Please Post Your Favorite Infinite Product (or Application Thereof)

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
Sign up Log in
20 replies · 2K views
benorin
Science Advisor
Insights Author
Messages
1,442
Reaction score
191
I've always had a fascination with infinite products. I like them, I do. To stimulate our ensuing conversation, I here post Knopp's two-way series-to-product (and vice-versa) "doorway" out of his book, Theory and Applications of Infinite Series pg. 226:

Knopp said:
1. If ##\prod_{n=1}^{\infty}(1+a_n)## is given, then this product, if we write
$$\prod_{\nu =1}^{n}(1+a_{\nu})=\mathfrak{p}_n \, ,$$
represents essentially the sequence ##(\mathfrak{p}_n )##. This sequence, on the other hand, is represented by the series
$$\mathfrak{p}_1 + ( \mathfrak{p}_2 - \mathfrak{p}_1) + ( \mathfrak{p}_3 - \mathfrak{p}_2) + \cdots = \mathfrak{p}_1+\sum_{n=2}^{\infty}(1+a_1)\cdots (1+a_{n-1})\cdot a_n$$
... [omitted text]
2. If conversely the series ##\sum_{n=1}^{\infty}a_n## is given, then it represents the sequence for which ##s_n=\sum_{\nu = 1}^{n}a_{\nu}##. This is also what is meant by the product
$$s_1\cdot\tfrac{s_2}{s_1}\cdot\tfrac{s_3}{s_2}\cdots = s_1\cdot \prod_{n=2}^{\infty}\tfrac{s_n}{s_{n-1}} = a_1\cdot \prod_{n=2}^{\infty}\left( 1+\tfrac{a_n}{a_1+a_2+\cdots +a_{n-1}}\right) \, ,$$
--provided it has meaning at all. ...

Maybe that'll break the ice... please post your favorite infinite product (or application thereof).
 
  • Like
Likes   Reactions: berkeman
Physics news on Phys.org
Consider a field ##\mathbb{F}## with ##q < \infty## elements. Consider the set of all monic polynomials in ##\mathbb{F}[X]## by ##\mathcal{M}##. Then

$$\sum_{N \in \mathcal{M}} z^{\deg(N)} = \prod_{P \in \mathcal{M}\setminus \{1\}: P \mathrm{\ irreducible}} \frac{1}{1-z^{\deg(P)}}$$ for ##0 \leq z < 1/q##.

Denote the amount of all irreducible polynomials in ##\mathcal{M}## of degree ##m## by ##i_m##. Then the above formula implies
$$q^k = \sum_{m | k} mi_m$$
and as a corollary of that ##i_m > 0## for all ##m##, which is highly non-trivial.

I really like this as this uses analysis to obtain algebraic results.
 
  • Like
Likes   Reactions: fresh_42
fresh_42 said:
$$
\sum_{n\in \mathbb{N}} \dfrac{1}{n^s} = \prod_{p\in \mathbb{P}}\dfrac{1}{1-p^{-s}}
$$

My formula has the same structure as yours! Moreover, both products range over irreducible elements. There must be some generalisation!
 
Math_QED said:
My formula has the same structure as yours! Moreover, both products range over irreducible elements. There must be some generalisation!
Maybe by taking the projective limit of ##\mathbb{F}_p## over all primes, possibly the algebraic closures of ##\mathbb{F}_p##. There is one crucial difference: Whereas Euler / Dirichlet have the variable in the exponent, your formula has the variable in the base and an invariant in the exponent. Hower, I would bet there is a connection. Do you know a formulation which uses characters instead of degrees?
 
fresh_42 said:
Maybe by taking the projective limit of ##\mathbb{F}_p## over all primes, possibly the algebraic closures of ##\mathbb{F}_p##. There is one crucial difference: Whereas Euler / Dirichlet have the variable in the exponent, your formula has the variable in the base and an invariant in the exponent. Hower, I would bet there is a connection. Do you know a formulation which uses characters instead of degrees?

What do you mean with characters?
 
fresh_42 said:
Group characters. I was wondering if the formula can be phrased as sum over functions ##\chi: \mathbb{F}_q \longrightarrow \mathbb{C}^*##, but this was just an idea, not that I had a solution in mind.

No, unfortunately I'm not aware of something like that. Maybe someone with expertise in analytic number theory can help here.
 
Does anybody know the geometric significance of infinite products? I get how multiplication of two complex numbers works graphically, polar form does this simply enough the modulus of the product is the product of the moduli and the argument of the product is the sum of the arguments but how does one extend this to infinitely many multiplications? Could we just start off with the partial product? Then apply some sort of limiting procedure? I've never seen this done in texts before, but seems like something they would have done along time ago
 
benorin said:
Does anybody know the geometric significance of infinite products? I get how multiplication of two complex numbers works graphically, polar form does this simply enough the modulus of the product is the product of the moduli and the argument of the product is the sum of the arguments but how does one extend this to infinitely many multiplications? Could we just start off with the partial product? Then apply some sort of limiting procedure? I've never seen this done in texts before, but seems like something they would have done along time ago

An infinite product is usually defined as

$$\prod_{n=0}^\infty a_n = \lim_{k \to \infty} \prod_{n=0}^k a_k$$

but some authors (e.g. Apostol) make conventions like such a product does not converge if it contains infinitely many zeros etc, though the partial products are all ##0## then.
 
Yes convergence of ##\prod_{k=1}^{\infty}a_n## will require that each term is non-zero and that ## | a_n|\rightarrow 1## as ##n\rightarrow \infty## and iirc (correct me if I'm wrong, I may be) ##\text{arg}\, a_n \rightarrow 2k\pi\text{ for some } k\in\mathbb{Z}^+##. These are necessary but not sufficient.
Edit: this is equivalent to requiring an infinite series general term go to zero for convergence and since the arg of the partial product is the partial sum of the arg's the general term must go to zero (up to multiples of ##2\pi##.
 
Last edited:
There is no reason why infinite products with infinite many factors unequal one shouldn't be considered, e.g. Euler's formula for the Riemannian zeta function. Or the distinction between direct sums and direct products in algebra.

I find the above remark about zeroes in a product more than strange, infinite or not.
 
iirc Knopp speaks of a product diverging to zero, but this is just convention. And for Euler's product for the zeta function I think you misread what I wrote because ##\left| \tfrac{1}{1-p^{-s}}\right| \rightarrow 1## as ##p\rightarrow \infty## through the primes for all s that the product converges for, no?

Edit: I re-read Knopp, he defines that no terms ##a_n## should vanish for all ##n>m## for some fixed m in order to converge and that the the partial product beginning immediately beyond this point, ##\prod_{\nu =m+1}^{n}a_{\nu}## tend as n increases to a definite limit.∏ν=m+1naν
 
Last edited:
fresh_42 said:
There is no reason why infinite products with infinite many factors unequal one shouldn't be considered, e.g. Euler's formula for the Riemannian zeta function. Or the distinction between direct sums and direct products in algebra.

I find the above remark about zeroes in a product more than strange, infinite or not.

I'm not saying they shouldn't be considered. I'm just saying that some authors do not consider such products as "converging". I do not have enough expertise in the matter to say this is justified or not.
 
benorin said:
The Euler product only converges for ##\Re s >1##.

I have't looked into the details. Wikipedia says:

One obtains a series identity first published by Konrad Knopp, which holds on ##{\displaystyle \mathbb{C} \setminus \{1+2\pi \mathrm{i} m/\log 2\mid m\in \mathbb{Z} \}}##

$$\zeta (s)={\frac {1}{1-2^{1-s}}}\sum _{n=0}^{\infty }{\frac {1}{2^{n+1}}}\sum _{k=0}^{n}(-1)^{k}\ {n \choose k}\ {\frac {1}{(k+1)^{s}}}$$
This has been proven by Helmut Hasse in 1930. There are therefore no further gaps or poles during the continuation. This finally results in holomorphy on ##{\displaystyle \mathbb{C} \setminus \{1 \}}.##