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6:00 AM
bleh...
I am not sure if it is even possible to sum that alternate series in this fashion:
$(((1-1)+1)-1)+1-\cdots$
One thing I am interested under this framework is to have some notation of measuring how diverging a divergent series is by checking if for each different value obtained due to nonassociativity, whether there is a correlation between some values and the pattern of the nesting of the brackets
 
@Secret no, it is not possible
btw that's the standard interpretation
 
Another thing I am interested in is whether there exists series that have countably infinite many diverging values corresponding to countably many ways of summing it
 
@Secret If a series is convergent but not absolutely convergent, then you can make it converge to anything you want by rearranging the terms
 
That's true, but nonassociativity does not allow rearranging terms, so the question might be a more restricted one
It only allow "rearranging brackets"
 
nonassociativity? of addition?
 
6:07 AM
yup, or more generally any binary operation
 
that wouldn't be "or," since addition is associative
 
not for infinite series
 
@Secret addition is not defined for infinite series
look up how sum to infinity is defined
 
what notion of associativity are you talking about for infinite series?
 
6:08 AM
@anon apparently (1-1)+(1-1)+...=0 and 1+(-1+1)+(-1+1)+...=1
 
that's a classical demonstration of why this series is divergent
 
no, that isn't a proof
 
What I am interested in is then to query how divergent a given series is, and whether it makes sense to treat the different ways of grouping terms together as some notation of path dependence in some space
hmm... actually now that I think about it, it appears that grouping terms in different ways for the above alternating series can give as any interger number as its value is controlled by how many 1s and -1s are left behind and where that infinite tail of 0 starts
e.g. $((1-1)+1)+((1-1)+1)+((1-1)+1)+(-1+1)+(-1+1)+\cdots = 3$
 
@Secret You have rearranged some terms there
 
ooops typo
 
6:21 AM
You can never get more than $1$ or less than $0$ by picking brackets, since this just amounts to picking subsequences of the sequence of partial sums.
 
I see, so there is indeed some kind of pattern in the divergence. Hmm, I am not sure if this property is significant enough to have a name, probably people will explain it in terms of subsequence like what you said
 
@Secret It might be interesting to consider some sort of "average" of the values in the sequence of partial sums (or of those subsequences)
 
Is that how they obtain what is called the Cesàro sum of the above series as $\frac{1}{2}$?
 
@Secret yes
> The Cesàro sum of a sequence is defined as the limit of the arithmetic means of the partial sums of that sequence.
 
@LeakyNun Ahh, I didn't know that was an actual thing. Cool
 
6:30 AM
In mathematics, 1 − 2 + 3 − 4 + ··· is the infinite series whose terms are the successive positive integers, given alternating signs. Using sigma summation notation the sum of the first m terms of the series can be expressed as ∑ n = 1 m n ( − 1 ) n − 1 . {\displaystyle \sum _{n=1}^{m}n(-1)^{n-1}.} The infinite series...
hmm...
(Obviously not realated to Cesàro sum). We cannot say that some given divergent series is equal to the average value of all its possible divergent values weighted by the number of ways it reached said divergent value by picking subsequences, because there are continuumly many thus the average will not work out.

I am not sure if it is even meaningful to sort of construct a space with points indexed by the ways subsequences were picked since at least in a natural sense, there seemed to be no obvious way on how to establish a partial ordering in such space to meaningfully talk about paths
My original idea is trying to came up some notion of topology generated by the different ways brackets are placed and arranged and see if that will give some ideas on how the different diverging values of divergent series depends on topology (if any), however it seemed to be a bit complicated as it now seems
 
7:34 AM
Hey @AlessandroCodenotti and @SteamyRoot!
 
@Daminark Hi
 
Hey @TobiasKildetoft, lol this is becoming a pattern :P
 
Morning
 
Hey everyone!
Hey @Daminark, @AlessandroCodenotti, @SteamyRoot!
 
8:02 AM
How's it going with all of you?
 
Learning how to make a Gantt chart in LaTeX
 
Busy studying for a computer science test tomorrow
 
What's that?
 
@Daminark A chart which describes what work will be done at what parts of the project
Fortunately, these are apparently standard enough that someone has made a package for them
 
Oh sick
 
8:06 AM
hi chat
 
Hai
 
Hi @BalarkaSen
 
I can't decide what I should be studying
I kind of want to do some differential geometry stuff, and then do topologically motivated things (Farb-Margalit say).
But differential geometry always scares the shit out of me
 
Insert Shia LaBeouf meme here
 
8:11 AM
Lol I just lost heart completely for diffgeo, ima work on algebra or k-theory instead
 
@Perturbative I'm gonna dab on you if you do that
 
@Balarka you may dab on your haters but what if they dab back?
 
@Daminark Roses are red / charcoals are black / what do I do if the haters dab back?
 
Yo @Alessandro!
 
8:15 AM
Ohhh shit I just looked up Dolan's channel to see if he fired anything else and I found this
 
Why is he angry at Canada if they allow his video?
 
Apparently that meme was extremely offensive to him.
:P
 
Meme?
 
The sponge-bob thing that got blocked I mean
but yeah I'm guessing he was poking fun at people who get triggered at stuff and hate on the wrong people
I still liked the previous version better, just for the hilarious background music
 
9:01 AM
@Secret That is obviously divergent. That's like one of those division by 0 things.
 
user136984
9:16 AM
Good morning
 
user136984
@BalarkaSen I just watched that out of curiosity... It was weird! :D
 
user84215
9:56 AM
Please visit Planning MSE University ; it has been updated with new posts and conversations.
 
Hi everyone.
Does any one know where I can buy a posters of Mathematicians?
For example, Grothendieck?
 
10:17 AM
If you post a picture of Grothendieck here, then you have become a poster of Grothendieck.
 
Hi, $$f\in C^3(\mathbb R) \text{ with }f' \times f'''<0.
\\\text{ Is it true that : }\forall a,b\in \mathbb R^2, |f(a)-f(b)| \leq |f'(\frac{a+b}{2})|\times |a-b| ?$$
Jensen + : $$g \in C^2([0,1])[/tex] \text{ with }f\in C^1([0,1],[0,1]) \text{ and }f(0)=f(1)=0.
\\\text{Is it true that : }|g(\int_0^1 f(x)\text{d}x) -\int_0^1g(f(x))\text{d}x | \leq \frac{1}{10}\max(|g''|)|\times \max(|f'|)^2 ?$$
 
10:39 AM
Grothendieck's relative point of view is a heuristic applied in certain abstract mathematical situations, with a rough meaning of taking for consideration families of 'objects' explicitly depending on parameters, as the basic field of study, rather than a single such object. It is named after Alexander Grothendieck, who made extensive use of it in treating foundational aspects of algebraic geometry. Outside that field, it has been influential particularly on category theory and categorical logic. In the usual formulation, the language of category theory is applied, to describe the point of view...
Lol, then I think like Grothendieck given how I tend to explosively overgeneralise things
 
11:29 AM
how do we know that sin(n)|n in N diverges
 
@Secret I tend to think that algebra tries to generalise more and more while analysis tends to specify more and more, going down to every epsilon and delta.
 
That might explain why I have a slightly better intuition on algebra, because not many special cases to worry about
 
@LeakyNun Is that a fraction?
 
@WillHunting no
that is a sequence
 
@LeakyNun It's going to oscillate between -1 and 1 forever regardless of what value I think
 
11:36 AM
a|b means a indexed by b
@Secret n in N
 
@LeakyNun Never seen that kind of notation.
 
the naturals is just a subset, so I think the same conclusion holds
 
@Secret argue that for the subset pi N
 
that will oscillate exactly between -1 and 1 forever
because sine is periodic with period 2pi
O wait
that's not cos
bleh, fine, so it does converge for some special values
4
Q: Sine of natural numbers

Matthew LevySo, I want to prove that $$\sup_{n \in \Bbb N}{\sin (n)} = 1$$ I was thinking of proving that some set related to $\pi$ is dense in $\Bbb R$ that will then imply there is some $n \in \Bbb N$ s.t. $\sin(n)$ is as close to $1$ as desired. ($\left( \frac m n \right) \times \pi\quad \forall m,n \in \...

This just showed how much I suck at analysis
 
You want to prove the sequence $(\sin(n))_{n \in \mathbb{N}}$ does not converge?
RIP my LaTeX skills
 
11:49 AM
@SteamyRoot yes
 
The easiest approach is most likely to take two subsequences
 
@SteamyRoot which two?
 
Doesn't matter that much, as long as they're "far enough apart" that they can't converge to the same value
 
Consider multiplying $\pi$ by various powers of $10$ and taking the floor
 
Like, for the first one, pick those natural numbers $n$ such that $n \in [\pi/2-1,\pi/2 +1]$
(modulo $2\pi$ of course)
And for the second one, the same with $-\pi/2$ everywhere
For the first sequence, you'll have $\sin(\pi/2 -1) \leq \sin(n) \leq 1$
and for the second $-1 \leq \sin(n) \leq \sin(-\pi/2 -1)$
Well, technically they're all strict inequalities but that doesn't matter. You then just have to note that $\sin(\pi/2 - 1) > 0$ and $\sin(-\pi/2 - 1) < 0$
So definitely those two subsequences can never converge to the same limit
 
12:27 PM
@WillHunting reminds me of a Terry Tao blog post, where he discusses ultra filters / non-standard analysis as a way to "automate" some of the minutiae of epsilon-delta arguments
(Though if memory serves there's a price, in that the results one gets in this way usually aren't as strong/precise as possible)
He also mentions a post of his discussing hard vs soft analysis which may be of interest
 
12:42 PM
hi chat!
I need to draw a decision tree with over 100 of leaves. Do you know some soft to comfortable do that?
@Mr.Xcoder so, as you are a master of the induction method now (I suppose :) ) and like unequalities, have a look at this: www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/…
 
@Kirill Master >_> - Don't think so
Sorry, don't really have time rn
Will take a look when I have time
Thanks!
 
@Mr.Xcoder sure, you have a good book to start with!
 
@Perturbative I like Shia La Beouf and his wife Mia Goth.
 
hi :)
 
ACTUAL CANNIBAL SHIA LABEOUF
 
12:57 PM
I like Mia's performance in A Cure For Wellness.
I think she looks a bit like me when I was younger. =)
 
1:19 PM
lol, that's the highest on the starboard, really.
 
Hi, $$P\in \mathbb C [x] \text{ with }P(\mathbb Q(\sqrt 2)) \subset \mathbb Q(\sqrt 2).
\\\text{Is it true that : }P \in \mathbb Q(\sqrt 2)[x] ?$$
 
ugh what
 
pick a basis for $\Bbb Q(p_n,\cdots,p_0,\sqrt{2})/\Bbb Q(\sqrt{2})$
@LeakyNun $f(A):=\{f(a):a\in A\}$
 
how to prove that $\displaystyle \lim_{x \to 0^+} \frac1x \exp \left( -\frac1x \right) = 0$?
@anon no, sorry
 
L'hopital.
 
1:32 PM
@Semiclassical no
 
$$\Bbb Q(p_n,\cdots,p_0,\sqrt{2})/\Bbb Q(\sqrt{2})$$
 
it will complicate the function
 
Depends what arrangement you do.
 
wait, it's currently in the form infty x 0
 
Though I think it's a lot easier if you first make the change of variable $u=1/x$.
(That's just a composition of limits thing.)
 
1:34 PM
@Semiclassical $\displaystyle \lim_{u \to \infty} u \exp (-u) = \lim_{u \to \infty} \frac u {\exp u} = \lim_{u \to \infty} \frac 1 {\exp u}$ that's better
 
thanks
 
how did i not think of it
how am i going to cope with university
 
@anon : the problem can be solved with one sentence
 
1:36 PM
I forget. What's the name of a mapping from a set to a subset of itself?
I want to say that $f:A\to B\subset A$ is an injection on $A$ but that seems wrong.
 
@Dattier $\Bbb Q(\sqrt2)$ is a field and a polynomial of degree $n$ is completely determined by its value on $n$ values
 
Bravo @LeakyNun
but it's true only for infinite fields
 
I guess "endomorphism" works.
 
user84215
How can I see that a property is related to intrinsic geometry of a surface or not?
 
user84215
How can I see that a property is related to intrinsic geometry of a surface or not?
 
1:42 PM
Generally a nonobvious question.
 
Another : $$P(x)=x^{2^{2017}}+x^{2^{2016}}+1 \\\text{How many rooth with } Im(z)\geq 0 \text{ and } Re(z)\geq 0 ?$$ Justify, your answer.
 
Even for Gaussian curvature, that's defined as the ratio of determinant of the Ist and the IInd fundamental form, it's not obvious that's a local isometry invariant (i.e. is an "intrinsic property").
 
That sounds like an argument principle problem.
 
Indeed, that is precisely the Theorema Egrergium
 
user84215
for example, why are the eigenvalues of the shape operator not intrinsic properties?
 
1:44 PM
i.e. the number of roots can be written as $\frac{1}{2\pi i}\int_\gamma \frac{P'(x)}{P(x)}\,dx$ where $\gamma$ winds around the first quadrant of the complex plane.
 
@Dattier counterexample?
 
of what ?
 
@Dattier you said it's true only for infinite fields
 
user84215
the above is the answer to my question?
 
Yes. As for your next question, come up with examples of locally isometric surfaces whose principal curvatures are different.
Sounds like a pretty good exercise to me.
 
1:48 PM
$x^{2^{2017}}+x^{2^{2016}}+1 = (x^{2^{2016}}+x^{2^{2015}}+1) (x^{2^{2016}}-x^{2^{2015}}+1)$ dios mio
 
Ah,@LeakyNun in $$\mathbb Z/2\mathbb Z, \alpha \in \mathbb F_4, P(x)=x(\alpha x+\alpha+1)=x$$
 
daaang
 
what is $\Bbb F_4$?
 
@LeakyNun Can you repeat that argument for the next factor down?
also, F4 is the finite field of order 4 (char 2 by necessity)
 
@Semiclassical ugh, no, 'coz like there's a $-x^{2^{2015}}$
 
user84215
1:50 PM
Why can we not conclude from the definition of the curvature of a surface that it is an intrinsic property?
 
sure, but what about the first factor?
 
@Semiclassical then how do we deal with the second factor?
 
one thing at a time, is my point
 
$$\mathbb F_4=\mathbb F_2[x]/(X^2+X+1)$$ @LeakyNun
 
and one might be able to argue that that factor has no roots in the first quadrant? (I may be entirely wrong on that though)
 
1:50 PM
@MathematicsA Because it depends on the 2nd fundamental form of the surface, which is an extrinsic object, not intrinsic?
 
user84215
Why?
 
Why what?
 
@Dattier ugh, finite fields
 
Can't you just solve for $x^{2^{2016}}$ first?
 
what is its additive group isomorphic to?
 
1:51 PM
yes @LeakyNun
 
oh duh
 
is "yes" a group?
 
If $u=x^{2^{2016}}$, then $x^{2^{2017}}=u^2$.
 
user84215
good question: "Why what?"
 
so yeah, that's just $u^2+u+1=0$, with roots $u=e^{\pm 2\pi i/3}$.
 
1:53 PM
Yes, you are adding a question mark after a word. That does not constitute a coherent question. If you want to get answers, ask precise questions.
 
user84215
I do not know enough math to ask my questions correctly.
 
So one just needs to figure out how many of the relevant primitive roots of $e^{\pm 2\pi i/3}$ land in the fourth quadrant
 
@MathematicsA Then you should learn the subject before asking the questions incorrectly.
 
which seems a bit tedious but straightforward.
 
I can recommend textbooks and notes.
 
user84215
1:55 PM
I am reading the Kuhnel's book.
 
@Semiclassical bravo
 
meh, wouldn't have done it without Steamy's hint
once you see that it falls apart
 
What about $$P(x)=x^{2^{2017}}+x^{2016}+1$$ ?
 
hi, can i ask?
 
> should learn the subject before asking the questions incorrectly
::Brain-explosion::
 
1:56 PM
@Dattier not a clue.
Though I know how to find it numerically (yay argument principle)
 
Explanation:
should learn the subject before (you can start) asking the questions incorrectly
vs
should learn the subject, before asking the questions incorrectly
 
With three sentences you can answer
 
$$(x^{2016})^2+x^{2016}+1$$
It's a pseudoquadratic equation
oops typo
 
uh, no
 
$(x^{2^{2016}})^2+x^{2016}+1$
ok nvm, I have no idea
 
2:03 PM
but the same method works for your first polynom (without power of 2)
@LeakyNun yes is a field
 
@Secret oh my god
how did I not notice it
 
No, that's a typo, there's a $2^{2016}$ power in Dattier's question.
 
@Dattier ugh, what
 
$$(x^{2^{2016}})^2=x^{2^{2017}}$$
 
But that's not a pseudoquadratic, because the $x^2$ term is off by the exponent base $2$
 
2:10 PM
@LeakyNun I answer to your old question for the first problem, where you are find the solution
 
This: $x^{2^{2017}}+x^{2^{2016}}+1$ will be a pseudoquadratic
 
@Secret it's not a clue, but a remark
I don't know what's a pseudoquadratic
 
I made up that term to refer to a polynomial equation that looks like a quadratic in disguise
 
@Secret i.e. it's a quadratic in x^p for some p
the generic question is: given distinct positive integers $a,b$, how many roots does $x^a+x^b+1$ have in the first quadrant?
 
Indeed indeed (I am always bad at terminology)
 
2:14 PM
the case of $a=2b$ is quadratic in $x^b$ and therefore can be resolved easily.
 
@Semiclassical I don't know answer to the generic question
 
uh, I have no idea for the generic question. It seemed to be a deg = max(a,b) diophataine equation
 
If $b|a$ then any particular case should reduce to tedious casework.
Otherwise, I have no idea.
in particular, I've no idea how one does the case of $a=2^{2016}$, $b=2016$ (i.e. Dattier's modified problem)
 
@All Do you want the answer or keep looking?
 
Any idea on why here math.stackexchange.com/questions/2403037/… I'm suggested to compute the local Lefschtez number with $I-d_pf$ and not $d_pf-I$?
 
2:18 PM
I suspect the answer had something to do with polynomial rings, which I know nothing about
 
No, the level is licence
 
I really don't much care either way, so you might as well.
 
a and b were even, btw
and indeed if just adding each term one by one, I get no roots
both $x^a$ and $x^b$ are concave up curves centered at zero when $a=2^{2016}$ and $b=2016$
the +1 then shift the whole thing up by one unit
so nothing should be touching the x axis
 
We're looking at the entire first quadrant, though
so that's not very illuminating.
 
a clue : not only even but divide 4
 
2:23 PM
well, all positive even cases (a,b) should have no roots since the $x^a$ and $x^b$ wil be simple concave up curves thus +1 will make the whole thing leave the x axis
The odd cases and case involving negatives is where it gets difficult
 
I guess what you're ultimately aiming is that if $u=x^{32}$ then the equation can be cast as $u^{2^{2012}}+u^{63}+1=0$.
Which simplifies things but not much.
 
do you want the answer ?
 
like I said, I really don't care either way.
 
yeah, too hard
 
You, must reasoning with action and orbit
 
2:27 PM
there's nothing periodic with polynomial powers unless you are in a finite ring?
 
The term 'Burnside's lemma' comes to mind, but I know basically nothing of it so
 
so what kind of action are we expecting?
If $x^a$ and $x^b$ are not elements of a group or even a ring, I don't see if there will be any nice patterns in terms of actions and orbits
 
hmm. actually. i don't suppose it's just as simple as $2^{2015}$ roots?
 
$a=5,b=1$ is a screwy case through
 
my logic being that it's a polynomial of degree 2^{2015} in $x^4$, and that $x\to x^4$ is a surjection from the first quadrant to the entire plane
 
2:30 PM
I'll let you look since you do not want me to give you the answer.
 
ehhh. we're not going to be able to find it numerically.
especially if it really is 2^2015
 
$$(x^{4})^{2^{2015}}+(x^4)^{54}+1$$
 
@Semiclassical But you have say : " I really don't care either way."
 
How does that makes things simpler?
 
@Secret Well, let $u=x^4$. Then that polynomial certainly has $2^{2015}$ roots in the entire complex plane.
 
2:32 PM
ok
 
If I take fourth powers of all those polynomials (in the sense of $(e^{i\theta})^{1/4}=e^{i\theta/4}$) then those 2^2015 roots all all land in the first quadrant.
 
It's simple if you want the solution just say it, otherwise I'll let you look
 
not sure that logic quite works but it's the best I've got.
 
Well, I have nothing to add, thus you can show us the solution
 
indeed
 
2:35 PM
well, if z is rooth then iz is rooth
so $$\text{cardrooth}=\frac{2^{2^{2017}}}{4}$$
because there are no reals solutions, ok ?
 
what is cardrooth? the roots in the 1st quadrant?
 
yes
 
That's basically the reasoning I had above, yeah.
assuming I divided by 4 right...
oh, bollocks.
$2^{2^{2017}}/4=2^{2^{2017}-2}\neq 2^{2^{2015}}$
so yeah. I had the right idea but couldn't do basic exponential arithmetic :/
 
yes, you had the good idea, but I don't have see it
 
yeah. I should've just written down $2^{2^{2017}}/4$.
and not tried to be fancy
 
2:45 PM
Thanks for your participation, see you later.
 
Hey, I'm wondering if there's a general term for algebraic structures where at least one of the underlying sets maps backwards in homomorphisms.
One example is Chu spaces. A homomorphism of Chu spaces S -> T maps states in T to states in S.
 
@TannerSwett What do you mean by one of the underlying sets maps backwards in homomorphisms?
 
Topological spaces are another example. A homomorphism of topological spaces S -> T maps open sets in T to open sets in S (although homomorphisms are usually not considered to explicitly contain such a mapping).
@WillHunting Well, I'm trying to describe what I mean by giving several examples.
 
Oh I see.
 
A homomorphism of locales S -> T maps the underlying set of T to the underlying set of S.
Finally, the simplest possible example, which is kind of a dumb example...
 
2:59 PM
The word 'pullback' sounds right but uh
don't trust me on that
 
Define an "anti-set" as being a set, except that a homomorphism of anti-sets S -> T is a function T -> S.
Yeah, I don't think pullbacks are related. A pullback is like a subset of a cartesian product.
 
fair enough.
I thought the language might be that (in the Chu case) a homomorphism of Chu spaces S-> T pulls states in T back to states in S.
 
Hmm. I don't actually know what the verb "to pull back" means.
 
yeah. it's more suggestive than anything. so caveat emptor
 
3:12 PM
The opposite of "to push forward"?
 
I don't know what that means either. :)
 
It's simple, you push or pull the door. =D
 
12 hours ago, by Semiclassical
YEAAHHH
 
Hey @Semiclassical I was wondering if you would like to share emails with me? If not it's OK.
 
maybe at some point, but not right now? in general I've kept my email away from MSE and for now I'd like to keep it that way
 
3:24 PM
Hmm OK, mine is jasperjloy@gmail.com, if you want to know.
 
kk
(tbh one could probably use the info I've said throughout the years here to figure out who I am off-line, and from there it's easy to get my address. but eh)
 
Oh I don't do that kind of thing.
 
nah, I just mean that I'm probably being a bit silly
given that if someone was really determined to email me off-line they could do so
 
Although I do know the futility of emails. I have had so many contacts that simply vanish after a while. They don't bother to respond for several months.
 
I'm terrible about email if I'm honest.
 
3:28 PM
how bad are these (bouncing ball)
elastic rebound
 
What is the question actually?
 
Probably what one should do is write the time in air for a given bounce as a function of the max rise of that bounce, which is itself proportional to the max potential energy (and thus max kinetic energy) in each bounce.
Since the time for an object of mass $m$ to descend from a height $h$ undergoing gravitational acceleration satisfies $h=\frac{1}{2}gT^2$, the total time for one bounce would be $2T=\sqrt{2h/g}$.
But the potential energy at max bounce is $U_{max}=mgh$, and $U_{max}=K_{max}$. So $2T= \sqrt{2h/g}=\sqrt{2/mg}\sqrt{mgh}=\sqrt{2K_{max}/mg}$. Hence the time in air for each bounce varies as the square root of the kinetic energy after rebound.
 
semi u talking to me?
@WillHunting asking if the graphs make sense, for my simulation
 
That should be enough info to deduce what fraction of the ball's energy is lost on each bounce.
yep
 
its elastic
why would it lose a fraction of energy
 
3:35 PM
If it's elastic, then the bounces would all be identical.
They're clearly not in yours.
 
it really is, is something to do with gravity
 
If no energy is lost on each rebound, then each bounce would have the same initial position and the same initial velocity. Hence each bounce would be identical.
This is not affected by gravity, since that's a conservative force.
So if your bounces are not identical, then you are not properly simulating a purely elastic rebound.
 
May I ask a new question?
 
@Kanishk Here in chat? Go for it.
I usually don't worry about interrupting other conversations. Anyone who's not interested in a new question can just ignore it.
 
3:53 PM
$$f(x)= \lim_{n\to \infty}\int_0^{f(x_0)} \tan\int_1^{f(x_1)} \tan \int_2^{f(x_2)}tan \int_3^{f(x_3)}\dots\int_n^{f(x_n)}\tan(f(x_n))\,dx_n\,dx_{n-1}\dots dx_0$$
In this question we have to find solutions of f(x)
f(x)=0 is one answer
I can't figure out the other possible solutions
Most important thing that f(x) is continuous and differentiable for all positive reals.
 
4:09 PM
@TannerSwett Exactly, though some folks here get irritated very easily, LOL.
 
4:48 PM
I've got another category theory question, and this time it's very broad.
How can categories of algebras be built out of Set?
One example of a "category of algebras" is the category of pointed sets.
This category is just the coslice category 1/Set.
Another example is the category of sets equipped with functions into themselves.
This seems to be a little more complicated to build. Start with the comma category id_Set/id_Set. Take the full subcategory of this where the left and right sets are the same.
 

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