Can you post the equations?

Are you getting an error message from pdsolve, or no result at all?

A simplified version of your plotting command:

plot(
     [zip(`[]`,X1, Y1), zip(`[]`, X2, Y2)],
     color= [blue, red],
     style= point, symbol= solidcircle, symbolsize= 8,
     view= [0 .. 2, 0 .. 1], scaling= constrained
);

A simplified version of your plotting command:

plot(
     [zip(`[]`,X1, Y1), zip(`[]`, X2, Y2)],
     color= [blue, red],
     style= point, symbol= solidcircle, symbolsize= 8,
     view= [0 .. 2, 0 .. 1], scaling= constrained
);

It is not clear what you mean by "solve", "solving for", "equation", and "plotting". Do you mean

to solve the equations x/(x+1) = .5, x/(x+1) = .9, etc., for x?

Or do you mean

to evaluate the expression x/(x+1) at x = .5, .9, etc.?

And, in either of the above situations, it is not clear what "plotting" means. We can plot expressions (but not equations) with one variable. We can plot equations with two variables. We can plot a sequence (not a series) by using the index number into the sequence as one of the coordinates.

@andy.zhou.nuaa I recommend you download the free DirectSearch package from the Maple Applications Center. It contains a solver similar to fsolve, but with more flexibility in the precision.

@Markiyan Hirnyk He means the Answer by Alec Milhailovs to which you referred.

@Markiyan Hirnyk He means the Answer by Alec Milhailovs to which you referred.

@Sergio Parreiras Yes, I see what you mean. You can use simplify (non symbolic) after simplify(..., symbolic). I'm sure someone would take a closer look at how to simplify the expression if you post the code for the expression.

@Sergio Parreiras Yes, I see what you mean. You can use simplify (non symbolic) after simplify(..., symbolic). I'm sure someone would take a closer look at how to simplify the expression if you post the code for the expression.

In the future, please post your code in plain text format or upload the worksheet. The "picture" of the code given in your Question is difficult to enter into Maple.

Also, please note the distinction between the Posts section and the Questions section of MaplePrimes. I moved this item from Posts to Questions.

Please post the equation eq that you used for the toy example. Plain text woud be preferable.

@Stavros The solutions are highly symmetric. Xk1 = Xk4 for all k. Xk3=0 for all k. So Xk1*Xk4=0 implies Xk1 = 0. The numerators of Xk can be obtained from the numerators of X1 by a simple permutation of [a,b,c,d].

Does that answer your questions?

@Stavros The solutions are highly symmetric. Xk1 = Xk4 for all k. Xk3=0 for all k. So Xk1*Xk4=0 implies Xk1 = 0. The numerators of Xk can be obtained from the numerators of X1 by a simple permutation of [a,b,c,d].

Does that answer your questions?

@Markiyan Hirnyk Thanks for finding it. So, to answer the original Question, f(x) = (2*x)!/x!/(x+1)!  Maple can verify this solution, even though rsolve won't find it:

f:= x-> (2*x)!/x!/(x+1)!:
g:= x-> sum(f(k)*(f(x-k-1), k= 0..x-1):
simplify(f(x) - g(x));

                                   0

And with this form, the conjectures are obvious.