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These are answers submitted by DJKeenan

I tried looking the series up in Hansen's Table, but no luck. Given how simple the definition, this seems like evidence that no one knows a closed form. Another approach, admittedly unlikely, is to use identify(evalf(sum(...))). But this too gave no answer.
If the ranges for r and zeta are, for example, -2..3 and -6..7, then how about something like this? with(Optimization): g:= zeta -> Maximize(a(r, zeta), r = -2 .. 3, method = 'branchandbound')[1]; Minimize(g, -6 .. 7, method='branchandbound')[1]; Note that method is used to get a global solution.
For some general background, it is worth noting that this is the “Monty Hall Problem”.
How about this? Optimization[Maximize](objective(x1, x2, x3, x4, x5, x6, x7, x8, x9), seq(x || i = 0 .. 7, i = 1 .. 9), initialpoint = [seq(x || i = some_point[i], i = 1 .. 9)]); [0.99999999999979594, [x1 = 0.249910963699999988, x2 = 2.07892373100000016, x3 = 2.25241025000000006, x4 = 5.71395350600000018, x7 = 3.14159233835433582, x8 = 3.14159233239993796, x9 = 3.14159261604398532, x5 = 0.347503411099999992, x6 = 1.57079632730711150 ]]
To add to Robert Israel's answer, if you want only solutions in R4, try RealDomain. use RealDomain in solve(F - [x1,x2,x3,x4]) end use;           {x3 = 0, x1 = 0, x2 = 0, x4 = 0}
Does changecoords help? This can do conversion to cartesian, from just about anything. changecoords([r, theta], [r, theta], 'polar')           [r*cos(theta), r*sin(theta)] changecoords([r, phi, theta], [r, phi, theta], 'spherical')           [r*sin(theta)*cos(`ϕ`), r*sin(theta)*sin(`ϕ`), r*cos(theta)]
… but about exp(Pi), this is transcendental, by the the Gelfond–Schneider theorem (which solved the seventh of Hilbert's problems). Gelfond also had a separate proof for exp(Pi), a few years earlier (cited here).
A good general guideline for GUIs is that (almost) anything that can be done via point-and-click should also be doable via some sequence of keystrokes.
Have you looked at Statistics:-PointPlot? If x and y are lists, then Statistics:-PointPlot(y,xcoords=x) should work well. I just tried it with 5000 points (in Maple 11), and it worked fine; response time was <2 seconds.
Here is a simpler example that illustrates the issue. fsolve(Int(t, t = 0 .. x) = 1); Error, (in fsolve) number of equations, 1, does not match number of variables, 2 But there is an easy solution: fsolve(x->Int(t, t = 0 .. x) = 1);
I have posted more about this in Product Suggestions.
Is this a homework assignment? (If so, does your teacher approve of asking on MaplePrimes?)
Brian, for some thoughts about using documents and the Standard Interface in Maple 11, see my blog entry related to this.
Another approach to the problem is to put the computation into its own section (or subsection). Then collapse the section.
There seems to be some terminological confusion. Yath is asking for all minimal cuts, whereas mincut finds a minimum cut. Maple does not have a built-in function for the "minimal cuts" problem. I googled a little and found some papers that have algorithms; here are two doi: 10.1007/BF01074775 10.1109/LCN.2000.891015 There might be others; I only spent a few minutes googling. Too, there might be different approaches that take advantage of some other functions in Maple.
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