Operator Inversion and Bauer's Series

The problem came from a post by Daniel Ángeles on X. The proposed series was

\sum_{n=0}^{\infty} \frac{4n+1}{(2n-1)^2(n+1)^2} \left(\frac{(2n-1)!!}{(2n)!!}\right)^3(-1)^n = \frac{8}{3\pi}.

The PDF linked above gives the full proof. Here I concentrate on the operator computation behind it.

I evaluated the series first and only later recognized an intermediate expression as Bauer’s series:

\mathcal B =\sum_{n=0}^{\infty}(4n+1) \left(\frac{(2n-1)!!}{(2n)!!}\right)^3(-1)^n =\frac{2}{\pi}.

The operator calculation below shows that the original sum is a rational multiple of $\mathcal B$ :

F(-1)=\frac{4}{3}\mathcal B.

The Reduction

Let $f_n$ denote the coefficient of $z^n$ in

F(z)=\sum_{n=0}^{\infty} \frac{4n+1}{(2n-1)^2(n+1)^2} \left(\frac{(2n-1)!!}{(2n)!!}\right)^3z^n.

Using $\frac{(2n-1)!!}{(2n)!!}=4^{-n}\binom{2n}{n}$ , its consecutive coefficients satisfy

\frac{f_{n+1}}{f_n} =\frac{(n+\frac54)(n-\frac12)^2(n+\frac12)} {(n+\frac14)(n+2)^2(n+1)}.

Since $f_0=1$ , this identifies $F$ as

F(z)={}_4F_3\left( \begin{matrix} \frac54,-\frac12,-\frac12,\frac12\\ \frac14,2,2 \end{matrix};z\right).

Set $\theta=z\frac{d}{dz}$ and

G(z)={}_3F_2\left( \begin{matrix} -\frac12,-\frac12,\frac12\\ 2,2 \end{matrix};z\right).

The quotient $(\frac54)_n/(\frac14)_n$ is $4n+1$ , so

F=(4\theta+1)G.

For the operator calculation, let

H(z)={}_3F_2\left( \begin{matrix} \frac12,\frac12,\frac12\\ 1,1 \end{matrix};z\right).

If $G(z)=\sum g_nz^n$ and $H(z)=\sum h_nz^n$ , then

\frac{h_n}{g_n} =\frac{(2)_n^2(\frac12)_n^2}{(1)_n^2(-\frac12)_n^2} =(n+1)^2(2n-1)^2.

Since $\theta z^n=nz^n$ , this coefficient identity may be written

H=BG, \qquad B=(2\theta-1)^2(\theta+1)^2.

The standard hypergeometric differential equation

\left[(\theta+1)^2\theta -z\left(\theta+\frac12\right) \left(\theta-\frac12\right)^2\right]G=0.

expands to $AG=0$ , where

A=(1-z)\theta^3+\frac{4+z}{2}\theta^2 +\frac{4+z}{4}\theta-\frac z8.

Inverting $B$ Modulo $A$

A Commutative Analogy

This example is not used in the proof. It isolates the inversion that will be needed below.

Consider the matrix

M=\begin{pmatrix}0&-1\\1&-1\end{pmatrix}, \qquad M^2+M+I=0.

Suppose $y$ is unknown and only

h=(M+I)y

is known. The task is to recover $y$ using the polynomial equation satisfied by $M$ .

Set

a(x)=x^2+x+1, \qquad b(x)=x+1.

Then $a(M)=0$ and $h=b(M)y$ . In $\mathbb Q[x]$ ,

a(x)-xb(x)=1,

so $b(x)^{-1}\equiv-x\pmod{a(x)}$ . Evaluating the identity at $M$ gives

a(M)-Mb(M)=I.

Since $a(M)=0$ and $b(M)y=h$ ,

y=-Mh.

All the operators in this example are polynomials in the same matrix $M$ , so they commute.

The Operator Computation

In the proof, $AG=0$ and $H=BG$ play the roles of $a(M)y=0$ and $h=b(M)y$ . The coefficients of $A$ and $B$ now depend on $z$ , and $\theta$ does not commute with them. I applied the extended Euclidean algorithm in the Ore ring

\mathcal R=\mathbb Q(z)\langle\theta\rangle, \qquad \theta f=f\theta+zf'(z),

writing coefficients to the left of the powers of $\theta$ . If $\delta f=zf'(z)$ , multiplication is determined by

(a\theta^m)(b\theta^n) =a\sum_{r=0}^{m}\binom{m}{r}\delta^r(b)\theta^{m+n-r}.

Right division is taken to mean

P=SQ+R, \qquad \deg_\theta R<\deg_\theta Q.

The divisor $Q$ occurs on the right of the quotient $S$ . With the arguments in the order $(B,A)$ , the extended algorithm computes operators $U,V$ and a rational function $g$ satisfying

VB+UA=g(z).

S.<z> = QQ['z']
R = S.fraction_field()
delta = (z * S.derivation()).extend_to_fraction_field()
Ore.<theta> = R['theta', delta]

A = (
    (1 - z)*theta^3
    + (4 + z)/2*theta^2
    + (4 + z)/4*theta
    - z/8
)
B = (2*theta - 1)^2*(theta + 1)^2

def right_xgcd(P, Q):
    r0, r1 = P, Q
    u0, u1 = Ore.one(), Ore.zero()
    v0, v1 = Ore.zero(), Ore.one()
    while r1:
        q, r = r0.right_quo_rem(r1)
        r0, r1 = r1, r
        u0, u1 = u1, u0 - q*u1
        v0, v1 = v1, v0 - q*v1
    return r0, u0, v0

def print_factored(name, op):
    print(f"{name} =")
    for i in reversed(range(op.degree() + 1)):
        c = op[i]
        if c:
            print(f"  theta^{i}: {factor(c)}")

g, V, U = right_xgcd(B, A)   # V*B + U*A = g

print("g =", factor(g[0]))
print_factored("V", V)
print_factored("U", U)

Factored output

g = (9/484) * (z - 1)^-4 * (z^2 + 13*z + 1)^2

V =
  theta^2: (-4/33) * (z - 1)^-2 * z^-1 * (z^2 + 13*z + 1)^2
  theta^1: (-38/363) * (z - 1)^-3 * z^-1 * (z - 3/19) * (z^2 + 13*z + 1)^2
  theta^0: (-8/363) * (z - 1)^-4 * (z - 13/8) * (z^2 + 13*z + 1)^2

U =
  theta^3: (-16/33) * (z - 1)^-3 * z^-1 * (z^2 + 13*z + 1)^2
  theta^2: (112/363) * (z - 1)^-4 * z^-1 * (z - 19/14) * (z^2 + 13*z + 1)^2
  theta^1: (20/363) * (z - 1)^-4 * z^-1 * (z^2 + 13*z + 1)^2
  theta^0: (2/121) * (z - 1)^-4 * z^-1 * (z^2 + 13*z + 1)^2

ok

The last nonzero remainder and the coefficient of $B$ have a common factor:

\begin{aligned} p&=z^2+13z+1,\\ g&=\frac{9p^2}{484(z-1)^4},\\ V&=-\frac{4p^2}{33z(z-1)^2}\theta^2 -\frac{2(19z-3)p^2}{363z(z-1)^3}\theta -\frac{(8z-13)p^2}{363(z-1)^4}. \end{aligned}

Since $BG=H$ and $AG=0$ , the Bézout identity acts on $G$ as

gG=(VB+UA)G=V(BG)+U(AG)=VH.

Equivalently, modulo the left ideal $\mathcal RA$ ,

(g^{-1}V)B=1-(g^{-1}U)A\equiv1\pmod{\mathcal RA}.

Set $C=g^{-1}V$ . Then

\begin{aligned} C={}&-\frac{176(z-1)^2}{27z}\theta^2 -\frac{8(z-1)(19z-3)}{27z}\theta +\frac{52-32z}{27},\\ G={}&CH. \end{aligned}

Since $F=(4\theta+1)G$ , we now have $F=(4\theta+1)CH$ . This operator has order three. On $H$ , it can be reduced modulo

D=\theta^3-z\left(\theta+\frac12\right)^3, \qquad DH=0.

Right division by $D$ , followed by replacing powers of $\theta$ with ordinary derivatives, gives

F(z)=16z(1-z^2)H''(z) +\frac{-88z^2-16z+56}{3}H'(z) -\frac{8z+4}{3}H(z).

The coefficient of $H''$ vanishes at $z=-1$ :

F(-1)=\frac43\bigl(H(-1)-4H'(-1)\bigr).

To identify the expression in parentheses, write

H(z)=\sum_{n=0}^{\infty}c_nz^n, \qquad c_n=\left(\frac{(2n-1)!!}{(2n)!!}\right)^3.

For $n\geq1$ , $nc_n$ decreases to zero, so the differentiated series converges uniformly on $[-1,0]$ by the alternating test. Termwise differentiation at $-1$ gives

H(-1)-4H'(-1) =\sum_{n=0}^{\infty}(4n+1) \left(\frac{(2n-1)!!}{(2n)!!}\right)^3(-1)^n =\mathcal B.

Hence $F(-1)=\frac43\mathcal B$ .

Evaluating Bauer’s Series

I will only summarize the rest of the evaluation. The differentiated transformations and the elliptic integral calculation are worked out in the PDF.

Clausen’s identity gives

H(z)=I(z)^2, \qquad I(z)={}_2F_1\left( \begin{matrix} \frac14,\frac14\\ 1 \end{matrix};z\right).

It follows that

\mathcal B=I(-1)\bigl(I(-1)-8I'(-1)\bigr).

Kummer’s quadratic transformation takes the form

I(4z(1-z))=T_K(z), \qquad T_K(z)={}_2F_1\left( \begin{matrix} \frac12,\frac12\\ 1 \end{matrix};z\right),

and the preimage of $-1$ needed here is

z_0=\frac{1-\sqrt2}{2}, \qquad 4z_0(1-z_0)=-1.

Differentiating Kummer’s identity at $z_0$ changes the expression for $\mathcal B$ to

\mathcal B =T_K(z_0)\bigl(T_K(z_0)-\sqrt2\,T_K'(z_0)\bigr).

Pfaff’s transformation sends $z_0$ to

z_1=\frac{z_0}{z_0-1}=3-2\sqrt2.

After differentiation and substitution at $z_0$ ,

\mathcal B =(6-4\sqrt2)T_K(z_1)^2 +(80-56\sqrt2)T_K(z_1)T_K'(z_1).

The derivative is eliminated with

\begin{aligned} T_E(z)&={}_2F_1\left( \begin{matrix}-\frac12,\frac12\\1\end{matrix};z\right),\\ T_K'(z)&=\frac{T_E(z)-(1-z)T_K(z)}{2z(1-z)}. \end{aligned}

Substitution at $z_1$ leaves

\mathcal B =-2T_K(z_1)^2+2\sqrt2\,T_K(z_1)T_E(z_1).

Set $k=\sqrt2-1$ , so that $k^2=z_1$ . The hypergeometric forms of the complete elliptic integrals are

K(k)=\frac\pi2T_K(k^2), \qquad E(k)=\frac\pi2T_E(k^2),

and the last expression becomes

\mathcal B =\frac{4}{\pi^2} \left(-2K(k)^2+2\sqrt2\,K(k)E(k)\right).

At $k=\sqrt2-1$ , the Landen transformation substituted into Legendre’s relation gives

2\sqrt2\,K(k)E(k)-2K(k)^2=\frac\pi2.

It follows that $\mathcal B=2/\pi$ and

F(-1)=\frac43\mathcal B=\frac{8}{3\pi}.

Conclusion

The extended Euclidean algorithm produces a left inverse for $B|_{\ker A}$ , turning $H=BG$ into $G=g^{-1}VH$ . I had not encountered this use of the algorithm before.

The inversion alone does not finish the reduction. The resulting formula for $F$ still contains $H''$ , with coefficient $16z(1-z^2)$ . This coefficient vanishes at $z=-1$ , leaving exactly $\frac43(H(-1)-4H'(-1))$ . Without that cancellation, the calculation would not reach Bauer’s series.