick

So that means there are two ways for a subset of a given set to break the least upper bound property:

1. Either the set B has no lower bound (thus there exists an infinitely decreasing sequence or the downward directing net is unbounded or there is a filter with no minimal element) resulting in L(B) to be empty

2. Or that L(B) has no least upper bound because any would be least upper bound will fall outside of L(B) or B (as in the rational counterexample)

Ted Shifrin

16:26

@Secret: You cannot break LUB in the real numbers, but you can in stranger ordered topological spaces.

Secret

1

Q: Not well-ordered sets satisfying the least upper bound property

Before proving that every closed interval in $\mathbb{R}$ is compact, James R. Munkres remarks that, in Section 27 entitled "Compact Subspaces of the Real Line" of Topology (2nd edition), Remark: We need only one of the order properties of the real line --- the least upper bound property. We ...

general-topology order-theory

> However, every linear order can be extended to a linear order which satisfies the least upper bound property by considering a construction similar to the construction of the real numbers by a Dedekind completion of the rationals.

and Re Ted: Ah I see, so in general I will need to go to things weaker than linear ordering to have more exotic counterexamples

Ted Shifrin

No, linear ordering, but not what's called a "linear continuum."

Of course, you can break the LUB axiom with $\Bbb Q$ ...

Secret

yeah, so is $\Bbb{R} - \{\text{insert sequence of numbers to take away}\}$ in general, those are the dense linear orderings but not linear continua

Ted Shifrin

right ...

Secret

But I am interested in more exotic ways to break least upper bound, if any

Ted Shifrin

16:35

So you've been reading Munkres?

Secret

Well, I read Munkres 6 months ago although I stopped halfway before continuous functions because then my chemistry PhD started

Ted Shifrin

Weird to tackle Munkres without doing serious analysis first :P

Secret

but failed to be mathematician is going through munkres recently, and when I read his recent question on proving lub, after explosive generalisation, lead me to this curiosity on what are the all possible classes of sets that break LUB look like

Ted Shifrin

Here's an example you should think about. Take a two-element set, say, $\{1,2\}$ and cross with $\Bbb Z$, in the dictionary order. So $(1,a)<(2,b)$ for all $a,b$, and $(i,a)<(i,b) \iff a<b$.

Can you give me a set that's bounded above but has no LUB?

Secret

16:53

$\{\forall a \in \Bbb{Z} : (1,a)\}$ or in general, any unbounded from above sequence of $a$ s forming a sequences of $(1,a)$. "The want to be least upper bound" would be the greatest lower bound of $\{\forall b\in \Bbb{Z} : (2,b)\}$ which there is no smallest $b$ thus such greatest lower bound does not exist

Ted Shifrin

There you go.

Although I don't like your set notation. You should write $\{(1,a): a\in\Bbb Z\}$, etc.

Secret

right, because the tuples are the elements

so the above example will be the class where $L(B)$ is unbounded from below in the LUB proof. Now I guess the harder question is: Can we prove that dense linear orderings which are not linear continua, and all sets $B$ such that $L(B)$ is unbounded from below the only two "classes" of LUB breaking sets, and how to approach such proof?

Ted Shifrin

I doubt one can prove such a thing.

What you said isn't right, anyhow, as any element of our set is a lower bound for $B$.

Secret

ah right, I mean the set $U(B)$ which is the set of all upper bounds of $B$, and $U(B)$ is unbounded from below

Ted Shifrin

Not having a least upper bound goes along with not having a greatest lower bound, so it's basically circular.

I thought $B$ was your set of upper bounds.

In our example, the set of upper bounds is bounded below. There's just no greatest lower bound.

That's always going to happen if the LUB axiom fails. If you poke a hole in a linear continuum, it's the identical phenomenon.

Secret

17:02

I see, so basically they are the same as the "poke a hole in the linear continuum" cases

hmm...

let me think a bit on how to construct a nonempty set $B$ such that $L(B)$ is empty...

Ted Shifrin

Well, conceptually, our example we've been talking about is very different. You can't add something to "fill in a hole," but the phenomenon is analogous — one "line" is the upper bounds for the other, and, vice versa, the other is the lower bounds for the second.

You can't. If $B$ is the set of upper bounds for a set $A$, then any $a\in A$ is a lower bound for $B$.

Alessandro Codenotti

Did someone say ordered set?

Secret

I am suspecting for that example, we can extend the set by adjoining an element $(1,\infty)$ such that $\forall a,b \in \Bbb{Z} : (1,a) < (1,\infty) < (2,b)$, then the hole should be filled in the extension

o wait, maybe not, because I can just pick

$\{(1,a) : a \in \Bbb{Z}\} \cup \{(1,\infty)\}$ and we will be screwed as before (because there is no smallest $b$)

hmm... this case is interesting...

Ted Shifrin

Heya @Alessandro. Here's our expert. :P

Alessandro Codenotti

Hi @Ted

Secret

17:17

Ted Shifrin

@Secret: Well, now $(1,\infty)$ is the LUB of our original subset.

But if you add it to the subset, then $(1,\infty)$ is the LUB and it's in the set.

This is like taking $(0,1)\cup (1,2)\subset\Bbb R$ and throwing in $1$.

Or you could try $(0,1)\cup (2,3)$ for fun. :P

Secret

yup, but we can play the same game again and we will found $\{(1,a) : a \in \Bbb{Z}\}
\cup \{(1,\infty)\}$ still have no LUB, thus technically the hole cannot be filled by adjoining elements

so it is as you said, very different from the "poke hole in a linear continuum case"

Ted Shifrin

Well, is it?

No, the set you just defined does have a LUB. It's $(1,\infty)$.

The LUB can certainly belong to the set.

Secret

ah right...

I almost forgot that LUB can also be the max

> However, every linear order can be extended to a linear order which satisfies the least upper bound property by considering a construction similar to the construction of the real numbers by a Dedekind completion of the rationals.

So by Asaf's comment, since the usual lexicographical order is a linear order, it means any holes can be filled in

so that means in order to have sets with holes so big that it can never be filled, we will be going into partial orders. But I will stop here for now as that's too much explosive generalisation in a day

Ted Shifrin

This probably gets into some interesting set theory if you proceed far enough. I leave you to @Alessandro.

Re the multivariable analysis stuff, you saw my answer earlier, @Secret? I can email you problem sets but mostly they refer to exercises in the book.

Secret

17:27

@TedShifrin I think it is silent that asked for the exercise?

Ted Shifrin

Oh, oops. You both start with S. My apologies.

hides in embarrassment

Alessandro Codenotti

Here's an interesting order on $Bbb R$ to think about: you can think about $\Bbb R$ as $\Bbb Z\times[0,1)$ (the stuff on the left and on the right of the comma), the usual order on $Bbb R$ is the lexicographic one. The antilexicographic one is a linear order on an uncountable set in which every element has a unique immediate successor and a unique immediate predecesser!

Ted Shifrin

hmm, what does antilexicographic mean?

Oh, I see, we go horizontally instead of vertically ...

Alessandro Codenotti

You compare the second coordinate first

And the first one if needed

Secret

This is why I love infinities. They are so cool (and they are one of the primary reasons I get into topology before analysis as I never can fit that into my undergrad which was packed with my chemistry and physics degree stuff)

Abcd

17:35

@TedShifrin Hi

Ted Shifrin

hi @Abcd

Abcd

@TedShifrin Comparing 2^pi and pi^2 , proof verification needed

Ted Shifrin

@Alessandro: Or you just write $[0,1)\times\Bbb Z$ for dopes like me ;P

Abcd

Define: $f(x) = x^{\frac 1x}$ for $x>0$

$f'(x)= \dfrac{x^{\frac1 x}(1-\ln x)}{x^2}$

Thus, $f(x)$ is strictly increasing till $x= e$ and strictly decreasing after $x=e$.

Now, which number is closer to $e$, $2$ or $\pi$?

$e\approx 2.7, \pi \approx 3.14$ , thus it's $\pi$ which is closer to $e$.

$\implies f(\pi)> f(e)$

$\implies \pi^{\frac 1\pi}> 2^{\frac 12}$

Raise both sides to the power $2\pi$ to get:
$$ \bbox[5px,border:2px solid black]
{
\pi^2 > 2^\pi
}
$$

Ted Shifrin

@Abcd: Personally, I would take logs first and look at $f(x)=\dfrac{\ln x}x$.

Abcd

17:39

@TedShifrin Do you see any loophole in my proof

Ted Shifrin

I don't think it's about who's closer to $e$. The graph isn't symmetric, after all.

That's a big loophole.

Abcd

@TedShifrin So how do I fix it

@TedShifrin ya, i know.

Ted Shifrin

I don't see how to fix it.

Maybe Taylor estimates on the $f(x)$ I gave?

Mike Miller

Or plug it into Wolfram eh

Abcd

Using physics the answer is easier

$\pi^2 = g$

$2^{\pi} = 8$

Alessandro Codenotti

17:42

Actually I messed that order up earlier I think

Can't think about two things at the same time

Mike Miller

Oh there is an elementary argument

2^(1/2) = 4^(1/4)

Abcd

@TedShifrin Is that OK as a Math.SE answer? (with equalities replaced by approx sign)

Mike Miller

Because x^(1/x) is decreasing after e...

Secret

Abcd

@MikeMiller Hmm yes.

Secret

17:44

confirmed alessandro's statement

Ted Shifrin

Ah, that's super clever, @MikeM.

Abcd

Now I am wondering about my Physics Answer.

@TedShifrin What are your thoughts on it?

Ted Shifrin

I don't know what you're talking about, @Abcd.

Abcd

@TedShifrin $\pi^2 = \text{gravitational acceleration}= g$ is well known.

Ted Shifrin

What?

Abcd

17:45

yes.

Ted Shifrin

If it's well-known, why is that I've never seen it?

Abcd

@Secret Isnt it well known?

@TedShifrin Maybe its been long since you did physics?

Ted Shifrin

No physicist I know has ever mentioned such a thing.

What's the proof, if it's "true"?

Secret

@Abcd no I am referring to an earlier comment while we are discussing about ordered sets

Abcd

Its mentioned in endless simple harmonic motion problems.

Ted Shifrin

17:47

Not in any course or text I've ever seen.

Abcd

@Secret No, I was asking you because you spend some time on h bar

Ted Shifrin

This must be some crazy Chinese thing.

3

Abcd

no

Secret

@Abcd how on earth is $\pi^2$ the same as the gravitational constant??

Abcd

@Secret i said gravitational acceleration not gravitational constant

Ted Shifrin

17:48

It's not even close as an approximation. This is garbage.

Abcd

9

A: Why is "$\pi^2= g $" where $g$ is the gravitational constant?

Maybe this helps: link Looks like some time ago the second was defined by $1/2$ of the oscillation time of a $1$ meter long pendulum. The oscillation time of a pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$. With $T = 2$ and $L = 1$ this gives $g = \pi^2$

Ted Shifrin

And this.

Abcd

wired.com/2013/03/what-does-pi-have-to-do-with-gravity/

quora.com/Why-does-π-2-g

Ted Shifrin

First of all, a pendulum is only an approximation of SHM. SMH.

I'm done with it.

It's a lame approximation. It's certainly NOT any sort of equality.

Abcd

I know

Ted Shifrin

17:50

Well, you're not acting like you know.

annoyed

Secret

It cannot be an equality, because
g 9.80665 (units)
but $\pi^2 \approx 9.86960440109$

that's way off on just the 2nd d.p.

Abcd

@Secret okay at least $\pi^2 \approx g$

Secret

I am not sure if you want to have a huge error when using that approximation

Abcd

I mean its OK to use it compare $2^\pi$ and $\pi^2$

Secret

well, Mike justified how that work abstractly, thus I am convinced

Abcd

17:54

@Secret I am really not telling any one to use it. My only point is that it can be used to compare 2^pi and pi^2

Secret

ok

Abcd

I am just wondering if I should add that as an answer to this question^

But I am also wondering if everyone will kill me if I do that

3

Q: How to compare $2^{\pi}$ and $\pi^2$ using calculus

How to compare $2^{\pi}$ and $\pi^2$ using calculus I guess $$f(x)=\frac{\ln x}{x}$$ wont help here since $2 \lt e \lt \pi$

calculus

Secret

if that is going to be an answer, you cannot use $\pi^2 \approx g$ because the real reason why it works is the asymptotic behaviour that Mike outlined

at least I will roast you if you put an imprecise answer lol