How Euler connected the zeta function to primes
Euler's product formula shows that the zeta function, a sum over all natural numbers, equals an infinite product of geometric series over the primes. The key insight is that when you multiply those prime-based geometric series together and expand all the terms, every natural number appears exactly once, because the fundamental theorem of arithmetic guarantees that every natural number has a unique prime factorization. This means the zeta function and the distribution of primes are encoding the same information in two different forms, which is why the Riemann zeta function is so deeply connected to prime numbers.
- The geometric series for a prime p takes the form 1 + 1/p + 1/p² + 1/p³ + ..., and sums to 1/(1 - 1/p) using the standard formula for an infinite geometric series.
- Euler arrived at this product formula in the 1730s, as an extension of his work on the Basel problem rather than as a targeted attack on prime numbers.
- The shared exponent s in the zeta function is what links both sides: raising the primes to the power s on the product side corresponds exactly to the same s in the sum 1/nˢ on the other side, making the formula work for any value of s.
Transcript
This chapter, from the episode video's captions · 1,230 words
15:51first number. So, the first number in this case is 1. You put that on the top 1 / 1 minus the common ratio. So, in this case, it would be 1 / 1 - 1/2 to get you two. And that's that's how you sum up a geometric series. Okay? So we're going to require this toolbox. All right. Now that we've understood that, this was Oiler's genius. He said, consider the geometric series for primes. Okay. We had just seen a geometric series for two. A geometric series for three would look like 1 + 1/3 + 1 9th + 127th because those are the powers of three. >> Mhm. >> We don't do four because four is not a prime. >> Mhm. We um we look at the geometric
16:32series for five >> which is 1 plus 1/5 plus 125th plus 1 125th [clears throat] and then there'll be 625 so on and so forth. All of those things are going to equal 1 / 1 - common ratio. So it's going to be 1 over 1 - 1 over2 over there. That's going to be 1 over 1 - 1/3. So that'll give you 2/3 actually over there. It's it's right or three halves sorry. And so this is the geometric series for powers of primes. >> Okay. [clears throat] >> Mhm. >> Oiler said, let's take a look at this and see if we can do something. All right. If we multiply all of these series together,
17:14so we multiply like the infinite sum up there, the infinite sum up there, the infinite sum up there, right? I'm going to really get um a multiplication of those individual Yes. representations, right? Those individual answers. And now you're already starting to see that secondhand side of Oilers's product formula. >> Yeah, I was going to say that it's the same. Yeah, there's the through line. >> Yeah, there's the through line. The the giant pi that means multiply. The giant sigma means means sum. The giant pi means multiply. And it's it's multiplying over the primes. >> Okay. for whatever power. In this case, the power is S equals 1. >> Mhm. >> Right. And so now we're seeing the right hand side. >> Yes.
17:55>> Now that's fine. Here's the real genius of Oiler. Why is he even doing this in the first place? Okay, this seems like a lot of stuff to do. >> Mhm. >> Without the punch line. Here is the punch line. What about the left hand side? >> The left hand side, as I said, it's a bunch of primes multiplied together, right? You've got 1 + 1/2 + 1/4 the powers of 1/2 the powers of 1/3 the powers of 1/5 you'll get the power of 17th later right? If I multiply those out and I do the foil method, if you remember like um you have to distribute each term has to multiply to each term, right? So if I if I multiply that out, the one can be
18:35multiplied to all the other ones to get a one, right? Then the 1/2 from the first one can be multiplied to all the other ones to get a 1/2. >> Mhm. >> Right. Um similarly, the 1/4 can be multiplied to all the other ones to get a 1/4. >> The 1/3 will be there, the 1/5 will be there. I could also get a 1/6th because the 1/2 times the 1/3 multiplied by all the other ones is going to get me the 1/6. I see >> the 1/8 is going to come from the the 1/2, right? The 1 nth is going to come from the 1/3 and then the 1112th can come from 1/4 multiplied by 1/3. Every single number is going to be represented in that infinite sum
19:15>> because primes make up the numbers. >> Right. >> Right. Right? Every single number that is out there, this is called the fundamental theorem of arithmetic, which means that every single number out there, a natural number, can be broken down into multiplications of primes, products of primes that are unique. There's there's only one way to do it. Okay? And there's only one way to choose all of the different terms in that sequence. And now you can notice, right? What are the ones that are missing? Right? >> 17th is missing because I haven't included that in this. Right? Mhm. >> Similarly, 11 1th is missing, but that's a prime. That'll be another that'll be another one on its own. Um 113th is
19:56missing, but also 114th is missing because 114th would be the 1 17th multiplied by the 1/2 multiplied by all the other ones. Right? So all of the other composite numbers are not there and the primes that I haven't included aren't there >> because they're just further down. >> Yeah, there's further down. I haven't included it. Right? If I added the if I added the 17th thing, then the 114th would show up and the 17th would show up, but the 111th would not show up, right? Because that would require another. >> As we expand the top row we have here, it will fill in these gaps that are currently present. And you could do this infinitely. >> Yes. And you can do this infinitely. And the key is you will get every single fraction ever. >> That's incredible. Okay. >> Right. >> Yeah. Yeah. Yeah. Yeah. >> This was Oiler's genius. >> Yeah. That
20:36>> and that's how he's showing this is called Oiler's product formula. And it's kind of crazy that like he set out to just solve the basil problem. >> And then and then he's like, "Oh, by the way, I also noticed this. [laughter] >> Just just have this one on the >> Yeah. Yeah. So this here you've already seen now the zeta function is on the left. That's the sum >> of the one over all the natural numbers, right? >> He has tied this now to a product over primes. Mhm. >> And now you can imagine if I do if I want to do um one the zeta function to the second power all I have to do is put the primes to the second power. >> Right. Right. Yes. Yes. >> And that's [clears throat] why the s is a common um exponent
21:18>> in in both across both. >> Oh that's so elegant. >> Right. And and this already shows you how the primes are so intertwined right >> with the zeta function. Right. They are one and the same. And so part of what we're building here is the connection between those two. >> Yes. I'm trying to I'm trying to justify why there's so much hype >> about this. Okay. >> It's because the remon zeta function has to do with primes >> in a very integral way. It's actually encoding the same information. >> Right. >> On one side there's a a product of primes. On the other side there's a sum over every natural number. >> This is really good. >> It's kind of cool. >> This is this is really good.
21:58>> Yeah. And this was this was in 1730s. >> Okay. >> No Wi-Fi then. >> No. Um now 1792
From What Claude Actually Did to the Riemann Hypothesis
Claude takes a real run at the Riemann Hypothesis, forcing us to ask what agentic AI can now do in mathematics, before we open the summer transfer window for America’s scientists.