If $k>1$, $k\equiv 1 \text{ mod 4}$, then $$\sum_{n=1}^\infty\frac{n^k}{e^{2\pi n}-1}=\int_{0}^\infty\frac{t^k}{e^{2\pi t}-1}dt=\frac{\zeta(k+1)\Gamma(k+1)}{(2\pi)^{k+1}}$$
Which can be interpreted nicely as, $$\int_{0}^\infty\frac{\lfloor{t}\rfloor^k}{e^{2\pi \lfloor{t}\rfloor}-1} dt=\int_{0}^\infty\frac{t^k}{e^{2\pi t}-1}dt$$
Though I still see no way to prove either in an elementry way, though there easily seen equivilent
I might pose the question in terms of asking to show the equivilence of those two integrals
As you can see though there is a simple pole at $x=y$ which would be a problem for an ordinary Mellin Transform, but here $Z$ indicates taking the finite part of the integral
@PeterTamaroff A pretty wide range. Basically, you take the Mellin Transform of a function and then figure out where in the complex plane it is analytic
So something like $\frac{1}{1+x^2}$ has a well-known mellin transform, but it is only analytic in the strip $0<Re(s)<2$ if $s$ is the variable we are Mellin Transforming to
It is mostly USEFUL for function where there is some kind of scaling symmetry. So in my example you can see that the argument of Bessel function is multiplied by $\beta$
Taking the mellin transform makes it so that $x$ and $\beta$ no longer appear as a product, which is useful for making an approximation like $\beta << 1$
Otherwise you have to worry about "well, $\beta$ is small but maybe x is large so . . ."
I'm trying to solve a problem in P&S now. Suppose that $\langle t_n\rangle $ is a sequence of numbers such that there exist another sequence of positive numbers $\langle \varepsilon_n\rangle $ converging to zero and $$ t_{n+1}>t_n-\varepsilon_n$$ Then $\langle t_n \rangle $ is everywhere dense between $\overline{\lim}_{n\to\infty}t_n$ and $\underline{\lim}_{n\to\infty}t_n$
@PeterTamaroff I don't mean to bore you with bessel functions and contours in the plane, but if you want to take a look at what I'm trying to get away with you can find it here at the bottom
@PeterTamaroff It's actually a program where I can collect all my notes and calculations and then it syncs to a server that can be accessed if you know the address
@KarlKronenfeld Haha indeed. Also don't forget that I am writing on a graphics table that is only 5" x 7" but it mapped to my entire laptop screen, so any wobble or mistake is magnified
@AlexYoucis Kudos to you. I couldn't solve most of those problem cold, right now. Thankfully I think it wouldn't take much study for me to get there but still
@AlecTeal It sounds like 'Steve' doesn't understand how the internet works...
@KevinDriscoll you'd think that Bear Grills or however you spell it guy would be upset, but I have a feeling that he wanted it all along. Dontcha just love happy endings?
@KevinDriscoll you didn't get that Barbra reference did you?
Have a look at the Barbra Strisand (again, unsure of spelling) effect. Basically a guy uploaded a picture of a house by the beach, it had 4 views, then 10 views (6 were lawyers) then they issued a take-down notice, then the Internet found out and made it hugely popular. @KevinDriscoll
@AlexYoucis Here they have done away with prelims for physics PhD students because they found that GRE scores were so highly correlated with passing the prelim that it was basically a wasted 6 months of study for the students
@KarlKronenfeld No, it's asking for a group such that $Z(G/Z(G))$ is non-trivial. The obvious example being any non-abelian simple group, such as $A_5$.
@KevinDriscoll another good example is Microsoft when talking about the XBone (XBox One) they'd just sort of mumble, the the Internet-shit storm that followed (and a 800% jump in Wii U sales, and trolling from Sony) kicked their ass, having said that I am still disgusted and don't want the first company to be involved with NSA's PRISM project to have a camera in my room. ->
Nor do I want to dance to turn the bugger on, having a remote control and the ability to turn the console off by the pad was the dream.
@KarlKronenfeld Yeah, I actually misread that on the exam too. I also languished over finding the $H$. Eventually I just said $F_2$, and showed that $F_2F_2'\supset F_2''\supset\cdots$.
Now for a maths question, when finding the intersection between a plane and a sphere, is there a "formal method" because it seems to pop up a lot on papers, I would simply: ->
Just solve the equations (the sphere and the plane) then tidy it up and parameterise if the Q asked.
I shall soon. This integral is part of the Mellin Transform of the integral equation that was the original form of the integral equation that you put the bounty on
The one you put the bounty on and this original one are related by a weird kind of transform themselves, called the Kontorovich-Lebedev Transform
@skullpatrol I still have not understood why the Kontorovich-Lebedev transform is useful in this problem. It was developed and usually appears in solving the Laplace Equation in polar coordinates over a wedge-shaped domain
@Twink You were suspended for 2 reasons, (1) being a bit rude to a couple people (you weren't INCREDIBLY rube, but you were a bit) and (2) frequently steering the conversation toward sexuality which is fine to discuss every now and then but not as an all the time thing
You were actually suspended by a mod from another stackexchange site
I asked a question I feel is quite important a few days ago and got an answer some believe is a good one, but I would like some second opinions or references before I mark the question as answered. You can find the question here: /questions/482726/would-proof-of-legendres-conjecture-also-prove-riemanns-hypothesis
@Twink You seem to have legitimate math questions and things to contribute which is great. The sooner you allow your sexuality to be as much of a non-issue as it is to all of us the smoother it'll be. I agree with @skullpatrol that you can feel free to ignore anyone you think is attacking you.
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A huddle of people running from city taxes all park their trailers on a pile of dust 45 minutes from any semblance of civilization and give it a name without incorporating.
Have fun. I'm out as well, unless someone knows of anything about the prime numbers that, if proven, would prove the RH?
I read |Li(x) - pi(x)| <= c(sqrt(x) log(x)) for some c is equivalent, but does this mean that, if it were proven, the RH would be too? This whole equivalence thing is a bugger, as I thought it meant that a proof on one would prove the other, either way.
the $x^{\theta}$ term comes from the contribution of the real part of the zeros of the reimann zeta function, and the logarithmic factor accounts for the number of them
I'll hereby translate an entry in the blog Gaussianos ("Gaussians") about Polya's conjecture, titled:
A BELIEF IS NOT A PROOF.
We'll say a number is of even kind if in its prime factorization, an even amount of primes appear. For example $6 = 2\cdot 3$ is a number of even kind. And we'll say...
actually i think it's L(x) = O(x^(1/2+e)) for all e > 0 not exactly the same as what i said
Sorry; had to put my kid to bed. So my problem, I take it, is that the "error term" approach is not the same as showing that the gap between consecutive primes is no bigger than the square root of the first, right?
maybe factoring can be done in polynomial time etc.
for example if prime gaps are in O(log(n)^k) for some k (much weaker than cramer's conjecture) then the next prime after n can be found in polynomial time because we can just test each one with the AKS algorithm
not quite the same as factoring but makes ya wonder
Either way, I don't think all numbers will be able to be factored in polynomial time given a proof of the RH; instead, I think highly composite numbers would be, moderately composite would be partially factorable in p, but those which are almost prime would not be factorable at all with the exception of a few.
That's my problem; can't get it with prime gaps, so you ave to go at it directly from the error term of the prime number theorem (or some similar formula).
As do I, though it's somewhat difficult algebraically. I like the O(root(p)) one better from that point of view, and I think it will be the first to be proven.
anyway i think it is interesting that it is seemingly consistent to be skeptical about RH and also believe prime gaps are O(log(n)^k) for some k
because all RH implies is the much weaker statement that prime gaps are O(sqrt(n) log(n)) and there isn't a very strong implication in the other direction either unless k is too small
What was throwing me is that it's the same "big O" for both the gap problem and the prime number theorem error term problem, which got me thinking the two problems were one and the same.
I'll hereby translate an entry in the blog Gaussianos ("Gaussians") about Polya's conjecture, titled:
A BELIEF IS NOT A PROOF.
We'll say a number is of even kind if in its prime factorization, an even amount of primes appear. For example $6 = 2\cdot 3$ is a number of even kind. And we'll say...
