My question is: how to find the coordinates of the vertices of a dodecahedron?
I can find the  coordinates of the vertices of a tetrahedron as the solutions of a certain polynomial system in 8 variables (see  tetrahedron.mw for details).
However, that approach seems not to work for a dodecahedron. A new idea is required.

PS. Of course, I have in mind a regular dodecadron.

Hello.

I am novice here and I have an question about export animations frames into sequence of postscript files. I don't googled nothing about this theme.
Here is my question:

Is there an option to convert the animated picture into  sequence of separate images and then (eg using the program cycle) save this images as separate PostScript images? I know the possibility of exporting to GIF file and then converting them into PS file. But I do not have a PS file with a sequence of bitmap images. I would sufficed for me to be able to see a separate frame of animation (through any index).

Thanx Jaroslav Hajtmar

eq1 := y=−26.21231979∗z+15.42332896+13.22411533∗e−.6786000000∗x

eq2 := y=−25.98077423∗z+14.81943362+13.53858145∗e−.6569000000∗x

# Comparing both equations, eliminating y

# Putting z= 0.5044

fsolve(eval(rhs(eq1) = rhs(eq2), z = .5044))

I even done manually as well

-26.21231979*(0.5044)+15.42332896+13.22411533*e^(-.6786000000*x) =
 -25.98077423*(0.5044)+14.81943362+13.53858145*e^(-.6569000000*x)

I cannot find x value?  

But it doesn't evaluate the value of x. Any other solution.

Edit : My main task is to calculate value of x by putting any value of z This is just an example

Say I have the following loops:

for C from 1 to 10 do
    r:=[]:
    for K from 2 to 10 do
    r:=[op(r),2*K+2*C-3];
    end do:
    print(r);
end do:

for C from 1 to 10 do
    r:=[]:
    for K from 2 to 10 do
    r:=[op(r),K*C+K+C-2];
    end do:
    print(r);
end do:

I wonder how could I write a procedure, say use expressions "2*K+2*C-3" and "K*C+K+C-2" as input arguments?

so I can call up like :

myfun(K*C+K+C-2) or myfun("K*C+K+C-2")

myfun(2*K+2*C-3)

I dont care whether the output(s) are lists, tables, or matrices.

My main difficulty is to get the expression to be procedure inputs.

Though if the output can be a  10 by 9 matrix, it's better.

Thanks,

casper

Using the fsolve commmand how does one solve for just the positive solutions and remove the dublicate values?

Thanks

Is there a way of increasing the spacing between the maple input and output?

Thanks

Hi, i was wondering how you would solve the equation

30+1144*r^4-832*r^2=0, 

exactly in maple.

Using the command

fsolve(30+1144*r^4-832*r^2);

-0.8301954535, -0.1950595709, 0.1950595709, 0.8301954535

Also, if possible how would you get maple to find the two exact positive solutions.

Thanks

when i got this error, i am confused i guess t is independant variable, x1,y1,z1 are dependant variables

x11 := [0.208408965651696e-3, -0.157194487523421e-2, -0.294739401402979e-2, 0.788206708183853e-2, 0.499394753201753e-2, 0.191468321959759e-3, 0.504980449104750e-2, 0.222150494088535e-2, 0.132091821964287e-2, 0.161118434883258e-2, -0.281236534046873e-2, -0.398055875132037e-2, -0.111753680372819e-1, 0.588868146012489e-2, -0.354191562612469e-2, 0.984082837373291e-3, -0.116041186868374e-1, 0.603027845850267e-3, -0.448778128168742e-2, -0.127561485214862e-1, -0.412027655195339e-2, 0.379387381798949e-2, -0.602550446997765e-2, -0.605986284736216e-2, -0.751396992404410e-2, 0.633613424008655e-2, -0.677581832613623e-2]:
y11 := [ -21321.9719565717, 231.709204951251, 1527.92905167191, -32.8508507060675, 54.9408176234139, -99.4222178124229, -675.771433486265, 42.0838668074923, -12559.3183308951, 5.21412214166344*10^5, 1110.50031772203, 3.67149699000155, -108.543878970269, -8.48861069398811, -521.810552387313, 26.4792411876883, -8.32240296737599, -1085.40982521906, -44.1390030597906, -203.891397612798, -56.3746416571417, -218.205643256096, -178.991498697065, -42.2468018350386, .328546922634921, -1883.18308996621, 111.747881085748]:
z11 := [ 1549.88755331800, -329.861725802688, 8.54200301129155, -283.381775745327, -54.5469129127573, 1875.94875597129, -16.2230517860850, 6084.82381954832, 1146.15489803104, -456.460512914647, 104.533252701641, 16.3998365630734, 11.5710907832054, -175.370276462696, 33.8045539958636, 2029.50029336951, 1387.92643570857, 9.54717543291120, -1999.09590358328, 29.7628085078953, 2.58210333216737*10^6, 57.7969622731082, -6.42551196941394, -8549.23677077892, -49.0081775323244, -72.5156360537114, 183.539911458475]:
a1 := Diff(x1(t),t) = k1*x1(t)+ k2*y1(t)+ k3*z1(t);
b1 := Diff(y1(t),t) = k4*x1(t)+ k5*y1(t)+ k6*z1(t);
c1 := Diff(z1(t),t) = k7*x1(t)+ k8*y1(t)+ k9*z1(t);
ICS:=x1(0)=x11[1],y1(0)=y11[1],z1(0)=z11[1];
sol:=dsolve({a1,b1,c1,ICS}, numeric, method=rkf45, parameters=[k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12]);
ans:=proc(p1,p2,p3) sol(parameters=[x1=p1,y1=p2,z1=p3]); end proc:
tim := [seq(n, n=1..27)];
FitParams:=Statistics:-NonlinearFit(ans, [x11, y11, z11], tim, initialvalues=<0.5,0.5,0.5>, output=parametervalues);

Hi, i am trying to solve the equations denoted eq1 and eq2 for x and r.

f:=r*(8 - 2*x^2);
g:=subs(x=f,f):
eq1:= g-x:
eq2:= expand(diff(g,x) + 1):

I am having a bit of trouble as these simultaneous equations have many solutions and using the command solve, just basically crashes maple. I just want the commands that would give the positive set of solutions only, ie. excluding all complex and negative solutions. 

Thanks in advance.

4th_order.mw Is it possible to solve this ODE with perturbation method using maple? If yes, please give the procedure.

Thanks.

Can one solve nonlinear ODE with perturbation method in maple? If yes, please the procedure.

I was looking at a question on another forum, and tried in Maple also, and Maple also have a problem with this integral. Here is a simple version. The problem is that int() gives different numerical answer from evalf(Int). Maple can't solve this analytically, so values have to be used for the integrand before calling int()

restart;
eq2:=(a/(a + c*z))^L*exp(-z)/sqrt(z);

L:=2:
a:=10^(0.1):
b:=10^(0.1):
c:=0.01*a:
int(eq2,z=0..infinity);   # 177.245
evalf(Int(eq2,z=0..infinity));   # 1.7551

I think the 1.7551 is the correct value. My question is: Why did maple give wrong answer from int()? Is it analytical reason, or purely numerical?

Maple 17.02, windows 7.

[*********************************************************************************]
[* DO NOT EDIT PROBLEMS TO REMOVE THE ORIGINAL QUESTION! -Carl Love as moderator *]
[*********************************************************************************] 

% use Euler formula to compute function y
for i = 1:N
    if i ==1

legend('numerical y','exact y','numerical g','exact g')

function g = f2g(a,b)
% f(x) = x
g = (b-a)/6*(a + 4*(a+b)/2 + b);

x11 := Vector([0.208408965651696e-3, -0.157194487523421e-2, -0.294739401402979e-2, 0.788206708183853e-2, 0.499394753201753e-2, 0.191468321959759e-3, 0.504980449104750e-2, 0.222150494088535e-2, 0.132091821964287e-2, 0.161118434883258e-2, -0.281236534046873e-2, -0.398055875132037e-2, -0.111753680372819e-1, 0.588868146012489e-2, -0.354191562612469e-2, 0.984082837373291e-3, -0.116041186868374e-1, 0.603027845850267e-3, -0.448778128168742e-2, -0.127561485214862e-1, -0.412027655195339e-2, 0.379387381798949e-2, -0.602550446997765e-2, -0.605986284736216e-2, -0.751396992404410e-2, 0.633613424008655e-2, -0.677581832613623e-2]):
y11 := Vector([ -21321.9719565717, 231.709204951251, 1527.92905167191, -32.8508507060675, 54.9408176234139, -99.4222178124229, -675.771433486265, 42.0838668074923, -12559.3183308951, 5.21412214166344*10^5, 1110.50031772203, 3.67149699000155, -108.543878970269, -8.48861069398811, -521.810552387313, 26.4792411876883, -8.32240296737599, -1085.40982521906, -44.1390030597906, -203.891397612798, -56.3746416571417, -218.205643256096, -178.991498697065, -42.2468018350386, .328546922634921, -1883.18308996621, 111.747881085748]):
z11 := Vector([ 1549.88755331800, -329.861725802688, 8.54200301129155, -283.381775745327, -54.5469129127573, 1875.94875597129, -16.2230517860850, 6084.82381954832, 1146.15489803104, -456.460512914647, 104.533252701641, 16.3998365630734, 11.5710907832054, -175.370276462696, 33.8045539958636, 2029.50029336951, 1387.92643570857, 9.54717543291120, -1999.09590358328, 29.7628085078953, 2.58210333216737*10^6, 57.7969622731082, -6.42551196941394, -8549.23677077892, -49.0081775323244, -72.5156360537114, 183.539911458475]):
a1 := Diff(x1(t),t) = k1*x1(t)+ k2*y1(t)+ k3*z1(t)+k4*u(t);
b1 := Diff(y1(t),t) = k5*x1(t)+ k6*y1(t)+ k7*z1(t)+k8*u(t);
c1 := Diff(z1(t),t) = k9*x1(t)+ k10*y1(t)+ k11*z1(t)+k12*u(t);
ICS:=x1(0)=x11[1],y1(0)=y11[1],z1(0)=z11[1],u(0)=7;
Zt := rhs(dsolve({a1,b1,c1,ICS},[x1(t),y1(t),z1(t),u(t)]));
Params := NonlinearFit(Re(Zt),<seq(k,k=0..N)>, C, [t], parameternames=[k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12], output=parametervalues):
A := eval(a, Params):
B := eval(b, Params):
a = A;
b = B;

i wait for a long time for dsolve still evaluating