Greetings. Not knowing what your equation looks like,
I will use an example. This is how I would check to see
if the answer that Maple gave me is correct.
> restart:
> eq1:=(x^3+x^2+x+1=0);
> sol:=solve(eq1);
This gives me 3 solutions for the above equation (eq1). Below I
check each of the answers in turn.
> subs(x=sol[1],eq1);
> subs(x=sol[2],eq1);
> subs(x=sol[3],eq1);
As expected, I get 0=0 for each.
Regards,
Georgios Kokovidis
Greetings. Not knowing what your equation looks like,
I will use an example. This is how I would check to see
if the answer that Maple gave me is correct.
> restart:
> eq1:=(x^3+x^2+x+1=0);
> sol:=solve(eq1);
This gives me 3 solutions for the above equation (eq1). Below I
check each of the answers in turn.
> subs(x=sol[1],eq1);
> subs(x=sol[2],eq1);
> subs(x=sol[3],eq1);
As expected, I get 0=0 for each.
Regards,
Georgios Kokovidis
y:=10+5*sin(theta+Pi/60);
plot(y,theta);
y2:= 2-3*cos( theta-Pi/4);
plot(y2,theta);
Remember that Maple computes using radians by default,
so I converted your degrees to radians first. One more
thing to remember is that lower case "pi" is not the
same and upper case "Pi". Make sure you use the upper
case version in your plot calls.
Type ?greek at the command prompt to bring up the help page
that shows you how to enter greek characters. ?plot will
show you many examples for plotting.
Regards,
Georgios Kokovidis
My simple 6x6 example gives an answer that is 271 pages long.
This does not seem to be practicle. Assuming you do get an
answer after waiting for a couple of weeks (again, not practicle),
what will you do with it. It will be thousands of pages long.
Is there a reason why you would like to do this. I am curious.
If you use numbers instead of symbolic variables, you will get
an answer in a short period of time. What is your application that
requires this? Again, just curious. I have not seen anything like
this before.
Regards,
Georgios Kokovidis
You are not getting and error because the program is running
just fine. The problem is with the size of your matrix. With
symbolic elements, as the size of your matrix grows, so does the
time in wich is takes to solve the problem, because of issues
related to storage. If my understanding is correct, the number
of elements needed to store a matrix of size N*N is N! With a
matrix of 6x6 elements, I got a solution in under a minute on
a laptop running a 2GHz intel with 512Meg ram. On the same machine,
if I increase the size to 7x7, after 5 minutes, still no answer.
For a 12x12 I doubt you will ever see a solution, and even if you do
it will be pages and pages worth of output that will not make any
sense. Below is what I used as my test case to measure this. When
x is equal to 6 I get an answer. When x is equal to 7, I end up stopping the process because it is taking to long.
> restart:with(LinearAlgebra):
> time1:=time():
> x:=6;
> [seq(s1||j,j=1..x)];
> symMatrix:=Matrix([seq([seq(a||i||j,j=1..x)],i=1..x)]);
> ans:=MatrixInverse(symMatrix):totaltime:=time()-time1;
Regards,
Georgios Kokovidis
The only way that I know that works for this would be to
create multiple sets of 3D plots that are continuous over
a given range, and then combine all of them using the "display"
command. Could you post the function that you are working with?
Regards,
Georgios Kokovidis
The only way that I know that works for this would be to
create multiple sets of 3D plots that are continuous over
a given range, and then combine all of them using the "display"
command. Could you post the function that you are working with?
Regards,
Georgios Kokovidis
> restart:
> with(plots):
> fx := cos(x)+2*sin(x):
> p0:=plot(fx, x = 0 .. 2*Pi, y = -3 .. 3,
xtickmarks =[0.785="p/4",1.57="p/2",2.36="3p/4",
3.14="p",3.92="5p/4",4.71="3p/2",5.50="7p/4",
6.28="2p"],axesfont=[SYMBOL,12]):
> p1 := plot([arctan(2), y, y = -3 .. 3]):
> display(p0, p1);
Regards,
Georgios Kokovidis
Below is an explanation given from ?sign at
the Maple prompt. Maybe it can shed some light.
plot(sign(x), x=-1..1);
Notice that the plot results in the line y=1.
This occurs because it computes the constant
sign(x)=1 and plots that. To get the expected plot,
enclose sign(x) in right-single quotes.
Regards,
Georgios Kokovidis
Try this below. Don't forget that the coefficients are
with respect to a certain variable, "x", for your case, so the
answers you get are with respect to this variable.
> restart:
> eq1:=(a+b*x)=(3+4*x);
> coefficients_right:=coeff(rhs(eq1),x);
> coefficients_left:=coeff(lhs(eq1),x);
Regards,
Georgios Kokovidis
Greetings. From the worksheet that you downloaded, if
you change the colons (:) at the end of each statement to
semicolons (;), then you will see the results of each of the
calculations. Do not change the semicolons at the end of the
worksheet for p1 and p2. If you do that the plot will not work.
Don't worry about the "Warning" message. You
will see what I mean when you get the full output from the
solve statement. The allvalues will display the actual solution
of the solve statement, and the evalf will convert it to floating
point notation. At that point, you cut and paste the output of
the evalf statement and plot it. The reason for solving the
equations is to give you the intersection and put them into a
a parametric form that can be plotted.
Hope this helps.
//Georgios
Greetings. From the worksheet that you downloaded, if
you change the colons (:) at the end of each statement to
semicolons (;), then you will see the results of each of the
calculations. Do not change the semicolons at the end of the
worksheet for p1 and p2. If you do that the plot will not work.
Don't worry about the "Warning" message. You
will see what I mean when you get the full output from the
solve statement. The allvalues will display the actual solution
of the solve statement, and the evalf will convert it to floating
point notation. At that point, you cut and paste the output of
the evalf statement and plot it. The reason for solving the
equations is to give you the intersection and put them into a
a parametric form that can be plotted.
Hope this helps.
//Georgios
Check out the following link below. Scroll down to section
7.1. There are examples there to get you started.
http://www.mhhe.com/math/advmath/rosen/r5/student/ch02/maple.html
Regards,
Georgios Kokovidis
This will plot and solve for both solutions. Just cut
and paste it into your worksheet.
> restart:
> with(plots):
> a:=implicitplot(4*(x-1)^2+(y-2)^2 = 5,x=-1..2,y=-1..1):
> b:=implicitplot(x^2+4*y^2 = 1,x=-1..2,y=-1..1):
> display(a,b);
> eq1:=4*(x-1)^2+(y-2)^2 =5;
> eq2:=x^2+4*y^2 =1;
> equations:={eq1,eq2};
> vars:={x,y};
> solution:=solve(equations,vars);
> sol:=allvalues({x = 2/3-1/3*RootOf(5*_Z^2-4*_Z-5,label = _L1),
y = 1/3*RootOf(5*_Z^2-4*_Z-5,label = _L1)});
> evalf(sol);
Regards,
Georgios Kokovidis
This will plot and solve for both solutions. Just cut
and paste it into your worksheet.
> restart:
> with(plots):
> a:=implicitplot(4*(x-1)^2+(y-2)^2 = 5,x=-1..2,y=-1..1):
> b:=implicitplot(x^2+4*y^2 = 1,x=-1..2,y=-1..1):
> display(a,b);
> eq1:=4*(x-1)^2+(y-2)^2 =5;
> eq2:=x^2+4*y^2 =1;
> equations:={eq1,eq2};
> vars:={x,y};
> solution:=solve(equations,vars);
> sol:=allvalues({x = 2/3-1/3*RootOf(5*_Z^2-4*_Z-5,label = _L1),
y = 1/3*RootOf(5*_Z^2-4*_Z-5,label = _L1)});
> evalf(sol);
Regards,
Georgios Kokovidis
