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MaplePrimes Activity


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Hello.

I would like to solve numerically highly nonlinear and cumbersome the second order differential equation. 

Applying the numerical procedure I got an error "Error, (in dsolve) found wrong extra argument(s): range = 0 .. 4*Pi, type = numerical". The similar problem has been described earlier here however I can't realize my problem. 

Below is my code

restart;

A1 := 8*Pi^3*R^2*n(x)^4*m+(2*Pi*sin((1/2)*x)*m*omega0*p+Pi*sin((1/2)*x)*m*omega0+3*Pi^2*(diff(n(x), x, x)))*n(x)^3+(-2*sin((1/2)*x)^2*m^2*omega0^2*p^2+2*cos((1/2)*x)^2*m^2*omega0^2*p^2-2*sin((1/2)*x)^2*m^2*omega0^2*p+2*cos((1/2)*x)^2*m^2*omega0^2*p)*n(x)^2+(-4*(diff(n(x), x, x))*sin((1/2)*x)^2*m^2*omega0^2*p^2-8*sin((1/2)*x)*(diff(n(x), x))*cos((1/2)*x)*m^2*omega0^2*p^2-4*(diff(n(x), x, x))*sin((1/2)*x)^2*m^2*omega0^2*p-8*sin((1/2)*x)*(diff(n(x), x))*cos((1/2)*x)*m^2*omega0^2*p)*n(x)+8*sin((1/2)*x)^2*(diff(n(x), x))^2*m^2*omega0^2*p^2+8*sin((1/2)*x)^2*(diff(n(x), x))^2*m^2*omega0^2*p;

R := 1; m := 1; p := 10; omega0 := 1000;

A2 := A1;

with(plots):
A3 := dsolve({A2, n(0) = n(4*Pi), (D(n))(0) = (D(n))(4*Pi)}, type = numerical, range = 0 .. 4*Pi):

odeplot(A3);


I appreciate for any help and suggestion.

 

Hello.

I'd like to extract numerical data from the plot of a implicit function and write it to the txt file. But, unfortunately due to a strong oscillating function and as a consequence extremely dense meshgrid I got a message that [Length of output exceeds limit of 1000000]. How to resolve this problem? Below is my code

restart;
R0 := 1+w_c*(sum((2*(-1)^(n+1)*Pi^2/n*n)*t*sin(2*Pi*n*mu/w_c)*cos(Pi*n)*exp(-2*Pi*n*G/w_c)/(w_c*sinh(2*Pi^2*n*t/w_c)), n = 1 .. 3000))/Pi-mu;
with(plots, implicitplot);
G := 0.; t := 0.1e-2;
R1 := implicitplot(R0, w_c = 0 .. 5, mu = 0 .. 3, gridrefine = 10):
data := getdata(R1);

Thank you in advance.

Good day.

I have a problem with the numerical approximation of incomplete elliptic integrals of the first kind in the case when the argument and the modulus are complex numbers. 

Let's consider an incomplete elliptic integral of the first kind EllipticF((a+I*b)*x, c+I*d), where a,b,c,d are arbitrary real numbers and x is very large number. Here is the code:

restart;
R0 := EllipticF((a+I*b)*x, c+I*d);
with(MultiSeries, series):
R1 := series(R0, x = infinity, 2);
a := .5; b := 1.75; c := 10; d := 12.5; x := 10^6;
evalf(R0); evalf(R1);

Below you can see that answers are different

.1680611942+.2058775337*I
-0.4563467782e-1-.3592094173*I+O(1/1000000000000000000)

While the same code for an incomplete elliptic integral of the second kind gives the correct approximation:

restart;
R00 := EllipticE((a+I*b)*x, c+I*d);
R11 := series(R00, x = infinity, 2);
a := .5; b := 1.75; c := 10; d := 12.5; x := 10^6;
evalf(R00); evalf(R11);
-1.687498740*10^7+2.374999011*10^7*I
-1.687501260*10^7+2.375000989*10^7*I+O(1/1000000)

What is the difference?
 

Hello everyone

I try to solve the system of nonlinear equations, where three variables are inside of integrals, for different values of the real parameter q.

It seems that  not for all values of q the system has a solution because I substitute different q but the function fsolve does not give any numerical answer or the special message even after the long evaluation. 

Here the code:

restart;
q := 70;
k0 := 100; n := 3; r := 1.1; t := 0.1e-2; Eg := n/((n+1)^(3/2)+1)^(2/3); kF2 := (1/((n+1)^(3/2)+1))^(1/3);
R1 := int(a^2*tanh(sqrt((a^2-mu)^2+d1^2)/(2*t))/sqrt((a^2-mu)^2+d1^2), a = 0 .. k0);
R2 := int(a^2*tanh(sqrt((a^2-mu+Eg)^2+d2^2)/(2*t))/sqrt((a^2-mu+Eg)^2+d2^2), a = 0 .. k0);
R3 := int(a^2*(1-(a^2-mu)*tanh(sqrt((a^2-mu)^2+d1^2)/(2*t))/sqrt((a^2-mu)^2+d1^2)), a = 0 .. k0);
R4 := int(a^2*(1-(a^2-mu+Eg)*tanh(sqrt((a^2-mu+Eg)^2+d2^2)/(2*t))/sqrt((a^2-mu+Eg)^2+d2^2)), a = 0 .. k0);
Eq1 := evalf(k0+k0*(r-1)-(1/2)*Pi*r*kF2*q-R1);           
Eq2 := evalf(k0-(1/2)*Pi*kF2*q-R2);
Eq3 := evalf(2/3-R3-R4);
fsolve({Eq1, Eq2, Eq3}, {d1, d2, mu});

I know that this system has the solution for some values of q but why I can't get the solution or the answer using fsolve?

Thank you in advance.

 

Update. I apologize, I made some errors in expressions R2 and R4. I have corrected the question and the code. But the problem is still unresolved.

Hello.

Regarding to my previous question I'd like to speed up calculations of the expression. 

restart;
tt := -0.689609e-3; T_c := .242731; mu := .365908; k := 1;
R1 := a*tanh((a^2-mu)/(2*T_c))*ln((2*a^2+2*a*q+q^2-2*mu-(I*2)*Pi*N)/(2*a^2-2*a*q+q^2-2*mu-(I*2)*Pi*N))/q-2;
R2 := Int(R1, a = 0 .. 10000);
R3 := q*ln((-q^2-k^2+mu+I*(2*N*Pi*T_c-(2*m+1)*Pi*T_c)+k*q)/(-q^2-k^2+mu+I*(2*N*Pi*T_c-(2*m+1)*Pi*T_c)-k*q))/(k*(tt+R2));
R4 := Sum(R3, N = -100 .. 100);
m := 1;
R5 := Int(R4, q = 0.1e-2 .. 10000);
R6 := evalf(R5);

Here I have integration procedure inside the expression R3, then the summation over the integer parameter N and then finally the integration again.

Is it possible to speed up calculations of this cumbersome expression? Or actually was I correct to write this simple code?

Thank you in a advance.

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