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I discovered a way to evaluate infinite products of even analytic functions with high accuracy.

$$ \prod_{k=1}^{\infty} f(k^2) \approx \prod_{k=1}^{\infty} \left(1-\frac{A_1}{k^2}+\frac{A_2}{k^4}-\frac{A_3}{k^6}+\cdots+\frac{A_n}{k^{2n}} \right) = \frac{\sin (\pi a_1) \sin (\pi a_2) \cdots \sin (\pi a_n)}{\pi^n a_1 a_2 \cdots a_n} $$

Here:

$$A_1=\sum^{n}_{j=1} a_j^2,~~~~~A_2=\sum^{n}_{j\neq l} a_j^2 a_l^2,~~~~~A_3=\sum^{n}_{j\neq l\neq m} a_j^2 a_l^2 a_m^2,~~~\cdots~~~A_n=\prod^{n}_{j=1} a_j^2$$

The idea is as follows:

$$ \prod_{k=1}^{\infty} \left(1-\frac{a^2}{k^2} \right)=\frac{\sin (\pi a)}{\pi a}$$

Since the product converges, we can multiply two or more products by term:

$$ \prod_{k=1}^{\infty} \left(1-\frac{a^2}{k^2} \right)\left(1-\frac{b^2}{k^2} \right)=\prod_{k=1}^{\infty} \left(1-\frac{a^2+b^2}{k^2}+\frac{a^2 b^2}{k^4} \right)=\frac{\sin (\pi a) \sin (\pi b)}{\pi^2 a b}$$

We can easily recognize the Vieta formulas:

$$a^2+b^2=A_1,~~~~~~a^2b^2=A_2$$

$$x^2-A_1x+A_2=0$$

The connection to the coefficients of the Taylor series in general is obvious and we can always find $a_j$ by solving the equation:

$$x^n-A_1 x^{n-1}+A_2 x^{n-2}- \cdots + A_{n}=0$$

$$a^2_j=x_j,~~~~~~j=1,\dots,n$$

Of course $a_j$ can be complex, the product will still be real.


Now some examples to see how it works. Consider the following two products:

$$\prod_{k=1}^{\infty} k \tanh \frac{1}{k} = \prod_{k=1}^{\infty} \left(1-\frac{1}{3k^2}+\frac{2}{15k^4}-\frac{17}{35k^6}+\frac{62}{2835k^8}-\cdots \right)$$

$$\prod_{k=1}^{\infty} \frac{\sqrt{\pi}}{2} k ~\mathbb{erf} \frac{1}{k} = \prod_{k=1}^{\infty} \left(1-\frac{1}{3k^2}+\frac{1}{10k^4}-\frac{1}{42k^6}+\frac{1}{216k^8}-\cdots \right)$$

Coincidentally, the second term is the same for both functions, so we have the first approximation:

$$1>\prod_{k=1}^{\infty} k \tanh \frac{1}{k}>\frac{\sqrt{3}}{\pi} \sin \frac{\pi}{\sqrt{3}}=0.535131$$

$$1>\prod_{k=1}^{\infty} \frac{\sqrt{\pi}}{2} k ~\mathbb{erf} \frac{1}{k}>\frac{\sqrt{3}}{\pi} \sin \frac{\pi}{\sqrt{3}}=0.535131$$

To get the following approximations we solve quadratic, cubic and fourth-order equations and obtain the roots.

For example, the first product gives us:

$$a,b=\sqrt{\frac{5\pm i \sqrt{65}}{30}}$$

$$\prod_{k=1}^{\infty} k \tanh \frac{1}{k}<\frac{\sin \pi a \sin \pi b}{\pi^2 a b}=0.620862$$

The second product:

$$a,b=\sqrt{\frac{5\pm i \sqrt{95}}{30}}$$

$$\prod_{k=1}^{\infty} \frac{\sqrt{\pi}}{2} k ~\mathbb{erf} \frac{1}{k}<\frac{\sin \pi a \sin \pi b}{\pi^2 a b}=0.649760$$

The sequence of partial products up to $50$ is compared to the numerical value of the infinite product and three approximants in the graphs below.

enter image description here

enter image description here


Another interesting product is connected to the number $e$ (see here):

$$\prod_{k=2}^{\infty} e \left(1-\frac{1}{k^2} \right)^{k^2}=\prod_{k=2}^{\infty} \left(1-\frac{1}{2k^2}-\frac{5}{24k^4}-\frac{5}{48k^6}-\cdots \right)=\frac{\pi}{e^{3/2}}=0.70098$$

It converges extremely slow. But the second approximant already does a very good job of approximating the exact value.

enter image description here

Finally, let's consider much simpler case:

$$\prod_{k=2}^{\infty} \cos \frac{1}{k}=\prod_{k=2}^{\infty} \left(1-\frac{1}{2k^2}+\frac{1}{24k^4}-\cdots \right)=0.719109$$

The first approximant is the same as in the previous product:

$$P_1=\frac{2\sqrt{2}}{\pi} \sin \frac{\pi}{\sqrt{2}}=0.7163756$$

And it's remarkably close to the numerical value. The third one already gives $6$ correct digits:

enter image description here

Is this method useful for evaluating infinite products? It's easy to implement in any algebra system. I know it can be generalized using gamma functions. Can you provide some reference on the topic?

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  • $\begingroup$ (+1) Nice observation. What is the connection between $a_k$ and $A_k$? Moreover, what is the description of $f$? $\endgroup$ – user98186 Feb 24 '16 at 20:54
  • $\begingroup$ $a^2_j$ are the roots of polynomial with $A_k$ as coefficients. $f$ is an even analytic function of $1/k$, defined by its Taylor series. It's all written above. $\endgroup$ – Yuriy S Feb 24 '16 at 20:58
  • $\begingroup$ Thank you. Do you have anything in mind to prove it? Even more, what do you think about the error term? A suggestion: write your proposition in canonical form, so the relation between the terms are easier to see. Log both sides of your main equation to see if there's something obvious (easily provable). If it turns out fruitless, try generating functions. $\endgroup$ – user98186 Feb 24 '16 at 21:07
  • $\begingroup$ Thanks for the comment. I don't think this needs any proof, since it's directly derived. It may be possible to estimate the error, by considering the log series. But please, note, the product of sines is stricktly finite. The limit can not be taken, so it's not an equivalence between two products, the right side just estimates the left side $\endgroup$ – Yuriy S Feb 24 '16 at 21:12

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