## Infinite product of complex numbers

### Set

context | $(z_i)$ … Sequence($\mathbb C$) |

definition | $\prod_{i=1}^\infty z_i\equiv \mathrm{lim}_{n\to\infty}\prod_{i=1}^n z_i$ |

#### Discussion

todo: Interestingly, I think I see the nLab doesn't want to allow e.g. $\prod_{n=1}^\infty (17-n)$ to be zero. (Reference below)

I recon infinite products may arise when x is written as $111\cdots11x$ and 1 is represented as something.

see the context of the “prod-step to exp-sum” formula blow.

can Euler products arise in this way?

And they may arise ans determinants of infinite dimensional operators, i.e. products of eigenvalues.

#### Theorems

##### Limit of a sequence

For nonzero $a_n$ and any $N,M,K$, we have

- $\lim_{n\to\infty}a_n=a_N\cdot\prod_{n=N}^\infty\frac{a_{n+1}}{a_n}$

- $\lim_{n\to\infty}a_n=a_M+\sum_{n=M}^\infty(a_{n+1}-{a_{n}})=a_M+\sum_{n=M}^\infty\left(\frac{a_{n+1}}{a_n}-1\right)\,a_n$

So both are equal. This might be useful for rewriting and computing $\prod_{n=d}^\infty f(n)$, or even find the limit. Mathematica can find the $a$-sequence with $a_{n+1}=f(n)\cdot a_n$, see (“RSolve”).

Also, from the sum formula with $a_n=\prod_{k=K}^{n-1}b_k$, we get

$\prod_{k=K}^\infty b_k=\lim_{n\to\infty}a_n=\prod_{k=K}^{M-1}b_k+\sum_{n=M}^\infty(b_n-1)\,\prod_{k=K}^{n-1}b_k$

$\prod_{k=0}^\infty b_k = \prod_{k=0}^{M-1}b_k + \sum_{n=M}^\infty(b_n-1)\,\prod_{k=0}^{n-1}b_k$ |
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b[n_] = 1 + 1/n!; Ks = 0; Ke = 100; Mid = 4; Product[b[k], {k, Ks, Ke}] Ks = 0; Ke = 100; Mid = 4; Product[b[k], {k, Ks, Mid - 1}] + Sum[(b[k] - 1) Product[b[k2], {k2, Ks, k - 1}], {k, Mid, Ke}]

todo: I think this formula holds also for finite products, i.e. when $\infty$ is replaces by some natural number $K_2>M>K$.

##### prod-step to exp-sum

Note that this isn't of the form as in this entry, because the terms depend on the upper limit $N$ that goes to infinite

$\lim_{N\to\infty} \prod_{n=0}^{N-1} \left(1+\frac{x_n}{N}\right)= \exp\left(\lim_{N\to\infty}\frac{1}{N}\sum_{n=0}^{N-1}x_n\right)$

I've seen this pop up in the derivation of the basic path integral. Where else does it pop up? Why, if $x_n$ doesn't depend on $N$, does this turn to an $\exp(\int something)$? I think they join some limits to a single one. I think that relates to what Ron said about quantization procedures.

h = (b - a)/m; f[X_] = c X^2; p = Product[ 1 + f[i] h , {i, a, b, h}] // simple a = 0; b = 7; c = 3; Limit[p, m -> \[Infinity]] Exp[Integrate[c X^2, {X, 0, 7}]]

##### Jacobi triple product

$\prod_{m=1}^\infty \left( 1 - q^{2m}\right)\left( 1 + w^{2}q^{2m-1}\right)\left( 1 + w^{-2}q^{2m-1}\right)= \sum_{n=-\infty}^\infty w^{2n}q^{n^2}.$

If the sum is a partition function, what does $\prod_{m=1}^\infty$ represent in terms of (algebraic) operations on state spaces?

Also

$\prod_{n=0}^{\infty}\left(1 + x^{2^n}\right) = \frac{1}{1-x}$