lemelinm

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14 years, 181 days

 

 

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Mario Lemelin
Maple 14.00 Win 7 64 bits
Maple 14.00 Ubuntu 10,04 64 bits
messagerie : mario.lemelin@cgocable.ca téléphone :  (819) 376-0987

MaplePrimes Activity


These are answers submitted by lemelinm

HI Robert,

 

When the delay will be over, I would be very interested by one exemple.  Could you send it directly to me?.

 

Mario Lemelin

mario.lemelin@cgocable.ca

Yes I understsand your point.  But at least it should be done if I ask to simplify symbolic.  Instead I receive something even worst.

But if I do a  plot in the region of the numerics solutions, I see that they will be 3 reals roots.  Is there a way to have the exact solution after knowingthat in a inexpensive way?

plot(x^3-6*x^2-7*x+58, x = -4 .. 6);

 

Mario Lemelin

mario.lemelin@cgocable.ca

This is very interesting.  If you do solve(x^3-6*x^2-7*x+58 = 0);, you obtain square root of a complex number.  I decided to take part by part the answer for the first root.  Of course, since I do an evalf, I am sure that they will be an approximation, not the right answer.  But as you can see, when I sum up part1+part2+2,  I get a real number (with 0. as imaginary part). 

What I am wondering is why Maple cannot simplify that fact immediately?

 

eq1 := (1/3)*(-378+(3*I)*sqrt(4701))^(1/3);

evalc(eq1);

part1 := evalf(%);

eq2 := 19/(-378+(3*I)*sqrt(4701))^(1/3);

evalc(eq2);

part2 := evalf(%);

tot := part1+part2+2;

 

Mario Lemelin

mario.lemelin@cgocable.ca

Here a way to do it in Maple 11:

with(Student:-VectorCalculus):

P:=<1,2>;

Q:=<1,-1>;

QP:=P-Q;

OR:=1/3*P;

and you see the result with;

plots[pointplot]([[1, 2], [1, -1], [0, 0], [1/3,2/3]], color = [red, blue, orange, black], view = [0 .. 1, -1 .. 2]);

 

Mario Lemelin

mario.lemelin@cgocable.ca

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