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3:57 AM
Anyone know how to typeset the symbols that look like $\lfloor$ and $\rfloor$ but shorter?
4:09 AM
Found them: $\llcorner$ and $\lrcorner$.
5:04 AM
Is "Thomas and Finney Calculus" good book for self-studying calculus?
5:43 AM
Q: Linear algebra and notation

maths student Assuming that $a \neq 0$ and $a d-b c \neq 0,$ perform the progression of row operations (in the order indicated) on the matrix $A=\left[\begin{array}{cc}{a} & {b} \\ {c} & {d}\end{array}\right]$. (a) $E_{1}\left(\frac{1}{a}\right)$ (b) $E_{21}(-c)$ (c) $E_{2}\left(\frac{a}{a d-b c}\right)$ (d...

6:15 AM
I'd appreciate an assist with this proof by induction problem I'm working through:
1 hour later…
7:15 AM
Using the definition of a compact set in a metric space, prove that in the Euclidean space R 2 the unit open ball centered at i is not compact.
8:07 AM
Consider any sequence (x_n) in the open disk that converges to a point on the unit circle. Then every subsequence of (x_n) converges to the same point on the unit circle. Hence no subsequence of (x_n) converges to a point in the open disk.
8:35 AM
Is $\aleph_{\aleph_1}$ singular under a model where $\aleph_1$ is singular?
It seems naively one can find that nonconstructive countable sequence that get to $\omega_1$ and then use replacement to map alephs to this sequence to go from $\aleph_0$ to $\aleph_{\aleph_1}$
$\aleph_1$ is always singular
and $\operatorname{cof}(\aleph_{\aleph_1}) = \aleph_1$
ah I see
8:57 AM
btw, it seems the finite cardinals are club wrt $\omega$ because $\sup (\omega \cap n) = n \in \omega$ and for every $n \in \omega$ there is an $m > n$
Mornin' all
actually no, the above is a mistake
finite cardinals are bounded
finite cardinals are not club wrt any cardinals because they each have a maximum meaning they are bounded despite closed.
w is also not club wrt any cardinal because it is bounded (there are no elements in w that are larger than any countable ordinal in w1).
The simplest example of a club set wrt w1 is the set of all countable limit ordinals
9:29 AM
The only club set wrt $\omega$ is itself
9:50 AM
no that is also wrong
the club sets wrt $\omega$ are the cofinal subsets of $\omega$
i.e. any countable sequence S with $\sup(S) = \omega$
10:09 AM
hey guys, good morning!
Lets say we have a linear subspace $F$ of a Hilbert space $\mathfrak H$.
How could I prove that $\bar F\subset (F^{\perp})^{\perp}$ (the closure of $F$ is a subset of the "double" orthogonal complement) implies $\bar F^{\perp} \subset F^{\perp}$?
I've made a type
actually the condition is $F\subset \bar F$ implies implies $\bar F^{\perp}\subset F^{\perp}$
10:50 AM
@RScrlli Just write the definition of what $A^{\perp}$ means. In general if $A \subset B$ the $B^{\perp} \subset A^{\perp}$
@LeakyNun what
@AlessandroCodenotti no?
Not in any ZFC model
ZFC proves that successor cardinals are regular
oh I must have mixed up singular and regular
I meant $\operatorname{cof}(\aleph_1) = \aleph_1$ in any case
Right, that's definitely true in ZFC (but it means that $\aleph_1$ is regular rather than singular!)
11:02 AM
ok thanks
It is consistent with ZF that its cofinality is $\omega$ however
is the main site down?
I cannot reach it
i can access the other stackexchange sites though
It's working fine for me
now its working for me too
11:27 AM
@AlessandroCodenotti Will in singular models $\text{cof}(\aleph_{\aleph_1})=\omega$ ?
because I should be able to find the cofinal sequence of $\aleph_1$ and then apply replacement $x \mapsto \aleph_x$ to turn them into alephs and hence producing a sequence for $\aleph_{\aleph_1}$?
Yes, $\aleph_\alpha$ and $\alpha$ have the same cofinality
When $\alpha$ is an infinite limit ordinal
Hey @Alessandro @Ultradark @s.harp etc.
Hey @ÍgjøgnumMeg
How's it hangin'
11:43 AM
Let me describe it this way:
Imagine a decreasing inconsistent function similar to y=-x, except the slope is zero everywhere
That is, I am failing without noticing I am failing
@ÍgjøgnumMeg it’s going well. Disk model of Euclidean geometry
The circular arcs are geodesics but I don’t know exactly why
You are basically mapping the euclidean plane onto the surface of a sphere while preserving many things such as right angles, hence straight lines become their noneuclidean counterpart: Geodesics
oh okay that makes more sense
Is it a compactification of the euclidean plane?
12:05 PM
Not sure, it certainly looks closed to me since any limit point is the boundary of that circle or its interior
so all those angles are right angles?
yeah, it is a conformal mapping, possibly related to the penrose mapping, at least visually, I don't know how to prove that
12:32 PM
Hi @ÍgjøgnumMeg how's the first week going?
Hi @ÍgjøgnumMeg @Alessandro
Hiya @Alessandro, it's going alright, haven't really done anything yet lol. First 2 weeks of ANT 1 are going to be elementary number theory, and for modular forms we're reviewing Möbius transformations so that's a bit slow too
Hey @Mathein :)
@Mathein Kasten is a really good lecturer though!
Hi @Mathei I saw the unit/counit definition of adjunction for the first time today
yes, he is
really enjoyed his enthusiasm lol
and him sliding around on the floors
Oh and, as it happens, we are doing Hecke theory in the modular forms vorlesung
12:36 PM
so the script is not complete yet?
He just hasn't written it in the Skript yet lol
we'll study Eisenstein series "instead of" theta series
or theta functions or whatever they're called
theta seris and theta functions are both a thing
Ah okey, well in any case Kasten said we'll just focus on Eisenstein series since they're somehow more interesting for him personally ol
Btw @Alessandro or @Mathein what is a Plenarübung?
@ÍgjøgnumMeg it's an optional lecture in which typically the assistant gives examples, background, answers questions etc.
12:51 PM
Ah okey thank you
1 hour later…
2:01 PM
@ÍgjøgnumMeg hi igjog^
2:14 PM
Hullo @s.harp :)
2:24 PM
how is heidelberg?
It's nice :P although I'm actually living in Mannheim, which is less nice hehe
do you know the manhattan metric? here in analysis courses we teach the mannheim metric instead :P
Hahaha nice
Thinking about the philosophy of infinity again (don't roast me, I have actually read the prerequiste text on this topic months ago) :
Let a be an object
Define some partial ordering so we can talk about $\leq$
Let b be an object
Then a potential infinite of $a$ is if for every $c \geq a$, there exists a $d$ such that $d \geq c$
(need to think about how to phrase this better, because $\forall c (c \geq a) \implies \exists d (d \geq c)$ is supposed to be a schema that is iterated until the condition is no longer met)
Thus a potential infinity $p$ is any mathematical object where it is preserved under the operation of upward extension.
2:52 PM
Let $w$ be euclidean space. Let $i$ be a sphere
$b$ is an actual infinity wrt $a$ if there exists at least one potential infinity $p$ of $a$ such that $b \geq p$
Let $Q$ be an embedding of $i$ into $w$
We can now denote the above schema be $\subseteq_{\to}$ which is a procedure that is interpreted this way:
@AlessandroCodenotti why do I feel like in ZF we still have $\operatorname{cof}(\aleph_1) = \aleph_1$?
Given any input $x$, $\subseteq_{\to} (x) \implies \subseteq_{\to} (y) \geq x$
2:59 PM
Let $Q$ also be isometric
This is a while loop, and will not exit until the implication fails to hold
Thus $p$ has the property that $\subseteq_{\to} (p)= p$
@AlessandroCodenotti never mind I found this: mathoverflow.net/a/147092/91375
Let $(NG)_p=Q$ be conformal
Now the program can give a couple of possibilities:
We knew from the above definition, a potential infinity $p$ is something that is left invariant by the program and the program does not terminate in any part of its run
acutally, screw that, starting over
oh so $p$ is sort of like a projection of infinity onto the sphere?
3:08 PM
@LeakyNun I don't know why you feel like that, but it's not necessarily the case :P
In fact ZF cannot prove that there is a single uncountable regular cardinal (but this is a result way harder than showing that $\aleph_1$ might be singular)
Let $a$ be an object
Now equip anything that is to be constructed with a partial order $\leq$
Let $C$ be a partial order with objects $c \in C$
Let $P$ be a program defined as follows:
1. $PC$ : Randomly pick $c \in C$, do $Pc$
2. $Pc \implies c \leq Pd$
3. Step 2 is terminated if no such $d \in C$
4. If step 2 terminates, repeat steps 1-3 by picking another $c$
5. If at any point $P$ runs without need to switch to another step, then output: $C$ contains a potential infinity and hence it is actually infinite
6. else if $P$ terminates after it has tried every $c \in C$, then output $C$ is finite
3:24 PM
How does one define a filtration on a graded module $A = \{A_n\}_{n \in \mathbb{Z}}$?
In my head I'm just thinking "Well it's just gonna be a collection of filtrations of each of the modules $A_n$" but I could be wrong
@Perturbative or more category-theoretically, a collection of graded submodules (indexed by a totally ordered set satisfying the order condition)
$P$ thus define a procedure for checking tarski finite partial orders
We can upgrade $P$ with a new step between 4,5 to tell between tarski finite and finite:
4.5. If for every $c$ tried it terminates at some value $m$, output $C$ is finite
Thus we now have a recipe for generating a variety of infinite sets:
Steps 1-2 will propagate a partial order starting from the randomly picked element $c$
Then the following checks will modify the behaviour of $P$
1. there is no $d \in C$: Output "maximal element encountered". This will loop back to steps 1-2 on another $c$
2. Steps 1-2 is never interrupted given the $c$ picked: Output "potential infinity wrt $c$ detected". This will loop back to step 1-2 on another $c$
3. All possible $c$ give condition 1: Output "$C$ is bounded"
4. condition 3 and the final value before condition 1 is encountered are the same: Output "maximum detected"
(typo: 3 should be Tarski finite)
3:59 PM
Q: Triple integrals with polar coordinates.

cdummieI have a following integral: $$\int_0^1 dx\int_0^{\sqrt{1-x^2}}dy \int_\sqrt{x^2+y^2}^\sqrt{1-x^2-y^2}z^2dz$$ Which i have to solve by introducing polar coordinates, which is, by itself, relatively simple: $$x=\rho\cos\theta\sin\phi \\ y=\rho\sin\theta\sin\phi \\ z=\rho\cos\phi$$ Besides this,...

How do you know u have to use spherical coordinates here not cylindrical
5. step 1-2 revists an element: Output "cyclic order of period <number of steps 1-2 for that c> detected"
6. step 1-2 never revists an element, but the elements obtained are distributed between a pair of lower and upper bound, Output "aperiodic bounded order of length <number of steps 1-2> detected"
ok screw that again, should have put a different order
1. If $Pc$ maintain its run all the way through without switching steps, output "potential infinity detected, $C$ is not strictly finite". Pick another $c$ and recompute $Pc$
2. If there is no $d \in C$ for a given $Pc$, output "maximal element detected". Pick another $c$ and recompute $Pc$
3. If Condition 2 occur for all $c \in C$, output "$C$ is tarski finite"
4. If Condition 3 occurs and the value that triggers Condition 2 are the same, output "maximum detected"
5. If $Pc$ maintain its run all the way through without switching steps, and some elements in that run are revisted, output "cyclic order of length <number of steps ran for $Pc$>" detected
6. If $Pc$ maintain its run all the way through without switching steps, no elements are revisited, and the whole output of $Pc$ has both a upper and lower bound, output "aperiodic cyclic order detected"
4:22 PM
@LeakyNun Actually I think that is a nicer perspective to view it, thanks
@Perturbative you're welcome
ok, need to try again, I messed up the definition of tarski finite
But really, not even ZF allow a set that is tarski finite but not finite?
hmm, I guess that might have something to do with how every infinite set must have a finite subset
Q: Proof of: $X$ is finite $\iff X$ is Tarski-finite

aerdna91I am self-studying Horst Herrlich, Axiom of Choice (Lecture Notes in Mathematics, Vol. 1876). In the fourth chapter, he deals with different definitions of finite set. Here is the classical one: Definition 1. A set $X$ is called finite if there exists a bijection $n \to X$ for some $n \in \ma...

Yeah, PA ensures that
I always found it redundant when the same concept is given many different names
and I always thought tarski finite is weaker than finite because "names are cool"
Let $C$ be a partial order of objects, with elements $c \in C$
Let $P$ be a program that does the following when $PC$ is executed:
1. Randomly pick a $c \in C$ that is not picked in the previous iteration, compute $Pc$
2. If $d \in C$ such that $c \leq d$, compute $Pd$
3. After a complete run for each $Pc$, refer to checklist.
4. Repeat Steps 1-3
(Very important assumption here. $P$ is not a turing machine. The best way to understand it is it can run for absolute infinite amount of time and steps, thus it can terminate for any transfinite value of time and steps thus there is no infinite loop in the usual sense of the term. I have yet to figure out a way to drop this)
1. If the current $Pc$ maintain its run without switching steps, output "potential infinity detected, $C$ is not finite"
2. If the current $Pc$ terminates due to inability to fulfill step 2, output "maximal element detected"
3. If all $Pc$ fulfills Condition 2, output "$C$ is bounded by maximal elements"
4. If all $Pc$ fulfills Condition 3 and the element that triggers termination is identical, output "$C$ has a maximum"
5. If the current $Pc$ fulfills Condition 1 but some elements are revisited, output "cyclic order of period <number of steps before each revisiting> detected"
6. If the current $Pc$ fulfill Condition 1, some elements are revisited irregularly, output "quasiperiodic cyclic order detected"
7. If the current $Pc$ fulfills Condition 1 and no elements are revisited, and that the output has both a upper and lower bound, output "aperiodic cyclic order detected"
8. If all $Pc$ fulfills Condition 1, and none of these have length equal to $C$, output "$C$ is dedekind finite". Otherwise output, "$C$ is dedekind infinite"
(actually no, I need to rule out countable subsets)
8. If all $Pc$ fulfills Condition 1 or 2, and none of these have the same length if one is a subsegment of another, output "$C$ is dedekind finite", else output "$C$ is dedekind infinite"
9. If all $Pc$ and its subsegments fulfills Condition 3, output "$C$ is tarski finite"
10. If all $Pc$ fulfills Condition 9 and each subsegment contain subsegment of length 1, output "$C$ is finite"
5:13 PM
Hello peeps, I'm interested in functions of form $\sum_1^n \frac{\exp(ak^2+bk)}{k+a}$
I found the Jacobi Theta functions (number 3 seems good) but it's for an infinite summation. Where should I be looking?
(Here I introduce a new kind of set call an indecomposable set. They are sets of some cardinality that taking its powerset only gives itself. Still need to work out how it will mess up set algebra. Bijections between these sets with other sets are done in an abstract fashion. Bijection can occur if they have the same cardinality, despite no element of such set can be demonstrated)
I can define my own function but I'd rather use something that also exists in the literature
@Secret there are no sets which don't get a larger cardinality from taking powersets
@1010011010 Why do you want to study those functions?
Anyway, it appears that starting with a computer that can compute indefinitely (and hence all transfinite steps will terminate), it is much easier to check for an infinite thing compared to a finite thing. Likewise, in real life computers, where the number of steps are bounded by a finite number, it is much easier to check for finite thing compared to an infinite thing
My suspicion is that, given any proof that start with an infinite thing, it is much easier to check whether something is infinite more than whether something is finite, and the converse is true if the proof has some notion of finiteness in it
hence there seemed to be some kind of correspondence between an infinite and a finite prover
NB An infinite prover is a program that proves things without assuming any in built notion of finite. In particular, it does not have a notion of natural numbers, only whether the job is completed with yes or no
In contrast, a finite prover is your typical turing machine, which has bounded memory and time, thus if it get stuck in an infinite loop, it cannot output anything
So it seems, even if in an alternate history, we humans knew more about infinity than finite mathematical objects, it seems to take quite an effort to define finiteness
analogous to how in this history where we are more familiar with the finite, it is very hard to prove most theorems involving infinity
5:43 PM
@TobiasKildetoft I'm studying the interacting 1d bose gas and this summation popped up in the context of wave packets
It's a physics thing, I'm not sure if there's any point in me explaining much further
not to me :) But some people here might have an idea also based on the physics.
go to h bar for physics questions, there are more physicists there

 The h Bar

General chat for Physics SE (physics.stackexchange.com). For M...
But the question is not physics related is it?
I'm asking about a summation identity
well you said your context is bose gas, so it seems more likely that statistical mechanics related users will knew it better. Otherwise I don't know of anyone right here who deal with summation identities
Fair, I've also posted there. Cheers
5:55 PM
Consider the equation $0=x+2y+z+e^{2z}-1$. If one solves this equation $z=z(x,y)$ at a point $P$ and tries to find the taylor expansion of $z$ at $P$, which will look like $z=Ax+Bx+Cx^2+Dxy+...$ (1), how does one solve for $x,y$ when plugging (1) in the equation? The $e$-term seems to cause some trouble here.
As a university student of Maths in England, is it useful to buy either Wolfram Pro or Mathematica? And if so, which?
don't they give you free access to mathematica or something
Don't buy any software as a student
Either use whatever your university has a licence for for you, or get an open source version
6:28 PM
Why is the "basis vector field" $d/(dx_i)$ in Minkowski space constant and covariant (I assumed the basis vectors in a tangent space are contravariant)? Or is there such a thing as “covariant constant”?
@ÍgjøgnumMeg Hey
I have a question about substitution principle in ring theory
6:55 PM
@eigenvalue: Yes, vector fields are contravariant tensors. "Covariant constant" is the phrase to say that the covariant derivative (i.e., the derivative using the connection) is zero. Nothing to do with co- and contra-variant.
7:46 PM
@AlessandroCodenotti you are familiar with measure theory, right?
Or were you a logician?
8:03 PM
There are things I'm way more familiar with but I know some measure theory, what's the question? @anakhro
I just wanted to work something out in words with someone familiar with the topic.
In particular if $\Delta = [0,1]\times [0,1]$ is the diagonal, then I want to compute $\int 1_\Delta(x,y)\,d\mu(x)$ where $\mu$ is the Lebesgue measure.
This is of course 0.
But I wanted to get the nitty gritty.
Is that $d\mu(x,y)$?
Just wrt x.
So you want to say that it is $0$ regardless of $y$?
Basically that integral is only being done across the x, so that means that $1_\Delta(x,y)$ is fixed in $y$. Meaning that it is basically the indicator function $1_{\{y\}}$, right?
8:07 PM
And since integrating the indicator function of something measurable is the same as taking its measure you're done
So all in one line it is:
$$\int 1_\Delta(x,y)\,d\mu(x) = \int 1_{\{y\}}(x)\,d\mu(x) = \mu(\{y\}) = 0.$$
That looks right?
That is, there are no notational problems, and I am thinking of it the right way?
Looks good to me
Great, thanks!
You're welcome!
@AlessandroCodenotti what sort of math have you been reading/doing lately?
8:13 PM
Incidentally some set theory which is somewhat measure theoretic. I'm going to give a seminar talk about a strong form of Fubini's theorem which is independent of ZFC
Also some coarse geometry
@TedShifrin thanks for the clarification!... this is what I suspected but the phrasing threw me off a bit.
Nice! Incidentally this calculation is part of showing Fubini-Tonelli fails for non-sigma-finite spaces.
Just using the counting measure for the y part.
Ah, makes sense
@AlessandroCodenotti so what exactly is this strong form of Fubini's theorem?
Suppose that $F\colon\Bbb R\times\Bbb R\to\Bbb R$ is such that $F_x$ and $F^x$ are measurable for almost all $x$ (those are $F$ with the first and second coordinate fixed) and such that $G(x)=\int F_xdx$ and $H(y)=\int F^ydy$ are both measurable. Must we have $\int G(x)=\int H(y)$?
Basically we don't ask that $F$ is measurable, but instead minimal assumptions to be able to write a double integral as in Fubini
8:23 PM
And so there are indeed a set of minimal assumptions?
Well this theorem is independent of ZFC
There are models where it is false (this is easy to show, it fails in every model of CH for example) and there are models where it's true (this is hard to show, it's a result published in 1980 by Friedman)
The "theorem" is that $\int G(x) = \int H(y)$?
Ah, I see.
Do measure theoretical things come up frequently in the context of logic?
Depends on the kind of set theoretical issues one wants to think about
But there are many measure theorical problems studied in set theory
8:35 PM
More so than other areas in analysis?
Yes I'd say so
Pretty interesting. But I guess that sort of makes sense. Topology also comes up in proof theory sometimes.
I was thinking more about set theory than logic (I actually know very little logic)
Set theory falls under "foundations" right?
8:43 PM
Okay, when I say "logic" that is what I have in my head that I mean.
Sorry, I guess not all of foundations is pure logic. :P
@MikeMiller Wow, there's a whole swath of these combinatorics/topology people who seem to like pictures in topology!
On that note, maybe you can explain the opening paragraph here to me
Hi @Balarka
Hi @Alessandro, @anakhro
Hi @BalarkaSen
Combinatorics makes the world go 'round.
@BalarkaSen What can I add t it
9:00 PM
1) What is the notion of $H_1$ that they speak of? It seems to be invariant under surgery? 2) What is an $L$-space? You told me once but I forgot
that's singular homology bro
If you take a knot in $S^3$ and do surgery with slope $p/q$, you get $H_1$ of the result is $\Bbb Z/p$; this follows from MV
$|H_1(S^3; \Bbb Z)| = 1$? Oh by $|.|$ they mean cardinality
@BalarkaSen In set theoretic topology an $L$-space is a regular, hereditarily Lindelöf but not separable space, but I guess there's another concept with the same name lol
I thought rank or something so didn't understand
@MikeMiller Yes, OK, this is clear. Should have guessed it from there.
2) The hat version of Heegaard Floer has $\chi(\widehat{HF}(Y)) = |H_1(Y;\Bbb Z)|$, where if that cardinality is infinite we return the number 0. So in particular we have the rank requirement $\dim \widehat{HF}(Y) \geq |H_1(Y;\Bbb Z)|$. An L-space is a rational homology sphere for which you have equality there.
Because $S^3$ is so simple its Heegaard Floer complex has 1 generator and no differential, so it is an L-space. Same with any $L(p,q)$, whence the name; but also connected sums of L-spaces are L-spaces.
The L-space conjecture provides a conjectural description of which 3-manifolds are L-spaces
9:06 PM
i see
I can add clarifying details if you want, but I don't know what to say
I do want to know more but it might be out of my league; I'll think and ask some questions later. Do you know a description of the Poincare homology sphere as a surgery along some knot in $S^3$?
(Should I be able to guess it?)
It's +1 surgery on one of the trefoils, I forget which one
i also forget why
Oh yeah I was wondering that
Trefoil has fundamental group $\langle x, y | x^2 = y^3 \rangle$, right?
I want to add $(xy)^5 = x^2$ as a relator, or something
The Wirtinger presentation is $\langle a, b | aba = bab \rangle$. Hm, what's $x$ and what's $y$
@BalarkaSen Hi topology guy , can you recommend an intro book to topology ?
9:19 PM
I think the meridian is $a$ in the second presentation but I forget in the first
Kinsey's "Topology of Surfaces" is a good introduction, barring that it gets the definition of a topology rwong
what does a conformal ball model of euclidean space look like? (euclidean space compactified into $D^3$)
@MikeMiller thanks
does wolframalpha.com/input/… mean there is never a real solution?
how can I solve $\lambda_1 e^{y - \lambda_1 e^y} (1 - \lambda_1 e^y ) + \lambda_2 e^{y - \lambda_2 e^y} (1 - \lambda_2 e^y ) = 0$ ? $\lambda_1, \lambda_2 > 0$
I am not sure where to start
wolfram alpha didn't seem to help or at least I didn't get it to help
My strong suspicion would be “you don’t”, except numerically
What is a Weyl chamber?
9:32 PM
If there’s any hope of an analytic solution, I’d start by substituting x=e^y
But I don’t hold out much optimism
@Semiclassical I really just want to know how many solutions it has. Is that easier to determine?
I believe it sometimes has one and sometimes has two
@MikeMiller Oh, ok. $x = aba$ and $y = ab$ works, because $x^2 = aba\cdot bab = (ab)^3 = y^3$. There should be a way to interpret this in terms of the $(2, 3)$-symmetry of the trefoil?
I just want to know under what criteria does it have one or two solutions
I kept trying stuff with inverses, good catch.
I’m dubious you’ll find a nice description
9:36 PM
@Semiclassical hmm...
ideally, though, what you seem to want is this: Consider the 2D plane of lambda1,lambda2 values
For what region of that plane do you get two solutions? One solution? No solution? etc
$x^2 = y^3$ should just be the center circle of the essential annulus in the complement of the trefoil (think of it as a torus knot)
I expect that can be studied numerically (the argument principle in complex analysis may provide a method)
So $y$ is the meridian and $x$ is the core circle of the torus the trefoil is embedded in, I think?
Yeah, that's what $y = ab$ is in Wirtinger.
9:54 PM
I think all of the generators of Wirtinger (a and b in this case) are a meridian of the knot up to unbased htpy, by defin
Good point
I should draw some picture tomorrow to understand if my interpretation is coherent and why $(xy)^5$ is the parallel curve to the knot
That's what we want to do, right? Glue a 1-handle in such that the meridian disks paste to the parallel curves to the trefoil?
These are the curves of intersection of a Seifert surface of the trefoil with an epsilon-torus around that knot, I mean to say
Any insights are appreciated: math.stackexchange.com/questions/3395278/…
10:11 PM
@schn WolframAlpha gives an answer in terms of the exponential integral; wolframalpha.com/input/…
Which is probably just a matter of the right choice of u-substitution
@Semiclassical Okay. Would there be a way to integrate f’(x) (as given in the question)? That is, f’(x) with the $e^{-x^3}$ terms. The range is supposed to be (0,ln 2), so maybe there is some power series solution to it as well...
10:35 PM
@schn what you get for f’(x) is this: wolframalpha.com/input/…
I’m not sure what you’re hoping to get tho
I guess what you’re after is the behavior here: wolframalpha.com/input/…
So if you can show that f’(x)<0 for all x>0, then f(x) is decreasing for positive x
And evidently f(x)->ln(2) as x->0
10:54 PM
@Semiclassical Cool. How does one evaluate, say Ei(x) for any x? It is somewhat unfamiliar territory, although the gamma function is a familiar function. Are they related in some way?
back in the day, you’d look up a table of numerical integrals
Nowadays WolframAlpha is a familiar enough resource
If you want to compute it yourself, you could use Riemann sums
I don’t know how the old tables would have compiled the values. Probably involved lots of slide rules
Okay, so there probably isn’t any explicit expression for f(x) then?
No one would ever use Riemann sums. Perhaps Simpson's rule ...
Not besides Ei itself, no
Yeah, using Riemann sums as such would be a waste
@Semiclassical Okay, so if one doesn’t have Wolfram, one would look Ei(-x^3) up in a table? There is no way that the gamma function could be of any help?
11:08 PM
Not really, no. That said, you don’t need to compute any values of Ei to show that f(0)=ln 2
@schn the gamma function is $\int_0^\infty$
@Semiclassical How do you conclude f(x)->ln(2) as x->0?

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