Items tagged with is

I want to solve for the roots of a polynomial, such as a x^2+b x + c = 0, for which the output is only the positive root. All coefficients/variables in the polynomial are positive. 

Recently, someone posted an answer to a question where at some point they performed this task and their solution was really slick. But I can't find it. The answer used either solve, or eval or something like that. (Yes, I did perform a search via the MaplePrimes search before asking this question.) 

 

Dear Friends, 

I would appreciate your help in resolving some issues. Let me describe my dummy code and the issues I am having. 

1. I have two functions f(x,a) and g(x,a) are well defined. 

2. I need to get roots of f(x,a). I am using the command soln := Roots(f(x,a)) which gives me a list of all possible values of x. 

3. I need to choose only one element -- say x* -- from the list "soln" such that x*=argmax{g(x,a) | g(x,a)>0 for x in soln}.

I am not able to find a technique (i) to evaluate g(x,a) for all elements of the list "soln", (ii) and select x* that maximizes g(x,a). 

Picking and choosing elements of the list one by one is difficult since the number of elements in the list can vary with parameter "a" and it would complicate the matter in numerical studies. 

I would sincerely appreciate any inputs in this regard. 

Thank you,

Omkar

 

im a huge fan of using the is boolean function maple has in my studies, and in as much as the CAS is a million times smarter than me, i just wanted to show an example of a situation where i am reluctant to use this feature. (see the uploaded worksheet)

 

perhaps someone with more  programming ability/experience than me can explain how it works, as so i can have a means of knowing whether or not it is appropriate to use in each situation.IS_function_problem_example.mw

Let

and f=

The elements of W are none zero. I want a procedure that return "true" if f is none zero w.r.t. W and return

"false" otherwise.

I am very confused about function is():

 

1. is(a-b=0) assuming a::real,b::real,a^2=b^2,a>0,b>0;

2. is(a-b=0) assuming a::real,b::real,a^2=b^2,a>=0,b>=0;

 

the first command gives me false, and the second command gives me true...

 

any thoughts?

Page 1 of 1