You mean that 3 + 4 cos t + cos 2t = 2 (1 + cos t)^2 >= 0 in one proof that the Riemann zeta function doesn’t vanish on the line Re z = 1?
That is indeed quite unmotivated and tricky. It is the math equivalent of looking at extremely optimized code, where more naive or alternate approaches are slower or take longer to work out. There are other proofs of the prime number theorem that don’t use this trig identity estimate, but they often use different tools (like harmonic analysis instead of complex analysis, or “elementary” methods) and the prime number theorem is often presented as a big goal of a complex analysis course, to students who are maybe OK with using random unmotivated tricks to reach a key step.
Well, holomorphic functions (like the Riemann zeta function) are harmonic, so it’s not unexpected to try to apply those methods to their study.
There are a couple of alternate approaches to the prime number theorem in this MathOverflow answer; a proof via harmonic analysis is detailed in this French paper (see Corollaire 2.6 for nonvanishing on the line Re z = 1). There’s also Selberg and Erdos’ “elementary” proof which doesn’t use complex analysis at all, but is very complicated (hence the scare quotes around elementary).
I’d be surprised if that was the same identity because the function n here is integer-valued. The coefficients involved look similar (modulo the extra factor of 2 in the last term) but the rest of this seems rather different, being deduced from the Fourier analysis of distributions.
I still can’t get over how the entire thing hinges on a random trig identity
You mean that
3 + 4 cos t + cos 2t = 2 (1 + cos t)^2 >= 0in one proof that the Riemann zeta function doesn’t vanish on the lineRe z = 1?That is indeed quite unmotivated and tricky. It is the math equivalent of looking at extremely optimized code, where more naive or alternate approaches are slower or take longer to work out. There are other proofs of the prime number theorem that don’t use this trig identity estimate, but they often use different tools (like harmonic analysis instead of complex analysis, or “elementary” methods) and the prime number theorem is often presented as a big goal of a complex analysis course, to students who are maybe OK with using random unmotivated tricks to reach a key step.
yes. Why would one use harmonic analysis when we are dealing with a complex function? Would be interesting to have a look at this proof.
Well, holomorphic functions (like the Riemann zeta function) are harmonic, so it’s not unexpected to try to apply those methods to their study.
There are a couple of alternate approaches to the prime number theorem in this MathOverflow answer; a proof via harmonic analysis is detailed in this French paper (see Corollaire 2.6 for nonvanishing on the line
Re z = 1). There’s also Selberg and Erdos’ “elementary” proof which doesn’t use complex analysis at all, but is very complicated (hence the scare quotes around elementary).the thing I wasn’t getting was since holomorphic is a stronger condition than harmonic, it would seem like using holomorphy would get you further.
thanks for the link, that seems to be the same proof if you notice the identify in lemam 4.4 3n(0) + 4n(a) + 2n(2a) 6 0
I’d be surprised if that was the same identity because the function
nhere is integer-valued. The coefficients involved look similar (modulo the extra factor of2in the last term) but the rest of this seems rather different, being deduced from the Fourier analysis of distributions.