How to prove that this iterated mean gives $\frac{2 \cdot 2^{7/8}}{\vartheta _2\left(0,\frac{1}{\sqrt{2}}\right)}$ for $a_0=2,b_0=1$?

Question

I tried a very simple iterated mean and got a very strange closed form for a particular value. The sequence in question goes like this:

$$a_{n+1}=\frac{2a_nb_n}{a_n+b_n}, \qquad b_{n+1}=\frac{a_{n+1}+b_n}{2}$$

Note that this looks like the Arithmetic-Harmonic mean (i.e. just the Geometric mean, or the method for computing square roots), however, it is different, since we substitute the current value of $a_n$ in $b_n$ instead of the previous value.

$$a_{n+1}=\frac{2a_nb_n}{a_n+b_n}, \qquad b_{n+1}=\frac{3a_nb_n+b_n^2}{2(a_n+b_n)}$$

Thus, this is similar in spirit to the Schwab-Borchardt mean, but this is a rational sequence. It's easy to show that it converges:

$$L(a_0,b_0)=\lim_{n \to \infty}a_n=\lim_{n \to \infty}b_n$$

What is truly suprising, the closed form I've got for the simple case:

$$L(2,1)=\frac{2 \cdot 2^{7/8}}{\vartheta _2\left(0,\frac{1}{\sqrt{2}}\right)}=\frac{2}{\sum_{k=0}^\infty 2^{-k(k+1)/2}}=$$

$$=1.2182994221324572310422292086114491025901998820372$$

I obtained the closed form in a roundabout way using Inverse Symbolic Calculator, OEIS and Mathematica.

This is one of the Jacobi elliptic functions. So far I haven't found any connection between the elliptic functions and this kind of iterated means. Except of course for the elliptic integrals and arithmetic-geometric mean, but that is a different case.

Neither was I able to find the general closed form for $L(x,y)$, or any other particular closed form.

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Some elementary considerations.

$$a_{n+1}-b_{n+1}=\frac{b_n}{a_n+b_n}\frac{a_n-b_n}{2}<a_n-b_n$$

$$\frac{b_{n+1}}{a_{n+1}}=\frac{3}{4}+\frac{1}{4}\frac{b_n}{a_n}=1+\frac{1}{4^{n+1}} \left(\frac{b_0}{a_0}-1 \right)$$

Thus, if $b_n/a_n=q_n$:

$$q_n=1+\frac{q_0-1}{4^n}$$

Furthermore, using the first definition of the sequence, we have:

$$b_n=2b_{n+1}-a_{n+1}=a_{n+1}(2q_{n+1}-1)=a_nq_n$$

$$a_{n+1}=\frac{q_n}{2q_{n+1}-1}a_n$$

$$\color{blue}{a_n=a_0 \prod_{k=0}^{n-1} \frac{1+\dfrac{q_0-1}{4^k}}{1+\dfrac{q_0-1}{2 \cdot 4^k}}}$$

Separating the numerator and denominator and taking the limit, we obtain well studied infinite products:

$$ \prod_{k=0}^\infty \left(1+\frac{q_0-1}{4^k} \right)$$

$$\prod_{k=0}^\infty \left(1+\frac{q_0-1}{2 \cdot 4^k} \right)$$

I can see the connection with Jacobi functions now. It's rather simple, but it seems that the closed form only exists for special cases, like the one above.

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Using q-Pochhammer symbols, we obtain:

$$L(x,y)=x \frac{(1-y/x;1/4)_{\infty}}{((1-y/x)/2;1/4)_{\infty}}$$

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Here is an interesting paper, relating Theta functions with iterative means: https://www.math.washington.edu/~morrow/papers/nate-thesis.pdf

Answer 1

$b_0/a_0=q_0$:

$$\color{blue}{L(a_0,b_0)=a_0 \prod_{k=0}^{\infty} \frac{1+\dfrac{q_0-1}{4^k}}{1+\dfrac{q_0-1}{2 \cdot 4^k}}}$$

$a_0=2$

$b_0=1$

$b_0/a_0=q_0=1/2$:

$$L(2,1)=2 \prod_{k=0}^{\infty} \frac{1-\dfrac{1}{2.4^k}}{1-\dfrac{1}{4 \cdot 4^k}}$$ $$L(2,1)=2 \prod_{k=0}^{\infty} \frac{1-\dfrac{1}{2^{2k+1}}}{1-\dfrac{1}{ 2^{2k+2}}}$$

$q_0=1/2$:

$$L(2,1)=2 \prod_{k=0}^{\infty} \frac{1-q_0^{2k+1}}{1-q_0^{2k+2}}$$

$$L(2,1)=2 \prod_{k=1}^{\infty} \frac{1-q_0^{2k-1}}{1-q_0^{2k}}=2 \prod_{k=1}^{\infty} \frac{1-q_0^{2k-1}}{(1-q_0^{k})(1+q_0^{k})}$$

$$L(2,1)=2 \prod_{k=1}^{\infty} \frac{1}{(1-q_0^{2k})(1+q_0^{k})}=2 \prod_{k=1}^{\infty} \frac{1}{(1-q_0^{k})(1+q_0^{k})(1+q_0^{k})} \tag 1 $$

The Jacobi triple product identity is: $$\prod\limits_{n=1}^{ \infty }(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1})=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} $$

$q=q_0^{1/2}$ $z=q_0^{1/2}$

$$\prod\limits_{n=1}^{ \infty }(1-q_0^{n})(1+q_0^{1/2}q_0^{(2n-1)/2})(1+q_0^{-1/2}q_0^{(2n-1)/2})=\sum\limits_{n = - \infty }^ \infty q_0^{n/2} q_0^{n^2/2} $$

$$\prod\limits_{n=1}^{ \infty }(1-q_0^{n})(1+q_0^{n})(1+q_0^{n-1})=\sum\limits_{n = - \infty }^ \infty q_0^{\frac{n(n+1)}{2}} $$

$$(1+q_0^{0})\prod\limits_{n=1}^{ \infty }(1-q_0^{n})(1+q_0^{n})(1+q_0^{n})=\sum\limits_{n = - \infty }^ \infty q_0^{\frac{n(n+1)}{2}} $$

$$2\prod\limits_{n=1}^{ \infty }(1-q_0^{n})(1+q_0^{n})(1+q_0^{n})=2\sum\limits_{n = 0 }^ \infty q_0^{\frac{n(n+1)}{2}} $$ We know that $q_0=1/2$: Thus,

$$\prod\limits_{n=1}^{ \infty }(1-q_0^{n})(1+q_0^{n})(1+q_0^{n})=\sum\limits_{n = 0 }^ \infty 2^{-\frac{n(n+1)}{2}} $$

If we put the result in Eq (1);

$$L(2,1)=2 \prod_{k=1}^{\infty} \frac{1}{(1-q_0^{k})(1+q_0^{k})(1+q_0^{k})} =\frac{2}{\sum_{k=0}^\infty 2^{-\frac{k(k+1)}{2}}} \tag 2 $$

$$L(2,1)=\frac{2 \cdot 2^{7/8}}{\vartheta _2\left(0,\frac{1}{\sqrt{2}}\right)}=\frac{2}{\sum_{k=0}^\infty 2^{-\frac{k(k+1)}{2}}}$$