Kitonum

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simplify(c*lambda*D(V)*V^(lambda-1) - lambda*c*D(V)*V^lambda/V);
                                                         
 0

@adel-00  

1. You have forgotten that now you have  epsilon1, not  epsilon.

2. In this new example, the system for  a  and  b  has solutions that are very close to 0. Therefore, it is necessary to specify ranges for unknowns  a  and   in  fsolve  command.

3. It is always useful to plot graphs for your equations (if of course it is possible) to visually verify the existence of solutions and their properties (see below).


 

restart;

epsilon:=epsilon1/(2*Pi): gamma1:=8*Pi: gamma2:=0.002*Pi: x1:=100*Pi: omega2:=200*Pi: delta1:=0.2*Pi: Delta:=25*Pi: G:=20*Pi: Omega:=0:

lambda1:=G*Omega*gamma1/(2*(0.25*gamma1^2+Delta^2)):
lambda2:=G*Omega*Delta/(0.25*gamma1^2+Delta^2):
lambda3:=gamma1+lambda1:
lambda4:=delta1-G^2*Delta/(0.25*gamma1^2+Delta^2):
lambda5:=2*x1^2*omega2/(omega2^2+gamma2^2):
c:=a^2+b^2:

f:=epsilon-(lambda1-I*lambda2)=(a+I*b)(lambda3+I*(lambda4+lambda5*(c)));
f:=convert(epsilon-(lambda1-I*lambda2)=(a+I*b)*(lambda3+I*(lambda4+lambda5*(a^2+b^2))),rational);
f1:=(lhs-rhs)(evalc(f));

B:=coeff(f1,I);    

A:=expand(f1-B*I);

(1/2)*epsilon1/Pi = a(25.13274123+I*(314.1592654*a^2+314.1592654*b^2-48.38248732))+I*b(25.13274123+I*(314.1592654*a^2+314.1592654*b^2-48.38248732))

 

(1/2)*epsilon1/Pi = (a+I*b)*(104703/4166+I*((940907/2995)*a^2+(940907/2995)*b^2-190627/3940))

 

(1/2)*epsilon1/Pi-(104703/4166)*a-(190627/3940)*b+(940907/2995)*b*a^2+(940907/2995)*b^3-I*(-(190627/3940)*a+(940907/2995)*a^3+(940907/2995)*a*b^2+(104703/4166)*b)

 

(190627/3940)*a-(940907/2995)*a^3-(940907/2995)*a*b^2-(104703/4166)*b

 

(1/2)*epsilon1/Pi-(104703/4166)*a-(190627/3940)*b+(940907/2995)*b*a^2+(940907/2995)*b^3

(1)

F:=t->fsolve({eval(A,epsilon1=t),B}, {a=-0.1..0.1,b=-0.1..0.1}):
F(1);
    
plot('eval(a^2+b^2,F(t))', t=0..5, labels=[epsilon1,a^2+b^2], view=[0..5,0..0.001], labelfont=[times,16],color=black,axes=boxed,font=[1,1,18],tickmarks=[4,2],thickness=2);

{a = 0.1345776972e-2, b = 0.2590582283e-2}

 

 

plots:-implicitplot([eval(A,epsilon1=1),B], a=-0.1..0.1,b=-0.1..0.1, color=[red,blue]);

 

 


 

Download plot1.mw

@acer  for the useful information.

@digerdiga 

Try this:

restart;
o:=[f1,f2,f3];
map(t->frontend(expand,[t]), o);

@Carl Love  I am sure, I meant "to solve symbolically for arbitrary parameters" and OP probably meant the same thing.

@shahid  You get the data as an array on a grid. You can convert this array to a table or something else. If the result below does not suit you, write exactly what table you want.

restart;
A:=plot3d(1/2*(q^(2*n)-5.*q^n+(2.*q-1)^n+3), q = 2 .. 15, n = 2 .. 20):
L:=plottools:-getdata(A);
convert(L[3], table);

 

@Rohith  I added another way to my answer. For me, both ways work.

@mehdi jafari  Your body is symmetrical about the plane  xOz , so it’s enough to calculate half of its volume. If you look at this body from the side of the y-axis, you will see this picture. In fact, this is the projection of the body onto the plane  xOz . Devide this projection into 3 regions that you see and then calculate volumes using double integrals.

    

 

@Carl Love In the third surface equation, I replaced  d1  with  d3 (OP probably just made a slip of the pen).

@Earl   See  emptysymbol  help page for this.

@vv  In fact, you have built a very thin ellipse. Once we realize the meaning of the problem, we can make it easier:
 plot([[1,1], [5,4]]);

 

 

@Rouben Rostamian  Thanks for pointing this out. Unfortunately, I do not know  output=...  option well and have never used it. However, I have never met an example where this option would be necessary. I think it is good that many tasks in Maple can be solved in different ways. We see that, for example, plots:-spacecurve command may well be replaced by  plots:-pointplot3d (with  style=line  option) or  plottools:-curve   commands.

@Carl Love  Thanks for the detailed explanation.

@jfreeman1212 

It's easy - you just need to solve the system:

solve({(x^2+y^2+2*y)^2=4*(x^2+y^2), y=x}, {x,y}, explicit);
       
   {x = 0, y = 0}, {x = sqrt(2)-1, y = sqrt(2)-1}, {x = -1-sqrt(2), y = -1-sqrt(2)}
                     

@Carl Love  Your procedure does not do what OP wants (look at  ListY  in the original post). In addition, it does not work in earlier versions than Maple 2018 (because of until).

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