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AppellF4

The AppellF4 function

	Calling Sequence
	AppellF4( $a, b, c_{1}, c_{2}, z_{1}, z_{2}$ )

Parameters

$a$	-	algebraic expression
$b$	-	algebraic expression
$c_{1}$	-	algebraic expression
$c_{2}$	-	algebraic expression
$z_{1}$	-	algebraic expression
$z_{2}$	-	algebraic expression

Description

•

As is the case of all the four multi-parameter Appell functions, AppellF4, 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, AppellF4 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 AppellF4 series and the corresponding domain of convergence can be seen through the FunctionAdvisor

>	$FunctionAdvisor (definition, AppellF4)$

$[F_{4} (a, b, c__1, c__2, z__1, z__2) = \sum_{_k1 = 0}^{\infty} \sum_{_k2 = 0}^{\infty} \frac{{(a)}_{_k1 +_k2} {(b)}_{_k1 +_k2} {z__1}^{_k1} {z__2}^{_k2}}{{(c__1)}_{_k1} {(c__2)}_{_k2}_k1!_k2!}, \sqrt{|z__1|} + \sqrt{|z__2|} < 1]$

(1)

A distinction is made between the AppellF4 doubly hypergeometric series, with the restricted domain of convergence shown above, and the AppellF4 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 AppellF4 variables subscripted with the numbers 1 and 2, the function remains the same; hence

>	$FunctionAdvisor (symmetries, AppellF4)$

$[F_{4} (a, b, c__2, c__1, z__2, z__1) = F_{4} (a, b, c__1, c__2, z__1, z__2), F_{4} (b, a, c__1, c__2, z__1, z__2) = F_{4} (a, b, c__1, c__2, z__1, z__2)]$

(2)

Note the existence of another symmetry, also visible in the double sum definition.

From the series' definition, AppellF4 is singular (division by zero) when the $c_{1}$ and/or $c_{2}$ parameters entering the pochhammer functions in the denominator of the series are non-positive integers, because these pochhammer functions will be equal to zero when the summation index of the series is bigger than the absolute value of the corresponding $c_{1}$ or $c_{2}$ parameter.

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 AppellF4 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, AppellF4 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 (parameters $c_{1}, c_{2}$ ) in the pochhammers 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

>	$FunctionAdvisor (singularities, AppellF4)$

$[F_{4} (a, b, c__1, c__2, z__1, z__2), (c__1 :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b :: (\neg ℤ^{(0, -)})) \lor (c__1 :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b :: (\neg ℤ^{(0, -)}) \land a < c__1) \lor (c__1 :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b :: ℤ^{(0, -)} \land b < c__1) \lor (c__1 :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b :: ℤ^{(0, -)} \land a < c__1 \land b < c__1) \lor (c__2 :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b :: (\neg ℤ^{(0, -)})) \lor (c__2 :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b :: (\neg ℤ^{(0, -)}) \land a < c__2) \lor (c__2 :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b :: ℤ^{(0, -)} \land b < c__2) \lor (c__2 :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b :: ℤ^{(0, -)} \land a < c__2 \land b < c__2)]$

(3)

The AppellF4 series is analytically extended to the AppellF4 function defined over the whole complex plane using identities and mainly by integral representations in terms of Eulerian integrals:

>	$FunctionAdvisor (integral_form, AppellF4)$

$[F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{Γ (c__1) (\int_{0}^{1} \frac{u^{b - 1} {}_{2}F_{1} (\frac{a}{2}, \frac{1}{2} + \frac{a}{2}; c__1; \frac{4 u^{2} z__1 z__2}{{(- 1 + (z__1 + z__2) u)}^{2}})}{{(1 - u)}^{b - c__1 + 1} {(1 + (- z__1 - z__2) u)}^{a}} &DifferentialD; u)}{Γ (b) Γ (c__1 - b)}, (\sqrt{|z__1|} + \sqrt{|z__2|} < 1 \land c__1 = c__2 \land 0 < ℜ (b) \land 0 < ℜ (c__1) \land 0 < - ℜ (- c__1 + b)) \lor (\sqrt{|z__1|} + \sqrt{|z__2|} < 1 \land c__1 = c__2 \land 0 < ℜ (b) \land 0 < ℜ (c__2) \land 0 < - ℜ (- c__2 + b))], [F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{\int_{0}^{\infty} \frac{u^{2 a - 1} {}_{0}F_{1} (; c__1; \frac{z__1 u^{2}}{4}) {}_{0}F_{1} (; c__2; \frac{z__2 u^{2}}{4})}{{&ExponentialE;}^{u}} &DifferentialD; u}{Γ (2 a)}, \sqrt{|z__1|} + \sqrt{|z__2|} < 1 \land b = a + \frac{1}{2} \land 0 < ℜ (a) \land ℜ (\sqrt{z__1} + \sqrt{z__2}) < 1 \land ℜ (\sqrt{z__1} - \sqrt{z__2}) < 1 \land - ℜ (\sqrt{z__1} - \sqrt{z__2}) < 1 \land - ℜ (\sqrt{z__1} + \sqrt{z__2}) < 1], [F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{Γ (c__1) Γ (c__2) (\int_{0}^{1} \int_{0}^{1} \frac{u^{a - 1} v^{b - 1}}{{(1 - u)}^{- c__1 + a + 1} {(1 - v)}^{- c__2 + b + 1} (1 - \frac{u {(1 + \frac{u (- α1 (z__1, z__2) + z__2)}{α1 (z__1, z__2)} - v α1 (z__1, z__2))}^{- c__1 - c__2 + a + 1 + b}}{{(\frac{α1 (z__1, z__2) - z__2}{α1 (z__1, z__2)})}^{- c__1 - c__2 + a + 1} {(1 - v α1 (z__1, z__2))}^{- c__1 - c__2 + b + 1}})} &DifferentialD; u &DifferentialD; v)}{Γ (a) Γ (b) Γ (c__1 - a) Γ (c__2 - b)}, \sqrt{|z__1|} + \sqrt{|z__2|} < 1 \land 0 < ℜ (a) \land 0 < ℜ (b) \land 0 < - ℜ (- c__1 + a) \land 0 < - ℜ (- c__2 + b) \land {α (z__1, z__2)}^{2} + (z__1 - z__2 - 1) α (z__1, z__2) + z__2 = 0], [F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{Γ (c__2) Γ (c__1) (\int_{0}^{1} \int_{0}^{1} \frac{u^{a - 1} v^{b - 1}}{{(1 - u)}^{- c__2 + a + 1} {(1 - v)}^{b - c__1 + 1} (1 - \frac{u {(1 + \frac{u (- α1 (z__2, z__1) + z__1)}{α1 (z__2, z__1)} - v α1 (z__2, z__1))}^{- c__1 - c__2 + a + 1 + b}}{{(\frac{α1 (z__2, z__1) - z__1}{α1 (z__2, z__1)})}^{- c__1 - c__2 + a + 1} {(1 - v α1 (z__2, z__1))}^{- c__1 - c__2 + b + 1}})} &DifferentialD; u &DifferentialD; v)}{Γ (a) Γ (b) Γ (c__2 - a) Γ (c__1 - b)}, \sqrt{|z__1|} + \sqrt{|z__2|} < 1 \land 0 < ℜ (a) \land 0 < ℜ (b) \land 0 < - ℜ (- c__2 + a) \land 0 < - ℜ (- c__1 + b) \land {α (z__1, z__2)}^{2} + (z__1 - z__2 - 1) α (z__1, z__2) + z__2 = 0]$

(4)

AppellF4 is the only one among the four Appell functions that has no single integral representation in the general case (all of its parameters arbitrary). These integral representations are also the starting point for the derivation of many of the identities known for AppellF4.

AppellF4 also satisfies a linear system of partial differential equations of second order

>	$FunctionAdvisor (DE, AppellF4)$

$[f (a, b, c__1, c__2, z__1, z__2) = F_{4} (a, b, c__1, c__2, z__1, z__2), [\frac{\partial^{2}}{\partial {z__1}^{2}} f (a, b, c__1, c__2, z__1, z__2) = - \frac{2 z__2 (\frac{\partial^{2}}{\partial z__1 \partial z__2} f (a, b, c__1, c__2, z__1, z__2))}{z__1 - 1} - \frac{{z__2}^{2} (\frac{\partial^{2}}{\partial {z__2}^{2}} f (a, b, c__1, c__2, z__1, z__2))}{z__1 (z__1 - 1)} + \frac{((- a - b - 1) z__1 + c__1) (\frac{\partial}{\partial z__1} f (a, b, c__1, c__2, z__1, z__2))}{z__1 (z__1 - 1)} - \frac{z__2 (a + b + 1) (\frac{\partial}{\partial z__2} f (a, b, c__1, c__2, z__1, z__2))}{z__1 (z__1 - 1)} - \frac{a b f (a, b, c__1, c__2, z__1, z__2)}{z__1 (z__1 - 1)}, \frac{\partial^{2}}{\partial z__1 \partial z__2} f (a, b, c__1, c__2, z__1, z__2) = - \frac{z__1 (\frac{\partial^{2}}{\partial {z__1}^{2}} f (a, b, c__1, c__2, z__1, z__2))}{2 z__2} - \frac{(z__2 - 1) (\frac{\partial^{2}}{\partial {z__2}^{2}} f (a, b, c__1, c__2, z__1, z__2))}{2 z__1} - \frac{(a + b + 1) (\frac{\partial}{\partial z__1} f (a, b, c__1, c__2, z__1, z__2))}{2 z__2} + \frac{((- a - b - 1) z__2 + c__2) (\frac{\partial}{\partial z__2} f (a, b, c__1, c__2, z__1, z__2))}{2 z__1 z__2} - \frac{a b f (a, b, c__1, c__2, z__1, z__2)}{2 z__1 z__2}]]$

(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 AppellF4. For example, the fourteen polynomial cases of AppellF4 are

>	$AppellF4 :- SpecialValues :- Polynomial ()$

$8, (a, b, c1, c2, z1, z2) \mapsto [[a :: ℤ^{(0, -)}, c1 :: (\neg ℤ^{(0, -)}), c2 :: (\neg ℤ^{(0, -)})], [b :: ℤ^{(0, -)}, c1 :: (\neg ℤ^{(0, -)}), c2 :: (\neg ℤ^{(0, -)})], [a :: ℤ^{(0, -)}, c1 :: ℤ^{(0, -)}, c2 :: (\neg ℤ^{(0, -)}), c1 \leq a], [a :: ℤ^{(0, -)}, c1 :: (\neg ℤ^{(0, -)}), c2 :: ℤ^{(0, -)}, c2 \leq a], [a :: ℤ^{(0, -)}, c1 :: ℤ^{(0, -)}, c2 :: ℤ^{(0, -)}, c1 \leq a, c2 \leq a], [b :: ℤ^{(0, -)}, c1 :: ℤ^{(0, -)}, c2 :: (\neg ℤ^{(0, -)}), c1 \leq b], [b :: ℤ^{(0, -)}, c1 :: (\neg ℤ^{(0, -)}), c2 :: ℤ^{(0, -)}, c2 \leq b], [b :: ℤ^{(0, -)}, c1 :: ℤ^{(0, -)}, c2 :: ℤ^{(0, -)}, c1 \leq b, c2 \leq b]]$

(6)

Likewise, the conditions for the singular cases of AppellF4 can be seen either using the FunctionAdvisor or entering AppellF4:-Singularities(), so with no arguments.

For particular values of its parameters, AppellF4 is related to the hypergeometric function. These hypergeometric cases are returned automatically. For example, for $c_{1} = c_{2}, z_{1} = - z_{2}$ ,

>	$(%AppellF4 = AppellF4) (a, b, c__2, c__2, - z__2, z__2)$

$F_{4} (a, b, c__2, c__2, - z__2, z__2) = {}_{4}F_{3} (\frac{a}{2}, \frac{b}{2}, \frac{a}{2} + \frac{1}{2}, \frac{b}{2} + \frac{1}{2}; c__2, \frac{c__2}{2}, \frac{c__2}{2} + \frac{1}{2}; - 4 {z__2}^{2})$

(7)

To see all the hypergeometric cases, enter

>	$FunctionAdvisor (specialize, AppellF4, hypergeom)$

$[F_{4} (a, b, c__1, c__2, z__1, z__2) = {}_{2}F_{1} (a, b; c__2; z__2), z__1 = 0], [F_{4} (a, b, c__1, c__2, z__1, z__2) = {}_{2}F_{1} (a, b; c__1; z__1), z__2 = 0], [F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{}_{2}F_{1} (a, a + \frac{1}{2}; c__1; \frac{z__1}{{(1 + \sqrt{z__2})}^{2}})}{2 {(1 + \sqrt{z__2})}^{2 a}} + \frac{{}_{2}F_{1} (a, a + \frac{1}{2}; c__1; \frac{z__1}{{(\sqrt{z__2} - 1)}^{2}})}{2 {(1 - \sqrt{z__2})}^{2 a}}, b = a + \frac{1}{2} \land c__2 = \frac{1}{2}], [F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{}_{2}F_{1} (a, a + \frac{1}{2}; c__2; \frac{z__2}{{(1 + \sqrt{z__1})}^{2}})}{2 {(1 + \sqrt{z__1})}^{2 a}} + \frac{{}_{2}F_{1} (a, a + \frac{1}{2}; c__2; \frac{z__2}{{(\sqrt{z__1} - 1)}^{2}})}{2 {(1 - \sqrt{z__1})}^{2 a}}, b = a + \frac{1}{2} \land c__1 = \frac{1}{2}], [F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{}_{2}F_{1} (b, b + \frac{1}{2}; c__1; \frac{z__1}{{(1 + \sqrt{z__2})}^{2}})}{2 {(1 + \sqrt{z__2})}^{2 b}} + \frac{{}_{2}F_{1} (b, b + \frac{1}{2}; c__1; \frac{z__1}{{(\sqrt{z__2} - 1)}^{2}})}{2 {(1 - \sqrt{z__2})}^{2 b}}, a = b + \frac{1}{2} \land c__2 = \frac{1}{2}], [F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{}_{2}F_{1} (b, b + \frac{1}{2}; c__2; \frac{z__2}{{(1 + \sqrt{z__1})}^{2}})}{2 {(1 + \sqrt{z__1})}^{2 b}} + \frac{{}_{2}F_{1} (b, b + \frac{1}{2}; c__2; \frac{z__2}{{(\sqrt{z__1} - 1)}^{2}})}{2 {(1 - \sqrt{z__1})}^{2 b}}, a = b + \frac{1}{2} \land c__1 = \frac{1}{2}], [F_{4} (a, b, c__1, c__2, z__1, z__2) = {(\frac{z__2 + 1 - z__1 - \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__2})}^{a} {(\frac{z__1 + 1 - z__2 - \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__1})}^{a} {}_{2}F_{1} (a, a - b + 1; b; \frac{{(- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}})}^{2}}{4 z__1 z__2}), c__1 = b \land c__2 = b], [F_{4} (a, b, c__1, c__2, z__1, z__2) = {(\frac{z__2 + 1 - z__1 - \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__2})}^{b} {(\frac{z__1 + 1 - z__2 - \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__1})}^{b} {}_{2}F_{1} (b, b - a + 1; a; \frac{{(- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}})}^{2}}{4 z__1 z__2}), c__1 = a \land c__2 = a], [F_{4} (a, b, c__1, c__2, z__1, z__2) = {}_{2}F_{1} (a, b; a - b + 1; \frac{(- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}) (z__1 + 1 - z__2 - \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}})}{2 z__1 (- z__2 - 1 + z__1 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}})}) {(\frac{z__1 + 1 - z__2 - \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__1})}^{a}, c__1 = a - b + 1 \land c__2 = b], [F_{4} (a, b, c__1, c__2, z__1, z__2) = {}_{2}F_{1} (a, b; a - b + 1; \frac{(- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}) (- z__2 - 1 + z__1 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}})}{2 z__2 (z__1 + 1 - z__2 - \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}})}) {(\frac{z__2 + 1 - z__1 - \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__2})}^{a}, c__2 = a - b + 1 \land c__1 = b], [F_{4} (a, b, c__1, c__2, z__1, z__2) = {}_{2}F_{1} (a, b; b - a + 1; \frac{(- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}) (z__1 + 1 - z__2 - \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}})}{2 z__1 (- z__2 - 1 + z__1 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}})}) {(\frac{z__1 + 1 - z__2 - \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__1})}^{b}, c__1 = b - a + 1 \land c__2 = a], [F_{4} (a, b, c__1, c__2, z__1, z__2) = {}_{2}F_{1} (a, b; b - a + 1; \frac{(- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}) (- z__2 - 1 + z__1 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}})}{2 z__2 (z__1 + 1 - z__2 - \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}})}) {(\frac{z__2 + 1 - z__1 - \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__2})}^{b}, c__2 = b - a + 1 \land c__1 = a], [F_{4} (a, b, c__1, c__2, z__1, z__2) = {}_{4}F_{3} (a, b, \frac{c__1}{2} + \frac{c__2}{2}, \frac{c__1}{2} + \frac{c__2}{2} - \frac{1}{2}; c__1, c__2, c__1 + c__2 - 1; 4 z__1), z__1 = z__2], [F_{4} (a, b, c__1, c__2, z__1, z__2) = {}_{4}F_{3} (\frac{a}{2}, \frac{b}{2}, \frac{a}{2} + \frac{1}{2}, \frac{b}{2} + \frac{1}{2}; c__1, \frac{c__1}{2}, \frac{c__1}{2} + \frac{1}{2}; - 4 {z__1}^{2}), z__2 = - z__1 \land c__1 = c__2]$

(8)

Other special values of AppellF4 can be seen using FunctionAdvisor(special_values, AppellF4).

By requesting the sum form of AppellF4, 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:

>	$FunctionAdvisor (sum_form, AppellF4)$

$[F_{4} (a, b, c__1, c__2, z__1, z__2) = \sum_{m = 0}^{\infty} \sum_{n = 0}^{\infty} \frac{{(a)}_{m + n} {(b)}_{m + n} {z__1}^{m} {z__2}^{n}}{{(c__1)}_{m} {(c__2)}_{n} m! n!}, \sqrt{|z__2|} + \sqrt{|z__1|} < 1], [F_{4} (a, b, c__1, c__2, z__1, z__2) = \sum_{k = 0}^{\infty} \frac{{(a)}_{k} {(b)}_{k} {}_{2}F_{1} (a + k, b + k; c__2; z__2) {z__1}^{k}}{{(c__1)}_{k} k!}, \sqrt{|z__2|} + \sqrt{|z__1|} < 1], [F_{4} (a, b, c__1, c__2, z__1, z__2) = \sum_{k = 0}^{\infty} \frac{{(a)}_{k} {(b)}_{k} {}_{2}F_{1} (a + k, b + k; c__1; z__1) {z__2}^{k}}{{(c__2)}_{k} k!}, \sqrt{|z__2|} + \sqrt{|z__1|} < 1]$

(9)

As indicated in the formulas above, for AppellF4 (also for AppellF2), and unlike the case of AppellF1 and AppellF3, the domain of convergence with regards to the two variables $z_{1}$ and $z_{2}$ is entangled, i.e. it intrinsically depends on a combination of the two variables, so the hypergeometric coefficient in one variable in the single sum form does not extend the domain of convergence of the double sum but for particular cases, and from the formulas above one cannot conclude about the value of the function when one of $z_{1}$ or $z_{2}$ is equal to 1 unless the other one is exactly equal to 0.

AppellF4 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,

>	$AppellF4 (a, b, c__1, c__2, z__1, z__2) = AppellF4 :- Transformations [Euler] [1] (a, b, c__1, c__2, z__1, z__2)$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{Γ (c__2) Γ (b - a) {(- z__2)}^{- a} F_{4} (a, a - c__2 + 1, a - b + 1, c__1, \frac{1}{z__2}, \frac{z__1}{z__2})}{Γ (c__2 - a) Γ (b)} + \frac{Γ (c__2) Γ (a - b) {(- z__2)}^{- b} F_{4} (b, b - c__2 + 1, b - a + 1, c__1, \frac{1}{z__2}, \frac{z__1}{z__2})}{Γ (c__2 - b) Γ (a)}$

(10)

Among other situations, this identity is useful when the sum of the square roots of the absolute values of $z_{1}$ and $z_{2}$ is larger than 1 but the same sum constructed with the arguments in the same position of AppellF4 on the right-hand side is smaller than 1. Another case where this identity is useful is when $z_{1} = 1$ , so that the two AppellF4 functions on the right-hand side will have the two main variables (last arguments) equal, in turn a special value of hypergeometric 4F3 type:

>	$eval (, z__1 = 1)$

$F_{4} (a, b, c__1, c__2, 1, z__2) = \frac{Γ (c__2) Γ (b - a) {(- z__2)}^{- a} {}_{4}F_{3} (a, a - c__2 + 1, \frac{a}{2} - \frac{b}{2} + \frac{c__1}{2}, \frac{a}{2} - \frac{b}{2} + \frac{1}{2} + \frac{c__1}{2}; c__1, a - b + 1, a - b + c__1; \frac{4}{z__2})}{Γ (c__2 - a) Γ (b)} + \frac{Γ (c__2) Γ (a - b) {(- z__2)}^{- b} {}_{4}F_{3} (b, b - c__2 + 1, \frac{b}{2} - \frac{a}{2} + \frac{c__1}{2}, \frac{b}{2} - \frac{a}{2} + \frac{1}{2} + \frac{c__1}{2}; c__1, b - a + 1, b - a + c__1; \frac{4}{z__2})}{Γ (c__2 - b) Γ (a)}$

(11)

This formula analytically extends to the whole complex plane the AppellF4 series when any of $z_{1} = 1$ or $z_{2} = 1$ (the latter using the symmetry of AppellF4 - see the beginning of the Description section).

A contiguity transformation for AppellF4

>	$AppellF4 (a, b, c__1, c__2, z__1, z__2) = AppellF4 :- Transformations [Contiguity] [1] (a, b, c__1, c__2, z__1, z__2)$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{F_{4} (b, a + 1, c__1, c__2, z__1, z__2) a}{a - b} - \frac{b F_{4} (a, b + 1, c__1, c__2, z__1, z__2)}{a - b}$

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The contiguity transformations available in this way are

>	$indices (AppellF4 :- Transformations [Contiguity])$

$[1], [2], [3], [4]$

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By using differential algebra techniques, the PDE system satisfied by AppellF4 can be transformed into an equivalent PDE system where one of the equations is a fourth order linear ODE in $z_{2}$ parametrized by $z_{1}$ . This linear ODE has four regular singularities, some of which depend on $z_{1}$ and the function's parameters. These singularities can be see directly from the function using the MathematicalFunctions:-Evalf:-Singularities command

>	$MathematicalFunctions :- Evalf :- Singularities (AppellF4 (a, b, c__1, c__2, z__1, z__2))$

$[0, \frac{(z__1 - 1) (a + b - c__1 + 1) (a + b - c__1 - 2 c__2 + 3)}{(c__1 - 1 - b + a) (- c__1 + 1 - b + a)}, z__1 + 1 - 2 \sqrt{z__1}, z__1 + 1 + 2 \sqrt{z__1}, \infty + \infty I]$

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You can also see a general presentation of AppellF4, organized into sections and including plots, using the FunctionAdvisor

>	$FunctionAdvisor (AppellF4)$

AppellF4

describe

$AppellF4 = Appell 2-variable hypergeometric function F4$

definition

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \sum_{_k1 = 0}^{\infty} \sum_{_k2 = 0}^{\infty} \frac{{(a)}_{_k1 +_k2} {(b)}_{_k1 +_k2} {z__1}^{_k1} {z__2}^{_k2}}{{(c__1)}_{_k1} {(c__2)}_{_k2}_k1!_k2!}$

$\sqrt{|z__1|} + \sqrt{|z__2|} < 1$

classify function

$Appell$

symmetries

$F_{4} (a, b, c__2, c__1, z__2, z__1) = F_{4} (a, b, c__1, c__2, z__1, z__2)$

$F_{4} (b, a, c__1, c__2, z__1, z__2) = F_{4} (a, b, c__1, c__2, z__1, z__2)$

plot

singularities

$F_{4} (a, b, c__1, c__2, z__1, z__2)$

$(c__1 :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b :: (\neg ℤ^{(0, -)})) \lor (c__1 :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b :: (\neg ℤ^{(0, -)}) \land a < c__1) \lor (c__1 :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b :: ℤ^{(0, -)} \land b < c__1) \lor (c__1 :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b :: ℤ^{(0, -)} \land a < c__1 \land b < c__1) \lor (c__2 :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b :: (\neg ℤ^{(0, -)})) \lor (c__2 :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b :: (\neg ℤ^{(0, -)}) \land a < c__2) \lor (c__2 :: ℤ^{(0, -)} \land a :: (\neg ℤ^{(0, -)}) \land b :: ℤ^{(0, -)} \land b < c__2) \lor (c__2 :: ℤ^{(0, -)} \land a :: ℤ^{(0, -)} \land b :: ℤ^{(0, -)} \land a < c__2 \land b < c__2)$

branch points

$F_{4} (a, b, c__1, c__2, z__1, z__2)$

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

branch cuts

$F_{4} (a, b, c__1, c__2, z__1, z__2)$

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

special values

$F_{4} (a, b, c__1, c__2, z__1, z__2) = 1$

$z__1 = 0 \land z__2 = 0$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = 1$

$a = 0$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = 1$

$b = 0$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = {}_{2}F_{1} (a, b; c__2; z__2)$

$z__1 = 0$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = {}_{2}F_{1} (a, b; c__1; z__1)$

$z__2 = 0$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{(1 + \sqrt{z__2})}^{- 2 a} {}_{2}F_{1} (a, a + \frac{1}{2}; c__1; \frac{z__1}{{(1 + \sqrt{z__2})}^{2}})}{2} + \frac{{(1 - \sqrt{z__2})}^{- 2 a} {}_{2}F_{1} (a, a + \frac{1}{2}; c__1; \frac{z__1}{{(1 - \sqrt{z__2})}^{2}})}{2}$

$b = a + \frac{1}{2} \land c__2 = \frac{1}{2}$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{(1 + \sqrt{z__1})}^{- 2 a} {}_{2}F_{1} (a, a + \frac{1}{2}; c__2; \frac{z__2}{{(1 + \sqrt{z__1})}^{2}})}{2} + \frac{{(1 - \sqrt{z__1})}^{- 2 a} {}_{2}F_{1} (a, a + \frac{1}{2}; c__2; \frac{z__2}{{(1 - \sqrt{z__1})}^{2}})}{2}$

$b = a + \frac{1}{2} \land c__1 = \frac{1}{2}$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{(1 + \sqrt{z__2})}^{- 2 b} {}_{2}F_{1} (b, b + \frac{1}{2}; c__1; \frac{z__1}{{(1 + \sqrt{z__2})}^{2}})}{2} + \frac{{(1 - \sqrt{z__2})}^{- 2 b} {}_{2}F_{1} (b, b + \frac{1}{2}; c__1; \frac{z__1}{{(1 - \sqrt{z__2})}^{2}})}{2}$

$a = b + \frac{1}{2} \land c__2 = \frac{1}{2}$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{(1 + \sqrt{z__1})}^{- 2 b} {}_{2}F_{1} (b, b + \frac{1}{2}; c__2; \frac{z__2}{{(1 + \sqrt{z__1})}^{2}})}{2} + \frac{{(1 - \sqrt{z__1})}^{- 2 b} {}_{2}F_{1} (b, b + \frac{1}{2}; c__2; \frac{z__2}{{(1 - \sqrt{z__1})}^{2}})}{2}$

$a = b + \frac{1}{2} \land c__1 = \frac{1}{2}$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = {(1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__2})}^{a} {(1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__1})}^{a} {}_{2}F_{1} (a, a - b + 1; b; \frac{{(- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}})}^{2}}{4 z__1 z__2})$

$c__1 = b \land c__2 = b$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = {(1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__2})}^{b} {(1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__1})}^{b} {}_{2}F_{1} (b, b - a + 1; a; \frac{{(- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}})}^{2}}{4 z__1 z__2})$

$c__1 = a \land c__2 = a$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = {(1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__1})}^{a} {}_{2}F_{1} (a, b; a - b + 1; - \frac{(- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}) (1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__1})}{2 z__2 (1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__2})})$

$c__1 = a - b + 1 \land c__2 = b$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = {(1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__2})}^{a} {}_{2}F_{1} (a, b; a - b + 1; - \frac{(- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}) (1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__2})}{2 z__1 (1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__1})})$

$c__2 = a - b + 1 \land c__1 = b$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = {(1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__1})}^{b} {}_{2}F_{1} (a, b; b - a + 1; - \frac{(- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}) (1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__1})}{2 z__2 (1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__2})})$

$c__1 = b - a + 1 \land c__2 = a$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = {(1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__2})}^{b} {}_{2}F_{1} (a, b; b - a + 1; - \frac{(- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}) (1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__2})}{2 z__1 (1 - \frac{- 1 + z__1 + z__2 + \sqrt{- 4 z__1 z__2 + {(- 1 + z__1 + z__2)}^{2}}}{2 z__1})})$

$c__2 = b - a + 1 \land c__1 = a$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = {}_{4}F_{3} (a, b, \frac{c__1}{2} + \frac{c__2}{2}, \frac{c__1}{2} + \frac{c__2}{2} - \frac{1}{2}; c__1, c__2, c__1 + c__2 - 1; 4 z__1)$

$z__1 = z__2$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = {}_{4}F_{3} (\frac{a}{2}, \frac{b}{2}, \frac{a}{2} + \frac{1}{2}, \frac{b}{2} + \frac{1}{2}; c__1, \frac{c__1}{2}, \frac{c__1}{2} + \frac{1}{2}; - 4 {z__1}^{2})$

$z__2 = - z__1 \land c__1 = c__2$

identities

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{F_{4} (b, a + 1, c__1, c__2, z__1, z__2) a}{- b + a} - \frac{b F_{4} (a, b + 1, c__1, c__2, z__1, z__2)}{- b + a}$

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

$F_{4} (a, b, c__1, c__2, z__1, z__2) = F_{4} (b, a + n, c__1, c__2, z__1, z__2) - \frac{b z__1 (\sum_{k = 1}^{n} F_{4} (a + k, b + 1, c__1 + 1, c__2, z__1, z__2))}{c__1} - \frac{b z__2 (\sum_{k = 1}^{n} F_{4} (a + k, b + 1, c__1, c__2 + 1, z__1, z__2))}{c__2}$

$z__1 \neq 1 \land z__2 \neq 1 \land c__1 \neq 0 \land c__2 \neq 0$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = F_{4} (b, a + n, c__1, c__2, z__1, z__2) - (\sum_{i = 0}^{n} \sum_{k = 0}^{n - i} \frac{F_{4} (a + i + k, k + i + b, i + c__1, k + c__2, z__1, z__2) (\binom{n}{i}) (\binom{n - i}{k}) {(b)}_{k + i} {z__1}^{i} {z__2}^{k}}{{(c__1)}_{i} {(c__2)}_{k}}) + F_{4} (a, b, c__1, c__2, z__1, z__2)$

$z__1 \neq 1 \land z__2 \neq 1 \land (c__1 :: (\neg ℤ^{(0, -)}) \lor n \leq |c__1|) \land (c__1 :: (\neg ℤ^{(0, -)}) \lor n \leq |c__2|)$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = F_{4} (a, b, c__1 - n, c__2, z__1, z__2) - a b z__1 (\sum_{k = 0}^{n - 1} \frac{F_{4} (a + 1, b + 1, c__1 + 1 - k, c__2, z__1, z__2)}{(c__1 - k) (c__1 - k - 1)})$

$z__1 \neq 1 \land z__2 \neq 1 \land (c__1 :: (\neg ℤ^{(0, +)}) \lor n - 1 < c__1)$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{Γ (c__2) Γ (b - a) {(- z__2)}^{- a} F_{4} (a, a - c__2 + 1, a - b + 1, c__1, \frac{1}{z__2}, \frac{z__1}{z__2})}{Γ (c__2 - a) Γ (b)} + \frac{Γ (c__2) Γ (- b + a) {(- z__2)}^{- b} F_{4} (b, b - c__2 + 1, b - a + 1, c__1, \frac{1}{z__2}, \frac{z__1}{z__2})}{Γ (c__2 - b) Γ (a)}$

$z__2 \neq 0 \land a :: (\neg ℤ^{(0, -)}) \land b :: (\neg ℤ^{(0, -)}) \land c__2 :: (\neg ℤ^{(0, -)}) \land (c__2 - a) :: (\neg ℤ^{(0, -)}) \land (c__2 - b) :: (\neg ℤ^{(0, -)}) \land (b - a) :: (\neg ℤ)$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{F_{1} (\frac{1}{2} + a - b, - b + a, b, 2 a + 1 - 2 b, \frac{4 \sqrt{z__2}}{{(\sqrt{z__2} + 1)}^{2}}, - \frac{4 \sqrt{z__2}}{- z__2 + z__1 - 2 \sqrt{z__2} - 1})}{{(\sqrt{z__2} + 1)}^{2 a} {(1 - \frac{z__1}{{(\sqrt{z__2} + 1)}^{2}})}^{b}}$

$c__1 = a \land c__2 = a - b + 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{F_{1} (\frac{1}{2} + b - a, b - a, a, 2 b + 1 - 2 a, \frac{4 \sqrt{z__2}}{{(\sqrt{z__2} + 1)}^{2}}, - \frac{4 \sqrt{z__2}}{- z__2 + z__1 - 2 \sqrt{z__2} - 1})}{{(\sqrt{z__2} + 1)}^{2 b} {(1 - \frac{z__1}{{(\sqrt{z__2} + 1)}^{2}})}^{a}}$

$c__1 = b \land c__2 = b - a + 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{F_{1} (\frac{1}{2} + a - b, - b + a, b, 2 a + 1 - 2 b, \frac{4 \sqrt{z__1}}{{(\sqrt{z__1} + 1)}^{2}}, - \frac{4 \sqrt{z__1}}{- z__1 + z__2 - 2 \sqrt{z__1} - 1})}{{(\sqrt{z__1} + 1)}^{2 a} {(1 - \frac{z__2}{{(\sqrt{z__1} + 1)}^{2}})}^{b}}$

$c__2 = a \land c__1 = a - b + 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{F_{1} (\frac{1}{2} + b - a, b - a, a, 2 b + 1 - 2 a, \frac{4 \sqrt{z__1}}{{(\sqrt{z__1} + 1)}^{2}}, - \frac{4 \sqrt{z__1}}{- z__1 + z__2 - 2 \sqrt{z__1} - 1})}{{(\sqrt{z__1} + 1)}^{2 b} {(1 - \frac{z__2}{{(\sqrt{z__1} + 1)}^{2}})}^{a}}$

$c__2 = b \land c__1 = b - a + 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{F_{2} (a, b, \frac{1}{2} + a - b, c__1, 2 a + 1 - 2 b, \frac{z__1}{{(\sqrt{z__2} + 1)}^{2}}, \frac{4 \sqrt{z__2}}{{(\sqrt{z__2} + 1)}^{2}})}{{(\sqrt{z__2} + 1)}^{2 a}}$

$c__2 = a - b + 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{F_{2} (b, a, \frac{1}{2} + b - a, c__1, 2 b + 1 - 2 a, \frac{z__1}{{(\sqrt{z__2} + 1)}^{2}}, \frac{4 \sqrt{z__2}}{{(\sqrt{z__2} + 1)}^{2}})}{{(\sqrt{z__2} + 1)}^{2 b}}$

$c__2 = b - a + 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{F_{2} (a, b, \frac{1}{2} + a - b, c__2, 2 a + 1 - 2 b, \frac{z__2}{{(\sqrt{z__1} + 1)}^{2}}, \frac{4 \sqrt{z__1}}{{(\sqrt{z__1} + 1)}^{2}})}{{(\sqrt{z__1} + 1)}^{2 a}}$

$c__1 = a - b + 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{F_{2} (b, a, \frac{1}{2} + b - a, c__2, 2 b + 1 - 2 a, \frac{z__2}{{(\sqrt{z__1} + 1)}^{2}}, \frac{4 \sqrt{z__1}}{{(\sqrt{z__1} + 1)}^{2}})}{{(\sqrt{z__1} + 1)}^{2 b}}$

$c__1 = b - a + 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{Physics :- Library :- Add (\frac{{(a)}_{k1 + k2} F_{2} (b, - k1, - k2, c__1, c__2, z__1, z__2)}{k1! k2!}, k1 + k2 \leq - a)}{{(−1)}^{a}}$

$a :: ℤ^{(0, -)}$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{(\frac{z__2 + 2 \sqrt{z__2} - z__1 + 1}{z__2 - 2 \sqrt{z__2} - z__1 + 1})}^{b} F_{3} (- b + a, b, \frac{1}{2} + a - b, \frac{1}{2} + a - b, 2 a + 1 - 2 b, \frac{4 \sqrt{z__2}}{{(\sqrt{z__2} + 1)}^{2}}, \frac{4 \sqrt{z__2}}{- z__2 + 2 \sqrt{z__2} + z__1 - 1})}{{(\sqrt{z__2} + 1)}^{2 a} {(1 - \frac{z__1}{{(\sqrt{z__2} + 1)}^{2}})}^{b}}$

$c__2 = a - b + 1 \land c__1 = a \land \frac{z__1}{{(\sqrt{z__2} + 1)}^{2}} \neq 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{(\frac{z__2 + 2 \sqrt{z__2} - z__1 + 1}{z__2 - 2 \sqrt{z__2} - z__1 + 1})}^{a} F_{3} (b - a, a, \frac{1}{2} + b - a, \frac{1}{2} + b - a, 2 b + 1 - 2 a, \frac{4 \sqrt{z__2}}{{(\sqrt{z__2} + 1)}^{2}}, \frac{4 \sqrt{z__2}}{- z__2 + 2 \sqrt{z__2} + z__1 - 1})}{{(\sqrt{z__2} + 1)}^{2 b} {(1 - \frac{z__1}{{(\sqrt{z__2} + 1)}^{2}})}^{a}}$

$c__2 = b - a + 1 \land c__1 = b \land \frac{z__1}{{(\sqrt{z__2} + 1)}^{2}} \neq 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{(\frac{z__1 + 2 \sqrt{z__1} - z__2 + 1}{z__1 - 2 \sqrt{z__1} - z__2 + 1})}^{b} F_{3} (- b + a, b, \frac{1}{2} + a - b, \frac{1}{2} + a - b, 2 a + 1 - 2 b, \frac{4 \sqrt{z__1}}{{(\sqrt{z__1} + 1)}^{2}}, \frac{4 \sqrt{z__1}}{- z__1 + 2 \sqrt{z__1} + z__2 - 1})}{{(\sqrt{z__1} + 1)}^{2 a} {(1 - \frac{z__2}{{(\sqrt{z__1} + 1)}^{2}})}^{b}}$

$c__1 = a - b + 1 \land c__2 = a \land \frac{z__2}{{(\sqrt{z__1} + 1)}^{2}} \neq 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{(\frac{z__1 + 2 \sqrt{z__1} - z__2 + 1}{z__1 - 2 \sqrt{z__1} - z__2 + 1})}^{a} F_{3} (b - a, a, \frac{1}{2} + b - a, \frac{1}{2} + b - a, 2 b + 1 - 2 a, \frac{4 \sqrt{z__1}}{{(\sqrt{z__1} + 1)}^{2}}, \frac{4 \sqrt{z__1}}{- z__1 + 2 \sqrt{z__1} + z__2 - 1})}{{(\sqrt{z__1} + 1)}^{2 b} {(1 - \frac{z__2}{{(\sqrt{z__1} + 1)}^{2}})}^{a}}$

$c__1 = b - a + 1 \land c__2 = b \land \frac{z__2}{{(\sqrt{z__1} + 1)}^{2}} \neq 1$

sum form

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \sum_{m = 0}^{\infty} \sum_{n = 0}^{\infty} \frac{{(a)}_{m + n} {(b)}_{m + n} {z__1}^{m} {z__2}^{n}}{{(c__1)}_{m} {(c__2)}_{n} m! n!}$

$\sqrt{|z__2|} + \sqrt{|z__1|} < 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \sum_{k = 0}^{\infty} \frac{{(a)}_{k} {(b)}_{k} {}_{2}F_{1} (a + k, b + k; c__2; z__2) {z__1}^{k}}{{(c__1)}_{k} k!}$

$\sqrt{|z__2|} + \sqrt{|z__1|} < 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \sum_{k = 0}^{\infty} \frac{{(a)}_{k} {(b)}_{k} {}_{2}F_{1} (a + k, b + k; c__1; z__1) {z__2}^{k}}{{(c__2)}_{k} k!}$

$\sqrt{|z__2|} + \sqrt{|z__1|} < 1$

series

$series (F_{4} (a, b, c__1, c__2, z__1, z__2), z__1, 4) = {}_{2}F_{1} (a, b; c__2; z__2) + \frac{a b {}_{2}F_{1} (a + 1, b + 1; c__2; z__2)}{c__1} z__1 + \frac{1}{2} \frac{a b (a + 1) (b + 1) {}_{2}F_{1} (a + 2, b + 2; c__2; z__2)}{c__1 (c__1 + 1)} {z__1}^{2} + \frac{1}{6} \frac{a b (a + 1) (b + 1) (a + 2) (b + 2) {}_{2}F_{1} (a + 3, b + 3; c__2; z__2)}{c__1 (c__1 + 1) (c__1 + 2)} {z__1}^{3} + O ({z__1}^{4})$

$series (F_{4} (a, b, c__1, c__2, z__1, z__2), z__2, 4) = {}_{2}F_{1} (a, b; c__1; z__1) + \frac{a b {}_{2}F_{1} (a + 1, b + 1; c__1; z__1)}{c__2} z__2 + \frac{1}{2} \frac{a b (a + 1) (b + 1) {}_{2}F_{1} (a + 2, b + 2; c__1; z__1)}{c__2 (c__2 + 1)} {z__2}^{2} + \frac{1}{6} \frac{a b (a + 1) (b + 1) (a + 2) (b + 2) {}_{2}F_{1} (a + 3, b + 3; c__1; z__1)}{c__2 (c__2 + 1) (c__2 + 2)} {z__2}^{3} + O ({z__2}^{4})$

integral form

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{Γ (c__1) (\int_{0}^{1} \frac{u^{b - 1} {}_{2}F_{1} (\frac{a}{2}, \frac{1}{2} + \frac{a}{2}; c__1; \frac{4 u^{2} z__1 z__2}{{(- 1 + (z__1 + z__2) u)}^{2}})}{{(1 - u)}^{b - c__1 + 1} {(1 + (- z__1 - z__2) u)}^{a}} &DifferentialD; u)}{Γ (b) Γ (c__1 - b)}$

$(\sqrt{|z__1|} + \sqrt{|z__2|} < 1 \land c__1 = c__2 \land 0 < ℜ (b) \land 0 < ℜ (c__1) \land 0 < - ℜ (- c__1 + b)) \lor (\sqrt{|z__1|} + \sqrt{|z__2|} < 1 \land c__1 = c__2 \land 0 < ℜ (b) \land 0 < ℜ (c__2) \land 0 < - ℜ (- c__2 + b))$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{\int_{0}^{\infty} \frac{u^{2 a - 1} {}_{0}F_{1} (; c__1; \frac{z__1 u^{2}}{4}) {}_{0}F_{1} (; c__2; \frac{z__2 u^{2}}{4})}{{&ExponentialE;}^{u}} &DifferentialD; u}{Γ (2 a)}$

$\sqrt{|z__1|} + \sqrt{|z__2|} < 1 \land b = a + \frac{1}{2} \land 0 < ℜ (a) \land ℜ (\sqrt{z__1} + \sqrt{z__2}) < 1 \land ℜ (\sqrt{z__1} - \sqrt{z__2}) < 1 \land - ℜ (\sqrt{z__1} - \sqrt{z__2}) < 1 \land - ℜ (\sqrt{z__1} + \sqrt{z__2}) < 1$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{Γ (c__1) Γ (c__2) (\int_{0}^{1} \int_{0}^{1} \frac{u^{a - 1} v^{b - 1}}{{(1 - u)}^{- c__1 + a + 1} {(1 - v)}^{- c__2 + b + 1} (1 - \frac{u {(1 + \frac{u (- α (z__1, z__2) + z__2)}{α (z__1, z__2)} - v α (z__1, z__2))}^{- c__1 - c__2 + a + 1 + b}}{{(\frac{α (z__1, z__2) - z__2}{α (z__1, z__2)})}^{- c__1 - c__2 + a + 1} {(1 - v α (z__1, z__2))}^{- c__1 - c__2 + b + 1}})} &DifferentialD; u &DifferentialD; v)}{Γ (a) Γ (b) Γ (- a + c__1) Γ (c__2 - b)}$

$\sqrt{|z__1|} + \sqrt{|z__2|} < 1 \land 0 < ℜ (a) \land 0 < ℜ (b) \land 0 < - ℜ (- c__1 + a) \land 0 < - ℜ (- c__2 + b) \land {α (z__1, z__2)}^{2} + (z__1 - z__2 - 1) α (z__1, z__2) + z__2 = 0$

$F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{Γ (c__2) Γ (c__1) (\int_{0}^{1} \int_{0}^{1} \frac{u^{a - 1} v^{b - 1}}{{(1 - u)}^{- c__2 + a + 1} {(1 - v)}^{b - c__1 + 1} (1 - \frac{u {(1 + \frac{u (- α (z__2, z__1) + z__1)}{α (z__2, z__1)} - v α (z__2, z__1))}^{- c__1 - c__2 + a + 1 + b}}{{(\frac{α (z__2, z__1) - z__1}{α (z__2, z__1)})}^{- c__1 - c__2 + a + 1} {(1 - v α (z__2, z__1))}^{- c__1 - c__2 + b + 1}})} &DifferentialD; u &DifferentialD; v)}{Γ (a) Γ (b) Γ (c__2 - a) Γ (c__1 - b)}$

$\sqrt{|z__1|} + \sqrt{|z__2|} < 1 \land 0 < ℜ (a) \land 0 < ℜ (b) \land 0 < - ℜ (- c__2 + a) \land 0 < - ℜ (- c__1 + b) \land {α (z__1, z__2)}^{2} + (z__1 - z__2 - 1) α (z__1, z__2) + z__2 = 0$

differentiation rule

$\frac{\partial}{\partial z__1} F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{a b F_{4} (a + 1, b + 1, c__1 + 1, c__2, z__1, z__2)}{c__1}$

$\frac{\partial^{n}}{\partial {z__1}^{n}} F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{(a)}_{n} {(b)}_{n} F_{4} (n + a, n + b, n + c__1, c__2, z__1, z__2)}{{(c__1)}_{n}}$

$\frac{\partial}{\partial z__2} F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{a b F_{4} (a + 1, b + 1, c__1, c__2 + 1, z__1, z__2)}{c__2}$

$\frac{\partial^{n}}{\partial {z__2}^{n}} F_{4} (a, b, c__1, c__2, z__1, z__2) = \frac{{(a)}_{n} {(b)}_{n} F_{4} (n + a, n + b, c__1, n + c__2, z__1, z__2)}{{(c__2)}_{n}}$

$f (a, b, c__1, c__2, z__1, z__2) = F_{4} (a, b, c__1, c__2, z__1, z__2)$

$\frac{\partial^{2}}{\partial {z__1}^{2}} f (a, b, c__1, c__2, z__1, z__2) = - \frac{2 z__2 (\frac{\partial^{2}}{\partial z__2 \partial z__1} f (a, b, c__1, c__2, z__1, z__2))}{z__1 - 1} - \frac{{z__2}^{2} (\frac{\partial^{2}}{\partial {z__2}^{2}} f (a, b, c__1, c__2, z__1, z__2))}{z__1 (z__1 - 1)} + \frac{((- a - b - 1) z__1 + c__1) (\frac{\partial}{\partial z__1} f (a, b, c__1, c__2, z__1, z__2))}{z__1 (z__1 - 1)} - \frac{z__2 (a + b + 1) (\frac{\partial}{\partial z__2} f (a, b, c__1, c__2, z__1, z__2))}{z__1 (z__1 - 1)} - \frac{a b f (a, b, c__1, c__2, z__1, z__2)}{z__1 (z__1 - 1)}$

$\frac{\partial^{2}}{\partial z__2 \partial z__1} f (a, b, c__1, c__2, z__1, z__2) = - \frac{z__1 (\frac{\partial^{2}}{\partial {z__1}^{2}} f (a, b, c__1, c__2, z__1, z__2))}{2 z__2} - \frac{(z__2 - 1) (\frac{\partial^{2}}{\partial {z__2}^{2}} f (a, b, c__1, c__2, z__1, z__2))}{2 z__1} - \frac{(a + b + 1) (\frac{\partial}{\partial z__1} f (a, b, c__1, c__2, z__1, z__2))}{2 z__2} + \frac{((- a - b - 1) z__2 + c__2) (\frac{\partial}{\partial z__2} f (a, b, c__1, c__2, z__1, z__2))}{2 z__1 z__2} - \frac{a b f (a, b, c__1, c__2, z__1, z__2)}{2 z__1 z__2}$

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 AppellF4 command was introduced in Maple 2017.

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

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