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.

@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.

The phenomenon that you describe is not strange to me. The fsolve most likely failed because it could not guarantee the desired amount of precision. Unlike solve, fsolve does not try to eliminate variables. You discovered a way to eliminate three of the variables with solve, leaving a single equation for fsolve. It is much easier to guarantee the precision numerically solving a single equation.

@Sergio Parreiras In your large expression, the thing superficially appears to be a common denominator is theta - 1. That is not a denominator of the overall expression; it a denominator of exponents. Since the exponents are exponents of terms in a sum, there's no obvious way they can be simplified.

@Sergio Parreiras In your large expression, the thing superficially appears to be a common denominator is theta - 1. That is not a denominator of the overall expression; it a denominator of exponents. Since the exponents are exponents of terms in a sum, there's no obvious way they can be simplified.