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12:00 AM
That's exactly what breaking up the sum as I said does for you.
GRR, stupid chat.
 
Yeah, but I was computing $A_{n-1}$ using the first recurrence formula and then pluging it in the general $A_n$ case, and then computing $A_{n-2}, A_{n-3}...$ etc.
 
Well, you still want to proceed with the induction as I framed it.
 
Yes, definitely.
 
@Shaun are you on & available for study?
 
@PenAndPaper The first question you can certainly ascertain yourself easily enough!
 
12:05 AM
By on I mean like around at home and then they'll receive a text ping and get back to me here
 
A text ping if you're not logged on this site?
I get pinged only when I log in here.
 
It was an abuse of language. Not sure how their tech works
We're trying to learn something from Sheaves in Geometry & Logic (Topos theory) by M&M
Hopefully, once we know enough of the basics we could one day understand what Johnstone's book is about
 
I wish we were able to change the ping sound. It gives me a scare
 
You could turn down your volume a smidgeon :)
 
$1 \ \iota$
 
12:17 AM
1 monomorphism?
 
Yes! One that is an inclusion
K, Shaun is probably busy. :>
 
Sorry, @PenAndPaperMathematics. It's 00:22 here. I'm quite tired.
 
G'night @Shaun.
 
12:50 AM
G'night @TedShifrin :)
 
 
3 hours later…
4:02 AM
Give the best real analysis lecture or book
I hope there are mentally retarded edition for people like me
 
4:25 AM
meth, i don't appreciate that phrase, as i think i have said before. i like the first half of walter rudin's book. chapters 1-7.
most of chapter 2 can be skipped.
there's a good part of chapter 1 that is also expressly flagged as optional.
 
Rudin is terrible unless for future mathematicians. Try Abbott or Wade.
 
4:43 AM
this is what george w bush, my personal hero, referred to as the tyranny of low expectations.
 
Mr C- average.
 
oh god, i looked it up, the phrase is 'soft bigotry of low expectations.' most of what is wrong with america can be traced to A+ students at schools that people like him went to, trying to outdo each other in what they think is debate class but is actually the lives of other people.
for more on my radical leftist views, consult my podcast, leslie speaks.
 
I don’t recommend my books for weak students. Gotta be realistic.
 
i was annoyed when i looked at your free book and saw that it was really good. i was hoping to needle you about at least something.
 
I’m sure you can succeed if you try.
That’s perhaps the most accessible of the four.
 
4:50 AM
i can take 20,000 potshots at rudin, and that's a book that some of which i like.
 
5:18 AM
How about Zorich analysis? I read it before and it was good.
 
thomas koerner's companion to analysis is also very good.
 
What about Spivak?
It seems to be a nice book for high-school students.
 
@leslietownes Google says it's GTM
 
Spivak is kinda hard
@Prithubiswas I would start with Stewart first
 
@mohan10216 I was talking about a good book for basic-real analysis.
 
5:23 AM
oh ok, oops.
Then Tao's analysis I seems good
it's the book i'm learning from now and I find it very nice and basic
 
love: i do think it is targeted at people who have thought about calculus before but it does not strike me as a graduate-level text.
exercises are broken into granular pieces. i think it is an undergraduate level book.
but i have never taught out of it.
 
You're a teacher?
 
i used to. i work elsewhere now.
 
I thought you were a student or smth.
 
i once taught out of ross 'elementary analysis: the theory of calculus.' it is OK for students who maybe only go on further to teach high school. it is written, i think, for that audience. which is not to say that it isn't rigorous, it just doesn't include a lot of pointers to higher levels of abstraction.
ajay i think my joie de vivre makes me seem younger than i am, or maybe even younger than i look. although i do look young.
if i buy a beer in a restaurant they ask for ID. maybe my wife pays them to do this to flatter me, i don't know.
 
5:31 AM
Don't call me Mohan please, it's my dads name, my name is Ajay.
 
i have corrected the record. :) thanks.
 
np :)
Quick question: What book do you think is the most basic for learning real analysis? In my mind it's tao's analysis.
 
tao's text is very good.
i met him once, he also seems like a nice human being. which you can't always say about authors, even good ones.
 
That's awesome! What did you guys talk about?
 
i think the furniture? i have a talent for finding the most extremely irrelevant thing and focusing on that.
 
5:35 AM
oh, I thought you guys talked about cool math stuff, lol.
 
one time in my current job we were visiting with the client and there was talk about a bird outside, and i went with a staff member to look at it and we spoke about it. it turns out the 'staff member' was the CEO. i had this conversation with my boss.
'what did you talk about?' '.. the bird that we saw.'
we got the job, however.
 
prime example of why it doesn't hurt to go off tangent once in a while.
 
i normally keep bird stuff kinda secret because birding marks you as a weirdo, but this guy really wanted to go see it.
he thought it was a rare morph of a kestrel but it was pretty obviously a merlin. we watched it eat a pigeon. this kind of stuff is super boring to people who aren't bird people.
 
There's nothing weird about that, bird watching seems fun, my Grandmother does it a lot. However i've never done it coz my mom is terrified of birds.
 
some people seem to be targeted by them. my best friend has been attacked by birds several times.
 
5:44 AM
Apparently a chicken pecked her eye when she was a baby but it turns out it was my moms older sister who got pecked in the eye and the trauma of seeing that stuck with her.
In my free time I write lot's of math problems for my schools math competition. The leader also gave me a whole chocolate bar because I wrote so much.
 
i've never been attacked by a bird, but they can be aggressive. i have had birds steal food from me.
it's very hard to write good problems.
 
I've seen that a lot on youtube, it's kinds funny as long as no-one gets hurt.
All of my problems mostly involve advanced logical reasoning and topics.
I've learned a lot of things from this site which helped me write them.
 
6:17 AM
the trick is to peck them in the eye first
gotta take that first move advantage
 
6:56 AM
> In a triangle $ABC$, a point $D $ is so taken on side $AB$ such that $CD ^ 2 = AD\cdot BD$. Prove that $\sin A \sin B \leq \sin^2 (\frac C2)$ Discuss the case of equality.
I have proved that $\sin A \sin B =\sin x \sin y $.
How to show $\sin x \sin y \leq \sin^2 (\frac C2)$ ?
From AM-GM,
I have got: $\sin x \sin y \leq \sin^2 (\frac C2) \cos^2 (\frac {x-y}2) $
 
an ostrich pecked me in the head once, it hurt.
i was not entirely lacking in culpability.
it bugs me when people delete a question when they get an answer.
 
 
2 hours later…
9:19 AM
0
Q: Proof verification: Given $\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$ and $\lim_{x\to 0} f(x)=0$, show that $\lim _{x\to 0}\frac {f(x)}{x}=0$.

KoroGiven that $\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$ and $\lim_{x\to 0} f(x)=0$, it is to be proven that $\lim _{x\to 0}\frac {f(x)}{x}=0$. Proof: Let $\epsilon\gt 0$ be fixed. \begin{align*} \frac {f(x)}x&=\sum_{k=0}^n\color{blue}{\frac{f(x/2^k)-f(x/2^{k+1})}{x/2^{k+1}}}(1/2^{k+1})+\frac{\color{gree...

Can anyone please review my proof shared in the above link? Thanks.
 
 
2 hours later…
10:55 AM
Hi all, regard a 3x3 matrix $M$ with $dim=2$ of the column/row vector space. Hence $det(M)=0$. This 2-dim vecor space spanned by the columns/rows corresponds to a plane with a well defined normal. My question is has this direction/norm vector a name and is there any "text book construction" (such as a wedge product of ...) that generates it?
I mean the $rank$ of $M$ is 2.
 
 
3 hours later…
2:24 PM
hi; i wanted to ask if can an algebraic expression of the form ax² + bx + c can always be factorised to (x + u)(x + v) form?
 
Hello I am curious how do you make 3D graphics on 2D screen?
 
2:57 PM
If earth wasn't a sphere (or spheroid etc.), what shape could it have?
 
why are some mathematics books so expensive? I want to purchase a copy of "Introductory Functional Analysis with Applications" by Kreyszig but most options are >100$
 
One reason could be popularity
You could try a less popular but also good book, and see if the prices change
also, if you're in university, they usually offer free books you can download
I know it's not a physical copy though
For example, Topology by Munkres is very overpriced iirc
 
@Jakobian ive been studying from a borrowed copy and i like it enough that i want to purchase my own copy. i was pretty surprised by the prices. maybe it is popularity since both these books are standard references ?
 
@MehdiSlimani I can find a copy for less than $70 on abebooks: abebooks.com/products/isbn/9780471504597
As to the price: it is not printed in large numbers (because there is not a huge market for it), so new copies are somewhat expensive, as publishing a book incurs some significant overhead which isn't being made up in bulk.
Used copies are similarly expensive, as the market will bear a higher price.
That being said, $70--100 for an advanced undergraduate level text isn't that bad. Think of it as, for example, lab fees (which are paid by chemistry or biology students).
And Kreyszig's book is good, so well worth the price to have on your shelf.
 
3:34 PM
@PyGamer0 yes
@Jakobian disk
 
Ah, yes. I thought about something more internal
 
@Jakobian geoid
@WilliamJohn You can make anything in 3d but don't make earth a sphere
@MehdiSlimani not everything is expensive if you are on right place and time
-library genesis
 
4:02 PM
I'm wondering how could life on our planet look like if it actually were a torus
 
4:19 PM
@Jakobian The physics which would permit a toroidal world would be so very different from real physics that I don't know what you could possibly say... :/
 
4:57 PM
on a related note, does any of you have experience using your office as a delivery address ? im wondering if this is accepted behavior as a phd student
 
5:14 PM
@RaphaelJ.F.Berger The only issue with your question is that the col. space and the row space may be different although they share the same dimension, so they may define two different planes. An example would be the following matrix: $$A = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix}$$, with col. space xy and row space yz
@RaphaelJ.F.Berger the orthogonal space to the col. space is the null space of the transpose. the orthogonal space to the row space is the null space. in your special case of a 3 by 3 rank 2 matrix, you can find these orthogonal spaces using cross products. if you are interested in the topic i recommend you this video: youtube.com/watch?v=nHlE7EgJFds
 
5:52 PM
can anyone please take a look at the following? Thanks.
9 hours ago, by Koro
0
Q: Proof verification: Given $\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$ and $\lim_{x\to 0} f(x)=0$, show that $\lim _{x\to 0}\frac {f(x)}{x}=0$.

KoroGiven that $\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$ and $\lim_{x\to 0} f(x)=0$, it is to be proven that $\lim _{x\to 0}\frac {f(x)}{x}=0$. Proof: Let $\epsilon\gt 0$ be fixed. \begin{align*} \frac {f(x)}x&=\sum_{k=0}^n\color{blue}{\frac{f(x/2^k)-f(x/2^{k+1})}{x/2^{k+1}}}(1/2^{k+1})+\frac{\color{gree...

 
@Fargle good morning
I am such a moron about forgetting about square matrix property
linear dependence
row or column matrix is linearly dependent then matrix is gonna give linearly dependent
sorry i write it in sloppy way
Suddenly struck in my mind why was I thinking something so hard about linear algebra
answer was 1 diagonally
fuk my stupid mind
 
@XanderHenderson Well, people on youtube say it's possible, even with our laws of physics. I'm just wondering how creatures would adapt to the environment.
 
My brain is atrophying too fast after a car accident I am officially retarded in IQ test
 
7:03 PM
How would I evaluate the integral $\int_{\gamma} \frac{1}{(z-z_0)}$ where $\gamma = {z \in \mathbb{C} : |z-z_0| = r}$ where $r >0$
I want to just write down the antiderivative to be $log(z-z_0)$ but apparently I am not allowed to do that because it doesn't exist on $\gamma$? I know I can parameterise it in terms of $\theta$ but why can I not just write down the former antiderivative
 
7:18 PM
there are many issues there. for one, $\log$ is not defined on the entire range of $\gamma$.
you could take $\gamma_\epsilon$ to be $\gamma$ with the little bit near $-1$ removed and then use $\log$. Taking limits (which would need to be justified) will get $2 \pi i$.
 
i would parametrize the thing and 'just do it.' as copper points out the result shows that you can't have an antiderivative on the whole curve.
 
it is more straightforward to parameterise $\gamma(t)=z_0 + r e^{it}$ on $[0, 2 \pi]$ and just compute.
bingo
 
this is a very deep question, govind. it is basically the heart of complex analysis.
i'm not joking.
 
Okay, I see - cheers guys :)
 
7:51 PM
@MehdiSlimani Thank you very much and I have meanwhile discovered that its also the kernel of the Matrix!
 
If $\lim_{x\to 0} f(x(\frac 1x-[\frac 1x]))=0$ then $\lim_{x\to 0} f(x)=0$?
$[.]$ denotes floor function.
Let $x\in (\frac 1{n+1}, \frac 1n)$, $n>1$ so that $f(x(\frac 1x -[1/x]))=f(x\{1/x\})=f(1-xn)$
I have no idea how to proceed further in this.
 
8:11 PM
Hint: Who says the limit exists?
 
I think it exists
 
I wonder the reason someone would give such an exercise.
Seems rather contrived.
 
What I did was, substitute $u = 1/x$ and then graph yourself the function $\{u\}/u$
 
are there reasons behind an exercise?
 
then I picked some big integer, and restricted to some $N<u\leq N+1$, and my argument works I think
 
8:18 PM
I'd say good exercises have reasons for giving them.
 
koro's course of study is eclectic because it is not tied to any particular course or textbook. if you are simply trying to understand a large number of things, it may be that in the middle, any reasons for posing one particular exercise are lost in the shuffle.
 
Probably there is some reason for the exercise and we don't know it (yet)?
 
so yeah, sometimes you just approach an exercise as a blank slate.
i don't know if there is a reason for anything. :)
 
leslie getting philosophical
 
reason for the exercise is for someone to do it
 
8:21 PM
Seems like a rather poor reason.
 
so I think it's more about, why are you doing the exercise
 
Leslie: this exercise is from a problems book.
 
and the answer is, entertainment for example
 
if you are preparing for an exam in a course taught out of textbook X, then you can begin to infer reasons from the structure of the textbook. i think koro's project is a little broader.
 
From problems on limits and this set of problems comes after problems on sequences and series.
 
8:22 PM
correct me if i'm wrong. but if you're taking an entrance exam or something, you can't know that the problems come from some source. they literally could come from anywhere.
 
Can someone ELI5 the reasoning behind this: There is a class of 30 students and 4 class committee positions (pres, vice pres, secretary, treasurer). I'm trying to find every combo of students and committee members. The only stipulation is that any student can serve any role, but they can't serve more than one role at once.

I thought the number of combinations would have been 30 choose 4, but apparently, it is 30! (factorial) I'm scratching my head as to why.
 
when i was preparing for grad school i worked on a large number of completely unrelated problems. it would have simplified my life considerably if i knew where my potential problems might come from.
 
@leslietownes I don't mean koro is pursuing low quality exercises in general. I just mean that this particular exercise seems (in particular) a bit contrived.
 
anakhro i agree. i'm struggling to think of any reason for posing this problem.
 
Leslie: true, I won't know the source in that case. But this problem is from a problems book called -Problems in Analysis amazon.in/…
I'm doing that because having learnt the concept of limit, I wanted to solve some problems. :)
 
8:26 PM
Um. Well, getting back to discussing the exercise. After substitution $u = 1/x$ and using the $\varepsilon$-definition of a limit, you can prove $f(1/u)\to 0$.
 
anakhro: like I said, there may be some reason for the exercise and we just don't know/realize it yet.
 
under: it seems to matter whether you distinguish between orderings of students who are not selected for any position. i wouldn't. the 30! seems to.
i don't think of that as a natural reading of the problem, but that may be what is going on here.
if you constructed a ledger whose first entry was pres, second vp, third was secretary, fourth was treasurer, and then an ordered list of the rest of people, there would be 30! ways of doing that.
 
Ooh, wait. The way I have it would indicate all of the ways of just choosing 4 people, I think.
Not paying attention to all of the others.
 
@Jakobian we'll have f({u}/u)$\to 0$. So by definition: given $d>0$ there is an $r>0$ such that $u>r\implies |f(\{u\}/u)|<d$.
 
I don't understand the point of that. I get why the answer makes sense from that perspective, but in reality, why would the orderings of the other people even factor in? Like, why would you want to know all of the ways you could list the class after selecting the committee?
 
8:32 PM
under: to me, it wouldn't make sense to pay attention to the others.
 
Exactly, hence why I chose the answer I did.
 
they don't affect who has what role.
you might be right and the answer wrong. it happens.
 
Maybe a bit more of an interesting limit question: let $a_n = \sum_{k=1}^n k!$ be the sum of the first $n$ factorials, and let $b_n = n!$. Find $\lim_{n\to\infty} a_n/b_n$. It's kind of annoying though because you have to pick a good inequality dealing with factorials.
 
Eh, alright. As long as I understand where the author's answer came from.
 
Well I guess you don't have to, but it makes it nicer.
 
8:36 PM
how full of a stirling do you need for that? :)
some applications of 'sitrlings inequality' can be solved with only the observation that (n/3)^n < n! < (n/2)^n for n large, which falls out of one of the standard definitions of e. you don't need to know any (n/e) stuff or the asymptotic corrections.
although in serious applications you definitely do.
koro what i just said is a problem solving hint of general application. feel free to use it. $5 royalty per use.
 
@Koro I think it means that $f(x) = 0$ for $x \in (0,2)$, but it says nothing about $x \le 0$ so with $f=1_{(-\infty, 0]}$ the condition is true but you cannot say anything about the limit of $f$ as $x \to 0$.
Look at the function $\phi(x) = 1-x \lfloor {1 \over x} \rfloor$.
another question deleted after a hint in a comment.
 
catching up on the starred comments re overpriced textbooks. gave me a happy memory of when some treatise on operator theory came out priced at over $1000 and my advisor and i had a long email chain making jokes about it.
 
I tried approaching this without Stirling's in mind, we can show that the limit exists so suppose that it equals $L$. Now, let's find $L$.
$\begin{align*}
\frac {a_n}{b_n}&=\frac {\sum_{k=1}^n k!-(k-1)!}{n!}+\sum_{k=1}^n \frac{(k-1)!}{n!}\\
&=\frac {n!-0!}{n!}+\frac{(n-1)!}{n!}.\frac {a_{n-1}}{b_{n-1}}\to 1+0=1
\end{align*}$
@copper.hat Hi, the question says "given so and so, prove that $\lim_{x\to } f(x)=0$". Are you suggesting that the question is wrong?
 
8:52 PM
what bugs me is inflated prices for books that are out of print. it seems a rude way to make money and the proceeds on used books have nothing to do with the author.
my white whale in grad school was gert pedersen's cstar algebras and their automorphism groups. it was trading so highly on ebay that someone stole it from the library and the university didn't have a copy.
google tells me there's been a second edition, which makes me happy.
too late for me, though.
 
@leslietownes The book has provided solution also but I didn't understand the solution also :(.
If I did, I would use the idea in other problems :).
 
@Koro unless i am missing something, yes. note that $x(\frac 1x-[\frac 1x]) \in (0,2)$ for $x \neq 0$ so, the limit can only 'say something' about $f$ on $(0,2)$.
 
@leslietownes My white whale was Linderholm's Mathematics Made Difficult for less than a thousand dollars.
 
@Koro you can get lost going up every maths side road.
 
I finally got an old library copy from an estate sale for a song.
 
8:55 PM
there are also ebooks of that for free from neerdowells.
 
@copper: please give me a while while I post the book's solution here.
 
pedersen's book had not attracted the attention of those who torrent math books.
 
@leslietownes ebooks don't count. :P
 
i would have settled for an ebook of pedersen :)
 
@leslietownes Our college had (still does?) a rule that if someone borrows a book from the library, they must return it before a certain date. If they didn't they'll be fined and won't get their grade card before clearing all dues.
someone stealing a book was never an option at my college.
 
8:57 PM
oh, i could steal a book from that library. watch me.
:)
whatever the policies are, there's no security, little monitoring, stupid obstacles. it's very much an honor system.
 
:). I'm afraid, if you don't have ID card, you won't be allowed to enter the library :(.
 
that would be the hardest part. but getting a book out of a library, if you are allowed into the library, is easy.
without checking it out.
my guess is some random undergrad had a business in seeing what was trading high on ebay and then spirited those books out of the library.
using their access to the shelves.
i'm not sure at the time that the library even bothered to check ID on the way in. once somebody can get their hands on the book it is over.
 
Haha, yeah that could happen. But not sure if they have now put some sensors at the entrance that detect bar-codes. :D
 
it depends on the system but you can zap RFID tags easy. and if all you need to do is toss the book over the sensors when nobody is looking, that's easy too.
i'll stop snitching on myself.
for the record, it's a bad idea to steal books from libraries, and antisocial to do so. but there is nothing stopping anybody from doing it, which i believe happened in this case.
was anyone else tutored in mischief as a child? my dad's thing was sneaking into museums without paying the fee. we'd done it dozens of times. never caught.
 
@copper: The solution given in the book is along the following lines:Given $\epsilon>0$, there is a $\delta, 0<\delta<1$, such that if $0<|x|<\delta$, then
$|f(1-x[1/x])|<\epsilon$. Taking $n$ so large as $1/n<\delta$. For $0<s<\frac 1{(n+1)}$, set $x=\frac {1-s}n$, then $\frac 1{n+1}<x<\frac 1n$. Thus $[1/x]=n$.
If $0<s<\frac 1{n+1}$, then $|f(s)|<\epsilon$. For $s<0$, one can proceed analogously.
I don't understand this.
 
9:07 PM
@Koro Ignore my comment, totally wrong, sorry.
Thank god for desmos
 
We have that $f(\{x\}/x)\to 0$ as $|x|\to\infty$, so fixing $\varepsilon > 0$, we get that there is $M$ such that $|f(\{x\}/x)|<\varepsilon$ for $|x|\geq M$, so now taking $N$ to be natural $\geq M$, we have $\{x\}/x = (N-x)/x$ for $N< x\leq N+1$. Now notice that $(N-x+1)/x$ takes values in $[0, 1/N)$, so for $x>N$ we have $|f(1/x)|<\varepsilon$
This is my solution after substituting $x\mapsto 1/x$
This is basically, observe how the graph of $\{x\}/x$ looks like, notice that between each two integers it continuously decreases to $0$!
 
the plot of $1-x\lfloor {1 \over x} \rfloor$ is mildly amusing, i should have plotted it first as that is my usual admonishment when commenting.
 
@Jakobian I think that in the second line, you meant $\frac{\{x\}}{x}=\frac{\color{blue}{x-N}}{x}$
 
yes, I meant that
I was wrong, it continuously increases to $1/(N+1)$, well, the reasoning is similar
 
Doesn't $\frac {x-N+1}{x}$ take values in $[0,\frac 2 N)$? (Just asking, although it doesn't affect the final conclusion that the limit exists and equals $0$).
 
9:17 PM
@anakhro $\frac1{n!}\sum\limits_{k=1}^nk!=1+\frac1n+\sum\limits_{k=1}^{n-2}\frac{k!}{n!}$ where $\sum\limits_{k=1}^{n-2}\frac{k!}{n!}\le(n-2)\frac1{n(n-1)}$
 
@Jakobian Thanks a lot. I understand how to solve this problem now.
 
@Jakobian you mean $\frac{\{x\}}{x}=\frac{x-N}{x}$?
 
I don't understand my mistake here https://math.stackexchange.com/questions/4363927/proof-verification-given-lim-x-to-0-fracf2x-fxx-0-and-lim-x-to/4363947?noredirect=1#comment9115580_4363947
Please let me know what I did wrong here.
 
9:34 PM
@robjohn yes
 
@Jakobian Can you please also take a look at my small question above?
I have almost reached the conclusion that the proof I presented in the OP is wrong.
I just wanted to know if my understanding is correct.
 
@Koro The way you align your answers makes them very awkward to read.
 
I have been advised to use align environment. So I have started using that more and more.
 
Well, having to scroll right & left just to look at a few lines of a proof makes it hard to read.
Also, this may be just me, but I have to read your proof many times to figure out what you are doing. I know there is a fashion in mathematics to write cryptic, Rudin like proofs, but personally I think explaining the derivation is a better idea.
 
9:50 PM
I have edited the post, copper.
 
@Koro For example, there is a jump after the line "There exists N such that' and I am still trying to figure out where that all came from.
 
@Koro It is hard to see the justification for the line starting with "There exists $N$ such that..."
It may be true, but it is hard to see why.
 
You introduced $h_n$ and then it disappeared?
 
For any $\epsilon\gt0$, there is an $x_\epsilon$ so that for $x\le x_\epsilon$, we have $\left|\frac{f(2x)-f(x)}{x}\right|\le\epsilon$. For any $x\le x_\epsilon$,
$$
\begin{align}
\left|\frac{f(x)-\lim\limits_{y\to0}f(y)}x\right|
&=\left|\sum_{k=0}^\infty\frac{f(2^{-k}x)-f(2^{-k-1}x)}{2^{-k-1}x}2^{-k-1}\right|\\
&\le\sum_{k=0}^\infty\epsilon2^{-k-1}\\[9pt]
&=\epsilon
\end{align}
$$
This seems to do it simply
 
I chose $N$ like this: For a fixed x, and a fixed πœ–>0, since $\lim_{n\to \infty} g_n(x)=0$, there is $N_1$ such that $n\ge N_1\implies |g_n(x)<\epsilon$ and since $\lim_{n\to \infty} h_n(x)=0$, there exists $N_2$ such that $n\ge N_2\implies |h_n(x)|<\epsilon|x|$. Choosing $𝑁:=\min(𝑁_1,𝑁_2)$, we get the $𝑁$ that was used in OP.
@copper.hat yes, because I bounded it with $\epsilon |x|$.
 
9:57 PM
@Koro I think you are missing my point. I can go figure it out, but I will only spend so much energy trying to solve an additional puzzle. It is better to be clear up front so that it is straightforward to follow.
 
No, I understood your point. I should have incorporated that in the post.
 
If you want cryptic: $|f(x)|=|f(x)-f({x \over2}) +f({x \over2}) -f({x \over 4 }) +\cdots| \le \epsilon|x|({1 \over 2}+{1 \over 4}+\cdots)=\epsilon|x|$
Sorry, I am grumpy today, so less patient than my usual impatient self.
and annoyed i didn't plot the 1-x[1/x] function earlier :-)
 
@robjohn i think that the first align line is wrong.
Because we can't assume convergence of the series from the get-go, I think.
@copper.hat that makes sense :) since $\lim_n f(x)=\lim_n(f(x)-f(\frac x{2^{n+1}}))$
The proof in my post indeed was wrong.
 
I still do not get the leap after "There exists N such that"
 
The leap relied on the fact that N is a constant, which as it turns out depends on x so the leap is invalid.
 
10:09 PM
So I cannot comment on correctness.
Ok, but it you explain the steps in the proof then it will be easier to unravel & check.
(I make lots of mistakes too, as demonstrated this morning.)
 
Sure, copper I can write the algo here if you like.
Let me explain what I did:
 
Well, no need. I know how to prove it.
You were asking for a solution verification, I was just pointing out that it is difficult to verify.
 
I know you do. But you said that my proof was not explanatory.
 
It is not.
Explaining it here is pointless.
 
Ok nvm. I understood the problem with the proof. :)
 
11:02 PM
h3y yall
 

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