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12:45 AM
we're running the air conditioner for the first time this year.
trying to get the second floor below 80.
 
I love almost integer stuff. It's interesting that there are lots of examples involving $e$, the most famous ones coming from the Heegner numbers. en.wikipedia.org/wiki/Almost_integer It's pretty easy to make quadratic ones, using Pell's equation.
 
oh, that's evil.
even by xkcd standards.
 
One of the pi guys (Borwein ?) did a similar thing involving cosh, IIRC. I should be able to find the anecdote on MO...
 
oh yeah, i've read that anecdote. it was one of the borweins.
 
1:05 AM
John Baez has an article about that stuff: Patterns That Eventually Fail
 
@leslietownes It's not that bad here.
 
it's muggy here.
 
Well, it's because munchkin is so full of hot air.
 
SD always has better weather. it's the best weather on earth.
 
I think the Borwein example involved the integral of a product of sinc terms, not cosh. More examples here: mathoverflow.net/questions/15444/…
 
1:15 AM
we have a coastal pine tree nearby that reminds me of SD.
 
We've had a couple of days with no rain (or little rain), but they've been very rare for months. At least the weather is starting to get cooler, so it's merely humid, not muggy.
Found it!
17
A: Integral of the Sinc product

ybeltukovBorwein integrals As Eckhard wrote in comments B[n] is the n-th Borwein integral. (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k])...

As a prank, Jonathan Borwein reported this to Maple, claiming there was a bug in the software. Maple computer scientist Jacques Carette spent 3 days trying to figure out the problem. Then he realized: There was no bug! That's what these integrals really equal!
 
that's the anecdote. you found it.
sadistic.
 
Perhaps. I like to think of it as an "interesting" teaching technique. ;)
 
1:38 AM
"Interesting," huh?
 
I'm trying to write a program to calculate the time of sunrise & sunset. Some of the equations I'm using come from this paper from 1989. doi.org/10.1093/mnras/238.4.1529 Unfortunately, their equation for sidereal time has a sign error. It took me a few days to fix that. :( But now I'm wondering if there are other errors in smaller terms that are harder to notice. I can check my results against results from JPL, but that doesn't really help much with the smaller terms.
@TedShifrin Well, I bet Jacques Carette now knows a lot more about those integrals than he otherwise would. :)
 
2:03 AM
@PM2Ring Back in the late 80s or early 90s, I wrote a program that would print out the sky including moon and planets. It was pretty accurate, but the moon position was a bit off since I used the Earth as a sphere.
I had about 880 stars, hand input from the Nautical Almanac and Astronomical Ephemeris.
I was wrong, it was written for the PET, so it was in the early 80s, maybe late 70s
 
loved the pet
was supposed to go to healdsburg winetasting with friends, but i met another friend yesterday who just texted that he tested +ve, so i had to cancel my end of the trip :-(
 
I got an extra graphics board that had 32K of RAM to use and 2 PROM sockets that I could use for some 3-D graphics software I'd written.
 
cool!
 
Nice work! This site has some amazing algorithms. neoprogrammics.com/index.html Accurately compensating for the effects of the Moon isn't easy. Here's the IAU2000 stuff for computing the effects of nutation on longitude & the obliquity of the ecliptic: neoprogrammics.com/nutations/nutations_1980_2000b I'm definitely not putting that stuff in my program!
 
@copper.hat It was that software, I think, that got me the job at Apple.
 
2:11 AM
@robjohn I can see why.
 
Lunar theory has improved significantly, especially in the last 20 years, thanks to refinements in Lunar Laser Ranging.
 
plus the 6502 was used in the II
 
@PM2Ring Yeah, I'm not real interested in trying to resurrect and improve that program. It lies on cassette tapes and printed on paper.
 
in Basic?
 
@copper.hat Yeah, I'd also written an assembler for the 68000 and upgraded to 68030+MMU
 
2:13 AM
i know its clunky by today's standards, but it was a great leaning vehicle
wow,
 
@copper.hat The PET software was in BASIC
 
JPL uses a fairly sophisticated model for the Moon's interior, but they also fit their ephemeris calculations to a lot of actual data.
 
@PM2Ring I'll leave that to them ;-)
 
the discovery that you could peek & poke opened up a whole world of exploration for me. we had no documentation at all, and coming across such stuff was practically impossible. byte magasine had some good stuff from time to time
 
It's still fun to compute orbits, but there's no way to compete with JPL. They integrate the equations of motion to the level where they model all the major bodies of the Solar system down to the top 340 asteroids as individual bodies, plus they have access to a huge amount of terrestrial & spacecraft observations for data fitting.
 
2:18 AM
@copper.hat That is how I wrote the 3-D software on the PET 6502 code POKEd in a BASIC program.
@copper.hat Loved BYTE
@PM2Ring If it were anyone but JPL, I'd say "get a life"
 
The best lunar theory from a couple of decades ago, ELP didn't bother trying to model actual observations directly, they took all their data from the then-current JPL ephemerides.
 
i wrote a little plot program that managed 'double resolution' plots. the pet had 16 characters with each quadrant on/off.
not terribly exciting, but i was so proud of that little routine
all sorts of things suddenly became interesting, scatter plots least squares, etc
 
I thought the PET was wonderful. I never owned one, but I played with them in the shop. A bit later, I had an Osborne
The Osborne 1 is the first commercially successful portable computer, released on April 3, 1981 by Osborne Computer Corporation. It weighs 24.5 lb (11.1 kg), cost US$1,795, and runs the CP/M 2.2 operating system. It is powered from a wall socket, as it has no on-board battery, but it is still classed as a portable device since it can be hand-carried when the keyboard is closed. The computer shipped with a large bundle of software that was almost equivalent in value to the machine itself, a practice adopted by other CP/M computer vendors. Competitors quickly appeared, such as the Kaypro II. ��2...
A few years later, I got an Amiga 2000. In fact, I still own one (not my original), which I use as a monitor stand. It might still work, (assuming its electolytics haven't dried up), but I no longer have an analog monitor for it. I guess I could get an adapter...
 
this is the best one.
The Sphinx (Russian: СФИНКС, romanized: SFINKS) was an experimental Soviet project for a home automation system, commissioned by the State Committee for Science and Technology and designed by Dmitry Azrikan, in collaboration with A. Kolotushkin and V. Goessen, in 1987. Sphinx, an acronym for Super Functional Integrated Communication System (Russian: Суперфункциональная интегрированная коммуникативная система), was intended to be an ensemble of modules that would allow consumers to easily interact with information systems. The home environment, as described in a 1987 issue of Soviet magazine Technical...
 
@copper.hat You could do a similar double res trick on the Osborne.
 
2:37 AM
Alessandro was the set theory fellow, right?
 
@TedShifrin I took a stats class with some math in it and i recall "manifold" and "chi squared test" being uttered in the same sentence ...
 
the reasons to use the chi square test are manifold
 
 
1 hour later…
3:48 AM
@copper.hat The graphics board I got (with the 32K RAM) had 320x200 pixel resolution, the full resolution of the screen!
I wrote a printer driver that used a three screen width buffer to print on the 960 dot wide dot matrix printer that the company my dad worked for made.
that was a 24K buffer (big on a 64K address space).
Thank heaven that the pixels were only 1 bit (black and white, no shades of gray)
 
4:08 AM
Is this ancient history being retold? :D
 
Some of those ancient ideas are still in use (eg, GIF is still very common for animations on the Web). Or are operating at a smaller scale on modern hardware, doing fun things with sub-pixels.
Modern high-speed printers are capable of much more accurate ink placement than the old printers. So some techniques that were rejected decades ago are now practical. Also, there are some superior image processing algorithms that were rejected years ago because they were too slow &/or consumed too much memory that are now practical on modern computers.
 
@robjohn impressive
 
4:23 AM
@TedShifrin yes, it's printed on a cave wall in dot matrix
 
Versatile printer!
 
4:45 AM
OTOH, there's a vast amount of incorrect image processing software. Even well-known proprietary software like Adobe Photoshop isn't immune.
 
4:57 AM
what??? software with bugs???
 
Spray for termites.
 
@copper.hat I wasn't talking about bugs. I'm talking about code that correctly performs incorrect calculations. ;)
 
and not a deliberate design choice? any examples easy to explain?
 
In some cases, yes, it's a deliberate design choice, sacrificing precision for speed. But in some cases, the bad algorithms result from inadequate understanding.
Admittedly, the mathematics of precision image processing is hard, and software developers may not have the necessary maths background. Also, important info related to colour hardware standards can be buried in technical articles &/or behind paywalls.
A simple example is the topic of gamma correction. Pixel colour values, eg in a PNG file use a non-linear encoding. But most image processing treats those values as linear. So any processes that rely on or modify the brightness of pixels end up with wrong values.
The wrong algorithm ends up becoming a "standard", and people using the software end up tweaking images by hand until they get something that looks good, not even realising that they're using bad software.
 
5:20 AM
hm. why are standards even behind a paywall.
 
Good question. Tradition! You put standards in the appropriate industry journal, which all professionals will have access to.
 
:(
i tried to check some iso thing at some point but didnt want to pay 100 swedish krona or whatever
was it iso....anyway.
 
But if you're a software developer, not a colour TV engineer, you probably don't read that journal. And you probably don't even realise that you're missing important info about how pixels on a screen actually behave.
IIRC, there's some thing related to gamma that recent versions of Photoshop can do properly. But by default, they do it the old way, and you have to go into an obscure menu to make it use the correct algorithm.
 
whoever had the bright idea to clearly label definitions and theorems was a genius
 
I guess that goes back to Euclid. We call him a geometer / mathematician, but he was actually employed as a librarian at Alexandria.
Here's an excellent article on gamma, and the errors associated with incorrect gamma handling: blog.johnnovak.net/2016/09/21/…
 
5:33 AM
> The popular graphics library <insert name here> I’m using handles gamma correctly.
lol
i guess they don't patch the older libraries for backwards compatibility?
or have i entered a cult of sacred knowledge
i think this explains why gimp's soft brush was uglier than photoshop's
or their gradient fills
 
Here's a great example from near the end of that page: incorrect gamma handling in 3D rendering is one of the main culprits behind the “fake plasticky CGI look”
 
eyyy theres a gradients section. totes called it
 
@CalvinKhor You generally can't fix this with a simple patch. You don't want old software to suddenly behave differently. But you could add new methods to supplement the old ones.
 
tho apparently ps is still wrong
 
> you have to go into an obscure menu to make it use the correct algorithm.
 
5:42 AM
right
thanks i hate image processing now
 
This is kind of a legacy thing. If we used 16 bits per channel instead of 8 bits, we could have gotten away with linear encoding instead of gamma-adjusted encoding (although gamma would still have advantages of not wasting bits on very dark & very bright colours).
Image processing is fun, if you do it in a colour space that's perceptually linear (eg Luv) and you use enough bits. Using algorithms that were devised when 32k bytes was a lot of RAM is madness.
 
haha im joking i have lots of fun with image editors and this explained a lot of things
 
5:59 AM
Oh, good. :) One of the refs at the end of that article is to poynton.com/GammaFAQ.html It's a bit old, with lots of stuff about analog TV & CRTs, but it's still mostly good, since modern display technology was designed to (sort of) mimic CRTs. Poynton's articles on gamma & colour are the de facto standards on this stuff for coders into image processing.
 
> Your connection is not private
Attackers might be trying to steal your information from poynton.ca (for example, passwords, messages or credit cards). Learn more
NET::ERR_CERT_DATE_INVALID
 
its still complaining. https such a mood killer
against my browsers better judgement im in
 
Yeah, the certificate expired 8 months ago. Oh well.
 
6:36 AM
Q is to represent a displacement of. 40km west of north and east Of north
When the angle is turned clockwise , it’s value and direction is said to be positive
In 1st case , it’s positive
min 2nd case , the direction of angle is anti-clockwise.
Should it value be -30?
 
in the 2nd case, the angle is clockwise
sure, though giving the direction of the arrow is the same as putting a minus sign IMO
$\Huge⏰$
 
Oh sorry
I did opposite
@CalvinKhor is my 1st case wrong ?
I have drawn according to the question but it’s direction of angle is opposite to that of 2nd one.
 
the choice of the minus sign is not right or wrong[ outside an exam environment ], as long as you're consistent. If anti-clockwise is positive then the diagram for the first case is correct but what you said is wrong
 
Yes
Anti is positive
Clock is negative
Should I consider to think about about +30 or-30?
 
What do you mean?
 
6:46 AM
In 1st case , direction is anti clockwise
We know that is positive direction
For 2nd case , direction is clockwise
Have I drawn a diagram of displacement of 40km , -30 or +30 east of north
 
you should follow what your text and teachers say. its clear enough to me but i dont know how they mark you
 
My text say nothing about the direction of angle
 
no solved examples?
 
Yes, there is but it does not have a drawn direction in it
 
add it if you want, then see what your teacher says when you recieve the work back. or ask directly. I can't help you decide what will ultimately be right for you
 
7:28 AM
$\Huge {(\ddot\smile)}$
 
$\Huge\vcenter{\stackrel{⦿{\large\triangle} ⦿}{\smile}}$
 
hahaha
$\Huge {(\ddot\smile)}^2+\Huge {(\ddot\smile)}+1=0\implies \Huge {(\ddot\smile)}=\omega,\omega^2$
 
8:07 AM
Is there anyone keeping track of what mathematical books are entering public domain?
 
8:25 AM
My teacher says degree is a scalar quantity
 
@TedShifrin thanks for the reply, right I see that, but what I was troubled by was that the second definition the OP states does not even seem to agree with the first one up to a constant factor
also, i have a burning question, this sounds weird to me and I'd like to know if its actually true.. so I seem to have a 'proof' that $H^2(\Omega)$ is finite dimensional, where $\Omega \Subset \mathbb{C}^n$ is a bounded domain, and the former symbol means the square integrable holomorphic functions $f : \Omega \rightarrow \mathbb{C}$
 
@porridgemathematics I know crazy things happen in $\mathbb C^n$ but are you claiming $n=1$ as well? that sounds very off
 
yeah that is my claim
 
but every polynomial should be in $H^2$ as $\Omega$ is bounded
I don't forsee a universe in which multivariate polys aren't "holomorphic", but I am reducing to $n=1$ out of pure fear of multivariate complex analysis
 
yeah that was exactly why it was why I think its wrong
 
8:38 AM
:) whats the proof?
 
so basically my proof went like this: Let $B$ denote the unit ball in $H^2(\Omega)$ (which is a hilbert subspace of $L^2$), then we claim $B$ is a locally bounded family, the reason is because for $f \in B$, $|f(x)| \leq \frac{||f||_{2}}{\sigma(x)}$ is satisfied, where $\sigma(x)$ is the volume of $B(x;d(x,\partial \Omega))$, so this should mean that $B$ is locally bounded
but then a locally bounded family of holomorphic functions is normal in the sense that any sequence has a locally uniformly convergent subsequence
in particular we can take some sequence $f_n \in B$, then take a subsequence $f_{n_k}$ converging locally uniformly to a holomorphic $f$. This $f$ is not apriori the $L^2$ limit of $f_{n_k}$, but I thought it should be because we can exhaust $\Omega$ with compact subsets, and then take limits
 
$B$ isnt a subspace though?
 
I mean $H^2(\Omega)$ is hilbert
 
oh
ill keep reading
ok, i think this is more or less alright, how do you get the dimension?
 
i thought this shows that $B$ is compact in $H^2(\Omega)$
and if $H^2(\Omega)$ was infinite dimensional, that would not be true
 
8:46 AM
hmmmmmmmmm
 
because we could embed $\mathcal{l}^2$ into it if it were infinite dimensional, and the unit ball in the former space is certainly non compact
just take $(1,0,...),(0,1,0,...),...$
 
must be the last bit, $L^2$ norm should blow up
or theres only finitely many polynomials and annals of math will be in contact shortly
 
this was the precise argument for the $L^2$ norm thing: Let $C \subset \Omega$ be compact, then $|f_{n_k}-f|^2 \mathbf{1}_{C} \rightarrow 0 $ pointwise, and $|f_{n_k} - f|^2 \leq 2^2 |f_{n_k}|^2 + 2^2 |f|^2 $, the RHS converges pointwise to $2^3 |f|^2$ on $C$, which has a finite integral because $C$ is compact, so by DCT $f_{n_k} \rightarrow_{L^2} f$ at least on compact subsets of $\Omega$
 
well locally uniformly already gives $L^2_{\text{loc}}$
but the uniform bound depends on distance to boundary
 
sure, but we should only need local boundedness to give that $B$ is normal by montels theorem
and at least locally, we can fix a small ball, and control $\sigma(x)$ for all $x$ in that small ball no?
 
8:55 AM
i have to google montel :') one sec
right ok, i agree that you can get an analytic limit for some subsequence
 
and you also agree that that analytic limit should be in $L^2_{loc}$
 
yes. but unless $\sigma^{-1}\in L^2$ I don't believe this gets you all the way to $L^2$
 
it shouldn't, but my argument was that on compact subsets of $\Omega$, $||f||$ should be less than or equal to $1$
because $f$ is the $L^2$ limit on this compact subset of functions in $B$
and those functions in $B$ have $L^2$ norm on all of $\Omega$ equal to $1$, so on some compact subset their $L^2$ norm should be at most $1$
 
hmmm.................ok you know what I agree that the limit is also in $L^2$. So I'm confused
do you want to write jointly to Inventiones
 
maybe $H^2(\Omega)$ is just not hilbert
im 99% sure im wrong haha
 
9:01 AM
oh im 200%....but I can't figure out the bug in the proof
 
im just revising for an oral on kahler geometry in two days, and this came up when looking at stuff related to the bergman metric on bounded domains
the claim maid there was that for such domains $H^2(\Omega)$ is a closed subspace of $L^2(\Omega)$, because of the mean value inequality I stated before
made*
 
I was happy enough to have proved that without asking you why its true
 
fair enough, so $H^2(\Omega)$ is definitely hilbert
hmmm, what about this then? This is a well established result in the literature: If $X$ is a compact complex manifold and $E$ is a holomorphic vector bundle on $X$, then the space of holomorphic sections of $E$ is finite dimensional
what if we take $X = \mathbb{C} \cup \{\infty \}$ and $E = X \times \mathbb{C}$
ah no, thats fine, these are just constant functions :)
 
hahaha
 
i thought for a second we can just use polynomials here too, but they blowup at $\infty$
 
9:06 AM
its possible for $f_n$ to be super nice and for the limit $f$ to be in $L^2$ without the convergence happening in $L^2$
 
im going to boil this down to , like you said, something wrong with my $L^2$ convergence argument
thats true
 
purely from bounded in $L^2$ they already converge weakly in $L^2$
 
you should ask them in your exam
hey, i gotta question for you
 
oh, i figured out the error
 
9:10 AM
oh, what is it?
 
to apply DCT on compact subsets $C$ as I wanted to, I need to show that $\int_{C} 2^2|f_{n_k}|^2 + 2^2|f|^2$ converges to $2^3 \int_{C} |f|^2$, but that means I need to show $f_{n_k} \rightarrow_{L^2} f$ on $C$, which is what I was applying DCT to try to show in the first place
 
tbh i didn't read that because $L^2_{\text{loc}}$ follows just from convergence locally uniformly
 
hmm, wait a sec, but if $f_{n_k}$ are locally uniformly bounded, they are uniformly bounded on compact subsets, so I could use DCT to show this, since $|f_{n_k}|^2 \leq M$, and $f_{n_k} \rightarrow f$ almost everywhere, this use of DCT should still go through
ugh, now im even more confused
yeah, $L^2_{loc}$ follows because the limit is continuous, we dont even need holomorphic
 
$\|f_n-f\|_{L^2(C)}\le |C|^{1/2} \|f_n - f\|_{L^\infty(C)} \to 0$ for fixed $C\Subset \Omega$
err....wrong power
but if you apply holder correctly
 
ah yeah, and holder also shows this, true
 
9:14 AM
and some people don't call the $L^\infty-L^1$ case Holder but wtv
 
okay, it has to be with the last bit then, so I think I can't say $||f_{n_k} - f||_{2} \rightarrow 0$ on $\Omega$ using DCT, because I don't have boundedness of $f_{n_k}$ on all of $\Omega$
and $f_{n_k}$ at least being dominated by something known to be in $L^2$ on $\Omega$ is necessary for that final application
 
yeah i think so
 
ugh, i feel so stupid spending so long on this lol
thanks for working thorugh it with me!
 
on the other hand, i have only ever seen finitely many polynomials....
 
9:18 AM
np :) it was fun
 
 
2 hours later…
10:52 AM
@CalvinKhor a Generalization of Hölder perhaps.
$\|f\cdot1\|_2\le\|f\|_\infty\|1\|_2$
 
right, in my head I did do $\frac12 = \frac12 + \frac1\infty$, so even this was a stretch :)
but you could just use holder and get there
$ \|f\|^2_{2} = \|f^2\|_1 \le \| f\|^2_\infty \|1\|_1$ then root
 
well, yes. But that would include the work that was done in proving the generalization. I guess we could say that Fermat's Last Theorem can be proven with basic arithmetic, if we don't mind doing all the intervening work ;-)
@CalvinKhor that is what was done in the linked answer.
 
:D the issue is not having a name for it...
also I can't quantify this but surely FLT should be further from basic arithmetic than the two versions of holder :P
 
sure, I was just extrapolating for emphasis.
 
this is only true for finite measure spaces though, no?
 
11:00 AM
yes
 
@porridgemathematics $\|1\|_2\lt\infty$
 
whereas holder works with whatever measure space you want
 
well ok. the application of holder is always true. whether or not the result is useful depends on if 1 has finite integral
 
sure, I like to see this result as essentially the statement : in a finite measure space, the $L^p$ 'averages' of a function converge upwards to $||f||_{\infty}$
i think this can be proven without holder
errr, well, the upwards part requires holder, so this is untrue
just convergence doesn't need holder though
 
on finite measure spaces, $L^\infty$ embeds into $L^1$, and on....atomic? spaces? or whatever collection $\ell^p$ falls in the reverse inequality holds
 
11:06 AM
yeah, this comes down to this problem in folland i think, basically for $p<q$ finite, you get $L^p \subset L^q$ if and only if the measure space has measurable sets of arbitrarily small positive measure
and I think the reverse inclusion holds with a similar sort of condition at least for finite $p,q$
oh whoops, *if the measure spaces does NOT have measurable sets of arbitrarily small positive measure
 
to match with $L^\infty\subset L^1$ you'll need $p>q$
right ok
 
so yeah, atomic seems intuitive
 
yes, that was what I meant by atomic. there is some constant $c$ so that $\mu(A)>0$ implies in fact $\mu(A)>c$
 
yeah thats precisely it
its an iff too
 
is this how this goes for the lower bound - say $0<C<1$, then $\|f\|_p^p \ge \int{|f|\ge
C \|f\|_\infty} |f|^p\ge C^p \|f\|_{\infty}^p \mu(|f|\ge
C \|f\|_\infty)$
take $p$th roots and $p\to\infty$ the measure goes to 1, then $C$ arbitrary less than 1 so result?
ok sounds good
....converges upwards though....hmm
I only get convergence with squeeze rule (above mentioned generalization of Holder for the upper bound)
the inequality chain should read
$$\|f\|_p^p \ge \int_{|f|\ge C \|f\|_\infty} |f|^p\ge C^p \|f\|_{\infty}^p \mu(|f|\ge C \|f\|_\infty)$$
 
11:57 AM
I'd like to sniff Heron.
 
12:47 PM
How do you feel about L'Huilier?
 
 
2 hours later…
2:27 PM
@porridgemathematics: Here is a solution to $\lim_{p\rightarrow\infty}\|f\|_p=\|f\|_\infty$ that based on Chabyshev's Markov inequality.
 
 
2 hours later…
4:56 PM
@CalvinKhor for convergence upwards i think its really just a matter of applying Holder to show that $(\frac{1}{\mu(X)} \int_{X} |f|^p)^{\frac{1}{p}}$ is increasing in $p$
@OliverDíaz nice, it makes sense to use markov there, i think most proofs essentially come down to using it in some form
 
5:10 PM
Why does Terence Tao call the Elliott-Halberstam conjecture a "sort of super-generalized Riemann Hypothesis" in a blog post? Does it imply the GRH? terrytao.wordpress.com/2007/06/05/…
 
good question. i dunno if anyone in the chat is a number theorist.
 
@porridgemathematics In this answer, I use Jensen's Inequality, but that is pretty close to Hölder. They are all convexity arguments, and so are closely related.
 
mathoverflow.net/questions/153078/… is the closest i could find from google.
 
@robjohn: For finite measure, it does make sense to use either Jensen's (or Holder's, which is in a sense an application of Jensen's inequality)
 
5:35 PM
Another approach is to use $$\|f\|_p=\left(p\int_0^\infty\mu\left\{x:|f(x)|\ge\alpha\right\}\alpha^{p-1}\,\mathrm{d}\alpha\right)^{1/p}$$
 
 
1 hour later…
6:46 PM
does $\sum_{n=1}^\infty e^{zn^{-s}}$ converge in a distributional sense?
 
 
1 hour later…
8:00 PM
@geocalc33 On what domain?
I am not sure any domain works, but it would be nice to know what domain you are looking at.
 

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