OK, this one seems to be true (at least as series: outside the circle of convergence, branch cuts may be chosen differently)..

> eq:=hypergeom([a,b],[c],z) 
         = (1-z)^(-a)*hypergeom([a,c-b],[c],z/(z-1));

> map(convert,eq,FormalPowerSeries,z);

Sum(pochhammer(b,k)*pochhammer(a,k)/pochhammer(c,k)/k!*z^k,k = 0 .. infinity) = Sum(pochhammer(b,k)*pochhammer(a,k)/pochhammer(c,k)/k!*z^k,k = 0 .. infinity)

To convert the right side into the left:

> convert(convert(rhs(eq),FormalPowerSeries,z),hypergeom);

hypergeom([a, b],[c],z)

The other way doesn't seem so easy. 

First, change variables: if u = z/(z-1), then z = u/(u-1).

> A:= eval(lhs(eq), z = u/(u-1));

A := hypergeom([a, b],[c],u/(u-1))

Multiply by (1-u)^(-a), then convert to Formal Power Series and then to hypergeom:

> A2:= A * (1-u)^(-a);

> convert(convert(A2, FormalPowerSeries,u), hypergeom);

hypergeom([a, c-b],[c],u)

Now divide by (1-u)^(-a) and change variables back to z.

> simplify(eval(%/(1-u)^(-a), u = z/(z-1)));

hypergeom([a, c-b],[c],z/(z-1))*(-1/(z-1))^(a)

And for the final bit of simplification:

> simplify(expand(%)) assuming z<1, a::posint;

(1-z)^(-a)*hypergeom([a, c-b],[c],z/(z-1))

Probably the simplest way, if all you want is to enforce x >= 0, is to replace x by t^2, and then when you get the result take x = sqrt(t).

Probably the simplest way, if all you want is to enforce x >= 0, is to replace x by t^2, and then when you get the result take x = sqrt(t).

OK, then.

> eqs:= {f=x*(1-g/y*(1-g))^2, g=x*(1-f/y*(1-f))^2};

Write g = f + d, and eliminate the variable f.

> eqs:= map(numer @ (lhs-rhs), subs(g=f+d,eqs));
  factor(resultant(op(eqs,f)));

x^4*d^4*(50*x^2*y^4*d^6+160*x^2*y^6*d^2+100*x*y^6*d^2+34*x^2*d^2*y^4+16*x^4*d^10*y-4*x^3*d^8*y^2-16*x^3*d^6*y^3+125*y^8+256*x^2*y^7+x*x^2*d^4*y^5+x^4*d^4-
68*x^2*d^4*y^4-32*x^3*d^4*y^3+4*x^3*d^6*y^2+16*x^2*y^5-128*x^2*y^6-12*x*y^6
-80*x*y^7+4*x^3*d^4*y^2-4*x^3*d^2*y^2-256*x^4*d^4*y^3+256*x^4*d^4*y^4
+64*x^3*d^4*y^4-48*x^4*d^8*y+48*x^3*y^3*d^2+96*x^4*d^8*y^2-16*x^4*d^4*y
+48*x^4*d^6*y-4*x^4*d^10+6*x^4*d^8-4*x^4*d^6-192*x^4*d^6*y^2+256*x^4*d^6*y^3
-176*x^2*y^5*d^2+96*x^4*d^4*y^2+256*x^3*d^2*y^5-192*x^3*d^2*y^4+x^4*d^12)

So you'll have a solution with d=0, and one or more other solutions where d is a root
of a rather unpleasant polynomial of degree 12.  Deciding where there are real solutions may not be easy.  However, since the coefficient of d^12 is x^4 and the
coefficient of d^0 is -128*x^2*y^6+256*x^2*y^7+16*y^5*x^2-12*x*y^6-80*x*y^7+125*y^8,
we can certainly say that when the latter is negative there will be real solutions with d <> 0.

I suspect your equations have a misprint, because as currently stated you certainly wouldn't expect F = G unless x = y.  Perhaps you meant

G:=(x,y)->x*(1-F(x,y)/y*(1-F(x,y)))^2;

Software Change Request, i.e. a bug report.

Software Change Request, i.e. a bug report.

In fact, on further investigation it seems that diff(H,x) - h = 3*Pi/4.

In fact, on further investigation it seems that diff(H,x) - h = 3*Pi/4.

It's not just a branch-cut problem affecting the definite integral: the antiderivative is wrong.

Consider:

> h := arctan( tan(x)^2 );
   H:= int(h, x);
   evalf(eval([h, diff(H,x)], x = 1));

[1.179746269, 3.535940759+.44e-9*I]

> plot([Re(H),h],x=0..Pi, colour=[red,blue], 
    discont=true,adaptive=false);

Clearly h is not the derivative of H on any of the three intervals..

I'm submitting an SCR.

It's not just a branch-cut problem affecting the definite integral: the antiderivative is wrong.

Consider:

> h := arctan( tan(x)^2 );
   H:= int(h, x);
   evalf(eval([h, diff(H,x)], x = 1));

[1.179746269, 3.535940759+.44e-9*I]

> plot([Re(H),h],x=0..Pi, colour=[red,blue], 
    discont=true,adaptive=false);

Clearly h is not the derivative of H on any of the three intervals..

I'm submitting an SCR.

First of all, BBP is for hexadecimal digits, not decimal digits.

You might look at <http://www.colab.sfu.ca/PiDay/3_14/Pi1.html>

which, among other things, contains a Maple implementation of the BBP algorithm.

First of all, BBP is for hexadecimal digits, not decimal digits.

You might look at <http://www.colab.sfu.ca/PiDay/3_14/Pi1.html>

which, among other things, contains a Maple implementation of the BBP algorithm.

The random number generator is the Mersenne Twister.  Its state consists of a table of 625 four-byte integers.  That is what is reset.  See ?RandomTools,MersenneTwister.

For more information on the Mersenne Twister you might try

<http://en.wikipedia.org/wiki/Mersenne_twister>

... just one more example of why 2D input should be avoided, especially by inexperienced users.