261 Reputation

8 Badges

21 years, 39 days

MaplePrimes Activity

These are questions asked by eclark

I solved two inequalities using

[solve]({f(theta) >0, f(theta) <=Pi}, theta)

for some complicated function f of theta, an solve gives me a list of intervals such as

[{theta <= Pi-arccos(-3/4+(1/4)*sqrt(13+16*sqrt(2))), arccos(3/4) < theta},

{arccos(-3/4+(1/4)*sqrt(13+16*sqrt(2)))+Pi <= theta, theta < 2*Pi-arccos(3/4)},

{theta <= 3*Pi-arccos(-3/4+(1/4)*sqrt(13+16*sqrt(2))), 2*Pi+arccos(3/4) < theta}]

I would like to convert each of the intervals to the range form a..b so I can plot the function over this range.

Is there a comand to convert something like {theta <=b, a < theta} to
a..b (assuming a and b are real numbers in some form)?  I can think of a rather complicated procedure to do it, but there should be something simpler.


Consider the following expression obtained from the solve command: Note that this uses the two variable arctan function.

p:=arctan((-cos(theta)^3-(1/2)*cos(theta)^2-(1/2)*cos(theta)*((2*cos(theta)+1)*(2*cos(theta)-3)*(cos(theta)+1)^2)^(1/2)+2*cos(theta)-(1/2)*((2*cos(theta)+1)*(2*cos(theta)-3)*(cos(theta)+1)^2)^(1/2)+3/2)^(1/2), -(1/2)*cos(theta)-1/2-(1/2)*((2*cos(theta)+1)*(2*cos(theta)-3)*(cos(theta)+1)^2)^(1/2)):

#ploting the expression shows a non-zero value at theta = Pi,  however if I convert p to a function using


# then f(Pi);  gives a value of 0

#On the other hand maximize(p,theta=3*Pi/4..5*Pi/4,location); shows a non-zero value of 4*Pi/5 at theta = Pi,  which agrees with the plot of p, namely, it returns:


{[{theta = Pi}, -arctan((10-2*5^(1/2))^(1/2)/(5^(1/2)+1))+Pi]}

Is this a bug? Or what?



I have just begun thinking of trying to make some mathematically defined objects using a 3d printer. I would be happy to hear from anyone who has done this using Maple to prepare input. Pointers for a novice in 3d printing would be appreciated.  I have access to a MakerBot Replicator 2. But the people who have it have only used it to scan objects and make 3d copies of them. 


I need a procedure to simplify rational functions r(x,y) = p(x,y)/q(x,y)  where p(x,y) and q(x,y) are polynomials in Z/(2)[x,y].

For example suppose r(x,y) = (1+x+y)/(x^2+x*y+y+1) then mod 2 this reduces to 1/(1+x) since 

            Factor(x^2+x*y+y+1) mod 2 = (1 + x + y) (1 + x)

I wrote a procedure to do this using Factors mod 2, but I wonder if there is a builtin procedure to make such simplifications that might be more efficient.  For now 2 variable is fine, but later I may want more than 2 variables.




The question of how to do matrix operations over a finite field with p^k elements k>1 was raised in this forum in April 27 2010. I wonder if there has been any progress since then for recent versions of Maple.

I am expecially interested in finding the rank of a matrix over a field of order p^k, k > 1.


1 2 3 4 Page 2 of 4