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AppellF1

The AppellF1 function

	Calling Sequence
	AppellF1( $a, b_{1}, b_{2}, c, z_{1}, z_{2}$ )

Parameters

$a$	-	algebraic expression
$b_{1}$	-	algebraic expression
$b_{2}$	-	algebraic expression
$c$	-	algebraic expression
$z_{1}$	-	algebraic expression
$z_{2}$	-	algebraic expression

Description

•

As is the case of all the four multi-parameter Appell functions, AppellF1, is a doubly hypergeometric function that includes as particular cases the 2F1 hypergeometric and some cases of the MeijerG function, and with them most of the known functions of mathematical physics. Among other situations, AppellF1 appears in the solution to differential equations in general relativity, quantum mechanics, and molecular and atomic physics.

Initialization: Set the display of special functions in output to typeset mathematical notation (textbook notation):

>	$Typesetting :- EnableTypesetRule (Typesetting :- SpecialFunctionRules) &colon;$

The definition of the AppellF1 series and the corresponding domain of convergence can be seen through the FunctionAdvisor

>	$FunctionAdvisor (definition, AppellF1)$

$[F_{1} (a, b__1, b__2, c, z__1, z__2) = \sum_{_k1 = 0}^{\infty} \sum_{_k2 = 0}^{\infty} \frac{{(a)}_{_k1 +_k2} {(b__1)}_{_k1} {(b__2)}_{_k2} {z__1}^{_k1} {z__2}^{_k2}}{{(c)}_{_k1 +_k2}_k1!_k2!}, |z__1| < 1 \land |z__2| < 1]$

(1)

A distinction is made between the AppellF1 doubly hypergeometric series, with the restricted domain of convergence shown above, and the AppellF1 function, that coincides with the series in its domain of convergence but also extends it analytically to the whole complex plane.

From the definition above, by swapping the AppellF1 variables subscripted with the numbers 1 and 2, the function remains the same; hence

>	$FunctionAdvisor (symmetries, AppellF1)$

$[F_{1} (a, b__2, b__1, c, z__2, z__1) = F_{1} (a, b__1, b__2, c, z__1, z__2)]$

(2)

From the series' definition, AppellF1 is singular (division by zero) when the $c$ parameter entering the pochhammer function in the denominator of the series is a non-positive integer because the pochhammer function will be equal to zero when the summation index of the series is bigger than the absolute value of $c$ .

For an analogous reason, when the $a$ and/or both $b_{1}$ and $b_{2}$ parameters entering the pochhammer functions in the numerator of the series are non-positive integers, the series will truncate and AppellF1 will be polynomial. As is the case of the hypergeometric function, when the pochhammers in both the numerator and the denominator have non-positive integer arguments, AppellF1 is polynomial if the absolute value of the non-positive integers in the pochhammers of the numerator are smaller than or equal to the absolute value of the non-positive integer (parameter $c$ ) in the pochhammer in the denominator, and singular otherwise. Consult the FunctionAdvisor for comprehensive information on the combinations of all these conditions. For example, the singular cases happen when any of the following conditions hold

>	$FunctionAdvisor (singularities, AppellF1)$

$[F_{1} (a, b__1, b__2, c, z__1, z__2), (c :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b__1 :: (\neg ℤ^{(0, -)}) \land a < c) \lor (c :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b__2 :: (\neg ℤ^{(0, -)}) \land a < c) \lor (c :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b__1 :: ℤ^{(0, -)} \land b__2 :: ℤ^{(0, -)} \land a < c \land b__1 + b__2 < c) \lor (c :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b__1 :: (\neg ℤ^{(0, -)})) \lor (c :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b__2 :: (\neg ℤ^{(0, -)})) \lor (c :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b__1 :: ℤ^{(0, -)} \land b__2 :: ℤ^{(0, -)} \land b__1 + b__2 < c)]$

(3)

The AppellF1 series is analytically extended to the AppellF1 function defined over the whole complex plane using identities and mainly by integral representations in terms of Eulerian integrals:

>	$FunctionAdvisor (integral_form, AppellF1)$

$[F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{Γ (c) (\int_{0}^{1} \frac{{(1 - u)}^{- 1 + b__1} {(- u z__1 + 1)}^{- c + a} {}_{2}F_{1} (a, b__2; c - b__1; z__2 u)}{u^{- c + b__1 + 1}} &DifferentialD; u)}{Γ (b__1) Γ (c - b__1) {(1 - z__1)}^{- c + a + b__1}}, 0 < ℜ (b__1) \land 0 < - ℜ (- c + b__1)], [F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{Γ (c) (\int_{0}^{1} \frac{{(1 - u)}^{b__2 - 1} {(- z__2 u + 1)}^{- c + a} {}_{2}F_{1} (a, b__1; c - b__2; u z__1)}{u^{- c + b__2 + 1}} &DifferentialD; u)}{Γ (b__2) Γ (c - b__2) {(1 - z__2)}^{- c + a + b__2}}, 0 < ℜ (b__2) \land 0 < - ℜ (- c + b__2)], [F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{Γ (c) (\int_{0}^{1} \frac{u^{a - 1}}{{(1 - u)}^{- c + a + 1} {(- u z__1 + 1)}^{b__1} {(- z__2 u + 1)}^{b__2}} &DifferentialD; u)}{Γ (a) Γ (c - a)}, 0 < ℜ (a) \land 0 < - ℜ (- c + a)], [F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{Γ (c) (\int_{0}^{1} \int_{0}^{1 - v} \frac{u^{- 1 + b__1} v^{b__2 - 1}}{{(1 - u - v)}^{- c + b__1 + b__2 + 1} {(- u z__1 - v z__2 + 1)}^{a}} &DifferentialD; u &DifferentialD; v)}{Γ (c - b__1 - b__2) Γ (b__1) Γ (b__2)}, 0 < ℜ (b__1) \land 0 < ℜ (b__2) \land 0 < - ℜ (- c + b__1 + b__2)]$

(4)

These integral representations are also the starting point for the derivation of many of the identities known for AppellF1.

AppellF1 also satisfies a linear system of partial differential equations of second order

>	$FunctionAdvisor (DE, AppellF1)$

$[f (a, b__1, b__2, c, z__1, z__2) = F_{1} (a, b__1, b__2, c, z__1, z__2), [\frac{\partial^{2}}{\partial {z__1}^{2}} f (a, b__1, b__2, c, z__1, z__2) = - \frac{z__2 (\frac{\partial^{2}}{\partial z__1 \partial z__2} f (a, b__1, b__2, c, z__1, z__2))}{z__1} + \frac{((- a - b__1 - 1) z__1 + c) (\frac{\partial}{\partial z__1} f (a, b__1, b__2, c, z__1, z__2))}{z__1 (- 1 + z__1)} - \frac{b__1 z__2 (\frac{\partial}{\partial z__2} f (a, b__1, b__2, c, z__1, z__2))}{z__1 (- 1 + z__1)} - \frac{f (a, b__1, b__2, c, z__1, z__2) a b__1}{z__1 (- 1 + z__1)}, \frac{\partial^{2}}{\partial z__1 \partial z__2} f (a, b__1, b__2, c, z__1, z__2) = - \frac{z__2 (\frac{\partial^{2}}{\partial {z__2}^{2}} f (a, b__1, b__2, c, z__1, z__2))}{z__1} - \frac{b__2 (\frac{\partial}{\partial z__1} f (a, b__1, b__2, c, z__1, z__2))}{z__2 - 1} + \frac{((- a - b__2 - 1) z__2 + c) (\frac{\partial}{\partial z__2} f (a, b__1, b__2, c, z__1, z__2))}{z__1 (z__2 - 1)} - \frac{f (a, b__1, b__2, c, z__1, z__2) a b__2}{z__1 (z__2 - 1)}]]$

(5)

Examples

Initialization: Set the display of special functions in output to typeset mathematical notation (textbook notation):

>	$Typesetting :- EnableTypesetRule (Typesetting :- SpecialFunctionRules) &colon;$

The conditions for both the singular and the polynomial cases can also be seen from the AppellF1. For example, the six polynomial cases of AppellF1 are

>	$AppellF1 :- SpecialValues :- Polynomial ()$

$6, (a, b1, b2, c, z1, z2) \mapsto'[[a :: ℤ^{(0, -)}, c :: (\neg ℤ^{(0, -)})], [a :: ℤ^{(0, -)}, c :: ℤ^{(0, -)}, c \leq a], [b1 :: ℤ^{(0, -)}, b2 :: ℤ^{(0, -)}, c :: (\neg ℤ^{(0, -)})], [b1 :: ℤ^{(0, -)}, b2 :: ℤ^{(0, -)}, c :: ℤ^{(0, -)}, c \leq b1 + b2], [b1 :: ℤ^{(0, -)}, c :: (\neg ℤ^{(0, -)})], [b2 :: ℤ^{(0, -)}, c :: (\neg ℤ^{(0, -)})]]'$

(6)

Likewise, the conditions for the singular cases of AppellF1 can be seen either using the FunctionAdvisor or entering AppellF1:-Singularities(), so with no arguments.

For particular values of its parameters, AppellF1 is related to the hypergeometric and elliptic functions. These hypergeometric cases are returned automatically. For example, for $z_{1} = 1$ ,

>	$(%AppellF1 = AppellF1) (a, b__1, b__2, c, 1, z__2)$

$F_{1} (a, b__1, b__2, c, 1, z__2) = {}_{2}F_{1} (a, b__1; c; 1) {}_{2}F_{1} (a, b__2; c - b__1; z__2)$

(7)

This formula analytically extends to the whole complex plane the AppellF1 series when any of $z_{1} = 1$ or $z_{2} = 1$ (the latter using the symmetry of AppellF1 - see the beginning of the Description section).

To see all the hypergeometric cases, enter

>	$FunctionAdvisor (specialize, AppellF1, hypergeom)$

$[F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{2}F_{1} (a, b__2; c; z__2), z__1 = 0 \lor b__1 = 0], [F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{2}F_{1} (a, b__1; c; z__1), z__2 = 0 \lor b__2 = 0], [F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{2}F_{1} (a, b__1; c; 1) {}_{2}F_{1} (a, b__2; c - b__1; z__2), z__1 = 1], [F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{2}F_{1} (a, b__2; c; 1) {}_{2}F_{1} (a, b__1; c - b__2; z__1), z__2 = 1], [F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{2}F_{1} (a, b__1 + b__2; c; z__1), z__1 = z__2], [F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{3}F_{2} (b__1, \frac{a}{2}, \frac{a}{2} + \frac{1}{2}; \frac{c}{2}, \frac{c}{2} + \frac{1}{2}; {z__1}^{2}), z__1 = - z__2 \land b__1 = b__2], [F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{1}F_{0} (b__1;; z__1) {}_{1}F_{0} (b__2;; z__2), c = a \land a \neq 0], [F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{z__1 {}_{2}F_{1} (1, a; c; z__1) - z__2 {}_{2}F_{1} (1, a; c; z__2)}{- z__2 + z__1}, b__1 = 1 \land b__2 = 1 \land z__1 \neq z__2]$

(8)

Other special values of AppellF1 can be seen using FunctionAdvisor(special_values, AppellF1).

By requesting the sum form of AppellF1, besides its double power series definition, we also see the particular form the series takes when one of the summations is performed and the result expressed in terms of 2F1 hypergeometric functions:

>	$FunctionAdvisor (sum_form, AppellF1)$

$[F_{1} (a, b__1, b__2, c, z__1, z__2) = \sum_{m = 0}^{\infty} \sum_{n = 0}^{\infty} \frac{{(a)}_{m + n} {(b__1)}_{m} {(b__2)}_{n} {z__1}^{m} {z__2}^{n}}{{(c)}_{m + n} m! n!}, |z__1| < 1 \land |z__2| < 1], [F_{1} (a, b__1, b__2, c, z__1, z__2) = \sum_{k = 0}^{\infty} \frac{{(a)}_{k} {(b__1)}_{k} {}_{2}F_{1} (a + k, b__2; c + k; z__2) {z__1}^{k}}{{(c)}_{k} k!}, |z__1| < 1], [F_{1} (a, b__1, b__2, c, z__1, z__2) = \sum_{k = 0}^{\infty} \frac{{(a)}_{k} {(b__2)}_{k} {}_{2}F_{1} (a + k, b__1; c + k; z__1) {z__2}^{k}}{{(c)}_{k} k!}, |z__2| < 1]$

(9)

As indicated in the formulas above, for AppellF1 (also for AppellF3) the domain of convergence of the single sum with hypergeometric coefficients is larger than the domain of convergence of the double series, because the hypergeometric coefficient in the single sum - say the one in $z_{2}$ - analytically extends the series with regards to the other variable - say $z_{1}$ - entering the hypergeometric coefficient. Hence, for AppellF1 (also for AppellF3), the case where one of the two variables, $z_{1}$ or $z_{2}$ , is equal to 1, is convergent only when the corresponding hypergeometric coefficient in the single sum form is convergent. For instance, the convergent case at $z_{1} = 1$ requires that $0 < Re (c - a - b_{1})$ .

AppellF1 admits identities analogous to Euler identities for the hypergeometric function. These Euler-type identities, as well as contiguity identities, are visible using the FunctionAdvisor with the option identities, or directly from the function. For example,

>	$AppellF1 (a, b__1, b__2, c, z__1, z__2) = AppellF1 :- Transformations [Euler] [1] (a, b__1, b__2, c, z__1, z__2)$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {(1 - z__1)}^{- b__1} {(1 - z__2)}^{- b__2} F_{1} (c - a, b__1, b__2, c, \frac{z__1}{z__1 - 1}, \frac{z__2}{- 1 + z__2})$

(10)

Among other situations, this identity is useful when both $z_{1}$ and $z_{2}$ have absolute values larger than 1 but one of the arguments in the same position of AppellF1 on the right-hand side has absolute value smaller than 1.

A contiguity transformation for AppellF1

>	$AppellF1 (a, b__1, b__2, c, z__1, z__2) = AppellF1 :- Transformations [Contiguity] [1] (a, b__1, b__2, c, z__1, z__2)$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{(c - 1) (F_{1} (a - 1, b__1, b__2 - 1, c - 1, z__1, z__2) - F_{1} (a - 1, b__1 - 1, b__2, c - 1, z__1, z__2))}{(- z__2 + z__1) (a - 1)}$

(11)

The contiguity transformations available in this way are

>	$indices (AppellF1 :- Transformations [Contiguity])$

$[1], [2], [3], [4], [5], [6], [7], [9], [8], [10]$

(12)

By using differential algebra techniques, the PDE system satisfied by AppellF1 can be transformed into an equivalent PDE system where one of the equations is a linear ODE in $z_{2}$ parametrized by $z_{1}$ . In the case of AppellF1 this linear ODE is of third order and can be computed as follows

>	$F1 (z__1, z__2) = AppellF1 (a, b__1, b__2, c, z__1, z__2)$

$F1 (z__1, z__2) = F_{1} (a, b__1, b__2, c, z__1, z__2)$

(13)

>	$simplify (op ([1, 2], PDEtools :- casesplit (PDEtools :- dpolyform (, no_Fn), lex)))$

$\frac{\partial^{3}}{\partial {z__2}^{3}} F1 (z__1, z__2) = \frac{((a + 2 b__2 + 4) {z__2}^{2} + ((- a + b__1 - b__2 - 3) z__1 - c - b__2 - 2) z__2 + z__1 (c - b__1 + 1)) (\frac{\partial^{2}}{\partial {z__2}^{2}} F1 (z__1, z__2)) + (((2 a + b__2 + 2) z__2 + (- a + b__1 - 1) z__1 - c) (\frac{\partial}{\partial z__2} F1 (z__1, z__2)) + F1 (z__1, z__2) a b__2) (b__2 + 1)}{(- 1 + z__2) (- z__2 + z__1) z__2}$

(14)

This linear ODE has four regular singularities, one of which is located at $z_{1}$

>	$DEtools [singularities] (subs (F1 (z__1, z__2) = F1 (z__2),))$

$regular = \{0, 1, z__1, \infty\}, irregular = \emptyset$

(15)

You can also see a general presentation of AppellF1, organized into sections and including plots, using the FunctionAdvisor

>	$FunctionAdvisor (AppellF1)$

AppellF1

describe

$AppellF1 = Appell 2-variable hypergeometric function F1$

definition

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \sum_{_k1 = 0}^{\infty} \sum_{_k2 = 0}^{\infty} \frac{{(a)}_{_k1 +_k2} {(b__1)}_{_k1} {(b__2)}_{_k2} {z__1}^{_k1} {z__2}^{_k2}}{{(c)}_{_k1 +_k2}_k1!_k2!}$

$|z__1| < 1 \land |z__2| < 1$

classify function

$Appell$

symmetries

$F_{1} (a, b__2, b__1, c, z__2, z__1) = F_{1} (a, b__1, b__2, c, z__1, z__2)$

plot

singularities

$F_{1} (a, b__1, b__2, c, z__1, z__2)$

$(c :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b__1 :: (\neg ℤ^{(0, -)}) \land a < c) \lor (c :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b__2 :: (\neg ℤ^{(0, -)}) \land a < c) \lor (c :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b__1 :: ℤ^{(0, -)} \land b__2 :: ℤ^{(0, -)} \land a < c \land b__1 + b__2 < c) \lor (c :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b__1 :: (\neg ℤ^{(0, -)})) \lor (c :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b__2 :: (\neg ℤ^{(0, -)})) \lor (c :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b__1 :: ℤ^{(0, -)} \land b__2 :: ℤ^{(0, -)} \land b__1 + b__2 < c)$

branch points

$F_{1} (a, b__1, b__2, c, z__1, z__2)$

$(a :: (\neg ℤ^{(0, -)}) \land b__1 :: (\neg ℤ^{(0, -)}) \land z__1 \in [1, \infty + \infty I]) \lor (a :: (\neg ℤ^{(0, -)}) \land b__2 :: (\neg ℤ^{(0, -)}) \land z__2 \in [1, \infty + \infty I])$

branch cuts

$F_{1} (a, b__1, b__2, c, z__1, z__2)$

$(a :: (\neg ℤ^{(0, -)}) \land b__1 :: (\neg ℤ^{(0, -)}) \land 1 < z__1) \lor (a :: (\neg ℤ^{(0, -)}) \land b__2 :: (\neg ℤ^{(0, -)}) \land 1 < z__2)$

special values

$F_{1} (a, b__1, b__2, c, z__1, z__2) = 1$

$z__1 = 0 \land z__2 = 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = 1$

$a = 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = 1$

$b__1 = 0 \land b__2 = 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{2}F_{1} (a, b__2; c; z__2)$

$z__1 = 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{2}F_{1} (a, b__2; c; z__2)$

$b__1 = 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{2}F_{1} (a, b__1; c; z__1)$

$z__2 = 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{2}F_{1} (a, b__1; c; z__1)$

$b__2 = 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{2}F_{1} (a, b__1; c; 1) {}_{2}F_{1} (a, b__2; c - b__1; z__2)$

$z__1 = 1$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{2}F_{1} (a, b__2; c; 1) {}_{2}F_{1} (a, b__1; c - b__2; z__1)$

$z__2 = 1$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{2}F_{1} (a, b__1 + b__2; c; z__1)$

$z__1 = z__2$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{3}F_{2} (b__1, \frac{a}{2}, \frac{a}{2} + \frac{1}{2}; \frac{c}{2}, \frac{c}{2} + \frac{1}{2}; {z__1}^{2})$

$z__1 = - z__2 \land b__1 = b__2$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {}_{1}F_{0} (b__1;; z__1) {}_{1}F_{0} (b__2;; z__2)$

$c = a \land a \neq 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{z__1 {}_{2}F_{1} (1, a; c; z__1) - z__2 {}_{2}F_{1} (1, a; c; z__2)}{- z__2 + z__1}$

$b__1 = 1 \land b__2 = 1 \land z__1 \neq z__2$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{arctanh (\sqrt{z__1}) \sqrt{z__1} - arctanh (\sqrt{z__2}) \sqrt{z__2}}{- z__2 + z__1}$

$a = \frac{1}{2} \land b__1 = 1 \land b__2 = 1 \land c = \frac{3}{2}$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{3 (\frac{arctanh (\frac{\sqrt{z__1} \sqrt{\frac{z__2 - z__1}{z__1}}}{\sqrt{1 - z__1}})}{\sqrt{\frac{z__2 - z__1}{z__1}}} - \arcsin (\sqrt{z__1}))}{\sqrt{z__1} z__2}$

$a = \frac{3}{2} \land b__1 = \frac{1}{2} \land b__2 = 1 \land c = \frac{5}{2}$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{2 a (\frac{arctanh (\frac{\sqrt{z__1} \sqrt{- 1 + \frac{z__2}{z__1}}}{\sqrt{1 - z__1}})}{{z__2}^{- \frac{1}{2} + a} \sqrt{- 1 + \frac{z__2}{z__1}}} - \frac{\sum_{_k2 = 0}^{- \frac{3}{2} + a} {(−1)}^{_k2} (\binom{- \frac{1}{2} + a}{_k2 + 1}) {(\frac{- 2 z__1 + z__2}{z__2})}^{_k2 + 1} (\sum_{_k1 = 0}^{_k2 + 1} \frac{(\binom{_k2 + 1}{_k1}) {z__1}^{_k1} ((\sum_{_k4 = 0}^{⌊\frac{_k1}{2} + \frac{1}{2}⌋ - 1} \frac{(\binom{_k1 - 1}{2_k4}) {(\frac{1}{2})}_{_k4} {(\frac{2 z__1 - z__2}{z__1})}^{_k1 - 1 - 2_k4} {z__2}^{2_k4} (2 \arcsin (\sqrt{z__1}) + \sqrt{1 - z__1} \sqrt{z__1} (\sum_{_k3 = 1}^{_k4} \frac{(_k3 - 1)! {(1 - 2 z__1)}^{2_k3 - 1}}{{(\frac{1}{2})}_{_k3}}))}{_k4! {z__1}^{2_k4}}) + (\sum_{_k4 = 0}^{- 1 + ⌊\frac{_k1}{2}⌋} 2^{2_k4 + 1} (\binom{_k1 - 1}{2_k4 + 1})_k4! {(1 - z__1)}^{_k4 + \frac{1}{2}} {z__1}^{- \frac{1}{2} -_k4} {(\frac{2 z__1 - z__2}{z__1})}^{_k1 - 2 - 2_k4} {z__2}^{2_k4 + 1} (\sum_{_k3 = 0}^{_k4} \frac{{(1 - 2 z__1)}^{2_k3}}{2^{2_k3} {(1 - z__1)}^{_k3} {z__1}^{_k3}_k3! {(\frac{3}{2})}_{_k4 -_k3}})))}{{(- 2 z__1 + z__2)}^{_k1}})}{2^{- \frac{1}{2} + a} {z__1}^{- \frac{1}{2} + a}})}{\sqrt{z__1}}$

$(- \frac{1}{2} + a) :: ℤ^{(0, +)} \land b__1 = \frac{1}{2} \land b__2 = 1 \land c - 1 - a = 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{3 (\sqrt{z__1} (z__1 - 2 arctanh (\sqrt{z__2}) (- 1 + z__1) \sqrt{z__2} - z__2) + arctanh (\sqrt{z__1}) (- 1 + z__1) (z__2 + z__1))}{2 (1 - z__1) \sqrt{z__1} {(- z__2 + z__1)}^{2}}$

$a = \frac{3}{2} \land b__1 = 2 \land b__2 = 1 \land c = \frac{5}{2}$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{5 (arctanh (\sqrt{z__1}) {(- 1 + z__1)}^{2} (3 {z__1}^{2} + 6 z__1 z__2 - {z__2}^{2}) + \sqrt{z__1} (5 {z__1}^{3} - 8 arctanh (\sqrt{z__2}) {(- 1 + z__1)}^{2} z__1 \sqrt{z__2} + {z__2}^{2} + z__1 z__2 (2 + z__2) - 3 {z__1}^{2} (1 + 2 z__2)))}{8 {(- 1 + z__1)}^{2} {z__1}^{\frac{3}{2}} {(- z__2 + z__1)}^{3}}$

$a = \frac{5}{2} \land b__1 = 3 \land b__2 = 1 \land c = \frac{7}{2}$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = (2 c - 2) ({(−1)}^{- \frac{5}{2} + c} {(1 - z__2)}^{- \frac{3}{2} + c} {z__2}^{\frac{1}{2} - c} \sqrt{\frac{z__2 - z__1}{- 1 + z__2}} arctanh (\sqrt{\frac{z__2 - z__1}{- 1 + z__2}}) + \frac{{(−1)}^{- \frac{3}{2} + c} {(1 - z__1)}^{c - \frac{1}{2}} (\sum_{_k1 = 0}^{- \frac{5}{2} + c} \frac{(_k1 + 1) 2^{_k1 - 2 c + 3} {z__1}^{\frac{_k1}{2} + 2 - c} (2 c - 5 -_k1)! (({(\sqrt{z__1} - \sqrt{\frac{z__2 - z__1}{- 1 + z__2}})}^{- 2 -_k1} + {(\sqrt{z__1} + \sqrt{\frac{z__2 - z__1}{- 1 + z__2}})}^{- 2 -_k1}) arctanh (\sqrt{z__1}) - (\sum_{_k2 = 0}^{_k1} \frac{{(−1)}^{_k2} ({(\sqrt{z__1} - \sqrt{\frac{z__2 - z__1}{- 1 + z__2}})}^{_k2 -_k1 - 1} + {(\sqrt{z__1} + \sqrt{\frac{z__2 - z__1}{- 1 + z__2}})}^{_k2 -_k1 - 1}) (\sum_{_k3 = 0}^{_k2} \frac{2^{_k2 - 2_k3} {(1 - z__1)}^{- 1 -_k2 +_k3} {z__1}^{\frac{_k2}{2} -_k3} (_k2 -_k3)!}{(_k2 - 2_k3)!_k3!})}{_k2 + 1}))}{(c - \frac{5}{2} -_k1)!})}{(- \frac{3}{2} + c)! (1 - z__2)})$

$a = 1 \land b__1 = - \frac{1}{2} \land b__2 = 1 \land (- \frac{3}{2} + c) :: ℤ^{(0, +)}$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \{\begin{array}{c} 1 & (b__1 = 0 \land b__2 = 0) \lor a = 0 \\ \frac{\sum_{_k2 = 0}^{a - c} \sum_{_k1 = 0}^{a - c -_k2} \frac{{(- a + c)}_{_k1 +_k2} {(b__1)}_{_k2} {(b__2)}_{_k1} {(\frac{z__1}{- 1 + z__1})}^{_k2} {(\frac{z__2}{- 1 + z__2})}^{_k1}}{{(c)}_{_k1 +_k2}_k2!_k1!}}{{(1 - z__1)}^{b__1} {(1 - z__2)}^{b__2}} & otherwise \end{array}$

$z__1 \neq 1 \land z__2 \neq 1 \land (a - c) :: ℤ^{(0, +)}$

identities

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {(1 - z__1)}^{- b__1} {(1 - z__2)}^{- b__2} F_{1} (c - a, b__1, b__2, c, \frac{z__1}{- 1 + z__1}, \frac{z__2}{- 1 + z__2})$

$z__1 \neq 1 \land z__2 \neq 1 \land (a :: ℤ^{(0, -)} \lor (b__1 :: ℤ^{(0, -)} \land b__2 :: ℤ^{(0, -)}) \lor \neg (1 < z__1 \lor 1 < z__2))$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {(1 - z__1)}^{- a} F_{1} (a, c - b__1 - b__2, b__2, c, \frac{z__1}{- 1 + z__1}, \frac{z__1 - z__2}{- 1 + z__1})$

$z__1 \neq 1 \land (a :: ℤ^{(0, -)} \lor (b__1 :: ℤ^{(0, -)} \land b__2 :: ℤ^{(0, -)}) \lor \neg (1 < z__1 \lor 1 < z__2))$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {(1 - z__1)}^{c - a - b__1} {(1 - z__2)}^{- b__2} F_{1} (c - a, c - b__1 - b__2, b__2, c, z__1, \frac{z__2 - z__1}{- 1 + z__2})$

$z__2 \neq 1 \land z__1 \neq 1$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{(- 1 + c) (F_{1} (- 1 + a, b__1, b__2 - 1, - 1 + c, z__1, z__2) - F_{1} (- 1 + a, b__1 - 1, b__2, - 1 + c, z__1, z__2))}{(z__1 - z__2) (- 1 + a)}$

$z__1 \neq z__2 \land a \neq 1 \land c \neq 1$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{{(a)}_{n} F_{1} (n + a, b__1, b__2, c, z__1, z__2)}{{(a - c + 1)}_{n}} - (\sum_{k = 1}^{n} \frac{{(−1)}^{k} (\binom{n}{k}) {(1 - c)}_{k} F_{1} (a, b__1, b__2, c - k, z__1, z__2)}{{(a - c + 1)}_{k}})$

$c \neq 1 \land ((a - c + 1) :: (\neg ℤ^{(0, -)}) \lor n \leq |a - c + 1|)$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{a F_{1} (a + 1, b__1, b__2, c, z__1, z__2) - b__1 F_{1} (a, b__1 + 1, b__2, c, z__1, z__2) - b__2 F_{1} (a, b__1, b__2 + 1, c, z__1, z__2)}{- b__1 - b__2 + a}$

$a \neq b__1 + b__2$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{(c - a) z__1 F_{1} (a, b__1, b__2, c + 1, z__1, z__2) - c F_{1} (a, b__1 - 1, b__2, c, z__1, z__2)}{c (- 1 + z__1)}$

$c \neq 0 \land z__1 \neq 1$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{(c - a) z__2 F_{1} (a, b__1, b__2, c + 1, z__1, z__2) - c F_{1} (a, b__1, b__2 - 1, c, z__1, z__2)}{c (- 1 + z__2)}$

$c \neq 0 \land z__2 \neq 1$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = F_{1} (n + a, b__1, b__2, c, z__1, z__2) - \frac{b__1 z__1 (\sum_{k = 1}^{n} F_{1} (a + k, b__1 + 1, b__2, c + 1, z__1, z__2))}{c} - \frac{b__2 z__2 (\sum_{k = 1}^{n} F_{1} (a + k, b__1, b__2 + 1, c + 1, z__1, z__2))}{c}$

$c \neq 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = F_{1} (a, b__1 + n, b__2, c, z__1, z__2) - \frac{a z__1 (\sum_{k = 1}^{n} F_{1} (a + 1, b__1 + k, b__2, c + 1, z__1, z__2))}{c}$

$c \neq 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = F_{1} (a - n, b__1, b__2, c, z__1, z__2) + \frac{b__2 z__2 (\sum_{k = 0}^{n - 1} F_{1} (a - k, b__1, b__2 + 1, c + 1, z__1, z__2))}{c} + \frac{b__1 z__1 (\sum_{k = 0}^{n - 1} F_{1} (a - k, b__1 + 1, b__2, c + 1, z__1, z__2))}{c}$

$c \neq 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = F_{1} (a, b__1 - n, b__2, c, z__1, z__2) + \frac{a z__1 (\sum_{k = 0}^{n - 1} F_{1} (a + 1, b__1 - k, b__2, c + 1, z__1, z__2))}{c}$

$c \neq 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {(\frac{z__1}{z__2})}^{b__2} F_{2} (b__1 + b__2, a, b__2, c, b__1 + b__2, z__1, 1 - \frac{z__1}{z__2})$

$z__1 \neq 0 \land z__2 \neq 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {(\frac{1}{1 - z__1})}^{b__1} F_{3} (a, b__1, b__2, c - a, c, z__2, \frac{z__1}{- 1 + z__1})$

$z__1 \neq 1$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {(\frac{1}{1 - z__2})}^{b__2} F_{3} (a, b__2, b__1, c - a, c, z__1, \frac{z__2}{- 1 + z__2})$

$z__2 \neq 1$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{Physics :- Library :- Add (\frac{{(a)}_{k1 + k2} F_{3} (b__1, b__2, - k1, - k2, c, z__1, z__2)}{k1! k2!}, k1 + k2 \leq - a)}{{(−1)}^{a}}$

$a :: ℤ^{(0, -)}$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {(\frac{z__1}{z__2})}^{b__2} F_{4} (a, b__1 + b__2, c, b__1 + b__2, \frac{{z__1}^{2}}{z__2}, \frac{(z__1 - z__2) (- 1 + z__1)}{z__2})$

$1 - c = 0 \land a - b__2 = 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{{(\frac{z__1}{z__2})}^{b__2} F_{4} (\frac{b__1}{2} + \frac{b__2}{2}, \frac{1}{2} + \frac{b__1}{2} + \frac{b__2}{2}, \frac{1}{2} + a, b__2 + \frac{1}{2}, \frac{{z__1}^{2} {z__2}^{2}}{{((- 1 + z__2) z__1 - z__2)}^{2}}, \frac{{(z__1 - z__2)}^{2}}{{((- 1 + z__1) z__2 - z__1)}^{2}})}{{(\frac{1}{2} - \frac{z__1}{2} + \frac{z__1}{2 z__2})}^{b__1 + b__2}}$

$c - 2 a = 0 \land - b__2 + b__1 = 0 \land \frac{1}{2} - \frac{z__1}{2} + \frac{z__1}{2 z__2} \neq 0$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {(\frac{z__1}{z__2})}^{b__2} {(1 + \sqrt{\frac{4 z__2 (z__2 + z__1) \sqrt{\frac{z__1}{z__2}} + {z__1}^{2} + 6 z__1 z__2 + {z__2}^{2}}{{(z__1 - z__2)}^{2}}})}^{2 b__1 + 2 b__2} F_{4} (a, b__1 + b__2, b__1 + b__2 - a + 1, c, \frac{4 z__2 (z__2 + z__1) \sqrt{\frac{z__1}{z__2}} + {z__1}^{2} + 6 z__1 z__2 + {z__2}^{2}}{{(z__1 - z__2)}^{2}}, \frac{2 z__1 ({(z__1 - z__2)}^{2} \sqrt{\frac{4 z__2 (z__2 + z__1) \sqrt{\frac{z__1}{z__2}} + {z__1}^{2} + 6 z__1 z__2 + {z__2}^{2}}{{(z__1 - z__2)}^{2}}} + (z__2 + z__1) (2 \sqrt{\frac{z__1}{z__2}} z__2 + z__1 + z__2))}{{(z__1 - z__2)}^{2}})$

$a - b__1 - \frac{1}{2} = 0 \land - b__2 + b__1 = 0 \land {ℜ (\frac{z__1}{z__2})}^{2} + {ℑ (\frac{z__1}{z__2})}^{2} \leq 1$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = {(\frac{z__1}{z__2})}^{b__2} {(1 + \sqrt{\frac{(- 4 z__1 + 8) \sqrt{1 - z__1} + {z__1}^{2} - 8 z__1 + 8}{{z__1}^{2}}})}^{2 b__1 + 2 b__2} F_{4} (b__2, b__1 + b__2, b__1 + 1, b__1 + b__2, \frac{(- 4 z__1 + 8) \sqrt{1 - z__1} + {z__1}^{2} - 8 z__1 + 8}{{z__1}^{2}}, - \frac{2 (\sqrt{\frac{(- 4 z__1 + 8) \sqrt{1 - z__1} + {z__1}^{2} - 8 z__1 + 8}{{z__1}^{2}}} {z__1}^{2} + (z__1 - 2) (z__1 - 2 \sqrt{1 - z__1} - 2)) (z__1 - z__2)}{{z__1}^{2} z__2})$

$a - b__1 - \frac{1}{2} = 0 \land c - 2 a = 0 \land {(ℜ (z__1) - 1)}^{2} + {ℑ (z__1)}^{2} \leq 1$

sum form

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \sum_{m = 0}^{\infty} \sum_{n = 0}^{\infty} \frac{{(a)}_{m + n} {(b__1)}_{m} {(b__2)}_{n} {z__1}^{m} {z__2}^{n}}{{(c)}_{m + n} m! n!}$

$|z__1| < 1 \land |z__2| < 1$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \sum_{k = 0}^{\infty} \frac{{(a)}_{k} {(b__1)}_{k} {}_{2}F_{1} (a + k, b__2; c + k; z__2) {z__1}^{k}}{{(c)}_{k} k!}$

$|z__1| < 1$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \sum_{k = 0}^{\infty} \frac{{(a)}_{k} {(b__2)}_{k} {}_{2}F_{1} (a + k, b__1; c + k; z__1) {z__2}^{k}}{{(c)}_{k} k!}$

$|z__2| < 1$

series

$series (F_{1} (a, b__1, b__2, c, z__1, z__2), z__1, 4) = {}_{2}F_{1} (a, b__2; c; z__2) + \frac{a b__1 {}_{2}F_{1} (b__2, a + 1; c + 1; z__2)}{c} z__1 + \frac{1}{2} \frac{a b__1 (a + 1) (b__1 + 1) {}_{2}F_{1} (b__2, a + 2; c + 2; z__2)}{c (c + 1)} {z__1}^{2} + \frac{1}{6} \frac{a b__1 (a + 1) (b__1 + 1) (a + 2) (b__1 + 2) {}_{2}F_{1} (b__2, a + 3; c + 3; z__2)}{c (c + 1) (c + 2)} {z__1}^{3} + O ({z__1}^{4})$

$series (F_{1} (a, b__1, b__2, c, z__1, z__2), z__2, 4) = {}_{2}F_{1} (a, b__1; c; z__1) + \frac{a b__2 {}_{2}F_{1} (b__1, a + 1; c + 1; z__1)}{c} z__2 + \frac{1}{2} \frac{a b__2 (a + 1) (b__2 + 1) {}_{2}F_{1} (b__1, a + 2; c + 2; z__1)}{c (c + 1)} {z__2}^{2} + \frac{1}{6} \frac{a b__2 (a + 1) (b__2 + 1) (a + 2) (b__2 + 2) {}_{2}F_{1} (b__1, a + 3; c + 3; z__1)}{c (c + 1) (c + 2)} {z__2}^{3} + O ({z__2}^{4})$

integral form

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{Γ (c) (\int_{0}^{1} \frac{{(1 - u)}^{- 1 + b__1} {(- u z__1 + 1)}^{- c + a} {}_{2}F_{1} (a, b__2; c - b__1; z__2 u)}{u^{- c + b__1 + 1}} &DifferentialD; u)}{Γ (b__1) Γ (c - b__1) {(1 - z__1)}^{- c + a + b__1}}$

$0 < ℜ (b__1) \land 0 < - ℜ (- c + b__1)$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{Γ (c) (\int_{0}^{1} \frac{{(1 - u)}^{b__2 - 1} {(- z__2 u + 1)}^{- c + a} {}_{2}F_{1} (a, b__1; c - b__2; u z__1)}{u^{- c + b__2 + 1}} &DifferentialD; u)}{Γ (b__2) Γ (c - b__2) {(1 - z__2)}^{- c + a + b__2}}$

$0 < ℜ (b__2) \land 0 < - ℜ (- c + b__2)$

$F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{Γ (c) (\int_{0}^{1} \frac{u^{a - 1}}{{(1 - u)}^{- c + a + 1} {(- u z__1 + 1)}^{b__1} {(- z__2 u + 1)}^{b__2}} &DifferentialD; u)}{Γ (a) Γ (c - a)}$

$0 < ℜ (a) \land 0 < - ℜ (- c + a)$

differentiation rule

$\frac{\partial}{\partial z__1} F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{a b__1 F_{1} (a + 1, b__1 + 1, b__2, c + 1, z__1, z__2)}{c}$

$\frac{\partial^{n}}{\partial {z__1}^{n}} F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{{(a)}_{n} {(b__1)}_{n} F_{1} (n + a, n + b__1, b__2, n + c, z__1, z__2)}{{(c)}_{n}}$

$\frac{\partial}{\partial z__2} F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{a b__2 F_{1} (a + 1, b__1, b__2 + 1, c + 1, z__1, z__2)}{c}$

$\frac{\partial^{n}}{\partial {z__2}^{n}} F_{1} (a, b__1, b__2, c, z__1, z__2) = \frac{{(a)}_{n} {(b__2)}_{n} F_{1} (n + a, b__1, n + b__2, n + c, z__1, z__2)}{{(c)}_{n}}$

$f (a, b__1, b__2, c, z__1, z__2) = F_{1} (a, b__1, b__2, c, z__1, z__2)$

$\frac{\partial^{2}}{\partial {z__1}^{2}} f (a, b__1, b__2, c, z__1, z__2) = - \frac{z__2 (\frac{\partial^{2}}{\partial z__2 \partial z__1} f (a, b__1, b__2, c, z__1, z__2))}{z__1} + \frac{((- a - b__1 - 1) z__1 + c) (\frac{\partial}{\partial z__1} f (a, b__1, b__2, c, z__1, z__2))}{z__1 (- 1 + z__1)} - \frac{b__1 z__2 (\frac{\partial}{\partial z__2} f (a, b__1, b__2, c, z__1, z__2))}{z__1 (- 1 + z__1)} - \frac{f (a, b__1, b__2, c, z__1, z__2) a b__1}{z__1 (- 1 + z__1)}$

$\frac{\partial^{2}}{\partial z__2 \partial z__1} f (a, b__1, b__2, c, z__1, z__2) = - \frac{z__2 (\frac{\partial^{2}}{\partial {z__2}^{2}} f (a, b__1, b__2, c, z__1, z__2))}{z__1} - \frac{b__2 (\frac{\partial}{\partial z__1} f (a, b__1, b__2, c, z__1, z__2))}{z__2 - 1} + \frac{((- a - b__2 - 1) z__2 + c) (\frac{\partial}{\partial z__2} f (a, b__1, b__2, c, z__1, z__2))}{z__1 (z__2 - 1)} - \frac{f (a, b__1, b__2, c, z__1, z__2) a b__2}{z__1 (z__2 - 1)}$

References

[1] Appell, P.; Kampe de Feriet, J. Fonctions Hypergeometriques et hyperspheriques. Gauthier-Villars, 1926.

[2] Srivastava, H. M.; Karlsson, P. W. Multiple Gaussian Hypergeometric Series. Ellis Horwood, 1985.

Compatibility

•	The AppellF1 command was introduced in Maple 2017.

•	For more information on Maple 2017 changes, see Updates in Maple 2017.

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