7. _{3}F_{2}[1] Rosetta stone

N_{g,h,i,j,k,n} := ( \begin{matrix}\frac{n + h}{n + 1} \frac{n + i}{n + g} \frac{n + j}{n + k} & 0 & n \\ 0 & \frac{n + h}{n + 1} \frac{n + i}{n + g} \frac{n + j}{n + k} & 1 \\ 0 & 0 & 1 \end{matrix} )

K_{g,h,i,j,k,n} :=

( \begin{matrix}\frac{(k − h)(k − i)(k − j) + hij}{k(k + g − h − i − j − 1)(k + n)} & \frac{hij}{(k + g − h − i − j − 1)(k + n)} & \frac{n(n + g − 1)}{k + g − h − i − j − 1} \\ \frac{1}{k + n} & \frac{k}{k + n} & 0 \\ 0 & 0 & 1 \end{matrix} )

J_{g,h,i,j,k,n} :=

( \begin{matrix}\frac{(j − k − g + h + i + 2)(j + n)}{(j − g + 1)(j − k + 1)} & \frac{hi(j + n)}{(j − g + 1)(j − k + 1)} & n\frac{n + g − 1}{j − g + 1} \frac{n + k − 1}{j − k + 1} \\ −\frac{(j − k − g + h + i + 2)(j + n)}{j(j − g + 1)(j − k + 1)} & \frac{j + n}{j}(1 − \frac{h}{j − g + 1} \frac{i}{j − k + 1}) & −\frac{n}{j} \frac{n + g − 1}{j − g + 1} \frac{n + k − 1}{j − k + 1} \\ 0 & 0 & 1 \end{matrix} )

This six-dimensional path invariant grid codifies all of the _{3}F_{2}[1] contiguity relations (and, as a limiting case, all the _{2}F_{1}[z] and _{1}F_{1}[z]). The three undisplayed matrices follow from symmetry: G_{g,h,i,j,k,n} = K_{k,h,i,j,g,n}, etc. The _{3}F_{2}[1] arises in element_{2,3} of ∏_{n≥0} N_{...,n}. Any contiguous sum can be reached in a finite number of multiplications by the parameter gunching matrices. The contiguity identity follows by closing the contour at n = ∞.

Contours that range indefinitely in indices other than n yield identities involving the general _{3}F_{2}[1] which are unsurprisingly resistant to conventional notation, due to the nontriangularity of the non-N matrices. But even if we never become facile with them, at least some such identities should be valuable for their rapid numerical convergence.

Perhaps more interestingly, by conceding linear constraints among the parameter indices, we can triangularize some of their matrices, and get all of the familiar _{3}F_{2}[1] identities (Dixon’s, Saalschütz’s, Whipple’s, Watson’s, etc.). When you change coordinates, Whipple’s becomes Watson’s, and Saalschütz’s becomes the very well poised _{5}F_{4}.

The imposition of a linear constraint entails the replacement of several indices by a new one. This corresponds to recoordinatization and then dimension reduction in the grid. An example is given in section 9 (“An application”).

As often happens when a system like the above is symmetric in several indices, one can find a corresponding system with those symmetric indices collapsed into one, and running at fractional speed, e.g. with (g, k) replaced by (\frac{k}{2} , \frac{k+1}{2}), or (h, i, j) replaced by (\frac{j}{3}, \frac{j+1}{3}, \frac{j+2}{3}). Another way to put it is that, in the (g, k) case, the matrix G_{k,h,i,j,k,n}K_{k+1,h,i,j,k,n} can be factored into the form M(2k)M(2k + 1). Such parameter specializations often simplify the quest for triangularizations, and lead to interesting identities upon subsequent coordinate changes.

Letting parameters, say h and g, blow up in a fixed relative ratio z yields a four-dimensional system for _{2}F_{1}[z], which is easier to strip mine for triangularizations, particularly when z is negotiable. Further specializing via the abovementioned “fractional parameter collapse” led me to

\begin{aligned}_{2}F_{1} [ \begin{matrix}−\frac{a}{2}, & \frac{1−a}{2} \\ & 2a + \frac{3}{2} \end{matrix} | \frac{1}{5}] & = \sqrt{π \frac{5 + \sqrt{5}}{5}} \frac{2^{5a+ \frac{3}{2}}}{5^{3a+1}} \frac{(2a + \frac{1}{2})!}{(a − \frac{1}{5})!(a + \frac{1}{5})!} \\ _{2}F_{1} [ \begin{matrix}−\frac{a}{2}, & \frac{1−a}{2} \\ & 2a + \frac{5}{2} \end{matrix} | \frac{1}{5}] & = \sqrt{π \frac{5 − \sqrt{5}}{5}} \frac{2^{5a+ \frac{7}{2}}}{5^{3a+2}} \frac{(2a + \frac{3}{2})!}{(a + \frac{2}{5})!(a + \frac{3}{5})!}. \end{aligned}

while strip mining the remaining degrees of freedom. (Professor P. Karlsson has lately informed me that these were dug up by W. Heymann in 1898 ([2],[3]), i.e., with his bare hands!)

I also have q-versions of the _{3}F_{2}[1] system. These have the annoying property of sometimes leaving factors of 1 − q instead of 0 when subjected to the transformations which triangularize their q = 1 cousins, which is one way to explain why we yet lack q-generalizations for some of our more exotic _{2}F_{1} identities.

8. Continued fractions

This is three N-K planes, j = −1, 0, 1, joined by two J matrices. In the top (j = 1) plane, the k direction computes a continued fraction, and on the bottom (j = −1), it’s the n direction.

N_{1,k,n} = ( \begin{matrix}(1 − n − k)z & a((k − 1)^{2} + b)z^{2} \\ 1 & a(n − k + 1)z \end{matrix} ) ,

K_{1,k,n} = ( \begin{matrix}((a − 1)k − (a + 1)n + 1)z & a(k^{2} + b)z^{2} \\ 1 & 0 \end{matrix} )

J_{0,k,n} := ( \begin{matrix}z^{n+k} & ((a + 1)n + k − 1)z^{n+k+1} \\ 0 & z^{n+k+1} \end{matrix} )

N_{0,k,n} := ( \begin{matrix}an & a(b − (a + 1)kn) \\ 1 & −n − (a + 1)k \end{matrix} ) , K_{0,k,n} := ( \begin{matrix}ak & a(b − (a + 1)nk) \\ 1 & −k − (a + 1)n \end{matrix} )

J_{−1,k,n} := ( \begin{matrix}z^{−n−k} & (1 − (a + 1)k − n)z^{−n−k} \\ 0 & z^{−n−k−1} \end{matrix} )

N_{−1,k,n} = ( \begin{matrix}((a − 1)n − (a + 1)k + 1)z & a(n^{2} + b)z^{2} \\ 1 & 0 \end{matrix} ) ,

K_{−1,k,n} = ( \begin{matrix}(1 − k − n)z & a((n − 1)^{2} + b)z^{2} \\ 1 & a(k − n + 1)z \end{matrix} )

Thus, to get a relation between two continued fractions, we must j-hop between planes before switching directions. If k runs from c to ∞ and n runs from d to ∞, and the corresponding continued fractions converge, then they are insensitive to whatever subsequent matrices accrue on the right by way of path closure. I.e., the two paths can be (j, k, n) = (0, c, d) → (1, c, d) → (1, ∞, d) and (0, c, d) → (−1, c, d) → (−1, c, ∞). This yields the identity

f(c, d) = f(d, c)

where

f(x, y) := ((a + 1)y + x − 1) z

+ ((a − 1)x − (a + 1)y + 1) z + \frac{a(x^{2} + b)z^{2}}{((a − 1)(x + 1) − (a + 1)y + 1) z + \frac{a ((x + 1)^{2} + b) z^{2}}{...}} .

Note that z can be simply canceled out of this identity, but it is handy to leave it in. Since we never travel in the j = 0 plane, you might consider discarding it and coalescing the two J matrices. My only objections to this are that it would destroy some symmetry, and more importantly, it would conceal the simple derivation, which just uses the technique in section 4 to force alternately the K and N matrices into continued fraction form.

Another application of this particular path invariant system is a direct interderivation of the series and continued fraction forms for arctan, in the form

1 + \frac{x^{2}}{3 + \frac{4x^{2}}{5 + \frac{9x^{2}}{...}}} = \frac{x}{arctan x} = \frac{\frac{2y}{1 − y^{2}}}{2 arctan y}

= \frac{1}{(1 − y^{2}) (1 − \frac{y^{2}}{3} + \frac{y^{4}}{5} − ⋯)},

where

x := \frac{2y}{1 − y^{2}} .

We merely remain in the j = −1 plane and get our series by choosing the parameters to annihilate the upper right element of K:

a := y,  b := 0,  z := −\frac{2}{1 − y^{2}} .

This gives

F_{n} := N_{−1, \frac{1}{2} ,n} = ( \begin{matrix}2n − 1 & n^{2} (\frac{2y}{1 − y^{2}} )^{2} \\ 1 & 0 \end{matrix} )

and

G_{k} := K_{−1,k,1} = ( \begin{matrix}\frac{2k}{1 − y^{2}} & 0 \\ 1 & −\frac{2ky^{2}}{1 − y^{2}} \end{matrix} ) .

Then the ratio of the lefthand elements of F_{1}F_{2}F_{3} … gives the continued fraction, while the corresponding ratio from G_{\frac{1}{2}} G_{\frac{3}{2}} G_{\frac{5}{2}} … gives the reciprocated series.

9. An application

\begin{aligned}J_{j,k,n} & := ( \begin{matrix}\frac{1 − q^{j−k+n+ \frac{1}{2}}}{1 − q^{2j+2}} \frac{1 − q^{j+k−n+ \frac{1}{2}}}{1 − q^{j−k−n+ \frac{1}{2}}} & \frac{1 − q^{2k}}{1 − q} \frac{1 − q^{2n}}{1 − q^{j−k−n+ \frac{1}{2}}} q^{j−k−n+ \frac{1}{2}} \\ 0 & 1 \end{matrix} ) \\ K_{j,k,n} & := ( \begin{matrix}\frac{1 − q^{k−n+j+ \frac{1}{2}}}{1 − q^{2k+2}} \frac{1 − q^{k+n−j+ \frac{1}{2}}}{1 − q^{k−n−j+ \frac{1}{2}}} & \frac{1 − q^{2n}}{1 − q} \frac{1 − q^{2j}}{1 − q^{k−n−j+ \frac{1}{2}}} q^{k−n−j+ \frac{1}{2}} \\ 0 & 1 \end{matrix} ) \\ N_{j,k,n} & := ( \begin{matrix}\frac{1 − q^{n−j+k+ \frac{1}{2}}}{1 − q^{2n+2}} \frac{1 − q^{n+j−k+ \frac{1}{2}}}{1 − q^{n−j−k+ \frac{1}{2}}} & \frac{1 − q^{2j}}{1 − q} \frac{1 − q^{2k}}{1 − q^{n−j−k+ \frac{1}{2}}} q^{n−j−k+ \frac{1}{2}} \\ 0 & 1 \end{matrix} ) \end{aligned}

This pretty, three-dimensional system, when path multiplied around an infinite rectangle based at j = a, k = b, n = 0 in either the k-n or n-j plane, gives Andrews’s q-gneralization of Bailey’s _{2}F_{1}[\frac{1}{2}] theorem ([4]). (And, by symmetry, there are four other ways to do it.)

By making the coordinate transformation n ← n + j + k + \frac{1}{2}, you get Andrews’s qgneralization of Gauss’s _{2}F_{1}[\frac{1}{2}] (also [4]). To preserve path invariance, this coordinate change requires that the J_{j,k,n} and K_{j,k,n} matrices be replaced by J_{j,k,n}N_{j+1,k,n} (or N_{j,k,n}J_{j,k,n+1}) and K_{j,k,n}N_{j,k+1,n} (or N_{j,k,n}K_{j,k,n+1}) respectively, followed by the actual substitution n ← n + j + k + \frac{1}{2}. The multiplication of J and K by N reflects that every incrementation of j or k must also increment what was formerly n.

A more interesting coordinate change, in greater detail, is

\begin{cases}J_{j,k,n} \\ K_{j,k,n} \\ N_{j,k,n} \end{cases}\} ← \begin{cases}J_{j,k+j,n+k+j} K_{j+1,k+j,n+k+j} N_{j+1,k+j+1,n+k+j} \\ K_{j,k+j,n+k+j} N_{j,k+j+1,n+k+j} \\ N_{j,k+j,n+k+j} \end{cases}\}

which gives

\begin{aligned}J_{j,k,n} & = ( \begin{matrix}−\frac{1 − q^{j−n+ \frac{1}{2}}}{1 − q^{2(j+1)}} \frac{1 − q^{j+n+ \frac{1}{2}}}{1 − q^{2(j+k+1)}} \frac{1 − q^{j+2k+n+ \frac{1}{2}}}{1 − q^{2(j+k+n+1)}} q^{3j+2k+n+ \frac{3}{2}} & −\frac{1 − q^{3j+2k+n+ \frac{1}{2}}}{1 − q} \\ 0 & 1 \end{matrix} ) \\ K_{j,k,n} & = ( \begin{matrix}\frac{1 − q^{2k+j+n+ \frac{1}{2}}}{1 − q^{2(k+j+1)}} \frac{1 − q^{2k+j+n+ \frac{3}{2}}}{1 − q^{2(k+j+n+1)}} q^{2j} & −\frac{1 − q^{2j}}{1 − q} \\ 0 & 1 \end{matrix} ) \\ N_{j,k,n} & = ( \begin{matrix}\frac{1 − q^{n+j+ \frac{1}{2}}}{1 − q^{n−j+ \frac{1}{2}}} \frac{1 − q^{n+j+2k+ \frac{1}{2}}}{1 − q^{2(n+j+k+1)}} & \frac{1 − q^{2j}}{1 − q} \frac{1 − q^{2(j+k)}}{1 − q^{n−j+ \frac{1}{2}}} q^{n−j+ \frac{1}{2}} \\ 0 & 1 \end{matrix} ) . \end{aligned}

Then a rectangle in the j-k plane, based at j = 0, k = (b − a)/2, n = a − \frac{1}{2}, gives

∏_{j≥0} ( \begin{matrix}−\frac{1 − q^{j+a}}{1 − q^{2j+2}} \frac{1 − q^{j+1−a}}{1 − q^{2j+b−a+2}} \frac{1 − q^{j+b}}{1 − q^{2j+b+a+1}} q^{3j+b+1} & \frac{1 − q^{3j+b}}{1 − q} \\ 0 & 1 \end{matrix} )_{1,2}

= \frac{\frac{b−a}{2}!_{q^{2}}}{(−\frac{1}{2})!_{q^{2}}} \frac{\frac{b+a−1}{2}!_{q^{2}}}{(b − 1)!_{q}} (1 + q)^{b},

where

z!_{q} := (1 − q)^{−z} ∏_{n≥1} \frac{1 − q^{n}}{1 − q^{n+z}} “ = ” \frac{1 − q}{1 − q} \frac{1 − q^{2}}{1 − q} … \frac{1 − q^{z}}{1 − q} .

Alternatively, writing q^{a} =: A, q^{b} =: B, and multiplying through by 1 − q,

∑_{j≥0}(1 − Bq^{3j})\frac{(A, q/A, B; q)_{j}}{(q^{2}, ABq, Bq^{2}/A; q^{2})_{j}} (−Bq^{(3j−1)/2})^{j} = \frac{(B; q)_{∞}}{(ABq, Bq^{2}/A; q^{2})_{∞}} .

Letting q → 1 in the penultimate equation, and dividing through by b,

4F_{3} [ \begin{matrix}a, & 1 − a, & b, & \frac{b}{3} + 1 \\ & \frac{b−a}{2} + 1, & \frac{b+a+1}{2}, & \frac{b}{3} \end{matrix} | −\frac{1}{8}] = \frac{\frac{b−a}{2}!}{(−\frac{1}{2})!} \frac{\frac{b+a−1}{2}!}{b!}2^{b}

This gives a rapidly convergent (three bits per term) series for the useful but very slowly (if at all) convergent _{2}F_{1}[a, b; c + 1|1]:

\frac{c! (c − a − b)!}{(c − a)! (c − b)!} = _{2}F_{1} [ \begin{matrix}a, & b \\ & c + 1 \end{matrix} | 1]

= \frac{_{4}F_{3} [ \begin{matrix}\frac{1}{2} + a + b, & \frac{1}{2} − a − b, & 2c − a − b + \frac{1}{2}, & \frac{2c−a−b}{3} + \frac{7}{6} \\ & c + 1, & c − a − b + 1, & \frac{2c−a−b}{3} + \frac{1}{6} \end{matrix} | −\frac{1}{8}]}{_{4}F_{3} [ \begin{matrix}a − b + \frac{1}{2}, & b − a + \frac{1}{2}, & 2c − a − b + \frac{1}{2}, & \frac{2c−a−b}{3} + \frac{7}{6} \\ & c − a + 1, & c − b + 1, & \frac{2c−a−b}{3} + \frac{1}{6} \end{matrix} | −\frac{1}{8}]}

= \frac{_{4}F_{3} [ \begin{matrix}c − 2b − a + 1, & −c + 2b + a, & c − a, & \frac{c−a}{3} + 1 \\ & c − b − a + 1, & b + \frac{1}{2}, & \frac{c−a}{3} \end{matrix} | −\frac{1}{8}]}{_{4}F_{3} [ \begin{matrix}c − 2b + 1, & −c + 2b, & c, & \frac{c}{3} + 1 \\ & c − b + 1, & b + \frac{1}{2}, & \frac{c}{3} \end{matrix} | −\frac{1}{8}]}2^{a}

= \frac{_{4}F_{3} [ \begin{matrix}c + a + 1, & −c − a, & c − a, & \frac{c−a}{3} + 1 \\ & c + 1, & \frac{1}{2} − a, & \frac{c−a}{3} \end{matrix} | −\frac{1}{8}]}{_{4}F_{3} [ \begin{matrix}c − b + a + 1, & −c + b − a, & c − b − a, & \frac{c−b−a}{3} + 1 \\ & c − b + 1, & \frac{1}{2} − a, & \frac{c−b−a}{3} \end{matrix} | −\frac{1}{8}]}2^{−b}.

Even if you neglect to use matrix products to evaluate the _{4}F_{3}s, these are notably cheaper than four invocations of the Γ function, especially when a, b, and c might be complex. Also, the (complex) Beta function is the special case

B(a, b) = \frac{_{2}F_{1}[a − 1, −b; a|1]}{b} = \frac{1}{(a + b − 1)_{2}F_{1}[1 − a, 1 − b; 1|1]} = \frac{π sin π(a + b)}{sin πa sin πb}_{2}F_{1} [ \begin{matrix}a, & b \\ & 1 \end{matrix} | 1].