On $q$-analogues Arising from Elliptic Integrals and the Arithmetic-Geometric Mean

TL;DR

This paper develops $q$-analogues of classical elliptic integral identities and the arithmetic-geometric mean's functional equation, extending derivatives to complex orders and expressing results as infinite products.

Contribution

It introduces new $q$-analogues for elliptic integrals and the AGM, connecting them through infinite product representations and complex extension of derivatives.

Findings

Established $q$-analogues of elliptic integral identities.

Extended derivatives of elliptic integrals to complex values.

Expressed $q$-analogues as infinite products.

Abstract

We prove $q$-analogues of identities that are equivalent to the functional equation of the arithmetic-geometric mean. We also present $q$-analogues of $F(\sqrt{k},\frac{\pi}{2})$, the complete elliptical integral of the first kind, and its derivatives evaluated at $k=\frac{1}{2}$. These $q$-analogues interpolate those $n$th derivative evaluations by extending $n$ to a complex variable $s$, and we prove that they can be expressed as an infinite product.

Equations

$$a_{n+1}=\frac{a_n+b_n}{2}\mathrm{and}{b}_{n+1}=\sqrt{a_n{b}_n}.$$

$$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{b}_n=M(a,b).$$

$$M(ca,cb)=cM(a,b)$$

$$M(a,b)=M(\frac{a+b}{2},\sqrt{ab}).$$

$$M(1,k)=\frac{1+k}{2}M(1,\frac{2\sqrt{k}}{1+k}).$$

$$\frac{1}{M(1,k)}=\frac{2}{\pi }{\int }_0^1\frac{dt}{\sqrt{1-t^2}\sqrt{1-(1-k^2)t^2}}.$$

$$F(k,\frac{\pi }{2})={\int }_0^1\frac{dt}{\sqrt{1-t^2}\sqrt{1-k^{2}t}}$$

$$F(x)=\frac{2}{\pi }F(\sqrt{x},\frac{\pi }{2})=\frac{2}{\pi }{\int }_0^1\frac{dt}{\sqrt{1-t^2}\sqrt{1-x{t}^2}}=\sum _{n=0}^{\infty }{a}_n{x}^n$$

$$a_n=(\prod _{j=1}^n\frac{2j-1}{2j}{)}^2.$$

$$F(1-k^2)=\frac{2}{1+k}F((\frac{1-k}{1+k}{)}^2).$$

$$[n{]}_q=\sum _{j=0}^{n-1}{q}^j=\frac{1-q^n}{1-q},$$

$$n{!}_q=\prod _{j=1}^n[j{]}_q$$

$${\left(\genfrac{}{}{0ex}{}{n}{m}\right)}_q=\frac{n{!}_q}{(n-m){!}_{q}m{!}_q}.$$

$$(\alpha ){!}_q=\frac{1}{(1-q{)}^{\alpha }}\prod _{n=1}^{\infty }\frac{1-q^n}{1-q^{\alpha +n}}$$

$${\left(\genfrac{}{}{0ex}{}{\alpha }{\beta }\right)}_q=\frac{(\alpha ){!}_q}{(\alpha -\beta ){!}_q(\beta ){!}_q}$$

$$\underset{q\to 1^{-}}{\lim }{\Gamma }_q(\alpha )=\Gamma (\alpha )$$

$$F(x)=\sum _{n=0}^{\infty }{a}_n{x}^n.$$

$$F(1-k^2)=\frac{2}{1+k}F((\frac{1-k}{1+k}{)}^2).$$

$$k=1-u$$

$$\frac{2}{2-u}F((\frac{u}{2-u}{)}^2)=F(u(2-u))$$

$$\frac{2}{2-u}\sum _{n=0}^{\infty }{a}_n(\frac{u}{2-u}{)}^{2n}=\sum _{n=0}^{\infty }{a}_n(u(2-u){)}^n.$$

$$(1+x{)}^a=\sum _{n=0}^{\infty }(\genfrac{}{}{0ex}{}{a}{n})x^n$$

$$\left(\genfrac{}{}{0ex}{}{-2n-1}{m}\right)=\left(\genfrac{}{}{0ex}{}{2n+m}{m}\right)(-1{)}^m$$

$$\frac{2}{2-u}\sum _{n=0}^{\infty }{a}_n(\frac{u}{2-u}{)}^{2n}$$

$$=\sum _{n,m=0}^{\infty }\frac{a_n}{2^{2n+m}}{u}^{2n+m}\left(\genfrac{}{}{0ex}{}{2n+m}{2n}\right)$$

$$=\sum _{k=0}^{\infty }\frac{u^k}{2^k}\sum _{n=0}^{\infty }{a}_n\left(\genfrac{}{}{0ex}{}{k}{2n}\right)$$

$$\sum _{n=0}^{\infty }{a}_n(u(2-u){)}^n$$

$$=\sum _{n=0}^{\infty }\sum _{m=0}^n{a}_n{2}^n{u}^n(-1{)}^m(\frac{u}{2}{)}^m\left(\genfrac{}{}{0ex}{}{n}{m}\right)$$

$$=\sum _{k=0}^{\infty }\frac{u^k}{2^k}\sum _{n=0}^{\infty }{a}_n{2}^{2n}(-1{)}^{k-n}(\genfrac{}{}{0ex}{}{n}{k-n})$$

$$\sum _{n=0}^{\infty }{a}_n\left(\genfrac{}{}{0ex}{}{k}{2n}\right)=\sum _{n^{\prime }=0}^{\infty }{a}_{n^{\prime }}{2}^{2n^{\prime }}(-1{)}^{k-n^{\prime }}(\genfrac{}{}{0ex}{}{n^{\prime }}{k-n^{\prime }}).$$

$$(-1{)}^{k-n^{\prime }}(\genfrac{}{}{0ex}{}{n^{\prime }}{k-n^{\prime }})=\sum _{j=0}^{\infty }(-1{)}^j\left(\genfrac{}{}{0ex}{}{2n^{\prime }+j}{2n^{\prime }}\right)\left(\genfrac{}{}{0ex}{}{k}{n^{\prime }+j}\right).$$

$$\sum _{n=0}^{\infty }{a}_n\left(\genfrac{}{}{0ex}{}{k}{2n}\right)=\sum _{n^{\prime },j\ge 0}{a}_{n^{\prime }}{2}^{2n^{\prime }}(-1{)}^j(\genfrac{}{}{0ex}{}{2n^{\prime }+j}{2n^{\prime }}\left)\right(\genfrac{}{}{0ex}{}{k}{n^{\prime }+j}).$$

Full text

1 Introduction

We present $q$-analogues arising from two closely related objects: the arithmetic-geometric mean and the complete elliptic integral of the first kind. We review these objects now.

We recall the definition of the arithmetic-geometric mean $M(a,b)$ of two real numbers $a$ and $b$: Let $a_{0}=a$ and $b_{0}=b$ and define

[TABLE]

Then

[TABLE]

For information about the arithmetic-geometric mean see D. A. Cox [6]. The properties

[TABLE]

and

[TABLE]

allow us to think of $M(a,b)$ as a function of one variable $k$ that satisfies the functional equation

[TABLE]

C. F. Gauss [7] proved that

[TABLE]

The integral

[TABLE]

is known as the complete elliptic integral of the first kind and we let $F(x)$ denote

[TABLE]

where

[TABLE]

Therefore the functional equation in terms of $F(x)$ is

[TABLE]

In Section 2 show that this functional equation is equivalent to a set of identities involving the $a_{n}$, and in Section 3 we prove $q$-analogues of those identities.

References that discuss the above relationship are [1], [4], [6], [8], [9]. The proofs we have found in the literature are the three of C. F. Gauss using integral substitutions, differential equations, and another also based on the power series coefficients $a_{n}$. These are discussed in [6]. There is also another proof using integrals by B. C. Carlson [5].

We now discuss how $q$-analogues enter the above discussion. We call our results “$q$-analogues” because they involve standard expressions from $q$-theory: the $q$-positive integers

[TABLE]

the $q$-factorial

[TABLE]

and the $q$-binomial coefficients

[TABLE]

We think of $q$ as an indeterminate in a formal power series or as a real number between 0 and 1. When $q=1$, the above expressions evaluate to the usual integers, factorials, and binomial coefficients. We also use the following generalizations

[TABLE]

and

[TABLE]

that recover the previous formulas when $\alpha$ and $\beta$ are integers. For complex numbers $\alpha$, the $(\alpha)!_{q}$ is referred to as the $q$-Gamma function $\Gamma_{q}(\alpha)$ and satisfies

[TABLE]

where $\Gamma(\alpha)$ is the Gamma function (see [2] for a proof). We will use this fact in Section 4. A $q$-analogue of trigonometric functions also appears in Section 4.

In Section 4 we present formulas that are $q$-analogues of $\frac{d^{n}}{dk^{n}}F(\sqrt{k},\frac{\pi}{2})$ evaluated at $k=\frac{1}{2}$. We prove that these formulas are equal to an infinite product which may be expressed using $\Gamma_{q}$. These equations also naturally allow the variable $n$ to take on complex values.

The $q$-formulas presented in this paper, then, may be viewed as seeking to define a $q$-analogue of the arithmetic-geometric mean, or rather a function or functions that satisfy a similar functional equation.

Another motivation is that the Jacobi theta functions are also closely connected to the arithmetic-geometric mean and elliptic integrals (Section 5). Furthermore, the theta functions are related to the Riemann zeta function and other Dirichlet series via the Mellin transform. Information about the arithmetic-geometric mean and elliptic integrals could thus be useful for understanding those Dirichlet series.

2 Identities for the Functional Equation of the Arithemtic-Geometric Mean

Let

[TABLE]

Suppose $F(x)$ satisfies the functional equation (1)

[TABLE]

We show this determines the $a_{n}$ and also evaluate the $a_{n}$ by setting $q=1$ in Theorem 4.

Let

[TABLE]

and the functional equation (1) becomes

[TABLE]

and as a power series becomes

[TABLE]

To the left side we now apply the binomial theorem

[TABLE]

and the fact

[TABLE]

to obtain

[TABLE]

where we have set $k=2n+m$.

Now the right side of (2) becomes

[TABLE]

where we have set $k=n+m$. Therefore for each integer $k\geq 0$

[TABLE]

Now we apply the following result which we call Identity 1:

For integer $n^{\prime}$ and integer $k$

[TABLE]

We prove this result in Theorem 1 using $q$-binomial coefficients. We get

[TABLE]

Setting $n^{\prime}+j=2n$ gives

[TABLE]

Setting $n^{\prime}+j=2n+1$ gives

[TABLE]

We refer to (3) and (4) as Identity 2. We present $q$-analogues of Identities 1 and 2 and prove them in Section 3.

3 Proofs of $q$-analogues of Identities 1 and 2

3.1 $q$-analogue of Identity 1

Theorem 1**.**

Let $k$ and $n$ be integers $\geq 0$. Then

[TABLE]

The sum has only finitely many non-zero terms if $k$ and $n$ are integers.

We prove two generalizations of this result. As noted, the terms in the sum are zero if $j>k-n$. We thus let $k=n+i$ and re-index $j\mapsto i-j$ to get

[TABLE]

We now allow $n$ and $i$ to be possibly non-integers $s$ and $b$ satisfying certain conditions in Theorems 2 and 3.

Theorem 2**.**

Suppose $b$ is an integer. Then

[TABLE]

Proof.

The statement is equivalent to

[TABLE]

This statement is proved in Lemma 1 for $\alpha=0$. ∎

Lemma 1**.**

Let $b$ be an integer. For all $\alpha$ and $s$:

[TABLE]

Proof.

Let

[TABLE]

The lemma statement is then

[TABLE]

We use induction on $b$. The lemma is true for $b=0$. Assume it is true for some $b\geq 0$ Consider $f(b+1,\alpha)$. Use

[TABLE]

to express $f(b+1,\alpha)$ as

[TABLE]

The first sum (5) is equal to

[TABLE]

and the second sum (6) is equal to

[TABLE]

Using the induction hypothesis we get that

[TABLE]

is equal to

[TABLE]

∎

Theorem 3**.**

Suppose $s+b=M$ is an integer $\geq 0$. Then

[TABLE]

Proof.

The statement is equivalent to

[TABLE]

This statement is proved in Lemma 2 for $\alpha=0$. ∎

Lemma 2**.**

Let $M$ be an integer $\geq 0$. For all $\alpha$ and $s$:

[TABLE]

Proof.

Let

[TABLE]

The lemma statement is

[TABLE]

We use induction on $M$. It is true for $M=0$. Assume it is true for $M\geq 0$. Use

[TABLE]

to express $f(M+1,\alpha)$ as

[TABLE]

The first sum (7) is equal to

[TABLE]

and the second sum (8) is equal to

[TABLE]

Use the induction hypothesis to get that

[TABLE]

this is equal to

[TABLE]

∎

3.2 $q$-analogue of Identity 2

We now present a $q$-analogue of Identity 2, the equations (3) and (4):

Theorem 4**.**

[TABLE]

We will use the following functions in a variable $c$:

Definition 1**.**

[TABLE]

Lemma 3**.**

For any $x,y$ and $z$,

[TABLE]

Proof.

This is proved by straightforward calculation. ∎

The following immediate corollaries describe two ways we will apply Lemma 3.

Corollary 1**.**

[TABLE]

Corollary 2**.**

[TABLE]

Definition 2**.**

For integer $m\geq 0$, define the function $I(c,m,q)$ by

[TABLE]

Theorem 5**.**

For integer $l\geq 0$,

[TABLE]

and

[TABLE]

Proof.

We use induction. The statement is true for $l=0$. Assume it is true for $l=L\geq 0$. Then we consider

[TABLE]

Now

[TABLE]

and

[TABLE]

Combining these we get

[TABLE]

This allows us to express (3.2) as

[TABLE]

Now for any $0\leq h\leq L$, let $S(h)$ denote

[TABLE]

For $h\geq 1$, we claim

[TABLE]

This follows from taking the $h$-th and $(h+1)$-th term in $S(h)$ and first applying Corollary 1 for $f_{h+1}(c,q)$ and $p_{L+1}(c,q)$; and then Corollary 2 for $a=2L+1$ and $k=2h$:

[TABLE]

By the same reasoning we check

[TABLE]

Now use

[TABLE]

to get

[TABLE]

This completes the part of the theorem for $I(c,2L+1,q)$. To this we add

[TABLE]

which completes the part of the theorem for $I(c,2L+2,q)$. ∎

Corollary 3**.**

Theorem 4 is true for the case of odd $m$.

Proof.

Theorem 5 shows that $I(c,2l+1,q)$ has a factor of $P_{l+1}(c,q)$. Evaluating at $c=l+\frac{1}{2}$ yields $P_{l+1}(l+\frac{1}{2})=0$. ∎

We introduce the variable $a$:

Definition 3**.**

For integer $l\geq 0$, define the function $G(c,a,l,q)$

[TABLE]

With this function we can express Theorem 5 as

[TABLE]

and

[TABLE]

We now evaluate $G(c,a,l,q)$ in terms of the $P_{i}(c,q)$:

Theorem 6**.**

[TABLE]

To prove this we first express the $F_{i}(c,q)$ in terms of the $P_{i}(c,q)$:

Lemma 4**.**

For integer $h\geq 0$,

[TABLE]

Proof.

We use induction. The statement is true for $h=0$. Assume it is true for an $h\geq 0$. We then multiply both sides by $f_{h+1}(c,q)$. To each $f_{h+1}(c,q)P_{h-v}(c,q)$ on the right side we apply Corollary 1 obtain

[TABLE]

Then we collect terms to equate the coefficient of $P_{h+1-v}(c)$ for each $0\leq v\leq h+1$ with the coefficient in the lemma. For $v=0$, we get

[TABLE]

For $1\leq v\leq h+1$, we get

[TABLE]

The above equation is implied by the following equation

[TABLE]

The above equation reduces to the following which is an instance of Corollary 2:

[TABLE]

This completes the proof. ∎

Now we can prove Theorem 6:

Proof.

We use induction on $l$. The statement is true for $l=0$. Assume it is true for some $l\geq 0$. Then we must show

[TABLE]

We apply Lemma 4 to $F_{l+1}(c,q)$ and equate the coefficient of $P_{i}(c,q)$ to the that in the Theorem to obtain for $0\leq i\leq l$:

[TABLE]

The above equation is implied by the following equation:

[TABLE]

This reduces to

[TABLE]

which is an instance of Lemma 3.

When $l+1=i$ the equation between the coefficients is

[TABLE]

This is implied by the following equation: when $l+1=i$

[TABLE]

also reduces to (10) for $l+1=i$. This completes the proof. ∎

We use Theorem 6 to evaluate $G(c,a,l,q)$ at $c=l$:

Theorem 7**.**

[TABLE]

Proof.

Theorem 6 expresses $G(c,a,l,q)$ as a function of $a$ using the Lagrange interpolation form of a polynomial. That is, for $0\leq i\leq l$ and $a=-2i$, each term in the sum is 0 except for the $i$-th term. Therefore we can easily evaluate $G(c,-2i,l,q)$ as a factored expression. After multiplying both sides of this theorem statement by

[TABLE]

both sides are polynomials in $q^{a}$ of degree at most $l$. Therefore if they agree at $a=-2i$ for $0\leq i\leq l$, then they are equal as functions of $a$. We get

[TABLE]

This simplifies to

[TABLE]

And

[TABLE]

simplifies to the same thing. This completes the proof. ∎

Now we can prove Theorem 4 in the case when $m$ is even:

Proof.

Let $m=2l$. Combining Theorems 5, 6, and 7, we evaluate $c=l$ and $a=2l+1$ to get

[TABLE]

This simplifies to

[TABLE]

which completes the proof. ∎

We include a result when $q=1$:

Lemma 5**.**

For integer $m\geq 0$, the following functions of $c$ are all equal:

[TABLE]

where

[TABLE]

Proof.

We prove that $H(c,m)=H(c,m+1)$ by showing

[TABLE]

The sum in the lemma for fixed $m$ and $c$ is absolutely convergent, as the product

[TABLE]

is convergent as $i\rightarrow\infty$ and

[TABLE]

as $i\rightarrow\infty$ where $C_{x}$ is a constant that depends on $x$. We have

[TABLE]

This follows from

[TABLE]

which can proved by induction. We denote

[TABLE]

From Corollary 1, we have

[TABLE]

This implies

[TABLE]

Using

[TABLE]

we obtain for any $N>0$

[TABLE]

Now as $N\rightarrow\infty$, the expression (12) goes to the right side of (11); expression (13) goes to 0; and expression (14) goes to 0 because the sum is convergent and $\displaystyle\lim_{N\rightarrow\infty}W(N)=0$.

∎

We note that this Corollary is sufficient to prove Identity 2 for $q=1$.

Corollary 4**.**

[TABLE]

Proof.

By Theorem 5, we have

[TABLE]

where

[TABLE]

For each $i>0$, $A(i,l)$ is an increasing function in $l$ and approaches 1 as $l\rightarrow\infty$. By the absolute convergence mentioned in the lemma for $H(c,0)$, we have that

[TABLE]

And

[TABLE]

because in the sum in the lemma, the term for $i>0$

[TABLE]

is decreasing in magnitude to 0 for fixed $c$ as $m\rightarrow\infty$ and remains the constant 1 if $i=0$. By the absolute convergence of the sum the limit is therefore 1.

∎

3.3 Trying to Reconcile Identities 1 and 2

Recall that the functional equation for the arithmetic-geometric mean is equivalent to

[TABLE]

for each integer $k\geq 0$. We set

[TABLE]

and therefore write

[TABLE]

where $f_{1}(n,k)$ and $f_{2}(n,k)$ are functions on $\mathbb{N}_{0}^{2}$ we will try to determine. To the above equation we apply Identity 1: for integer $n^{\prime}$ and integer $k$

[TABLE]

We get

[TABLE]

Setting $n^{\prime}+j=2n$ gives

[TABLE]

Now with $m=2n$, Identity 2 is

[TABLE]

Therefore we have

[TABLE]

For $k=4$, we therefore have a system of nine equations that come from the nine possible values for $(n,n^{\prime})$ such that $0\leq n^{\prime}\leq 2n$ and ${k\choose 2n}\neq 0$:

[TABLE]

These nine equations are in the eight variables

[TABLE]

and we check that the system has no solution.

Setting $n^{\prime}+j=2n+1$ gives

[TABLE]

so

[TABLE]

Alternatively we can we can start from Identity 2 and set $2n=n^{\prime}+j$ and see what formula results that corresponds to Identity 1:

[TABLE]

The above formula is equal to $\displaystyle(-1)^{n^{\prime}-k}{n^{\prime}\choose k-n^{\prime}}$ at $q=1$, but for other $q$ it in general does not factor and is not equal to $(-1)^{n^{\prime}-k}{n^{\prime}\choose k-n^{\prime}}_{q}$ times some power of $q$. However, when $k=n^{\prime}+1$ we do get

[TABLE]

which actually follows from (16). That is, what (19) is missing to make it coincide with (16) is a factor of $q^{\frac{j(j-1)}{2}}$. Therefore perhaps (19) can be written as a sum of $q$-binomials, for example, to give another $q$-analogue of Identity 1.

If we start from Identity 1 again and set $f_{2}(n,k)$ to be all 0, we get the sum for Identity 2 to be

[TABLE]

which does not completely factor either.

If we try to bypass Identity 1 and compare the coefficients of $u^{k}$ directly, we get the identity: for each $k\geq 0$

[TABLE]

We attempt a $q$-analogue of the above equation for $k=3$ with

[TABLE]

where we have let $2^{2n}$ become $\displaystyle\prod_{j=1}^{n}(1+q^{j})^{2}$. It can be shown that this equation as a function of $q$ is not true for any real values of $a,b,$ and $c$. The same holds if we try to let $2^{2n}$ become $\displaystyle(1+q)^{2n}$ or just $2^{2n}$.

4 $q$-analogues and the Complete Elliptic Integral of the First Kind

Recall

[TABLE]

Therefore

[TABLE]

We present two $q$-analogues of the above formulas. In Section 4.1 titled “$q$-analogue of the Sum”, we give a $q$-analogue of (21), which is actually phrased as a $q$ analogue

[TABLE]

In Section 4.2, titled “$q$-analogue of the Integral”, we give a $q$-analogue of (20). Despite the title of Section 4.2, we are actually giving a $q$-analogue of another sum that is obtained from that integral.

4.1 $q$-analogue of the Sum

We define a $q$-analogue of the function $1+\sin(\pi s)$ which we will use in Theorem 9.

Definition 4**.**

[TABLE]

Theorem 8**.**

The function $(1+\mathrm{SinPi})(s,q^{2})$ is 2-periodic in $s$ and

[TABLE]

Proof.

The $2$-periodicity follows from

[TABLE]

Now we prove the limits $q\rightarrow 1^{-}$. We express

[TABLE]

where

[TABLE]

As $q\rightarrow 1^{-}$,

[TABLE]

which is equal to

[TABLE]

where we have used

[TABLE]

And we have

[TABLE]

by Lemma 6.

Finally we have

[TABLE]

from standard trigonometric identities. ∎

Next we prove the limit in the previous lemma. It is a $q$-analogue of a product similar to the Wallis product for $\pi$.

Lemma 6**.**

[TABLE]

Proof.

We have

[TABLE]

for all $x$. Setting $x=\frac{1}{4}$ and then taking the reciprocal gives

[TABLE]

Let $L$ denote the limit

[TABLE]

We claim that $L=\frac{\pi}{\sqrt{8}}$. We claim that for each integer $n\geq 1$

[TABLE]

is an increasing function of $q$ for $q\in(0,1)$. That $\frac{\partial}{\partial q}f(q,n)\geq 0$ for $q\in(0,1)$ is equivalent to

[TABLE]

for $q\in(0,1)$. The above expression is equal to

[TABLE]

for $q\in(0,1)$, where we have used

[TABLE]

Since

[TABLE]

we can bound the limit $L$ between

[TABLE]

for any $N$. This completes the proof. ∎

Now we can prove the $q$-analogue of (21).

Theorem 9**.**

For any $s\in\mathbb{C}$ and $q\in(0,1]$,

[TABLE]

where

[TABLE]

and $C_{2}(1)=\frac{1}{\sqrt{2\pi}}$.

Proof.

We first prove the theorem for $q\in(0,1)$. From the definition of the $q$-binomial coefficient for non-integer $s$, we use

[TABLE]

to see that the theorem is equivalent to

[TABLE]

To this equation we multiply both sides by $\displaystyle\prod_{j=1}^{\infty}(1-q^{4j})$ and set $x=q^{-2s}$ to get

[TABLE]

Let

[TABLE]

Then

[TABLE]

We prove that

[TABLE]

Let

[TABLE]

so

[TABLE]

We calculate that

[TABLE]

We claim that

[TABLE]

We prove this claim by induction on $N$. It is true for $N=0$. Assume it is true for some $N\geq 0$. Then the induction step is implied by the identity

[TABLE]

Therefore

[TABLE]

and

[TABLE]

Iterating gives

[TABLE]

and

[TABLE]

This proves the theorem for $q\in(0,1)$.

To prove it for $q=1$, note that if $s$ is a non-negative integer, all sums and products become finite, so we may take the limit $q\rightarrow 1^{-}$ and we are done. If $s$ is a negative integer, then each term in the sum is 0 and the right hand side is also 0.

If $s$ is not an integer, we follow the same procedure for $q<1$, but, instead of multiplying by $\displaystyle\prod_{j=1}^{\infty}(1-q^{4j})$ at (26), we divide by $\displaystyle\prod_{j=1}^{\infty}(1-q^{2j}x)$. With $x=q^{-2s}$, we let $q=1$ and set

[TABLE]

Lemma 7 proves the convergence of the sum $f(s)$.

We prove that

[TABLE]

By the same reasoning for $q<1$, we have

[TABLE]

The right side of the above equation goes to 0 as $N\rightarrow\infty$ by the same reasoning we give for the bounds of $f_{n}(s)$ in Lemma 7. Therefore

[TABLE]

proving

[TABLE]

Iterating we have

[TABLE]

and

[TABLE]

because for $N>\mathrm{Re}(s)$, we have

[TABLE]

from the proof of Lemma 7. This proves the theorem for $q=1$.

Now

[TABLE]

and

[TABLE]

Therefore we can express the right side as

[TABLE]

where

[TABLE]

Now

[TABLE]

because $C_{2}(q)^{2}$ is a $q$-analogue for the Wallis product of $\pi$; the limit follows from similar reasoning in Lemma 6 by taking $x=\frac{1}{2}$ in the product for $\sin(\pi x)$. ∎

Lemma 7**.**

For $s$ not a positive integer, the sum $f(s)$ is convergent, where

[TABLE]

Proof.

The sum on the right is convergent because if $\mathrm{Re}(s)>0$ then we may bound $|f_{n}(s)|$ by

[TABLE]

where $p_{s}(n)$ is a polynomial in $n$ whose coefficients and degree depend on $s$. To see this, we have for $n>\mathrm{Re}(s)$

[TABLE]

Now the product at (27) is convergent as $n\rightarrow\infty$; the product at (28) is bounded by a polynomial in $n$ depending on $s$, and (29) is

[TABLE]

which is bounded by 1.

If $\mathrm{Re}(s)<0$, then by the above reasoning we may bound $f_{n}(s)$ for $n\geq 1$ by

[TABLE]

where $K$ is a constant independent of $s$. ∎

We include this lemma which be useful elsewhere.

Lemma 8**.**

Let $t,s,q\in\mathbb{R}$ such that $t\geq s$. Let

[TABLE]

Then for fixed $t$ and $s$, $f(t,s,q)$ is an increasing function of $q$ on $(0,1)$.

Proof.

Taking $\frac{\partial}{\partial q}f(s,t,q)$, we see that the lemma is equivalent to

[TABLE]

for $q\in(0,1)$. This is equivalent to

[TABLE]

being a decreasing function of $t$ for $t\in\mathbb{R}$ and for fixed $q$. To prove that $g(t,q)$ is a decreasing function, choose $\alpha>0$ and write by the binomial expansion

[TABLE]

Therefore

[TABLE]

For $t\in(-\infty,\alpha)$, each term in the above sum is positive and, for $n\geq 2$, decreases in magnitude as $t$ increases to $\alpha$, while the $n=1$ term remains constant. Therefore $g(t,q)$ is decreasing on $(-\infty,\alpha)$ for any $\alpha>0$. ∎

4.2 $q$-analogue of the Integral

Now we prove a $q$-analogue of (20). We first show how to obtain a sum from the integral.

Lemma 9**.**

[TABLE]

Proof.

We make the change of variable $t\mapsto\sqrt{t}$ and express $\displaystyle\frac{1}{\sqrt{2-t}}(\frac{t}{2-t})^{s}$ as a binomial series in $1-t$ to obtain

[TABLE]

To this we apply

[TABLE]

and

[TABLE]

to obtain

[TABLE]

which is equal to

[TABLE]

∎

We present a $q$-analogue of the above sum and its evaluation as a product in the following theorem:

Theorem 10**.**

Let

[TABLE]

Then for $s\in\mathbb{C}$ not a negative integer and $q\in(0,1]$,

[TABLE]

where

[TABLE]

and $C_{3}(1)=\sqrt{\pi}$ and $f(0,1)=\frac{1}{\sqrt{2}}$. That is,

[TABLE]

Proof.

We first prove theorem for $q\in(0,1)$. We multiply the left side of the theorem by

[TABLE]

and set $x=q^{2s}$ to obtain

[TABLE]

where

[TABLE]

We prove

[TABLE]

We claim

[TABLE]

where

[TABLE]

We prove (31) by induction on $N$. It is true for $N=0$ and $1$. Assume it is true for $N\geq 1$. Then

[TABLE]

Now

[TABLE]

To this we apply the identity

[TABLE]

Therefore

[TABLE]

This completes the induction step.

Because of the $q^{N}$ in $R_{N}(x,q)$, we have

[TABLE]

so

[TABLE]

and thus

[TABLE]

Iterating we obtain

[TABLE]

Now we divide both sides by

[TABLE]

which completes the proof for $q\in(0,1)$.

For $q=1$, we follow the same procedure for $q<1$, but do not divide by

$\displaystyle\prod_{j=1}^{\infty}(1-q^{2j})(1-q^{2s+2j})$ at (30). We let

[TABLE]

In Lemma 10 we prove that the sum on the right is convergent. We now prove

[TABLE]

From the above reasoning for $q<1$ we have

[TABLE]

The limit of the above sum as $N\rightarrow\infty$ is 0 because the product on the right converges as $N\rightarrow\infty$ for any $s$ not a negative integer. This proves (32). Iterating we have

[TABLE]

where $\displaystyle\lim_{N\rightarrow\infty}f(s+N)$ is determined in Lemma 10.

The expression of the product using $q$-factorials follows from their definition. $C_{3}(1)=\sqrt{\pi}$ follows from the same reasoning used for the limit of $C_{2}(q)$. And we can determine

[TABLE]

by comparison with the evaluation found in Theorem 4.1. ∎

Lemma 10**.**

The sum

[TABLE]

is convergent and

[TABLE]

Proof.

First we prove that

[TABLE]

is convergent for $s$ not a negative integer. We group the $2n$-th and $(2n+1)$-th terms together to express the sum as

[TABLE]

We compare (33) to the sum when $s=0$:

[TABLE]

Using Stirling’s approximation

[TABLE]

we have

[TABLE]

Therefore (34) is convergent. If $s<0$ and $s\in\mathbb{Z}+\frac{1}{2}$, then the sum (33) is finite. For other $s$, we apply the limit comparison test to to sums (33) and (34) to get

[TABLE]

This infinite product is convergent to a non-zero number because the sum

[TABLE]

is convergent. Therefore (33) is convergent for any $s\in\mathbb{C}$ not a negative integer.

We claim

[TABLE]

First we have that if $a,b\in\mathbb{R},j\in\mathbb{Z}$ with $a$ and $j>0$, then

[TABLE]

Therefore in (35), using $a+bi=2s+2N$, we assume that $N$ is so large that $\mathrm{Re}(N+s)>0$. Next, the sum on the right of (35) is convergent using Stirling’s approximation again, so for any $\epsilon>0$ we can choose $n_{1}$ such that

[TABLE]

for all $m>n_{1}$ and also such that

[TABLE]

for all $n>n_{1}$ with $\mathrm{Re}(N+s)>0$. Thus we have

[TABLE]

This proves the claim (35). As mentioned in Theorem 4.2 the sum $f(0)=\frac{1}{\sqrt{2}}$ by comparison with Theorem 4.1. This completes the proof. ∎

5 Further Work

• •

See if there are $q$-analogues of other proofs of the arithmetic-geometric mean functional equation.

• •

See if $q$-analogues can be found for the arithmetic-geometric mean applied to complex numbers.

• •

Find $q$-analogues for generalizations of the geometric-mean such as the cubic counterpart in [3].

• •

Try to reconcile Identities 1 and 2 to construct a $q$-analogue of the functional equation itself, possibly using more than one function.

• •

Use $q$-analogues of $F(x)$ to determine $q$-analogues of $k(x)^{2}$ and thus $\theta_{3}(e^{\pi x})^{2}$.

For this point, $k(x)^{2}$ is the function

[TABLE]

Now $k(x)^{2}$ is also determined by the properties

[TABLE]

and

[TABLE]

That is, those two properties imply

[TABLE]

where $\overline{\kappa}(n)$ is the sequence $\{1,6,104,3024,130176,7831296,...\}$. Therefore a $q$-analogue of $F(x)$ can by used to define a $q$-analogue of $k(x)^{2}$ via (36) and (37). Then a $q$-analogue of $k(x)^{2}$ can be used to define a $q$-analogue of $\theta_{3}(e^{\pi x})^{2}$ by

[TABLE]

We note that $\theta_{3}(q)$ can itself be viewed as arising from a $q$-analogue of $\sin(\pi x)$, so above we are talking about a $q$-analogue of a function that is a specialization (at $q=e^{-\pi}$) of a $q$-analogue of another function ($\sin(x)$).

We also note that the Mellin transform of $\theta_{3}(e^{\pi x})^{2}-1$ is a $\Gamma$ function factor times

[TABLE]

where

[TABLE]

Therefore considering $\theta_{3}(e^{\pi x})^{2}$ directly may be easier than considering $\theta_{3}(e^{\pi x})$ and would contain information about $\zeta(s)$ and its zeros. Studying the coefficients of $k(x)^{2}$ or its $q$-analogues could yield information of the generalized Turán inequalities for (39) or an expression of the coefficients as elementary-symmetric polynomials.

• •

The Mellin transform (39) follows from a Lambert series for $\theta_{3}(e^{\pi x})^{2}$. Find a combinatorial proof of this identity.

• •

Equation (38) is actually combinatorial identity. Find an explicit combinatorial proof of this identity and see if it has a $q$-analogue.

• •

See if $q$-analogues and infinite product evaluations exist for elliptic integrals of the second kind.

• •

The coefficients $a_{n}$ are

[TABLE]

where we may interpret $\displaystyle{2n\choose n}^{2}$ as the number of lattice paths on a square grid that start at one corner and go to the opposite corner and then return. Find out how Identity 2 translates into operations on these lattice paths.