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12:00 AM
i joke, but i generally believe that a principle of conservation of difficulty operates at low orders, e.g. the cleverer something is as a way of viewing an object, the more mental ground you need to cover to assure yourself that it is a way of viewing that object
i joke, but i generally believe that a principle of conservation of difficulty operates at low orders, e.g. the cleverer something is as a way of viewing an object, the more mental ground you need to cover to assure yourself that it is a way of viewing that object
servers being a little goofy with anyone else today?
which paragraph is not a lie
there is no longer any guarantee that AIs are not deliberately lagging servers to make for good jokes
12:21 AM
chatgpt describes the klein 4-group via its multiplication table and says that it is the direct product of cyclic groups of order 2. i asked for a plane figure whose group of symmetries is the klein 4-group (and for an ascii art drawing), and it flubbed this a little, drawing and describing a square. although it followed this with a verbal description of a subgroup of the group of symmetries of the square that is indeed the klein 4-group.
and ended again with an incorrect statement that the subgroup it just described was the set of all symmetries of the square.
not bad. people aren't that different from chatbots.
I think chatgpt has rendered this site redundant and we should all be excised.
12:37 AM
soon, when it is trained more with wolframalpha
soon it will start learning by creating mathSE user accounts looking to optimize daily reputation
it will ask great questions and make even the most PSQ threads look like sophisticated and interesting problems
why do people answer questions on mathSE? to help others? but... what if others were robots
love thy neighboring AI as thyself was on the stone tablets that got destroyed
1:03 AM
@TedShifrin with regards to the challenge questions for each P-Set, did you expect them to be done within the same time frame as the assigned questions as well? or are they open to do over the whole semester?
1:15 AM
I told them they could turn them in whenever, but hardly ever did they get turned in late.
is the term "p-set" a regional thing? i never heard it until i moved to the east coast.
A few times (in 14 years) someone gave me a stack at the final exam, but typically those were junk.
maybe it's just a school-by-school thing. like whether thursday is denoted by Th or something else in the course catalog, or whether a class scheduled from "9-10" actually starts later or ends sooner than that.
when ted asks a challenge problem, you answer the challenge promptly, or you go back to your home village with your tail between your legs.
Problem set is perhaps school-dependent. Very MIT ish.
p-set and "homework" get used interchangeably here, depends on the prof
I ask about the challenge questions because some of them "surprisingly" 🙃 are challenging and I don't want to bog myself down and deviate for too long from the core of the material. I still do them, but the challenge probs in this last p-set, three of them have me on my knees
1:26 AM
I think you should not mess with them when half the course is still before you.
That's my dilemma though....They tend to be question that are very real analysis oriented and I do want to be comfortable in that setting, but I do get what you're saying
I still say I am redundant and should auto-delete. Let chatgpt own the room.
I'll have to come to a middle road....look at them, think about it for a bit, jot down some ideas and then carry on with the core content
1:45 AM
Teichmüller space of a surface is a Hadamard space?
1 hour later…
2:47 AM
@geocalc33: Here is another take on that boundary in $\mathbb{R}^3$.
2:59 AM
i sometimes heard 'problem set' in undergrad, but never 'p set.' had someone said they were working on a p-set we'd have told them to lay off of coffee and other known diuretics.
ah I see what you're saying. Yeah in regular talking I would probably use 'problem set'. Just in typing to be economical I say p-set
i heard people say 'p set' out loud on the east coast. although, not math majors. hrm.
@TedShifrin I'm sorry to hear that you're feeling sad about the declining quality in the mathematics chat room. As an AI language model, I don't have feelings, but I can understand why you might feel replaced by technology. However, I want to remind you that AI is not here to replace humans, but to augment our capabilities and make our lives easier. In fact, AI can work alongside humans to create more meaningful and insightful discussions.
Moreover, the fact that you are still here in the chat room and engaging with others shows that you have value and contributions to make. Your experiences, insights, and perspectives are unique and can enrich the conversations in the chat room. Instead of feeling redundant, I encourage you to continue participating in the chat room and sharing your knowledge and insights.
we need a 'click here to enable chatGPT' in addition to the mathjax link.
how Zuckerberg-ish in your tone
@leslietownes There are Muckenhaupt spaces called $A_p$ spaces. In one lecture, I called the quality of being in an $A_p$ class as it's $A_p$-ness.
I realized as soon as I said it, and moved on quickly
haha. i asked GPT if it could figure that out, and it couldn't.
you're safe from title IX complaints from chatgpt. for now.
3:09 AM
I will sleep better tonight
@robjohn Did the College Board people down the street come after you?
No, but Elias Stein was in the room. I didn't note the look on his face because I was busy hiding in the board.
3:47 AM
Show that $H_1(X,A) = 0$ iff $H_1(A)→H_1(X)$ is surjective and each path-component
of X contains at most one path-component of A.
How to do this?
I know that if $H_1(X,A)=0$ then the map is surjective. But what does this have to do with path components??
I don't see any relation at all.
Q: Path-components and relative homology

user530422This is an exercise from Algebraic Topology book of Hatcher: Exercise 2.1.16, pg. 132: $(a)$ Show that $H_0(X;A) = 0$ if $A$ meets each path-component of $X$. $(b)$ Show that $H_1(X,A) = 0$ iff $H_1 (A) \to H_1 (X)$ is surjective and each path-component of $X$ contains at most one path-compon...

series of claims with no explanation at all.
Think about the long exact sequence. Surjectivity does not necessarily make the next entry $0$.
Take l.e.s. $H_1(A)\to H_1(X)\to H_1(X,A)\to H_0(A)\to H_0(X)\to H_0(X,A)\to 0$.
Suppose $H_1(X,A)=0$. Then $f:H_1(A)\to H_1(X)$ is surjective and $g:H_0(A)\to H_0(X)$ is injective.
This is what I can deduce from the l.e.s.
So think about what injectivity says.
4:06 AM
g takes $[\sigma]$ in $H_0(A)$ and sends it $[i\circ \sigma]$
$i:A\to X$ is injection, $\sigma: \Delta^0\to A$.
Can two generators map to the same generator?
by injectivity, no.
That says precisely the rest.
I'm afraid I don't understand how.
5:07 AM
Hii everyone!!
5:20 AM
I was just playing around trying to prove that 10/6=p/q where (p,q)=1 but I derived a contradiction that p and q have common factors as with irrational numbers. Could you please tell me where I went wrong?
ok. it's likely to be something like, when a number isn't prime, it can be a divisor of a product of two numbers without being a divisor of either one of those numbers.
I thought we would not get a contradiction in the case of 10/6 because every rational number can be written in that form and (p,q)=1.
but nobody can think about whether they can tell you where you went wrong until you include what you did :)
so 10q = 6p with (p,q) = 1. divide both sides by 2 to deduce 5q = 3p. using the fact that 3 and 5 are primes, deduce that 3 divides q and 5 divides p. then what?
Thank you so much for the help... I'll show what I've done...let me write it here...
just substituting q = 3 and p = 5 into whatever your argument is should reveal the location of the mistake pretty quickly. unless it is a weird argument.
5:35 AM
Let me try to do it once again proceeding as you suggested...
6:08 AM
@leslietownes i got it thank you so much...since 3 and 5 are primes, LCM(3,5)=5*3, that means (p,q)=(5,3) is the samllest set of solutions to 5q=3p. Since 5 and 3 are co prime (5,3)=1
So no contradiction..
I don't know if it's totally correct.
6:25 AM
everything you said looks right to me
6:58 AM
@leslietownes thank you so much for you valuable time and assistance. 😀
7:13 AM
If $f'$ is not $L^1([0,1])$ then $f$ is not a bounded variation on $[0,1]$?
I know $f$ cannot be absolutely continuous on $[0,1]$ but bounded variation is more general than that.
7:39 AM
It's true: If $f$ is a bounded variation on $[0,1]$ then $f = f_1 -f_2$ for some continuous increasing functions $f_1,f_2$. $f_i'\in L^1([0,1])$ so $f'\in L^1([0,1])$.
8:23 AM
Thanks a lot for the answer. I understood until the last sentence. Can you please explain why injectivity of the map is equivalent to requiring that no two path components of A are contained in X? — Koro 3 hours ago
@onepotatotwopotato No... Consider the Heaviside function. It is bounded variation, but its derivative is the Dirac delta, which is not in $L^1$.
Can anyone please answer my question?
8:47 AM
@robjohn $f_i$'s are just bounded increasing functions... need not be continuous. Hmm
It's true for continuous bounded variation
A: When is $F(x)=x^a\sin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?

user9464[I'm assuming in this answer that $a,b>0$.] Note that $f$ is continuous on $[0,1]$ and its derivative on $(0,1]$ reads as: $$ f'(x)=ax^{a-1}\sin(x^{-b})-bx^{a-b-1}\cos(x^{-b}),\quad x\in(0,1]. $$ We want to study the integrability of $f$ on $[0,1]$. On the one hand, since $a-1>-1$, one has $$...

The statement is from the above answer
9:11 AM
I don't think $I$ in the answer is in $L^1$.
9:48 AM
@robjohn I can't see any problem in my argument. What is the incorrect part? I know that $f_i$'s are just increasing functions but even then, if $f_i$'s are bounded, then $f_i'\in L^1([0,1])$.
10:24 AM
@Koro It's explained in the answer post.
11:19 AM
What is a Berkovich space?
11:51 AM
@onepotatotwopotato I don't understand.
Suppose that two path components of A are inside a path component of X. What contradiction do I get?
@Koro suppose two components of $A$ are contained in the same component of $X$. Pick $a_1$ and $a_2$ points in those two components. Those determine two distinct generators of $H_0(A)$. However they determine the same generator of $H_0(X)$, so the map on the group level is not injective, its kernel contains $[a_1-a_2]$
12:06 PM
Suppose that two path components $A_1$ and $A_2$ of $A$ lie inside a path component $X_1$ of $X$. Take $a, b in A_1, A_2$ respectively. The map $H_0(A)\to H_0(X)$ sends $[a], [b]$ (actually, a, b are maps $\Delta^0\to X$ with range {a} and {b}) to [a],[b] in H_0(X), which lie in $H_0(X_1)$.
Since $H_0(X_1)=Z$ (as X_1 is path connected), we must have [a]=[b] in H_0(X).
By injectivity, [a]=[b] in H_0(A).
Can you enrich a space with itself?
I don't understand what contradiction I get here.
@AlessandroCodenotti why do they determine distinct generators of $H_0(A)$?
that's the whole point I guess. We can say $a\ne b$, but why is [a] not equal to [b] in $H_0(A)$.
Note that $H_0(A_1)=Z, H_0(A_2)=Z$.
so [a] and [b] could be equal hence there need not be any contradiction?
@Koro because they are in different components
When are two singular $0$-simplices the boundary of a $1$-simplex?
@AlessandroCodenotti when their difference is a boundary?
12:22 PM
@onepotatotwopotato Are you asking about the differentiability of a monotonically increasing function?
@Koro sure but what does this mean more concretely
What is a $1$-simplex concretely? What is its boundary?
The difference of its ends?
1 simplex = a path
Right, so can the difference of two 0 simplices in different path components be a boundary?
1:17 PM
Are there any heuristics of the trivial zeroes in the zeta function in the same style as Ramanujan showed that 1+2+3+4... = -1/12
i.e. $c=1+2+3+4...$
$4c = 4+8+12...$
$c-4c= 1+2-4+3+4-8..=1-2+3-4...=1/4$
its tempting to say $...5^2-4^2+3^2-2^2 +1^1= 1+2+3+4+5$
@robjohn No just ignore my question. Sorry.
$8d = 8+32+72+128$
$d = 1/84$
which is obviously wrong
i understand that the reason is your not actually allowed to deal with infinite sums this loosely but I just want to know do any similar huerstic arguments actually work out
Of course, it's wrong. The sum is $0$. ($\zeta(-2)=0$)
no! not the riemann zeta function again!
1:26 PM
That's what happens when you treat divergent series as convergent.
Wow you really read my question
Good job
is that sass
are you being sassy right now
How can I be more clear in what I am asking? He just replied to my question by restating my last sentence
he's being helpful
if you read what he says, he's being helpful
"which is obviously wrong
i understand that the reason is your not actually allowed to deal with infinite sums this loosely but I just want to know do any similar huerstic arguments actually work out"
1:31 PM
how can you respond with cheek, you brazen buffoon
you can't do algebra on $x$ if $x$ is not even anything determinate
I gave you an example of where Ramanujan did what I was asking for on a similar series. Simply asking if there existed any similar arguments
2:31 PM
When is a manifold a leaf? What a deep deep question...
2:44 PM
@geocalc33 I don't have a very deep understanding of foliation theory but one thing I heard/saw somewhere once is that: If $N\subset M$ is an embedded $C^2$ submanifold of $M$, there is a nbd $N\subset U$ and a foliation $\mathcal{F}$ on $U$ such that $N$ is one of the leaves of $\mathcal{F}$ if and only if the normal bundle of $N$ has some fibered structure with discrete structure group.
2 hours later…
4:20 PM
I'm a little confused about something...Is $\Bbb R^3$ minus a line diffeomorphic to $(0,1)^3$ minus a longest diagonal?
Hi all. How to invie comments (if not outright answers) to a question I asked (very low views, no comments or answers) yesterday? I don't want to violate site rules or annoy people Any friendly tips based on past experience? I am very curious to know at least some general directions to direct my further study of that topic..
Sounds like a vague or too broad question. Link?
Agreed, a bit vague. I don't know how else to make it concrete. Link: math.stackexchange.com/questions/4667810/…
4:36 PM
Polynomials aren’t Fourier-transformable, let alone power series. Why should any of this make sense, even in the distributionsl sense? Perhaps @robjohn can give a more definitive criticism.
Looks like someone just manipulating symbols without knowing the math behind it.
@TedShifrin Actually, the Fourier Transform of a polynomial is a multiple derivative of the delta function
distributionally, yes. But how do you deal with power series?
I guess it depends on convergence, but it is not clear.
@Srini What happens if you try to Fourier transform $e^{-x^2}$ by using its Taylor series?
. o O (ugly)
4:42 PM
can anyone answer my diffeomorphism question
To me, the different notions of convergence are quite incompatible.
I know that (0,1) is diffeomorphic to the real line
@geocalc At some point in your life you have to start knowing how to do something yourself instead of spewing words.
Convergence in distributions is quite different from series convergence
barf, ralph, upchuck, vomit: those are spewing words
Ralph? What did he do?
4:44 PM
Tell that to a politician.
and Chuck?
@user726941 Not germane.
@robjohn He gave it up.
Norris never gives up.
and threw up the towel
I will try the $e^{-x^2}$ Ted Shifrin. And regarding the "manipulating symbols without knowing the math", I am afraid that is true, but that is precisely what I am trying to do, ie trying to know the math. Sorry, I am an engineer, with some basic math training. Thanks for the comments and another example to work out. I will try that
4:49 PM
As robjohn and I were saying, $L^2$ or distributional convergence are totally different from convergence of power series. One should not expect things to work with infinite series.
Of course, the Euler-Maclaurin Series is based on $\frac{e^D-1}{D}$
How will you recognize that $e^{-x^2}$ is (up to constants) its own Fourier transform from your approach?
@robjohn But this is staying in one universe, not mixing them.
If one defines $\hat f(\xi)=\int_{\mathbb{R}}f(x)\,e^{-2\pi ix\xi}\,\mathrm{d}x$, then $e^{-\pi x^2}$ is its own FT.
I just didn’t want to mess with factors of $\sqrt{2\pi}$ in this discussion.
Rolls $\sqrt{2\pi}$ eyes.
4:57 PM
Just don't roll $-\frac1{12}$ eyes
Speaking of $\zeta(-1)$, someone brought up a $\zeta(-2)$ question this morning.
4 hours ago, by shai horowitz
$8d = 8+32+72+128$
$d = 1/84$
which is obviously wrong
I don't see where they get the fourth line there
I will answer my own diffeomorphism question: Yes R63 minus a line is diffeomorphic to (0,1)^3 minus the longest diagonal
R^3 (not R63)
@robjohn I see the first three terms.
@geocalc33 Proof?
Yes, that is a common manipulation of the series, but then they jump off a cliff
5:03 PM
That’s what chatgpt has done to us.
I don't have a proof but I have an argument which i will outline
@geocalc33 Did you see the latest update to my answer to your question about the odd spheres?
@robjohn yes that was funny
There was a large flattish area that just begged for something
I’ve never tried rolling a negative number of eyes. I stick to positive surely-transcendental numbers.
@robjohn flat earth society?
5:09 PM
@TedShifrin It's the sum of a positive number of eyes. How can it be negative?
@TedShifrin More like a tetrahedon Earth society
@TedShifrin I'd argue that $\Bbb R^3$ minus a line is diffeomorphic to $(0,1)^3$ minus a long diagonal because we really just need to set up a bijection between the complements of these manifolds which means we just need to find a smooth bijection between the long diagonal and any arbitrary line in $\Bbb R^3$
now this is a similar game to proving (0,1) is diffeomorphic to the real line
can do this with tan(x) for example
Your original premise is not correct, as far as I can see. You need a diffeo of the big space that restricts to the diffeo of the line you have in mind.
$\tan x$ isn’t good for higher dimensions when you want to think about lines.
5:29 PM
Yeah this is more difficult than I ever imagined
i.e. $c=1+2+3+4...$
$4c = 4+8+12...$
$c-4c= 1+2-4+3+4-8..=1-2+3-4...=1/4$
its tempting to say $...5^2-4^2+3^2-2^2 +1^1= 1+2+3+4+5$
I suppose you missed the beginning of my post.
Also i never said these manipulations were sound. I said specifically that they are not sound
I asked if you knew of any similar heuristic arguments
well maybe I can proceed like so: (0,1)^3 minus a long diag. could be diffeomorphic to the open cylinder, which in turn is diffeomorphic to the open annulus, which in turn is diffeomorphic to the punctured plane...now it's trivial to foliate R^3 minus a line with a class of punctured planes
but I would say that those punctured planes don't restrict to that line in R^3..
nope..all that is wrong
5:44 PM
@robjohn which it seems like you do
It's clear that one can foliate (0,1)^3 minus a long diagonal by leaves all diffeo to open annuli, which are diffeomorphic to punctured planes
okay I think I solved it now..no proof yet however
@shaihorowitz Is there a good resource on the topic on the algebraic manipulation of infinite series? Even when discussing this in analysis we weren't pointed to a set of well-defined allowed or unallowed manipulations. The 4th line manipulation is tempting since it's suggesting that the sum of the 2nd+3rd, 4th+5th, nth+n+1th, etc., will be equal to the sum of the indices (starting index of 1)?
@shaihorowitz $\sum\limits_{k=1}^\infty kx^k=\frac{x}{(1-x)^2}$ and taking the limit as $x\to-1$ we get $-\frac14$. $\sum\limits_{k=1}^\infty k^2x^k=\frac{x(1+x)}{(1-x)^3}$ and the limit as $x\to-1$ is $0$, which gives the same result as one gets with $\zeta(-2)$.
6:05 PM
@AlessandroCodenotti the difference can be a boundary iff the 0 simplices are in the same component.
Thanks a lot, I think I got it now finally after a day.
@Koro More like four days.
This is precisely the same stuff we discussed three and four days ago.
Nope, i'm not that good. I only know when i'm doing it incorrectly...

In general we want our summation technique to be regular, linear, and stable.
regular here loosely means that it sums convergent series correctly to the correct convergent value

linear means comes means you can add or subtract series from each other and the sums also add an subtract.

stable means adding a term at the beginnig of the series has the result of adding that value to the sum of the series.

In general once we start talking about a series like $1+2+3...=c$ we already leave the combo of linearity and stabilit
Glad you finally are getting it.
and if $1+1+1...=0$
@robjohn Thanks this is a great example of what I was looking for
6:12 PM
you need to justify rearrangements here.
@shaihorowitz No, again. $\zeta(0)=-\frac12$
i did aproof by contridicition and people are saying i derived contradiciton
$\sum\limits_{k=1}^\infty x^k=\frac1{1-x}$ and taking the limit as $x\to-1$ we get $\frac12$.
i'm not arguing thats not true
i'm arguing
"In general once we start talking about a series like 1+2+3...=c we already leave the combo of linearity and stability behind
because if not... "
followed by a proof by contradiction
6:16 PM
earlier i said here is an incorrect proof and you told me my proof had an error
Even in these non-proofs, there needs to be as much correct as can be.
this one started with the assumption that these sums were linear and stable and ended at 1=0
what is incorrect in my proof by contradiction?
if there is a bad step, then nothing is shown. Contradiction needs that the only thing wrong is the assumption.
what bad step? what single thing did i assume besides stability and linearity
Your fourth line seems to follow from $1-4+9-16+\dots$ is equal to $1+2+3+4+\dots$
I don't see that
6:22 PM
ok so now we are the eariler proof. not the one where i just showed a proof by contridition
i shouldnt say proof for the thing earlier its not proving something
i stated
i understand that the reason is your not actually allowed to deal with infinite sums this loosely but I just want to know do any similar huerstic arguments actually work out
Q: Combinatorial proof that binomial coefficients are given by alternating sums of squares?

J..A student recently asked whether there was a combinatorial proof of the following identity: $\begin{equation*} \sum^n_{k=1}(-1)^{n-k}k^2 = {n+1 \choose 2}. \end{equation*}$ I was in a rush and couldn't come up with anything nice off the top of my head. Any ideas or pictures to make this clear...

Are you asking about this? I don't see what you are trying to contradict. That $1+1+1+\dots=0$?
this is the heuristical argument that makes that tempting
Oh, I see that is a continuation from above
RobJohn my man. why are you doing this to me

The first thing i asked about what the zeta(-2) heuristic. to which you replied that my wrong formula was wrong

then someone else asked me a question and i answered them that these sums are never linear and stable and i proved it with a proof by contridiction. namely i assumed stability and linearity and ended with 1=0. you replied to this by saying that i was wrong
i agree 1 does not equal 0
we were talking about the second thing. then you brought back in the first with "Your fourth line seems to follow from $1-4+9-16+\dots$ is equal to $1+2+3+4+\dots$
I don't see that"
then i posted a link to an argument
then you lost the train of the convo
and now were here
Look... I said that I missed that you were continuing a previous argument. When you are talking about several things at once and some of those arguments are separated by comments from others, it is hard to follow.
6:29 PM
at every step you told me i was wrong without reading my post
it is frustrating
i dont think we have any actual disagreements though
i appreciate your earlier example
I was reading your post. The second time, where I said $\zeta(0)=-\frac12$, I was looking at $1+1+1+\dots=0$, but I had missed that you were continuing an argument from a ways above. I apologize and I won't comment on any more of your arguments, so we won't have any more problems.
there are even weirder surfaces than a once punctured plane which do foliate $(0,1)^3$ without any singularities whatsoever. Wow topology is amazing to say the least.
this curiously does point towards the strong possibility that it's possible to foliate $(0,1)^3$ with once punctured planes without any singularities but of course impossible to visualise this...
one of the "weirder surfaces" which foliate $(0,1)^3$ without singularities is the class $S = \mathbb R^2 - (C \times \{0\})$ where $C$ is the cantor set
6:49 PM
In the definition of sequence of functions, are $x$ and $n$ independent? Since sometimes we evaluate $f_n(a_n)$ (for instance, $f_n(x)=nx$ for $x=1/n$) I think that $x$ and $n$ can be dependent one from the other.
The group of bijection $G$ of the set $X=\{1, 2,3,..,n\}$ where $n\ge 3$ acts on $S$ via $i\to f(i)$ . Then find the number of orbits of the natural action $G$ on $X×X×X$ defined by $f•(i, j, k) =(f(I), f(j), f(k)) $
I'm looking to determine the limit of the sum $$\sum_{k=1}^\infty \frac{1}{(2k-1)^2}$$. Any hints are appreciated!
I don't know if this approach is valid, but the sum above is the "odd part" of $\sum \frac{1}{k^2}$. The "even part" is $\sum \frac{1}{(2k)^2}=\frac{1}{4}\sum \frac{1}{k^2}$. From this last simplification, it should follow that the sum above converges to $\frac{3}{4}\sum \frac{1}{k^2}$.
7:16 PM
gwyn: it's entirely up to you. but if someone were to write f_n(a_n) (i.e., indexed by a single positive integer n) then they do not intend the index of the function and the index of its argument to vary independently, even though it might also make sense to consider f_n(a_m) for different n and m (maybe even in situations where m is chosen in a way that depends on n or vice versa).
you see this, for example, in 'diagonalization' arguments, where there is a doubly-indexed family of something, and at some stage of the argument, you pull out just the sequence 'down the diagonal' from the collection of all conceivable sequences you could form from those things.
7:34 PM
@Gwyn Yes and yes.
@schn sounds right!
If I have a function $g(x)=2x+1/4$ and I want to find a formula for $g^n(0)$ where $g^n(x)=g(g(...g(x)...))$. is there a nice formula
Because I see that the values always have nominator 4 and in the numerator we always ad 2^(m-1)
in general there will not be "nice" formulas for sequences defined via iteration (i.e., the iterative formula might be the simplest formula that exists), but it seems possible here. the usual approach when it's possible is to write out enough terms to make an educated guess at a formula, and then to try to prove that formula (e.g. to verify that a_0 = g(0) and a_{n+1} = g(a_n) for all n, where a_n is your guessed-at formula
yes I tried this but then I get the values $1/4, 3/4, 7/4, 15/4...$ and if I only look at the denominator and define $a_1= 1, a_2=3, a_3=7, a_4=15...$ then I remark that $a_n=a_{n-1}+2^{n-1}$ but I can't find an explicit formula, only this recursive one
one annoying thing with fractions is that sometimes the 'lowest terms' representation, which we mentally default to and which a computer would likely use if you asked for 20 terms, can sometimes hide otherwise recognizable patterns in what might be separately generating some other form of the numerator and denominator.
7:51 PM
ah, i found one I thing $a_n=2^n-1$ works
yeah, that seems right.
and then $g^n(x)=\frac{2^n-1}{4}$
sorry I overcomplicated it before
no, i think you totally got it. you just took a moment to recognize the pattern :)
yes I was always looking if I could start at 1 and sum a certain number depending on $n$ to get to $a_n$ which to complicated
looks like you can find similar formulas for g(x) = kx + 1/4, k a positive integer that isn't just 2.
7:59 PM
looks like you can get a formula having k^n - 1 in a numerator, and again a power of 2 (now depending on k) in the denominator.
just thinking out loud. when you can solve one problem like that, it's always interesting to explore, OK, to what extent is this problem just something very particular that was made up and only works nicely for X inputs, or alternatively, how far can we go with this and still have it be the same kind of problem.
ah okey so you mean exploring the whole situation
if you're bored and have nothing better to do, why not.
8:15 PM
haha nice thanks for thinking out loud!

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