That raises a good point. Isn't $\sqrt x$ real analytic on $(0,1)$?

I'm reading about square summable sequences here and why the product of two square summable sequences is summable. I don't understand why the answerer introduces $C>0$ when Lemma 1 only contains $\frac12$.

Glad to hear you're well on the road to recovery, bionic @copper.

@TedShifrin Much appreciated!

@psie Who knows. He makes it sound like $C$ can be any arbitrary positive number. It's just wrong.

right, maybe it should be $C\geq \frac12$

Who cares. Just use $1/2$, as you said.

ok

There are lots of people who write posts on this site and get lots of votes despite errors.

hmm yeah, and editing such typos is discouraged I believe

You know about Cauchy-Schwarz, anyhow, don't you? Don't you know that $|\sum a_k\bar b_k|^2 \le \sum |a_k|^2\sum |b_k|^2$?

ah yes, familiar with Cauchy-Schwarz, but didn't know it applied here, neat

@copper.hat Yeah, hash tag me too (when I lived in Reno). I didn't bike at all in Riverside (there was nowhere where I felt safe on a bike), and my bike has fallen into disrepair. The nearest shop is in Flag, so I really need to just take it in and get it fixed up so I can ride it around town.

@TedShifrin yeah... weird huh

(Just the extended binomial theorem, isn't it?)

yeah. Or $\exp((1/2)\ln(z))$

Sure.

@leslietownes right, thank you for sharing how one could look at it! I can follow it to a certain extent. I think for now I will leave it then, as my main intent with this exercise was to understand the material in Conway better, but as you already guessed, not much from Conway seems to yield a straightforward solution for this problem.

@JohnZimmerman What do you know about this problem? I'm not an expert, is it well-defined, smooth? What do you know?

If I'm going to help I want some background

number theory is outside of my scope

@XanderHenderson i know i have hit some generational barrier when i start complaining about new tech. i used to be able to maintain my bike, now the best i can do is part replacement and even that can be challenging unless you do it frequently (like the screw orientation on l & r pedals). disk brakes are powerful and nice to ride, but a total pain to maintain (so i don't have them).

plus any visit to the bike shop now costs as much as my bike did, typically $2-300 and that is pre mad inflation.

@copper.hat When I bought my bike, it was around $3k (I have a pretty nice Trek bike with a carbon frame). While living in Reno, I tended to put 10-20k miles on it every year (so the frame has about 60k miles on it right now). I am pretty good with changing tires and breaks and tightening things and whatnot, but I haven't been on it in ages, and it needs a lot of work. I'm happy to pay someone else to do that.

@Jakobian I know $J(x)$ is increasing. I also know that $f(0)=0$ and $f(1)=1.$

@XanderHenderson my bike was $350 (my mom gave me a present :-), so it has a lot of sentimental value). i typically (at least used to be) put 5-10k miles on trail, but typically fairly hilly. i fix punctures on trail, but beyond that & replacing brake pads it has just gotten too complicated/memory/tool intensive.

don't get me started on the ebike...

@Jakobian I suspect that the integration "smooths" the jumps in $\pi(k)$ sufficiently perhaps making $J(x)$ strictly increasing and possibly even infinitely differentiable. I also suspect $J(x)$ is a linear transformation of $\pi(k)$. I have checked indeed that $f(0)=0$ but have to check the linearity condition.

@copper.hat Oh, I don't do trails. Trails suck. I ride on pavement!

pavement sucks, or rather, bay area drivers are dangerous. plus the trails are less crowded and the views better :-)

While living in Reno, I'd do 12 miles at least 5 days per week to get to campus, and I'd ride to the California border (or some similar distance) on the weekends while the weather held (there were some nice, paved bike trails which roughly paralleled the 80; about 40 miles, round trip).

@copper.hat Yeah, I buy that.

Sometimes, I'd ride to Carson City (about 70 miles, round trip), if I felt like making a day of it.

for a while i was working in a research lab in Berkeley and used to cycle the 3 miles back & forth. I would say that one out of three days there was a close dooring or reversing in front of me incident. luckily i was younger and more aware. but theft in Berkeley was a problem, wheels, saddles, etc, disappeared until i decided to skip formalities and bring the bike into the office (it was a corporate sort of thing, so fairly formal, certainly for Berkeley).

@JohnZimmerman so did you prove its well-defined?

@Jakobian yes it's a well-defined function

@copper.hat UNR had bike lockers, so I rented one for \$5 per semester. I could lock my bike in a box all day. It was great.

Once I started my masters, I could probably have kept my bike in my office, but there as no need.

Normally one is interested in functions that are "asymptotic" to $\pi(k)$ in the sense that $\lim_{k \to \infty} f(k)/\pi(k)=1$

Prove uniform convergence first, to show continuity

Is it differentiable? Maybe. Is it twice differentiable? I doubt it

Don't expect things from this - not sure why would you, $\pi$ isn't even continuous

I doubt it is differentiable.

Same but it might be

Before you pass to the limit, you need $r$ and $rx$ both to be nonintegers. Both integrals diverge, so some careful analysis is required even to start.

By the way I mean local uniform convergence*

@AlessandroTerminiello I got $1+x+\frac{x^2}2-\frac{2x^3}3-\frac{19x^4}{24}+O\!\left(x^5\right)$

We were only doing degree 3.

cuz the ultimate problem was a limit $/x^3$.

I missed that the $x^3$ term was in two pieces.

then theirs matches what I got

Yeah. I checked he was right …. See what I meant about algebra weakness …

@XanderHenderson a few places in the bay area have Bike link lockers. It's a great deal, 5cent/hr with some card key thing. Very handy when I commuted to the south bay.