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12:03 AM
I think that is a fairly straightforward generalization. The point is that if you have a natural isomorphism F=>G (a natural monomorphism would actually suffice), then that natural isomorphism and G already determine F.
You should, if I'm not mistaken, obtain a functor from the indiscrete category whose underlying set is $\coprod_{f\colon\mathrm{Ob}(\mathcal{C})\rightarrow\mathrm{Ob}(\mathcal{C})}\prod_{x\in\mathcal{C}}\mathrm{Iso}(F(x),x)$ to the slice category $[\mathcal{C},\mathcal{C}]/1_{\mathcal{C}}$, which is an embedding onto the image, which is precisely the full subcategory on the
@TedShifrin Hopefully they sort your record out.
Q3 is too open-ended for me to have any worthwhile idea. Given the remark I just made, I feel like it comes down to asking something akin to "what's the next best thing to a monomorphism" and I don't have an answer to that.
Do you mean $\coprod_{F\colon\mathrm{Ob}(\mathcal{C})\rightarrow\mathrm{Ob}(\mathcal{C})}\prod_{x\in\mathcal{C}}\mathrm{Iso}(F(x),x)$ right?
You know what. I'm already satisifed with this combo. With the last generalization.
I know I did half of the work, but you translated that and made some points more clear to me. Q3 can wait. I'll work it myself if it is possible to jist use something more complex than the indiscrete category.
When i'll have more maturity I'll try to study a solution for it my self and see if there is some adjoint. And as long as Q1 goes. when I'll feel ready I'm sure I'll discover the general thing in some lost n-lab page.
@Thorgott So to reward you for this chat and your time, I hope you can copypaste those two constructions and you useful remark into an answe to q2+opinion on q3. I'll give you the bounty.
12:17 AM
Hello, can I ask if $AA^{\intercal} = A^{\intercal}A$?
In general, totally not. They might not even have the same shape.
I've never seen intercal before. I use ^\top.
@copper.hat When the website traffic subsides, I know what I have to do. Thanks.
@TedShifrin Ok, thank you. I thought that they are equal but no.
Probably you are interested in the square matrices case.
Even in the square case, they are not generally equal. Think about the entries.
12:32 AM
Mhh, ye. In general we can't conclude that $A_{(i)}\cdot A_{(j)}=A^{(i)}\cdot A^{(j)}$
I posted an answer, I hope that it's clear
:thumbs up:
Now I'll steal from you answer the power to draw commutative diagrams on MSE mwhuahahaha >:D
it's a bit awkward, but sadly MSE doesn't support tikz
All theft of TeX is normal.
hahah... @TedShifrin that's LITERALLY how I learnt TeX... by stealing here on MSE.
To bad I had to steal some more English skill xD, well nvm
12:46 AM
I feel like that's how almost everyone learns
And I bought several Latex books once I started writing books.
Well maybe long ago that was necessary. I wonder if it is still so. With so many online resources.
That was long ago, but I still use them occasionally.
1:06 AM
hi everyone!
Hey I have a question for you. A curiosity I had back when I tried to take my first courses. In your opinion "functor people", I'm gonna stick with the joke, are more geometric ppl or algebraic ones? In your experience.
I used to be at home with abstraction and algebra, even abstraction for it's own sake. But I started to really appreciate and understand CT when I discovered Vector spaces and linear algebra.
I ask because I had a weird experience at UNI. I was already well exposed to all sort of set theory, foundational issues and the basic definitions of category made very sense to me.
Back then we had three first year courses: ALG 1(set-theory+ ring theory), GEO 1 (linear algebra+affine spaces+eigentheory) and ANALYSIS.
Everyone was good at analysis, expect for me. Most were amazing doing computations of basis, subspaces and things like that... but, expect few geniuses, most were bad with the Algebra course, and most of them had some problems understanding proofs and the logic behind. I.e. mo
Category theory is fundamentally more algebraic in nature. I think that's not even controversial to claim.
1:22 AM
No argument from me. Formalism is more algebraic.
Hi, @Alexandru
Doesn't seem controversial to me. I heard a professor telling me that ZFC is pretty algebraic in nature, so I'm not gonna be shoked by your claim.
i agree with thorgott, which also means i agree with ted.
this is such a bad day for me.
Indeed. But you're my attorney to do battle with amWhy.
I do not think there's an inherent discrepancy between geometry and formalism
Yes, there is, if geometry is taught properly. Sadly, very few algebraists teach algebra unformally.
1:27 AM
@Thorgott I'm still surprised by the big divide I felt among first year students. I felt that too since to me convergence of series and asymptotics was like doing random black magic nonsense ... while, to me, algebra was like clean fresh water.... but to many it was like the most difficult thing on earth
a formal perspective and an intuitive perspective can coexist..
@TedShifrin are you claiming that for teaching geometry, properly=unformally?
Dos somebody know intutive proof of convergence of babylonian mathod?
It making me scratch my head on how to prove convergence related to itération.
nope, I'm sorry :/
Please tag me if somebody knows about it :`(
1:38 AM
@MphLee thor and I are talking about graduate level geometry, not high school. But I hate formalism in high school geometry.
So, given $N\in\mathbb N$ and $x_0=q$ a first approximation of $\sqrt{N}$ We need to prove that the sequence defined recursively by $x_{n+1}:=(x_n+\frac{N}{x_n})\frac{1}{2}$ converges to $\sqrt{N}$. Is that what your are asking @Omniman ?
What is graduate level geometry? When I asked you about "geometric" I was only referring to vector spaces, affine spaces, projective spaces...
You and Thor were talking about differential geometry or schemes?
Oh sorry
Schemes is algebraic geometry. I meant differential. There’s too much formalism in algebraic algebraic geometry.
I know now how to do it sorry I was quite confused with guess
grr, no schemes
1:44 AM
I was stupidly imaginating it to alternate lol
OK good to know, so when educated people say "garduate level geometry" they really mean differential g.
No, not necessarily. But Thor and I have a context here.
The term is very broad.
2:01 AM
Hi Ted. Just got back to analysis today. Worked out the algebra from the cone and tangent plane question and it is a lot nicer. After the algebra and solving it in terms of $x$ being the value I'm looking at everything being proportional to, I ended up with

$y = \frac{bx}{a}$ and $z = \frac{cx}{a}$

I know it's closer, but it still doesn't feel right. If I were to express it in terms of a line I would have $(x, \frac{bx}{a}, \frac{cx}{a})$....I mean I guess this does say that lines of this form are the ones that are shared between the tangent plane and the cone.....So reflecting on it, t
Thanks to everyone. Cya.
@TedShifrin I just made a nice geometric discovery (not applied or numerical for once...)
@dc3rd It's right, but avoiding denominators avoids worrying about things being 0.
@Alexandru I like applied when I understand it.
do you remember when we talked about clothoids briefly?
Oy. Very vaguely.
2:09 AM
well I found an analytic expression for the spherical equivalent
I will do a write up soon
Link me when you do.
I've been looking for this for a year and a half (with numerical analysis projects on the side)
my daughter just told my cat to "shut up"
now she's blowing bubbles in a glass of water and responded to a request to stop with "i do it at school"
She takes after you.
One other quick question. When I was figuring out the proportions I had the expression $x^2(c^2 - a^2) - 2bacy + y^2(c^2-b^2) = 0$. So through "fiddling" around I was able to arrive at the expression $(bx-ay)^2 = 0$. And I only knew this was what I wanted to get to because you had told me. So it was really a combination of shot in the dark/luck and what you told me. There was no "purpose" to my attempt.

So after being all wordy I want to ask should I think about things through the lens of:

"try to simplify things as much as possible without losing any information AND if information does
2:28 AM
You have to use $a^2+b^2=c^2$, of course. That’s to be expected. Your middle term is just wrong.
its all about behaviour modification
typed it wrong....oops
$2abxy$ was meant
Then it’s right. My comment stands.
I know the solution is right...I was thinking more in general terms when I am not blessed to have you tell me "the form should be xyz"
so when I'm out in the rough and tumble real world and nobody can hold my hand..... :(\
Copper came in with the brief but informative answer....lol
math just is so damn humbling, has me always second guessing myself......love and hate for the attrition of this discipline.
now the arm is starting to feel sore from my first jab today.....:(
Of course you want to simplify those difference of squares coefficients!
2:52 AM
If $|\lambda_{1}|<1$ and $|\lambda_{2}|<1$ then can $1 - (\lambda_{1} + \lambda_{2}) + \lambda_{1}\lambda_{2} < 1$ ?
yeah why not
3:13 AM
what's up
my daughter just went to bed. her last act before doing so was walking around with a blanket on her head shouting "i'm being a ghost."
i told her, ghosts don't tell you that they're ghosts, and if they did, they'd probably just say "i'm a ghost"
"i'm being a ghost" sounds too costume-y
maybe Halloween came early.
seems like it. earlier when my wife came home from being out, my daughter said "it's a monster in the house!" i don't know where she picked this up but she's really into it.
3:20 AM
did you gave her a hard math problem that's what my friend's daughter says as well.
we be like uhhh its friendly problem.
she can count to about 20. if i ask her to count higher she just repeats some of the lower numbers.
she's advanced, she's counting with multiplicity.
that's awesome.
teach her basic number theory haha.
i don't want to set her down the path of being a juvenile delinquent
hahaahha that's funny.
is number theory cool?
If $|\lambda_{1}|<1$ and $|\lambda_{2}|<1$ then can $1 - (\lambda_{1} + \lambda_{2}) + \lambda_{1}\lambda_{2} < 0$ ?
3:35 AM
i think it's too hard. elementary number theory is fun though.
sorry there wa s atypo to my above question
If $|\lambda_{1}|<1$ and $|\lambda_{2}|<1$ then can $1 - (\lambda_{1} + \lambda_{2}) + \lambda_{1}\lambda_{2} < 0$ ? thinking to prove
1-(a+b)+ab = (1-a)(1-b) if this helps you think about it, the sign of the product being negative only if the two factors have opposite sign.
i'm not sure if a and b are real or complex here. "lambda" makes me suspicious.
Inequalities with complex numbers are a no-no.
Negative numbers will make the desired inequality completely impossible!
Thats true
hey man, i'm just throwin' ideas out there. some functional analysts will talk about A-B being positive even if A and B aren't self adjoint.
i'm using the sarcastic method, which i learn mostly by reading your comments to OPs on main.
i may have made my own modifications to the technique
3:47 AM
May, yes.
4:06 AM
Is that Mai oui?
Mai is gone. Maïs is in season.
4:26 AM
Gnome bad
2 hours later…
6:07 AM
@BAYMAX Expand $(1-\lambda_1)(1-\lambda_2)$.
A: limit superior of real value function (Exercise 9.3.4 in Tao Analysis I)

Koro\begin{equation} \limsup _{x\rightarrow x_{0} ;x\in E} f(x)=\inf\Bigl\{\sup \bigl\{f(x):x\in E\land |x-x_{0} |< \delta \bigr\} :\delta \in \mathbf{R}^{+}\Bigr\} \tag{A} \end{equation} We'll prove that the following are equivalent: $\displaystyle \limsup _{E\ni x\rightarrow x_{0}} \ f( x) =L$ Fo...

I recently proposed an answer to this question
But it seems that the proposed definition is not working for the case when limsup is $-\infty$
And I can't believe this fact. The definition should be able to accommodate $-\infty$ case easily.
I have one idea for this: if limsup f(x) is $-\infty$ then every $a\in \mathbb R$ is an upper bound for $\sup \{f(x): x\in E\land |x-x_0|\lt \delta \}$ for every $\delta \gt 0$.
So if $L=-\infty$, then the limit of $f(x)$ as $x\to x_0$ is of course $-\infty$. What do you need to show?
This will ensure that $-\infty$ is inf of the set $\sup \{f(x): x\in E\land |x-x_0|\lt \delta \}$ is $-\infty$.
Hi Ted. My last comment was in addition to my last comment, not a reply to your comment.
No problem. I just don't have the patience to read the post. I'm wondering what the issue is.
Ok,not a problem. The problem is to give an equivalent definition of $\limsup f(x)$ as $x\to x_0$, $x\in E$. Equivalent definition using sequences.
And limsup f(x) is as defined above in the linked post (the definition is visible here), you don't have to go to the post.
I proposed that following two are equivalent: 1. $\displaystyle \limsup _{E\ni x\rightarrow x_{0}} \ f( x) =L$

2. For every $\displaystyle \epsilon >0$ and every sequence $\displaystyle ( x_{n}) \subset E-\{x_{0} \}$ converging to $\displaystyle x_{0}$, there exists $\displaystyle N\in \mathbb{N}$ such that $\displaystyle n\geq N\Longrightarrow $$\displaystyle f( x_{n}) < L+\epsilon $ and $\displaystyle f( x_{m}) >L -\epsilon $ for infinite $\displaystyle m$.
In case $L$ is finite.
6:24 AM
I'm asking specifically what needs to be done in the $-\infty$ case.
If $L=+\infty$, then also this definition works, I tried to show that at the end of my answer. What bothers me is that this definition does not seem to work in case $L=-\infty$.
Ted, in case of $L=-\infty$ case, I want to prove equivalence of the above two statements $(1)$ and (2).
Please let me know if my question doesn't make sense yet.
Epsilonics don't make sense with $-\infty$.
Right. So I used the following idea for $L=\infty$.
If $L=\infty$, $\displaystyle f$ is not bounded from above in any neighborhood of $\displaystyle x_{0}$. So for every sequence $\displaystyle x_{n}\rightarrow x_{0} ,\ x_{n} \neq x_{0}$ for any $\displaystyle n$, we can find a subsequence $\displaystyle x_{n_{k}}$ such that $\displaystyle f( x_{n_{k}})\rightarrow +\infty $
Oh I think, I'll need to modify the proposed definition for $\infty$ and $-\infty$ cases
The existing proposed def. will probably not work as it is.
$-\infty$ should be way easier. There are no possible subsequential limits other than $-\infty$.
I'm thinking
6:37 AM
Sometimes I find it useful to define of an $\epsilon$ neighborhood of a finite $x_0$ to be $(x_0-\epsilon,x_0+\epsilon)$ and an $\epsilon$ neighborhood of $+\infty$ to be $(1/\epsilon,\infty)$ and an $\epsilon$ neighborhood of $-\infty$ to be $(-\infty,-1/\epsilon)$.
The key fact is that these neighborhoods shrink as $\epsilon\to0$ ($\epsilon\gt0$).
Hi Robjohn, in Rudin's book in continuity chapter nbd of $\infty$ was defined similarly
nbd of $\infty$ is set of all $x\in \mathbb R$ such that $x>c$ for any $c\in \mathbb R$
similarly for negative $-\infty$
yes, but this lets us look at $\epsilon\to0$ in all cases.
right :-)
$N_\epsilon(x)$ for $x\in\overline{\mathbb{R}}$
7:04 AM
Is there any pattern like this in mathematics?
All patterns exist in mathematics. How to describe them is the problem.
you can get that first one by standing near a fence and looking at it the right way
7:21 AM
The next one by standing near a group of fences
yeah, i'm just not sure the fences are parallel in #2 or #3. might have to do some work to figure out how to slant them and place them.
and probably a finite . . . until they run into each other. i'm assuming fences in straight lines.
at least #3 makes me nervous about that.
i like it as artwork
I am sure we can arrange a set of fences at various places with various sizes to look like those.
finding someone to do it might be an issue. my dad needed a guy to a build a fence in just one straight line, he did about half of it and then didn't come back for like three weeks
@leslietownes contractors can be variable
7:37 AM
Q: Dirichlet's Theorem and $\dfrac{d}{c} \Bbb{Z} + \dfrac{a}{b}$ the fractional cosets...

OlympicComputerChairSitterDirichlet's Theorem on arithmetic progressions is meant for the cosets $d\Bbb{Z} + a$ where $\gcd(a,d) = 1$. However, what if you worked with fractional coefficients and thus "cosets" of the form $\dfrac{d}{c}\Bbb{Z} + \dfrac{a}{b}$, where the two fractions are given as reduced? I think this woul...

Late night number theory problem for you all
@leslietownes can you solve it?
@AndrewMicallef what's up?
Nm just chilling before goodbye drinks
My collegues are wishing me fare thee well
Just reading about fences it seams
Do you guys know about Dirichlet's Theorem?
on arithmetic progressions
Hi, @Koro
are you good at NT?
7:57 AM
@OlympicComputerChairSitter: I am new to NT so don't have much expertise on it, I'm afraid.
Dirichlet's theorem is that there are infinitely many primes in the coset $a \Bbb{Z} + d$ where $\gcd(a,d) = 1$.
I'm saying, take fractions as $a,d$ instead
This might give a whole new theorem
which might be easily solved since Dirichlet's is
@Koro don't be afraid of NT, only be afraid of the prime numbers :)
I fell victim to the primes long ago :|
8:32 AM
@Prithubiswas There's a round version of your diagram at en.wikipedia.org/wiki/Ordinal_number
@PM2Ring Oh ok. This looks interesting :)
8:48 AM
@Prithubiswas If you'd like a gentle introduction to transfinite set theory, check out Infinity and the Mind by mathematician and science fiction author Rudy Rucker.
9:01 AM
@PM2Ring I am currently studying "Theory of Sets" by E.Kamke and reading about transfinite cardinals and ordinals from it.
3 hours later…
12:14 PM
@TedShifrin You owe me no apology Ted although I appreciate it.
12:28 PM
Any help to accomodate $\infty$ and $-\infty$ in definition proposed in $(2)$ in the answer please?: math.stackexchange.com/questions/4141057/…
@robjohn I tried with this but I have no clarity/confidence with the result I got using this..
That is, for $\infty$ case for example, for any sequence $(x_n)$ and for any $c\in \mathbb R$, indeed it is true that $\exists N$ such that $n\ge N\implies f(x_n)\gt c$ but do I do with the "and" situation in $(2)@?
1 hour later…
1:37 PM
@Koro why not simplify with $$\limsup_{x\to x_0}f(x)=\inf_{\lower{2pt}{\delta\gt0}}\sup_{x\in N_\delta(x_0)}f(x)$$
using the $N_\epsilon$ I gave earlier. This should work for $x_0\in\overline{\mathbb{R}}$
You can do the same for $N_\epsilon(L)$
1 hour later…
2:57 PM
don't waste space
@RussianBotWhoKnowsYourIP This user has trolled at least three sites today with the same: Anti vaxers are morons. The user may need a healthy break.
3:17 PM
i am waiting for said user to tell me my ip address :-)
3:43 PM
Hi @Koro
@robjohn Hi Rob, this is the problem: $(2)$ has two statements connected by “and” one is: for every $\epsilon\gt 0$ and sequence $x_n$ cnvgng to $x_0$ there exists N such that $n\ge N\implies f(x_n)\lt L+\epsilon$ and second is $f(x_m)>L+\epsilon$. Using the $N(x_0)$ suggested earlier, I’m worried that only one is satisfied at a time :’(
Hi @copper.hat. How are you?
You already told your ip to the said user once:)
Good thanks! Getting ready for some morning exercise :-)
@copper.hat I think it's actually possible to get the IP addresses of everyone in a room by uploading your image, let me find the meta question about it
It is not actually that hard to find one of my public ip addresses, I was curious if that user would do so :-)
If I recall correctly , I may be wrong though:-)
3:46 PM
@user certainly SE/SO has access, but I doubt that they 'out' ip addresses.
@user what image are you referring to?
No, there's an actual vulnerability afaik if someone uploads an image without using i.stack.imgur
Without using stack imgur how will one even upload image here ?
I think by default image gets uploaded to that server. No?
@copper.hat Not any image in particular
Q: Allowing images from external sources opens doors to serious security exploits

Aaron YodaikenNote: Please see this answer for more info on how this poses a security risk. I'm sure you guys already know this, but allowing users to put images from any source in questions means that askers can get all the analytical data you can about their question. So a malicious user could theoreticall...

If you're hosting an image yourself, you can tell which IPs access it. Nothing major, but still
@copper.hat Great! Quite the opposite here. It’s nighttime :-)
@user that makes sense, but would require me to access a suitably trojaned question.
however, thanks for pointing it out, i certainly did not think of that.
3:53 PM
Is ip exposure that serious? What will anybody achieve by that even if they have it?
I have seen authorities using ip to track down the unruly
I don’t think there will be a financial loss to me if anybody knows my ip.
Probably not yeah
4:25 PM
IP exposure is generally not that huge of an issue for most people. Especially if you put it in a question, you'll get a huge list of IPs which tells you nothing.
However, you can drop it into chat rooms and use set theory to identify specific users' IPs. You can also get moderators' IPs though I'd rather we don't discuss how.
@hyper-neutrino i think it is more the hypothetical rather than any serious impact. however it means that it is possible with some uncertainty.
but getting a vpn address is not much use to someone
It's not that big of a deal but it's still a security exploit and also it violates GDPR. Also yes, I agree with copper, it's mostly a hypothetical. This has been known and unfixed for 11 years and SE hasn't burnt to the ground (yet).
It doesn't stop at IP grabbing though; the linked post also has other potential issues listed.
i used to be a strong privacy advocate/hack but over the years have mollified my stance considerably.
the irony of gdpr & ilk is that for many normal functions it requires exposing more data.
(the requirement of agreeing to certain terms.)
I am angry with VPNs
I'll never forget the day when I had a VPN to view content not available in my region.
Guess what, VPN was detected!
Since then, I don't care about VPNs.
4:44 PM
You can probably get the vast majority of the users to fall for a simple ip grabber so I don't think it matters that much
you also need your browser to not leak a ton of info even through the vpn
@Koro Most likely the provider has a list of well known VPN address ranges. I have a few VPNs, including one I run myself, so have never had an issue.
People are generally unaware of just how much potential exposure they have and where they are placing trust (such as browsers).
It does not matter in the 'real world' because physical proximity is required, but not so in the big bad interweb.
5:01 PM
yes if you want a vpn you should get a vps and run it yourself I think
5:25 PM
Should Mathematica return True for all four evaluations if my solution is correct, or am I inputting something incorrectly? i.imgur.com/HQDhDB8.png
(first time doing separation of variables and first time using Mathematica)
It's been so long since I saw @MikeMiller, @BalarkaSen here
5:58 PM
@user10478 Perhaps, if Mma could compute those sums.
@robjohn As in, I entered the sums wrong, or it just literally won't compute infinite sums?
@user If the browser requesting the image supplies other information, which they don't need to, but often do, then, more information than simply the IP can be obtained.
@user10478 It can compute some sums, but not all.
So is there a way to have it check my answers to these problems? Checking by hand looks to be very time consuming.
@user uploading an image here puts it on imgur.com. I don't think there is anything that can be done with that.
@user10478 You could evaluate them numerically and see if they agree to a large number of decimal places. That is not a proof, but it would be very unlikely that they would accidentally match to 20 places.
6:15 PM
Like change the infinite sums to sums going up to 20 (or 1000 or w/e it will compute in a reasonable time)?
since later terms in Fourier Series are always less significant
@robjohn Oh, I see, that's good to hear
@user10478 Instead of $\left((-1)^{n+1}+1\right)a_n$, why not use $2a_{2n+1}$ to sum twice the odd indexed terms?
Mma might have an easier time with the sums then.
all of the $n$'s need to be changed to $2n+1$
@user However, if you use <img src="...">, then your server gets the request. Of course, that exposes your server to everyone.
@user10478 it's the false statement of the day! :)
definitely true for those sums though
Hi Leslie
6:22 PM
@leslietownes if they are not less significant, the series does not converge, or am I missing your point?
Rob, I'll think more on the nbd
of infinity
I strongly believe the statements can be shown to be equivalent
That might be what the answer in the book did. There are some $2n + 1$'s floating around in there and their summation starts at $0$. I just wasn't quite able to manipulate one into the other, hence trying Mathematica.
my point is that a truncation up to some fixed number of terms is not guaranteed to capture the 'most significant' terms
nothing more than that
6:26 PM
Hmm, I thought that was a property of Fourier approximations, albeit not partial sums in general, that they get closer and closer to the target function the more terms you add.
here the terms are decreasing very rapidly, maybe monotonically (?), but that isn't guaranteed either
@user10478 You should also use Sin[x] for $\sin(x)$ and E for $e$ or Exp[x] for $e^x$
user i'm referring to a situation where maybe the first 20,000 coefficients of a function happen to be kinda small, then coefficient 20,001 is for some reason enormous, then the rest are zero. this is not the series you are talking about and not a common thing to see but it is possible.
i'll stop talking about my joke, it wasn't funny
i feel mean sometimes when i point out that someone should know certain facts, like what closed or bounded mean.
if they're asking a question with those words in it, it isn't mean
6:28 PM
@robjohn Okay, those are easy fixes.
more when you have been interacting over the years and basic stuff is still missing.
i guess it is frustration on my part.
u[x_, t_] := 8/Pi^3 Sum[Sin[(2n+1) Pi x]/((2n+1)^3 Exp[4 (2n+1)^2 Pi^2 t]), {n, 0, Infinity}]
writing name with first letter small is cool
i wouldn't think it's mean then either. some people do approach math like you can just breeze by technicalities every time and that half of the words don't matter. that works in some fields because the technicalities and word choices don't matter so much.
a friend once had to share a workspace with a programmer who left his desk one day and he left up a screen where he was googling up the syntax for something that any programmer in whatever language it was 'should know.' same kinda thing.
i suspect the person is working in earnest, but i suppose my intent it to help them move along rather than just to answer a question
6:32 PM
@leslietownes I guess you would find a function like that only by reverse engineering it? That any problem using exponents, trig, rational functions, etc., will have the nice monotonic behavior?
umm, i would consider myself moderately experienced in a few languages, but always have to look up syntax etc, especially if i switch regularly (as i do).
c++ is a lawyer's dream.
in this context there was no switching regularly, they had been working on one project with one language for several months. i don't know the specifics, but to my friend it felt very out of place.
@copper.hat because everything can be overridden?
@user10478 there are theorems that do supply decay information for families of functions, and usually better 'regularity' (roughly speaking) guarantees better decay. but even many of those theorems would not guarantee that the 100Nth coefficient is smaller than the Nth, for example, just because the decay is in a limit.
@copper.hat I never heard that before
6:35 PM
but yes for commonly combined functions would not have series with behavior like the example i hypothesized.
@robjohn well, if you do anything with templates/move constructors/lambdas, etc, you need to know the details of the constructs and the libraries that use them.
@leslietownes I was not talking about taking a finite sum, but getting a numerical estimate of the whole sum. This often takes the form of a finite sum, but not always.
@robjohn if you read the comment my initial comment was responding to, it was limited to that
and the previous remark about going up to 20,000
i.e., purely truncation of the series up to some N
@robjohn i like the abstractions of lisp (esp. scheme), it lets me get the ideas down in a fairly compact manner. with c++ one has to be aware of many, many details.
@leslietownes Okie, thanks
6:39 PM
@leslietownes I believe that I replied to that.
However, I think that most of the discussion was about this comment
a series can (1) be convergent, (2) have a 1001st term that is big, (3) have a first 1000 terms that are not. in this case, it would be a mistake to say, with reference to the 1000th partial sum, that the later terms are 'always less significant'
that's all
are you series?
@copper.hat he is very Sirius!
let the Sirius B
we may be formalizing my joke and the comments i was responding to in different ways. in any case, i admit that my remark was at cross purposes with the actual series in that problem
the paddle shaft of the Sirius was kept in my home town of passage west for years
6:43 PM
did they remove it?
not sure where it is, might be in a nearby park
because i'm booking plane tickets if it's still there
i went to high school with the grandson of captain roberts
why my brain is full of useless sh*t
there was a period in my life where i was very into books about adventure on the high seas
the wikipedia page for the sirius includes a photo of the paddle shaft :)
@robjohn Where did you get the factor of $2$ that made the $4$ an $8$?
6:48 PM
$(-1)^{n+1}+1$ is $2$ for odd $n$ and $0$ for even $n$; so I restricted to odd $n$ by substituting $n\mapsto2n+1$ and multiplied by $2$
wow, that is the river ferry in the background just down the road. pretty cool.
cobh, touted as the last landfall of the titanic, is just across the water.
Okay, your answer is closer to the book's than mine, but the book has a $4$ there.
8 mins ago, by robjohn
let the Sirius B
That is Sirius
i have starred your message
however, the text input line of the chat window in chrome obliterates the bottom part of the starred messages.
maybe the expectation is that laptops are much bigger than mine
@copper.hat unstar that one and star the source line then
6:54 PM
@Koro where is that?
@copper.hat that is what I see
That is view from my cabin from my new office @copper.hat
They transferred me to the bank of sea ...
@robjohn here is what i see: imgur.com/a/YVPXPZR
US embassy is also close-by.
@Koro nice
6:57 PM
@user10478 somehow, a factor of two is missing. Mine is directly from yours. Did you leave anything out?
@copper.hat Bombay
Someday I will get there :-)
Oh you haven't been?
I want to visit Kolkata someday
unfortunately i had only a month there, too many places to see.
@robjohn Yeah, I see the extra factor now. I guess I need to force Mathematica to check the full answer, or maybe taking the derivatives by hand won't be as bad as I thought since the variables are separated.
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