Using calculus to do number theory • Hacker News Board

Re: Using calculus to do number theory

#21

Post by wbl » Thu, Mar 19, 2026, 12:56 AM UTC

arch1t3cht wrote: post #4

joshuaissac wrote: post #2

The mathematical field of tackling number theory problems in this way is called analytic number theory. https://en.wikipedia.org/wiki/Analytic_number_theory The prime number theorem, on how prime numbers are distributed amongst the integers, was first proved using analytic techniques.

Analytic number theory exists and involves calculus, but it's not what the linked post is about. The article talks about Hensel's lemma, which is a purely algebraic statement with a purely algebraic proof, which, however, is inspired by techniques from calculus. This is typically still categorized as algebraic number theory.

Hensel's lemma is an analytic fact about the radius and speed of convergence of Newtons method in the p-adics.

Re: Using calculus to do number theory

#22

Post by articulatepang » Thu, Mar 19, 2026, 1:01 AM UTC

krackers wrote: post #19

articulatepang wrote: post #9

I love complex analysis, and that's the branch of calculus that is most associated with number theory. For example, it was critical in the original proof of the prime number theorem and Dirichlet's theorem on primes in arithmetic progressions. Today, all kinds of number theoretic functions are studied using complex analysis, like the famous Riemann zeta function, Dirichlet L-functions, theta functions, and so on. So…

>for any y >= n with f(y) = 0 (mod n), there's some x between 0 and n-1 There's a simpler way to see this, any such y can be represented as y = n k + x where i,j are divisor & remainder. Then f(y) = f(n k + x) = f(x) modulo n since by binomial theorem all other terms other than those with just x will be divisible by n.

Thank you and yes, I agree! It's neat to use the binomial theorem to see this, because that's the tool the article uses for the main trick/insight it's explaining.

Re: Using calculus to do number theory

#23

Post by krackers » Thu, Mar 19, 2026, 3:51 AM UTC

articulatepang wrote: post #22

krackers wrote: post #19

Earlier quoted context omitted.

>for any y >= n with f(y) = 0 (mod n), there's some x between 0 and n-1 There's a simpler way to see this, any such y can be represented as y = n k + x where i,j are divisor & remainder. Then f(y) = f(n k + x) = f(x) modulo n since by binomial theorem all other terms other than those with just x will be divisible by n.

Thank you and yes, I agree! It's neat to use the binomial theorem to see this, because that's the tool the article uses for the main trick/insight it's explaining.

>neat to use the binomial theorem to see this,

Yeah as far as I understand this basically _is_ where the connection to calculus comes in. As the article points out, it's very similar to computing the derivative via infinitessimals, e.g. (x+dx)^3 = x^3 + 3x^2 dx + O(dx^2), and from that you recover the term linear in dx as the derivative, 3x^2 . For the derivative you consider epsilon as a nilpotent where dx^2 = 0, and for hensel lifting lemma any term with a factor of p^2 gets zeroed (modulo p^2).

I'm just a layperson but I'm not really convinced that this is differential calculus in a meaningful sense. The other commenter mentioned "formal derivative" which seems fitting. There might be a connection to "umbral calculus" (something i know nothing about fwiw) though in the way you use the formal computations of derivatives for identities on polynomial equations, even without there being any differentials actually involved.