Recent content by keebs

Andrew Wiles and Fermat...

Ah, ok. Because if either one of those is true then it implies that pi(x+1)~x/lnx.

What about with the prime counting function? Is pi(x+1)~x/lnx?

Ahhh, ok. Thank you.

t may be small, but e1/t is large. What do you do in physics then? Where do you work?

I have a few questions about asymptotic functions, and was wondering if anyone could help... If h(x)~g(x), is h(x+1)~g(x)? And, if h(x)~g(x), is h(x)h(x+1)~g(x)g(x+1)? Thanks in advance for any help...

I found this on arxiv...is this guy a loon or do the proofs seem reasonable? Proofs

And to calculate an integral of Pi(n) you will need many more terms than sum(1,n) if you're going to use integrals, because calculating integrals numerically requires splitting up the curve into little tiny rectangles, finding the area of all of the rectangles, and then summing up all of the...

But as t increases the volume increases, thus a computer will require more time to compute the integral.

A simple solution for FLT would arise if you could prove the abc conjecture...

...that's the impression I've always gotten. The factorization part of Shor's algorithm can be done on a classic computer, but it's when you get to the order-finding problem that Shor's algorithm takes advantage of the quantum technology (I don't remember where I read this, but once I do I'll...

It doesn't always require the same number of operations because you are integrating up to variable t, which can vary (hence the term "variable"). In fact, because you will need to integrate numerically it may take even more operations to get a good approximation than a summation would.

Basically Riemann gave a formula for the prime counting function that includes a sum over all zeros of the zeta function (well, not exactly, it's actually a sum of x to the power of all zeros of the zeta function), and if all of the zeros lie on the critical line than we can get a good estimate...

Hmmm...interesting. I didn't know that, thanks. It makes sense because if it doesn't converge to 0 then "towards the end" (I guess you could say that) of the summation you'd just be adding values very close to a certain constant (or adding diverging terms) over and over and over again, but...

Yeah, but when n approaches infinity Pn approaches the set of all natural numbers because Pn is the set of all numbers generated by all primes up to n, so when we include all of the primes it should generate all of the natural numbers. Why does the terms have to go to zero in order for the...