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00:56
Where can I learn more about computing series and conversely what their partial sum formulas are?
Life is pain relying on Wolfram Alpha
See I previously asked about what the use of a generating function for a sequence is until I realized you can use it directly to compute that value if you have a partial sum formula for it.
Furthermore, my research has led me to conclude that there is nothing further I can do to compute the integer coefficients I've been looking for without a relation for the integer and fraction parts of the logarithm. It must exist since, as previously noted, the fractional part contains the identity of the odd part.
It's easy enough to see that for some odd number $x$ normalized by some integer number of factors of an integer $p^{\lfloor\log_p(x)\rfloor}$ as $y = p^{-\lfloor\log_p(x)\rfloor} x$, it's easy enough to see that $$p^q = \sum_{n=0}^{\lfloor\log_p(x)\rfloor} \lceil\{p^n y\}\rceil$$
Pain. I changed plan in the middle of writing and forgot to update things
$$\lfloor\log_p(x)\rfloor = 1 + \sum_{n=1}^{\lfloor\log_p(x)\rfloor + k} \lceil\{p^n y\}\rceil \land k\geq 0$$
For some reason I thought that just works with $n=0$, but that would only be the case if the number of factors were one more than $\lfloor\log_p(x)\rfloor$.
Literally just adds one while the fractional part of $y$ is non-zero.
 
2 hours later…
03:02
@HashNuke if you let $E_{a,b}$ denote the (bounded) linear functional on $C[0,1]^2$ given by evaluation at $(a,b)$, see if you can show that $S = \bigcap_{(a,b) \in [0,1]^2} \ker(E_{a,b} + E_{0,0} - E_{a,0} - E_{0,b})$
03:18
Wow. Isn't that sneaky.
03:38
@leslietownes any idea for L^2 functions?
i dunno. certainly this C[0,1] idea doesn't work anymore, because we don't have "evaluation at (a,b)". i wonder if you can express the condition in terms of averages or something though. like lebesgue's differentiation theorem.
I expected L^2 to have similar answer
process
03:56
beats me. can you "identify" in some sense the orthocomplements of each "L^2[0,1]" in L^2[0,1]^2?
Does the functional $\| f(x, y) + f(0, 0) - f(x, 0) - f(0, y)\|_2$ work?
well, i don't know what "f(x,0)" is for an element of L^2. you can change what f is on the axes without changing its L^2 equivalence class.
but, per my vague notion above, maybe you can write down some things that would average to what the result would be if you have an element of L^2[0,1]^2 that is a sum of L^2[0,1] functions.
If $f(x)\leq g(x)$ in an interval, $[a,b]$ such that both $f,g$ are differentiable in this interval, then, $\int_a^bf(x)\leq \int_a^b g(x)$ : Is this assertion/lemma true in general?
This is because, I used this lemma ( I actually, figured this out intutively ) to solve a problem...
shouldn't this come from definition of Riemann integral/Lebesgue integral?
yes. franklin, i think we've even discussed this result specifically before.
04:10
@HashNuke might be, but I dont know, about them. So, I can't answer this...
Mar 9 at 6:36, by leslie townes
early in riemann integration, yoou might learn the definition in terms of riemann sums, which makes clear that if h(x) >= 0 on [a,b] then the integral from a to b of h(x) dx is >= 0
@leslietownes exactly! To be elaborate, I felt and I am sure, we did. I used to note down everything in my notepad, but now the device is damaged unfortunately. So, I couldn't access it. Can you provide me a way, to access those chats ?
there should be a clickable link in the line above [the timestamp]
Ok, trying...
"well, i don't know what "f(x,0)" is for an element of L^2." could you please elaborate it?
04:22
mm, maybe start with single variable functions. i don't know what f(0) is for an element of L^2[0,1]. L^2 is variously defined, but is often realized as some kind of quotient. roughly speaking, you need to 'mod out' by functions having zero 'norm,' or the 'norm' you'd like L^2 to have isn't a norm. this means identifying some functions when they disagree on a set of measure zero, and so you lose the ability to evaluate an element of L^2 at points, which are examples of sets of measure zero.
now, once you start "evaluating" a function at multiple places and adding up, eventually you can get to things that have meaning for elements of L^2, for example, integrals of f against other L^2 functions make sense.
Oh ..
and right now that idea you're having is somewhere in the middle.
hence my vague thoughts about averaging, integrating functions over small sets. that vibe might point toward some recipe that will kick out an 'f' and 'g' for which h(x,y) = f(x) + g(y), at least when h is assumed to have that form.
but it's more complicated than in the continuous case, where we could just write that out and have it make sense in terms of point evaluations.
sorry if this is vague, im trying not to be too precise because my wife and daughter both have persistent coughs right now, to the point that it's impossible for me to think straight. it's just coughing, coughing, coughing.
the cat was even meowing a minute ago, i think because the coughing confused her.
things never come alone
thank you so much for your time
its a cool question. i'm wondering when i get the cough.
in a way this is revenge, usually it's me who is persistently slightly sick.
How do you get creative? How are people creative in general?
04:31
@leslietownes Oy.
@leslietownes Thanks a lot! You're a lifesaver! Problem resolved...but how did you get that time-stamp so quickly? But again: Thank you!
I announced earlier that I now have Covid, after 3 years …
ted! no!
2
The new normal …
we got munchkin boosted yesterday.
04:36
@leslietownes Ohh...I got the trick! There's an option here "Search for message containing"....
shhh, don't tell everyone about that.
04:53
It seems that $\| f(x, y) - f(x, 0) - f(0,y) + f(0, 0)\|_2$ does not work at all. I cannot show this is continuous at $0$ afterall.
@AMDG you can view the Zeta function as a mellin transform of a certain theta function. This theta function satisfies a functional equation which allows you to get a functional equation for the Zeta function that has a symmetry about the critical line. The gamma factor $\pi^{-s/2}\Gamma(s/2)$ in the functional equation comes from the mellin transform of $e^{-\pi*x^2}$ (which is it's own fourier transform). Tate's thesis shows that the gamma factor is in some sense an optimal choice for this fact
@TedShifrin Get well soon
*factor
05:08
Get well soon :)
 
1 hour later…
06:27
@TedShifrin Get well soon Professor
Is there a language that is useful other than English? Especially for math
something like yaiphah's idea ought to work. the "L^2" subspaces are not quite orthogonal, but the orthocomplements of their intersection in each one of them are, unless i'm missing something. which is the kind of relative position that feels like it oughta be enough.
one potato, "useful" is such a subtle concept, haha. for printed math, historically, german, french, and russian were 'big' languages, and still are, although which one is the most useful might depend on your subfield.
or more historically, latin. :D
and that's historical, i dunno about today. even within math, depending on field, mandarin or japanese might be more important.
outside of math, mandarin is probably the most useful language by many metrics.
06:48
I see. Thanks
How to solve this problem?
I've drawn venn diagrams and got that conclusion II is correct. But answer is given that Neither conclusion I nor II follows. Why?
in depicting the a "general" relationship venn diagrams aren't always good at depicting all possibilities, in particular, where some of the intersections are empty, or something is strictly contained in something else.
they're also fiddling with your mind by using names for real life things that you might have intuition about, in place of abstractions like "some A are B" "some B are C"
although i think they want you to answer as though they're asking about abstractions.
those two statements would be true in a world with exactly one tablet, that also happens to be the world's only laptop and the world's only computer. and II isn't true in that world. so it isn't an abstract consequence of those two statements.
i haven't seen the diagram, but if you do draw one, II is probably an assertion that one of the 'regions' depicted has to have something in it. when it fact it could be empty.
I've drawn these two venn diagrams.
According to the first venn diagram, conlusion I is wrong and conclusion II is right.
And according to the second venn diagram, both conclusions are right.
the second venn diagram isn't a picture of everything that can possibly happen consistent with the given information. or at least, not without the understanding that some of those shaded regions could have nothing in them.
in that second diagram, the center - the part that all the circles cover - could be everything. there could be nothing else in any of the other regions, even though they appear in your drawing.
07:04
This is also possible
yes. and note that this diagram is either implicitly assuming that there are T's that aren't L's, and L's that aren't C's, or we have to look at it and remind ourselves that maybe there's nothing in those outer two loops, though the diagram is able to capture that possibility.
Yes.
So what should be an apt way to solve this problem?
your picture does it. the first piece of information tells you that the blue region is not empty. the second piece of information tells you that the yellow region is not empty. now visualize what conclusions I and II are asking you to say about the diagram.
"some tablets are computers"? well, that certainly could happen, if the green region isn't empty. but i wasn't told that. and it could be there's some stuff in the pure blue, and the pure yellow, but nothing in the green.
"some laptops are not tablets"? that's asking me to deduce that the pure-yellow region is not empty. but i wasn't given that either. it could be there's only stuff in the green region, or some stuff in the green and some stuff in the pure blue, and nothing else.
Some laptops are not tablets -> It is true I think. There are some laptops which are not a part of tablets.
it's consistent with the given information that the only nonempty region is the green one. that was my first example, the world where there's only one laptop, one tablet, and one computer, all the same one thing.
07:14
Ohh yes.
But it is still confusing.
the venn diagram, in being capable of expressing all of the possibilities, is sometimes not that good at suggesting differences between different possibilities.
you have to mentally imagine some of those regions being empty, or collapsing, or swallowing up the whole diagram.
Okay
i don't think of venn diagrams as very useful general tools. they can sort of work when you have two or three conditions, but as we're seeing here, even then there's some mental effort. at least, i'm not a very visual thinker.
and once you have five or six conditions, forget it.
08:02
@leslietownes 😂😂😂...
 
4 hours later…
12:17
@robjohn I think I have figured it out. You change the order of summation and then get the telescoping sum from the previous series (i.e. from $\sum_{k=1}^\infty\left(\zeta(2k)-1\right)$), however, with a $(-1)^j$ factor in front. This is also a telescoping sum. The answer turns out to be $1/4$.
I had doubts about the change of summation though, since our terms have different signs. I guess this is only possible if $\sum_{I} |a_{jk}|<\infty$, where $I$ in our case is $\{1,2,\ldots\}\times\{2,3,\ldots\}$. Luckily we have $\sum_{I} |a_{jk}|<\infty$...we showed this in the previous sum (i.e. $\sum_{k=1}^\infty\left(\zeta(2k)-1\right)$), so all is good.
@schn How does that turn out to be $\frac14$? There is something missing.
Give me a second.
$$\begin{align}
\sum_{k=1}^\infty\sum_{j=2}^\infty\frac{(-1)^j}{j^{2k}} &=\sum_{j=2}^\infty\frac{(-1)^j}{j^2-1}\\ &=\frac12\sum_{j=2}^\infty(-1)^j\left(\frac1{j-1}-\frac1{j+1}\right)\\ &=\frac12\left(1-\frac12\right)\\
&=\frac14
\end{align}$$
@robjohn
that looks okay
which is $$\sum_{k=1}^\infty (1-\eta(2k))$$
12:49
~figured it out~
13:12
in my local community, people keep discussing 'whether is there any relation between (undergraduate) GPA and research ability?' Well... 'yes almost surely' is the correct answer isn't it?
correlation, completely since research admission is constructed so as to make it so (with reason)
Full list of statements:
Some tablets are laptops.
Some tablets are not laptops.
Some laptops are computers.
Some laptops are not computers.

The conclusions here form syllogisms with the above. Not sure what the diagrams are for.
Though I'm curious to know what a dialectics question is doing in mathematics.
13:36
Are there any good calculus 2 textbooks where I can practice sequences and series?
@PrithuBiswas have you checked stewart?
@shintuku No.
his textbooks are used a lot in calculus classes
@PrithuBiswas e.g. in the 2020 edition of Calculus: Early Transcendentals, there's an intro to sequences and series at chapter 11
it's only an introduction, so the sort of stuff you'd see in a calculus class
13:51
@shintuku Thanks for your suggestion =)
14:40
I have quick question, if $X$ is TVS and and $M$ is subspace, does $M$ need to be closed for $X/M$ to be TVS?
uh, might be a matter of definitions, do you require TVSs to be Hausdorff?
yeash ...
but in general does it require?
15:00
how do I proceed further
Mad
Mad
15:24
Good day. a physics student here, i am trying to learn some basic Wave theory, and i am trying to learn it mathematically. it is very hard to come by any material that does not skip over the mathematics . i have been trying to define the "Wavelenght" of a periodic function, mathematically
Suppose F is a periodic function with period T . how can we "mathematically" define the Wavelenght of this Function?
@Mad you could try distance between two extrema
Mad
Mad
i have been thinking about using some metric
@Mad I am not a physician (I took a semester of physics in high school and a quarter of quantum mechanics in grad school), but "wavelength" is usually the "fundamental period" of a periodic function.
That is, the wavelength of $f$ is $\min\{ T : f(x+T) = f(x)\}$, assuming that this minimum exists.
Mad
Mad
Suppose that a function F fullfills F(x) = F(x+T). then in this case it woul dbe true that $\lambda = T$
@Mad This is not quite right. The wavelength of $\sin$ is $2\pi$, but $4\pi$ satisfies your criterion.
Mad
Mad
15:27
However
the example you mention is not accurate
Unless you add the criteria (For all x)
Then it would fit i believe
@Mad Who are you replying to, here?
Mad
Mad
To you
Since for example, the Sinus function repeats itself before one period.
Then according to your defintion the wavelenght will be less than 2pi
@Mad ?
I was lazy, and left out a quantifier.
If you want to be pedantic, $\lambda = \min\{ T : \forall x, f(x+T) = f(x)\}$, assuming that this limit exists.
Mad
Mad
15:30
This makes alot of sense
And i was thinking something in this direction
however is the word limit accurate here
@Mad I don't understand...
What do you mean by "is the word limit accurate here"?
Mad
Mad
I fail to see the "limes" in your statement. but i suppose you mean if the Value exist.
@Mad No, there is no limit in my statement. There is not meant to be a limit.
It is a minimum, should that minimum exist.
Mad
Mad
Yes.
There are periodic functions which do not have a fundamental period.
Mad
Mad
15:34
what is meant by fundamental period?
But this has nothing to do with limits.
@Mad I gave you the definition above. It is $\min\{ T : \forall x, f(x+T) = f(x) \}$.
Mad
Mad
so this is called "fundamental". What is "none fundamental"?
@Mad What do you mean "none fundamental"? I haven't used this phrase, nor have I seen it used elsewhere.
Mad
Mad
That is true. I deducted that from your statement
You mentioned there exists functions that do not have a fundamental period but are periodic
Mad
Mad
15:37
I am not aware of such objects, can you perhaps deliver an example ?
For example, the characteristic function of the rational numbers. This function is defined by $$\chi_{\mathbb{Q}}(x) = \begin{cases} 1& \text{if $x\in \mathbb{Q}$, and} \\ 0 & \text{if $x \not\in\mathbb{Q}$.}\end{cases}$$
This function is $T$-periodic for any positive rational number $T$.
Mad
Mad
Thats very interesting example
Thus $\min\{T : \chi_{\mathbb{Q}}(x + T) = \chi_{\mathbb{Q}}(x) \}$ does not exist.
Mad
Mad
Thank you for explaining!
@XanderHenderson Does the constant function have a fundamental period?
15:43
@PrithuBiswas No.
Not per my definition.
@XanderHenderson Oh ok.
16:06
@TedShifrin: Sad to hear that. Get well soon!
16:18
@TedShifrin i sent you positive energies but they are powerful, beware
@Koro @shin Thanks. It’s not too bad (not like @leslie’s family).
@XanderHenderson That’s a terrible definition. Then every function has period $0$ (or $-\infty$?).
You mean $\inf$ of the set of positive periods :)
@TedShifrin Positive $T$.
And I mean minimum, not $\inf$.
So your char fn of the rationals has no period.
Fine.
@TedShifrin It has not fundamental period.
It has period $T$ for any positive rational $T$.
1 hour ago, by Xander Henderson
This function is $T$-periodic for any positive rational number $T$.
Right. So fundamental applies only when you have a discrete subgroup.
16:39
@TedShifrin Yes. That is the way I prefer to define it.
I’ve actually never given such a definition, but it’s fine with me.
Is it correct to say that if $a_n>0$ and $\sum_{n=1}^\infty a_n$ converges, then $\sum_{n=1}^\infty \frac{1}{a_n}=+\infty$? I think yes: the convegence of $\sum a_n$ implies that $\lim_{n\to+\infty} a_n=0$ and so if $n$ is big then $|a_n|<1$. Hence, since $a_n>0$, if $n$ is big then $\frac{1}{a_n}>1$ and so, since $\lim_{n \to +\infty} a_n$ exists, we have that $\lim_{n\to+\infty} \frac{1}{a_n} \ge 1$. So $\frac{1}{a_n}$ does not tend to $0$ with $a_n>0$, hence $\sum \frac{1}{a_n}=+\infty$.
how much measure theory do i need for: (1) functional analysis (2) probability theory (3) fractals? should i just do the entire course?
Or even: $\frac{1}{a_n}>1$ implies $\sum_{n=1}^k \frac{1}{a_n} > k$ and so $\sum_{n=1}^\infty \frac{1}{a_n}=\lim_{k\to+\infty} \sum_{n=1}^k \frac{1}{a_n} \ge \lim_{k\to+\infty} k=+\infty$.
I like how everything in math has useless names
The more important it is, the more useless its name gets
16:46
The most important theorem should be one letter anyways
by compression
Yes, the A theorem. Because it's the first and foremost.
This will prove useful to future generations hoping to grasp a thing without having to recall an ancient, dead man's unpronounceable last name, or worse, a triple from multiple different countries having equally unpronounceable names.
Also, clearly it should be called Euler's second constant, not Oiler-Macaroni
@CowperKettle hahaha
@CowperKettle if sin x wore pants?
Like imagine calling it something useful like the logaro-harmonic difference constant
16:51
@AMDG Most things in mathematics are named after people.
@CowperKettle :) If Dirichlet function wore pants?
Those names are, in my opinion, useful, as they connect us to the history of our field. Even when Stigler's law rears its head.
@XanderHenderson Yeah, that's what I'm getting at. It annoys me because they tell nothing of the essence of the thing to which they are named.
@AMDG Sure.
I mean maybe, but also... you could just mention "it was first discovered by '...'"
16:53
And, to give a counterpoint, I have sometimes argued (kind of half-seriously) that we should remove the names from theorems, and give them useful names.
@AMDG Again, Stigler. Nothing is named after the original author. :D
Which presumably is not Stigler's law, but simply Stigler's law because yes.
@AMDG Indeed. I think that the original statement is due to Plato or something.
I'm sad to say that boolean in my own field is irredeemably ubiquitous to change at this point.
It's become a word in its own right
Wikipedia claims that Merton should get credit for Stigler's law: "Stigler himself named the sociologist Robert K. Merton as the discoverer of "Stigler's law" to show that it follows its own decree, though the phenomenon had previously been noted by others."
@AMDG I mean, it could be worse. At least it isn't, for example, a part of your body.
Like a fallopian tube.
It could be worse, but it really doesn't get much worse than this. For things that are serial in nature, arbitrary names are justified. For things representing genera and species, its name should tell something of the thing itself. $\gamma$ ain't just "another avenue off the boulevard of another street". It's a fundamental and ubiquitous transcendental constant like $e$ is.
(cold take of the year)
I mean I assume everyone believes by default that $\gamma$ is conjecturally transcendental
17:01
By $\gamma$, do you mean the Euler-Macheroni constant?
(however that's spelled)
Yes, the Euler-Mascheroni constant
Well... it does have a name. :D
Yeah, but at least we in computer science don't go off naming self-balancing binary trees something silly like... Chomsky Hierarchy or something (as you can see, my own field is not impervious to this disease).
@AMDG I don't see it as a "disease". It is the way in which we honor the people in our fields. It is common across the sciences. Again, there are likely good reasons to give things more descriptive names, but I don't regard naming things out of people as a fundamentally bad thing.
Like, bruh. Just call it the grammar hierarchy.
@XanderHenderson It's a disease in my eyes insofar as it is a deviation from the rule I just gave, though I don't think I'd be able to live well if we named every common number after some random person, however important his work may have been.
Someone somewhere thought the idea of associating gold with a ratio dubbed $\phi$ was a good idea, and then someone discovered the rest of the "metallic" numbers which have a very convenient property for computing their reciprocals.
I just don't like it because the names are so far removed from the reality they name.
17:09
AKA the Xander principle.
I think I've heard of that guy before
While we're at it, the AMDG principle is incredibly important. I bet you'd like to know what it means without having to look it up, but too bad because I named it after myself because honor before utility.
Tell me, is there a platonic relationship between the regular polyhedra, or is there a more intimate relationship between them?
Oh, you've done it yourself!
What is a "platonic relationship"? :P
$\text{platonic relationship} \superset \text{Friendzone}$
I can't latex smh
17:14
\supset
Annoyingly supsetting that I've been defeated by something which deviates from my naming standard (no abbreviations).
17:26
@geocalc33 Ok, but what is a Mellin transform exactly, and what is a theta function? Sounds quite abstract. Gamma is easy to see: beginning with factorial, it's the number of ways of permuting $n$ distinct objects. Generalizing, Gamma can be understood in essence as the solution of the recurrence relation $f(x + 1) = f(x) x$ and intuitively as just the generalization of the factorial, so relating it to other things isn't too hard to see.
If you look at Zeta on the other hand, its wikipedia page lists only one application that's very specific and nuanced as far as I can see.
@AMDG The gamma function is not the only possible extension of the factorial, though.
Why that extension, rather than some other extension?
Yeah, but you can consider it a necessary extension as opposed to arbitrary.
@AMDG Why?
What is special about gamma?
It's a distinct relation in its own right. That's special enough.
I don't understand. That doesn't make any sense to me.
17:32
Well then what are you asking exactly?
I am asking you to explain why gamma is the natural extension of the factorial function.
Without handwaving.
I mean, there are good reasons, but they seem no less arbitrary than the zeta function, which is the unique meromorphic extension of the series $\sum 1/n^s$ to most of the complex plane.
a constant feature of amdg's reasoning is confusing potentiality with actuality (pew pew)
Well I don't know all the properties of Gamma as opposed to factorial, so perhaps there exists a better, more necessary extension. If the properties of both hold equally with factorial being a strict subset, then I would ask: what else can be called its natural extension?
@AMDG What do you mean by "necessary extension"?
For example, Hadamard's gamma extends the factorial function to a function which is entire (no poles). That seems like a nice property.
Metaphysically necessary, i.e., Gamma as an analytic continuation of the factorial is what must be the simplest analytic continuation possible, but I don't actually know what the full definition of analytic continuation is. I'm only aware of the concept as a vague notion of extending the domain of a relation such that it is possible to compute or holds beyond the domain of the original relation.
17:39
@AMDG What do you mean by "simplest"?
itt: attempt to do philosophy in mathematics
The Hadamard gamma does a better job of being an analytic continuation of the factorial function, in that it is entire. Doesn't that mean that it is "simpler"?
Salut @Ted
Well I'm interested here in the essence of factorial. What relation continues the factorial to all the integers in such a way that it can be reconciled with its definition for the positive integers? The rationals? The reals?
I have a question for you when you have time
17:42
@AMDG There are infinitely many such functions.
What does it mean to take the factorial of -1?
Salut @Astyx …. Tu as remarqué que Hippa est revenu?
Ah non je n'avais pas vu
What question?
@AMDG Well... you can't really answer that question with the usual gamma function.
17:44
So I have a (smooth) variety X, and a closed subvariety Z of codimension 2. Let Y be the blowup of X along Z. Let $\alpha$ and $\beta$ be in $H_{n-2}(Z)$. These give rise to cycles $A$ and $B$ which are $\alpha\times\mathbb P^1$ and $\beta\times\mathbb P^1$
What can I say about the intersection product $A.B$? I'm convinced it's $\alpha.\beta$ times the self intersection of the exceptional divisor, but I barely know any of this stuff
The normal bundle is likely nontrivial, so how are you getting a global product with $\Bbb P^1$?
Ok, but if you take after it for what it is in itself, then what relation best describes that? is $(-n)! = (-n)(-n-1)(-n-2)... = -(n)(n+1)(n+2)...$, or something else? As I understand it, it is the product of all possible integers up to $n$, but if we include all the negative integers, then suddenly we'd need the product of all negative integers also if we interpret this rule literally, but that isn't possible strictly speaking, so it cannot be the correct interpretation.
@TedShifrin This is what I meant by "I barely know any of this stuff" :p what's the normal bundle?
it is whatever you define it to be, it has no inherent meaning
Shin, shut up.
17:48
Putting the sign at the front actually was a mistake on my part there.
@AMDG The usual gamma function has poles at the negative integers. So it doesn't really extend the factorial there.
@TedShifrin now ted don't be rude
Stop being an ass.
that's an unfortunate interpretation, and no need to be rude
@XanderHenderson Well we can't get at $\Gamma(x)$ yet. We have to look at $n!$ and extend it to the integers.
17:49
@AMDG What?
I don't understand.
I thought we were determining why Gamma is the best analytic continuation or not
@Astyx What is the blow-up? You replace $Z$ with its projectivized normal bundle.
or necessary continuation
@AMDG I asked you to justify the claim that gamma is the "best" extension, or the most natural extension, or whatever.
I have no dog in the fight.
But I don't understand why you believe it to be the "best".
And I don't understand why you think that $\Gamma$ is more natural than $\zeta$.
Oh ok, in my setting Z is the intersection with a linear subspace of codim 2
17:51
(which is what started my inquiries).
Based on what little I know, it seems to be the best extension because it appears to directly extend the domain of $n!$ without any addition or subtraction of properties.
Also, be careful with real/complex dimensions. What is $n$?
@AMDG There are infinitely many functions which do this.
$n$ is the complex dimension of $X$
And there's also infinitely many ways of representing the number 1, yet the best representation of 1 is 1.
17:53
Codimension is also complex codimension
@AMDG This is false equivalence.
So $\alpha$ is middle-dimensional in $Z$?
Yes
And A in X
and even if it was an adequate equivalence, 1 is nothing beyond it's algebraic properties. there's no reason it couldn't be represented by anything else that had the same actions
@XanderHenderson and if I said instead "infinitely many antiderivatives for any relation that has one"?
17:54
no reason why it should be the best either
So we know the normal bundle. If you intersect (generically) with a hyperplane, the normal bundle is the normal bundle of the hyperplane, which is $\mathscr O(1)$.
What are you intersecting?
$X$ with a hyperplane. Then intersect again, and the normal bundle is the direct sum.
@AMDG Okay, but now we are saying something, which is why we are generally careful about how we define antiderivatives. If a function on an interval has an antiderivative, then it has infinitely many antiderivatives. But the difference between any two is a constant. So the distinction is relatively unimportant.
@TedShifrin exactly
17:58
But there are infinitely many "nice" functions which extend the factorial function to a larger set of complex numbers, and not all of these are simply related. Again, the usual gamma has a lot of poles; Hadamard's gamma does not.
So why should we prefer the usual gamma?
So how do you recover the intersection product from there?
@XanderHenderson So let me ask, however, why it is so that there is a constant difference between each when, considering it relates the area of the function being integrated, would not some constant $C$ added to the result of integrating imply that our function $f$ has also some extra constant area and therefore some constant offset away from the $x$ axis proportional to $C$?
The intersection theory depends on the normal bundle of the exceptional divisor. If you blow up a point in a surface, the exceptional divisor has self-intersection $-1$.
@AMDG This is a consequence of the mean value theorem, and is a statement about antiderivatives, not a statement about Riemann integrals.
An antiderivative has no a priori relation to an area.
The fact that areas and antiderivatives are related is a theorem (a fairly important one... fundamental, even).
The thing to think through, Astyx, is blowing up a $\Bbb P^1$ in $\Bbb P^3$.
I no longer have all my books, notes, or brains, but Hartshorne has a paper on the cohomology of blow-ups. There’s also some stuff in Griffiths/Harris.
18:07
@XanderHenderson So then would you say antiderivatives have a better natural definition as the surface satisfying the position of a point over time as it continuously changes in vector based on $f$? I can understand it in that sense then as being a posteriori to the area.
@TedShifrin thanks I'll have a look
@AMDG I would say that the notion of differentiation exists, and once you differentiate a function, you would like to know how to un-differentiate that function.
That is what an antiderivative is... nothing more, nothing less.
I’m a little bothered by the fact that $n-2$ might be odd, in which case we’re not looking at intersections of complex things.
You can assume $n$ to be even
I suppose that's fair, but even the notion of derivative being something that gives the vector $f'(x)$ of a point $(x, f(x))$ seems to be rooted in a geometric reality as the limit of a set of secants.
18:12
@AMDG That's one way of attacking it. It isn't the only way.
That said, it seems more natural to consider it as such, and that's actually been an idea I've had for a general antiderivative formula, but again, what else could it be in essence?
E.g. it can be thought of in terms of the "best" linear approximation of a function (and Taylor series give you the "best" approximation by a polynomial of specified degree).
@AMDG I don't understand your question.
again it has no inherent meaning
why do you keep supposing this?
We define an operation: differentiation. We note that many functions have derivatives. We then ask if we can reverse the process: given some function, can we know that it is the derivative of some other function? can we classify all such functions?
and why would it's essence be hidden from the actual use (and only use) we make of it?
we define objects and define their behaviour, then we follow the consequences of our definitions
18:19
The derivative is its own essence, and the act of differentiating is also its own essence; but there are necessarily certain things, for instance, geometric essences like the circle and its relation to $\sqrt{1 - x^2}$ and the hyperbola hiding in there as well, which relate by necessity, such as the positive integers with the quantity of objects I have, or how many boxes can fit in a truck assuming all are rectangular prisms and the space is itself a rectangular prism.
@AMDG This is not math.
One can see that the properties of the circle correspond perfectly with this algebraic expression, and that, by necessity, it must be its own inverse: considered algebraically, because it must; and considered geometrically, because a circle has the same appearance for any orientation.
temperature here is 29 to 36 deg celsius with 75% humidity.
It's been years since I lived in a hot environment like this.
@XanderHenderson Well it's a matter of the philosophy of math, but I guess the set theory people would try to disagree and say it is therefore math because "essence" is anything definable by a set, and a set is really equivalent to "genus", while subset is equivalent to "species".
@AMDG This is only correct if you are very careful to take only 1/4 the circle.
18:23
Yes, thank you for the correction :)
So the circular symmetry really is destroyed.
this is worse than philosophy, it's magical thinking, and it's used to guide mathematical investigation. each and every single one of those conclusions come from somehow supposing things must magically be that way
Dude...
Yeah man, cuz ontology and set theory are magic.
That definitely makes sense.
Finding face maps in cellular homology required 'degree' which I don't know.
is it true that cellular homology = simplicial homology ?
ontology is fine as long as it's actually not just statements of opinions, or repeatedly concluding that there are some magical skeletons making things be the way they are'
18:28
some of the examples seem to say yes.
While you're at it, Shin, you may as well say that $5 + 7 \neq 12$.
@Koro yes. Hatcher proves that somewhere.
Oh I see.
I remember from one of Professor Albin's lectures that nth homology is about detecting 'n-dimensional' holes in the space.
It's such a wonderful way to put it. Highly intuitive also.
I am not entertaining your absurd nonsense Shin.
Take a circle $S^1$. One takes a 0 cell and attached to it a 2 cell.
Let's find $H_2(S^1)$. Intuitively, find 2 dimensional holes in $S^1$. Are there any? No.
So it is $0$.
18:33
@XanderHenderson In any case, my efforts were an attempt to see how best to consider differentiation and integration in themselves besides just "operations that have definitions". I could define logical negation in essence as setting all ones to zeros, and all zeros to ones, or I suppose I could just give a table of inputs and outputs too, except that even the table itself implies its necessity of existing as an operation and that binary operations can be implicitly defined as a single number.
e.g., xor == 6 because you have 0110 for inputs 00, 01, 10, and 11.
@AMDG tangents and area respectively? It is also used in constructing buildings.
@AMDG I am a mathematician, not a philosopher. I want to do math, not philosophy.
That may be so, but the question of essence is inevitable. Two things are equal if they are in fact equal, but if they are equal, then they must be the same thing. Doesn't matter to me if you represent these operations as numbers and see to what things it equals or if you just consider them as essences directly and compare them.
If you say so.
Considering that math pretty heavily revolves around the equals sign, I'd say it's quite important :)
18:43
Suppose I'd like to calculate $\sum_{j=2}^\infty\sum_{k=2}^\infty\frac{1}{k^{j}}$. Then I can only switch the order of summation if $\sum_{I}^\infty \frac{1}{k^{j}}$ converges (absolutely), where $I=\{2,3,\ldots\}\times\{2,3,\ldots\}$. How can you show that $\sum_{I}^\infty \frac{1}{k^{j}}$ converges?
If it were a "single series", then I would try to show convergence by using comparison theorems such as limit comparison. However, in double series we have two indices and this makes me very uncertain.
@AMDG Not to a working mathematician.
I mean ontology isn't a practical science in itself, but I'm just saying it's important enough that the practice wouldn't be possible without a correct philosophy to guide it (here constituted by the theorems of mathematics).
Why isn't 1 prime?
@AMDG Because, in most cases, it is convenient to define the word "prime" in such a way that 1 does not meet that definition.
Well we don't avoid things because they're inconvenient in math, surely...
@AMDG I don't understand your comment.
We can define words to mean whatever we like.
18:54
What's so inconvenient about 1 being prime?
@AMDG If 1 were prime, then we would have to start a lot of theorems with "Let $p$ be prime, $p\ne 1$..."
Why would 1 be an exception, and why would it be inconvenient? 0 is already an inconvenience that shows up everywhere.
@AMDG Go the other direction: what is gained by defining the word "prime" so that 1 meets the definition?
Well that's the thing: if by definition a number is prime if its only factors are 1 and itself, then 1 must be prime, so I mean... is reality an inconvenience to anyone? I think not.
Again, we can define words to mean whatever we like. There is nothing magical about these definitions. The consensus of the mathematical community (remember, mathematics is created by people) is that we should not define the word prime so that it includes 1.
@AMDG That isn't the only definition.
18:58
Well if it has multiple definitions, then there are multiple kinds of primality.
The more grown up definition is that a non-unit $p$ is prime if, whenever $ab = p$, either $a$ or $b$ is a unit.
The number 1 is the same idea in everyone's head regardless of what symbol we give it.
1 is a unit, so it still holds
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