Carl Love

Carl Love

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12 years, 132 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

MaplePrimes Activity


These are replies submitted by Carl Love

@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.

@Stavros Okay. I'd prefer if you post any followup questions right here, or in a new Question if they can stand alone.

@Stavros Okay. I'd prefer if you post any followup questions right here, or in a new Question if they can stand alone.

@acer I thought that someone might catch that, thinking that I meant "assume" in the mathematical sense. I regret not editing it. What I meant was "In order for Maple to perform the expansion, it is necessary to assume that they (the factors) are positive." (Hoping that you understand the distinction that I make with boldface.) But you do bring up two other possibilities for assumptions that I hadn't considered: If you assume that a is integer, or that all factors except for one are positive, it will also perform the expansion:

expand((A*B*C)^a) assuming a::integer;

expand((A*B*C)^a) assuming A > 0, B > 0;

@acer I thought that someone might catch that, thinking that I meant "assume" in the mathematical sense. I regret not editing it. What I meant was "In order for Maple to perform the expansion, it is necessary to assume that they (the factors) are positive." (Hoping that you understand the distinction that I make with boldface.) But you do bring up two other possibilities for assumptions that I hadn't considered: If you assume that a is integer, or that all factors except for one are positive, it will also perform the expansion:

expand((A*B*C)^a) assuming a::integer;

expand((A*B*C)^a) assuming A > 0, B > 0;

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