Dirichlet theorem and the PNT

[Rami5743]

About the author

My name is Rami Aizenbud. I am a mathematician working in the Weizmann Institute of Science. You can find more information about me on my website

Quick Summary

This is about 2 related topics The Dirichlet theorem and the PNT.

My first choice is probably Dirichlet theorem, but I am flexible.

Target medium

video/videos in the stile of 3b1b

More details

I have written a Wikipedia article about the Dirichlet Theorem in the Hebrew Wikipedia I think this is a very exiting topic which is very suitable for animation. The prove involves lots of ideas from all over math, and many of these ideas can be illustrated. On the other hand it is rather deep topic so it is quite challenging to make it accessible. Nevertheless, I have explained the proof to my 10 years old kid, and I think he understand a big part of it.

Contact details

you can e-mail me on aizenr@gmail.com, and we can set a zoom meeting.

[3b1b]

Hi Rami,

Dirichlet's theorem is indeed a very interesting topic, but getting it to be accessible might be a fun challenge. Can you add any details on exactly what approach you had in mind? Maybe even an outline or storyboard describing some of the core mini lessons that would have to be involved along the way?

[Rami5743]

Here is a possible plan:

• Video 1. general introduction to the topic:

  1. formulation + motivation
  2. recollection of Euclid's proof 
  3. optional: discuss its (in)effectivity.
  4. optional: Discus Euclid-like proof for special cases of the Dirichlet theorem.
  5. From Euclid's proof to (simplified) Erdos' proof (i.e. to count the number of all numbers from 1 to n assuming there are only k primes in this interval)
  6. optional: discuss its (relative) effectivity.
  7. Euler's product and diversion of $\sum \frac{1}{p}$.

  This video can serve also as a joint introduction to PNT and Dirichlet theorem. In this case, item 4 is less relevant, but items 3, and 6 become more relevant. Also, in this case, one can add Mertens'  (very simple) proof that $$\sum \frac{ln(p)}{p} \sim \sum \frac{1}{n}$$ which serves as a heuristic for PNT. If we add all the optional points we might have to split the video into 2.

• Video 2.   Diriclet argument for simple special case(s): Let $d=3$. Let $A= \sum \frac{1}{p}$ for $p=1 \mod 3$ and $B= \sum \frac{1}{p}$ for $p=2 \mod 3$. We know $A+B=\infty$. We need to prove $|A-B|<\infty$. This can be deduce from $0<L(1,\chi)<\infty$, where $\chi$ is the non-trivial character of $(\mathbb Z/ 3\mathbb Z) ^\times$. The later is very easy computation. Cases $d=4,6$ are similar, case $d=8$ is not much harder. But other cases (for example $d=5$ or $d=10$) are a bit more problematic (see item 4 in the next video).

• Video 3.   Reduction to $L(1,\chi)=0$: 1. Fourier transform on finite abelian groups -  there are several possible arguments, one has to choose. The proof of Fourier transform is a computation for a given modulus, so maybe we can postpone it to another video (not necessarily part of this sires).

  2. The decomposition $$ \sum{p=a \mod d} \frac{1}{p^s}=\sum\chi \bar \chi(a) ln(L(s,\chi)) +O(1)$$
  3. The convergence of $L(1,\chi)$. Here also there are several possible arguments, I have a favorite one, for which I made an animation. 
  4. There is one more delicate point that often is swept under the rug, one has to decide whether to do it here (I prefer not swept under the rug).
  5. emphasize that now we have an argument that reduces the theorem for a particular progression to a computation, 
  6. optional - we can give the formula $$L(1,\chi)=-\frac{1}{m}\sum{k=1}^{m-1} \ln(1-e^{2\pi i k/m})\sum{l=1}^{m}\chi(l)e^{2\pi i kl/m}$$  It is not useful for the general proof but useful for special cases

We can stop at this point, declaring that we won, since we made a very hard problem, into a computation that one can easily perform with a computer till, say, $d=1000$. We also can frame the problem as if the aim is to understand the case $d=10$ (last digit analysis) or the special cases discussed in the "prime spirals" video, which are all follows from the above explanation.

On the other hand, we can also finish the proof:

• Video 4.   completion of the solution:

  1. Dirichlet (simple) argument for the non-real case. I made a nice animation here to explain the asymptotic of $\zeta$ near 1. The most elegant way is to compare the sum to the integral, however, one can do it in a more elementary way. One has to choose whether to mansion it and to what extent.

  2. Optional - we can squeeze this lemon to its end. Namely, we can:

     • deduce the Dirichlet theorem for many progressions ($a$ is non-square modulo any factor of $d$).
     • deduce the Dirichlet theorem for all powers of a given number (e.g. 10) depending on one computation.
     • deduce the Dirichlet theorem for all progressions except those whose difference is a multiplier of one number.

     If we add all these points we might have to postpone the next ones to another video.

  3. optional - non-elementary, but very short Landau's proof.

  4. The elementary Merten's analytic proof. Here, one has to study the asymptotes of the defining sires of $\zeta(1/2)$. I made another animation for it. One also uses here the Dirichlet Hyperbola method.

  5. Optional - discussion of the more algebraic proofs

  6. Optional - Explanation that in all of the above approaches, one can avoid using item 1.

I'll be happy to discuss any of it on zoom.

[InigoMontoya314]

Hi, I'm from #124 and #136. From my perspective(someone who just graduated from high school but have relatively good math background, so I've heard of Dirichlet's theorem), I think one way to make it work for a student like me is to go for special cases like d=1, and even that proof is quite challenging for me. There could be easier proofs that I'm not aware of, so please let me know if that's the case.

[Rami5743]

@ajayberriman, Thank you very much for your interest. It seems that we have rather different things in mind, so I do not think that it will be fretful to work on this together. However, if you have any mathematical question on the subject or on a related topic, I'll be happy to try to answer it.

[Rami5743]

@InigoMontoya314, It depends on what do you mean by Dirichlet theorem. The standard meaning is that there are infinitely many primes in the sequence. In this case, d=1 is just Euclide theorem, though it is very butiful I'd hardly call it challenging.

Dirichlet actually proved a stronger statement: the diversions of $$\sum\frac{1}{p}$$ when p is in the progression. The case p=1 becomes now Euler result, and ofcores any farther discussion of Dirichlet theorem have to start with it. There are more quantitative version of the Direchlet theorem, the strongest of which being the PNT for arithmetic progression, here the case d=1 is just the PNT, which is at list as chalnging as the Dirichlet theorem and deserve a treatment by its own.

[Rami5743]

@all, Let me explain what do I mean by simplified Erdos argument: Assume that there are only finitely many primes: $p_1,\dots pk$, then since any number can be decompose into prime factors (note that we do not use the uniqueness here) there are no more then $\prod \log{p_i}n$ numbers up to $n$. This means that $n \leq (log_2 n)^k$ which implies $log_2 n \geq n^{\frac{1}{k}}$, this leads to a contradiction.

This in fact gives the following bound: $$\pi(n) \geq \frac{\ln(n)}{\ln(log_2 n)}.$$

This is a slightly simplified version of an argument by Erdos that gives the slightly stronger bound $$\pi(n) \geq \frac{\log_2(n)}{2}.$$

Theas bounds are much weaker then the reality (given by the PNT) however they are way better then the bounds that can be obtained from original Euclid argument. This argument attributed to Erdos, so if Euler was awer of it he did not publish it, however it seems to me natural to consider this argument as transitional stage between Euclid and Euler. Namely, While in this argument we count all numbers below n, in Euler's product we use the same method to count all numbers but giving the number $k$ weight $k^{-s}$.

[InigoMontoya314]

@Rami5743 I see, I'm concerned with proof of proposition 3 in this paper, so I didn't realize the simple Euclidean argument. Thank you!