Advanced Search

gulliet

281 Reputation

7 Badges

20 years, 5 days

Social Networks and Content at Maplesoft.com

Contact gulliet

MaplePrimes Activity


These are replies submitted by gulliet

You can use convert() to go...

gulliet 281
March 01 2009
0 0
You can use convert() to go from one form of an expression to another form that is mathematically related to the original one. For instance, you can express the result as a product of factorials in place of the Beta() function. 97_intbetaconvertgammafactorial.jpg Regards, -- Jean-Marc

You can use convert() to go...

gulliet 281
March 01 2009
You can use convert() to go from one form of an expression to another form that is mathematically related to the original one. For instance, you can express the result as a product of factorials in place of the Beta() function. 97_intbetaconvertgammafactorial.jpg Regards, -- Jean-Marc

John, Thank you for...

gulliet 281
February 28 2009
0 0
John, Thank you for clarifying this point. Of course you are right: the assuming clause does not change anything in my first three examples, as I could have checked by myself, 97_solveineqreabydefault.jpg or pay more attendion to the fine print of the documentation that states:
In general, variables and parameters will automatically be assumed to be real-valued in the context of one or more inequalities.
(solve/ineqs -> Description -> 2nd line) Regards, -- Jean-Marc

John, Thank you for...

gulliet 281
February 28 2009
John, Thank you for clarifying this point. Of course you are right: the assuming clause does not change anything in my first three examples, as I could have checked by myself, 97_solveineqreabydefault.jpg or pay more attendion to the fine print of the documentation that states:
In general, variables and parameters will automatically be assumed to be real-valued in the context of one or more inequalities.
(solve/ineqs -> Description -> 2nd line) Regards, -- Jean-Marc

I use Filtered HTML, copy...

gulliet 281
February 27 2009
0 0
I use Filtered HTML, copy and past my code in < pre > and < /pre > directives and use the file manager to include some copy of the screen in jpeg format (full size, the code for medium and small seems to be broken). Regards, --Jean-Marc

I use Filtered HTML, copy...

gulliet 281
February 27 2009
I use Filtered HTML, copy and past my code in < pre > and < /pre > directives and use the file manager to include some copy of the screen in jpeg format (full size, the code for medium and small seems to be broken). Regards, --Jean-Marc

Does Maplesoft care?...

gulliet 281
February 27 2009
0 0
[ why is the Maple tag not working? ]
The real question might be, "Does Maplesoft care at all about the issue?" The problem (and possibly some related issues about links to images being broken) has been reported in this very forum several weeks ago and ... nothing, not a word from Maplesoft (at least to confirm whether the problem exists). Regards, --Jean-Marc

Does Maplesoft care?...

gulliet 281
February 27 2009
[ why is the Maple tag not working? ]
The real question might be, "Does Maplesoft care at all about the issue?" The problem (and possibly some related issues about links to images being broken) has been reported in this very forum several weeks ago and ... nothing, not a word from Maplesoft (at least to confirm whether the problem exists). Regards, --Jean-Marc

You might be interested in the function ...

gulliet 281
February 27 2009
0 0
Also, you might be interested in the function time(). Below, we have added a counter for the number of iterations and we display the overall cpu time taken so far. 
S := []: idx := 0: tm := time():
for alpha from 0 to 3 do 
  for beta from 0 to 2 do 
    idx := idx + 1;
    printf("Iter #%d, alpha = %d, beta = %d", idx, alpha, beta);
    S := [op(S), fsolve({eq1,eq2},{x,y})];
    printf(", %f seconds.\n", time() - tm); 
  end do; 
end do; 

Iter #1, alpha = 0, beta = 0, 0.602000 seconds.
Iter #2, alpha = 0, beta = 1, 1.120000 seconds.
Iter #3, alpha = 0, beta = 2, 1.648000 seconds.
Iter #4, alpha = 1, beta = 0, 2.283000 seconds.
Iter #5, alpha = 1, beta = 1, 2.925000 seconds.
Iter #6, alpha = 1, beta = 2, 3.523000 seconds.
Regards, --Jean-Marc

You might be interested in the function ...

gulliet 281
February 27 2009
Also, you might be interested in the function time(). Below, we have added a counter for the number of iterations and we display the overall cpu time taken so far. 
S := []: idx := 0: tm := time():
for alpha from 0 to 3 do 
  for beta from 0 to 2 do 
    idx := idx + 1;
    printf("Iter #%d, alpha = %d, beta = %d", idx, alpha, beta);
    S := [op(S), fsolve({eq1,eq2},{x,y})];
    printf(", %f seconds.\n", time() - tm); 
  end do; 
end do; 

Iter #1, alpha = 0, beta = 0, 0.602000 seconds.
Iter #2, alpha = 0, beta = 1, 1.120000 seconds.
Iter #3, alpha = 0, beta = 2, 1.648000 seconds.
Iter #4, alpha = 1, beta = 0, 2.283000 seconds.
Iter #5, alpha = 1, beta = 1, 2.925000 seconds.
Iter #6, alpha = 1, beta = 2, 3.523000 seconds.
Regards, --Jean-Marc

Writing the main sequence as...

gulliet 281
February 27 2009
0 0
Writing the main sequence as follows should help you to het an idea of what the program is doing and how much computations (loops) have been done. 
restart:with(plots):
mu:=0.1:
rs:=((x+mu)^2+y^2)^0.5:
re:=((x+mu-1)^2+y^2)^0.5:
Q:=beta/10*(1-mu)*(cos(alpha/100))^2:
S:=(x+mu)*cos(alpha/100)+y*sin(alpha/100):
T:=-(x+mu)*sin(alpha/100)+y*cos(alpha/100):
eq1:=(1-mu)*(x+mu)/rs^3+mu*(x-mu+1)/re^3-x=Q*S/rs^3:
eq2:=(1-mu)*y/rs^3+mu*y/re^3-y=Q*T/rs^3:
S := []: 
for alpha from 0 to 9 do 
  for beta from 0 to 2 do 
    printf("Computing alpha = %d and beta = %d\n", alpha, beta);
    S := [op(S), fsolve({eq1,eq2},{x,y})]; 
  end do; 
end do; 
pts:=seq(subs(i, [x,y]), i=S):
P:=pointplot({pts},axes=boxed,symbol=circle, color=black):
P;

Computing alpha = 0 and beta = 0
Computing alpha = 0 and beta = 1
Computing alpha = 0 and beta = 2
Computing alpha = 1 and beta = 0
Computing alpha = 1 and beta = 1
Computing alpha = 1 and beta = 2
Computing alpha = 2 and beta = 0

[.... deleted rest of the the output and the resulting plot ....]
Regards, --Jean-Marc

Writing the main sequence as...

gulliet 281
February 27 2009
Writing the main sequence as follows should help you to het an idea of what the program is doing and how much computations (loops) have been done. 
restart:with(plots):
mu:=0.1:
rs:=((x+mu)^2+y^2)^0.5:
re:=((x+mu-1)^2+y^2)^0.5:
Q:=beta/10*(1-mu)*(cos(alpha/100))^2:
S:=(x+mu)*cos(alpha/100)+y*sin(alpha/100):
T:=-(x+mu)*sin(alpha/100)+y*cos(alpha/100):
eq1:=(1-mu)*(x+mu)/rs^3+mu*(x-mu+1)/re^3-x=Q*S/rs^3:
eq2:=(1-mu)*y/rs^3+mu*y/re^3-y=Q*T/rs^3:
S := []: 
for alpha from 0 to 9 do 
  for beta from 0 to 2 do 
    printf("Computing alpha = %d and beta = %d\n", alpha, beta);
    S := [op(S), fsolve({eq1,eq2},{x,y})]; 
  end do; 
end do; 
pts:=seq(subs(i, [x,y]), i=S):
P:=pointplot({pts},axes=boxed,symbol=circle, color=black):
P;

Computing alpha = 0 and beta = 0
Computing alpha = 0 and beta = 1
Computing alpha = 0 and beta = 2
Computing alpha = 1 and beta = 0
Computing alpha = 1 and beta = 1
Computing alpha = 1 and beta = 2
Computing alpha = 2 and beta = 0

[.... deleted rest of the the output and the resulting plot ....]
Regards, --Jean-Marc

Not sure whether this is...

gulliet 281
February 27 2009
0 0
Not sure whether this is what you are looking for: 
plots:-implicitplot((x^2-y^2)/(x^2+y^2), x = -.5 .. .5, 
y = -.5 .. .5, grid = [30, 30]);
97_rationalinplicitplot.jpeg Regards, --Jean-Marc

Not sure whether this is...

gulliet 281
February 27 2009
Not sure whether this is what you are looking for: 
plots:-implicitplot((x^2-y^2)/(x^2+y^2), x = -.5 .. .5, 
y = -.5 .. .5, grid = [30, 30]);
97_rationalinplicitplot.jpeg Regards, --Jean-Marc

Factorization over the complex field...

gulliet 281
February 27 2009
To factorize a polynomial, you can use the function factor() and specify the field extension over which to factor in its second argument. For instance, 
> factor(x^2+36, complex);

> factor(x^2+36, I);

> help("factor");
Regards, --Jean-Marc
2 3 4 5 6 7 8 Last Page 4 of 15