asceduardo

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I have the following trigonometric expression:

detA:= -(-((-16*cosh(beta*`ℓ`*xi[2])*sinh(beta*`ℓ`)+16*sinh(beta*`ℓ`*xi[2])*cosh(beta*`ℓ`))*sinh(beta*xi[1]*`ℓ`)+(-16*sin(beta*`ℓ`*xi[2])*cos(beta*`ℓ`)+16*cos(beta*`ℓ`*xi[2])*sin(beta*`ℓ`))*sin(beta*xi[1]*`ℓ`)-(16*(-sin(beta*`ℓ`*xi[2])*sin(beta*`ℓ`)+cosh(beta*`ℓ`*xi[2])*cosh(beta*`ℓ`)-sinh(beta*`ℓ`*xi[2])*sinh(beta*`ℓ`)-cos(beta*`ℓ`*xi[2])*cos(beta*`ℓ`)))*(cos(beta*xi[1]*`ℓ`)-cosh(beta*xi[1]*`ℓ`)))*sin(beta*`ℓ`*xi[2])+16*cos(beta*`ℓ`*xi[2])*((sin(beta*`ℓ`*xi[2])*sin(beta*`ℓ`)-cosh(beta*`ℓ`*xi[2])*cosh(beta*`ℓ`)+sinh(beta*`ℓ`*xi[2])*sinh(beta*`ℓ`)+cos(beta*`ℓ`*xi[2])*cos(beta*`ℓ`))*sin(beta*xi[1]*`ℓ`)+(sin(beta*`ℓ`*xi[2])*cos(beta*`ℓ`)-cos(beta*`ℓ`*xi[2])*sin(beta*`ℓ`))*(cos(beta*xi[1]*`ℓ`)-cosh(beta*xi[1]*`ℓ`))))*sinh(beta*`ℓ`*xi[2])+16*cosh(beta*`ℓ`*xi[2])*((sin(beta*`ℓ`*xi[2])*sin(beta*`ℓ`)-cosh(beta*`ℓ`*xi[2])*cosh(beta*`ℓ`)+sinh(beta*`ℓ`*xi[2])*sinh(beta*`ℓ`)+cos(beta*`ℓ`*xi[2])*cos(beta*`ℓ`))*sinh(beta*xi[1]*`ℓ`)+(sinh(beta*`ℓ`*xi[2])*cosh(beta*`ℓ`)-cosh(beta*`ℓ`*xi[2])*sinh(beta*`ℓ`))*(cos(beta*xi[1]*`ℓ`)-cosh(beta*xi[1]*`ℓ`)))*sin(beta*`ℓ`*xi[2])-(16*(cos(beta*`ℓ`*xi[2])*cosh(beta*`ℓ`*xi[2])-1))*((sin(beta*`ℓ`*xi[2])*cos(beta*`ℓ`)-cos(beta*`ℓ`*xi[2])*sin(beta*`ℓ`))*sinh(beta*xi[1]*`ℓ`)+sin(beta*xi[1]*`ℓ`)*(sinh(beta*`ℓ`*xi[2])*cosh(beta*`ℓ`)-cosh(beta*`ℓ`*xi[2])*sinh(beta*`ℓ`)))

where:

xi1_val:= 0.05:
xi2_val:= 0.10:
ell_val:= 12:

I want to determine the roots "beta[1]..beta[N]" of this expression. For this, I have tried:

evalf(subs([xi[1] = xi1_val, xi[2] = xi2_val, `ℓ` = ell_val], detA)):
sol_beta:= Student[Calculus1][Roots](%, 1e-5..15);

which yields:

sol_beta:= [10.4719755]

However, I'm pretty sure there is another root before this, as we can see by the plot of the function:

plot(subs([xi[1] = xi1_val, xi[2] = xi2_val, `ℓ` = ell_val], detA), beta = 5.2..5.3,
    axes = boxed,
    gridlines = true,
    labels = [typeset(beta), typeset(`det A`)],
    size = [0.5,0.5]
    );

it is clear that there is a root about beta = 5.2359, which is not captured by "Roots" function.

Can someone help me build an algorythm that will get all roots wihtin a given interval? Particularly, I need it to be an efficient routine, because I will later vary parameters "xi[1]" and "xi[2]" and make a surface with the solutions. So I will run it several times.

I need to solve a system of differential equations, which have been obtained from the method of multiple scales. However, for some reason Maple returns an error and will not give a solution. Could someone help me find a solution? The associated filed is attached.

File.mw

I am developping an algorythm in which I need to obtain a 3D plot and its respective colorbars. As far as I know, Maple does not have a function with includes such a colorbar automatically (like Matlab, for example). So, I have had to get 3D plot and colorbar separatelly, as independent plots. It has worked fine, but now I need to export such graphics as EPS files, which imply merging them as a single graphic.

Here is a small example, where I try to put such graphics inside a table for then exporting it as EPS:

restart:

with(plots):
with(DocumentTools):

graph:= plot3d(x^2+y^2, x = -10..10, y = -10..10, colorscheme = ["zgradient", ["Red","Blue"]]):

grad_min, grad_max:= (min, max)(op([1,3], indets(graph, specfunc(anything, GRID)))):

colorbar:= densityplot(axis_z, axis_x = 0..0.1, axis_z = grad_max..grad_min, style = patchnogrid, size = [100,250], axes = boxed, tickmarks = [0,10], labels = ["",""], colorscheme = ["zgradient", ["Red","Blue"]]):

tab:= Tabulate([graph, colorbar], exterior = none, interior = none, weights = [4,1]):

 It has not worked, though. Can anyone suggest a solution for that? I need to export both graphics as one single file, preferably EPS, because of its high resolution.

I am developing an algorythm which returns some differential equation, which I want to simplify. Here is an example:

eqq:= k[t]*(`ℓ`^2)*(diff(q[3](tau), tau, tau)+(5*alpha-sigma+2*theta+1)*q[3](tau)+(-4*alpha+sigma-theta)*q[2](tau)+q[1](tau)*alpha) = -(sqrt(m*(1/k[t]))*`ℓ`*k[t]*`Δθ`*(q[3](tau)-q[2](tau))*sin(sqrt(Lambda*k[t]*(1/m))*sqrt(m*(1/k[t]))*tau)+2*xi*sqrt(lambda*k[t]*m)*(diff(q[3](tau), tau)))*`ℓ`*(1/sqrt(m*(1/k[t])))

I want the parameters to be associated to the the variables, q[1](tau)q[2](tau)q[3](tau) and their derivattives. So, I have used "collect" command, as below:

vars:= {q[1](tau),q[2](tau),q[3](tau),diff(q[1](tau),tau),diff(q[2](tau),tau),diff(q[3](tau),tau),diff(q[1](tau),tau$2),diff(q[2](tau),tau$2),diff(q[3](tau),tau$2)}:
collect(eqq,vars);

The problem is that the equations remain with non-simplified terms, such as the terms inside the "sine" functions and the term "k[t]*ell^2". The command "simplify" does not have any effect. Ideally, I would like to have something like this:

(diff(q[3](tau), tau, tau))+alpha*q[1](tau)+(-4*alpha+sigma-theta)*q[2](tau)+(5*alpha-sigma+2*theta+1)*q[3](tau)+2*xi*sqrt(lambda)/`ℓ`*(diff(q[3](tau), tau))-`Δθ`*sin(sqrt(Lambda)*tau)*q[2](tau)+`Δθ`*q[3](tau)*sin(sqrt(Lambda)*tau) = 0;

Does anyone know how to solve that?
 

I would like to plot a graph whose legend has a Greek character with a numeric subscript. Below, there is a small example, where I show what I am trying to do:

restart:

# Simplified example, which is not working:

omega0:= 10:
plot(omega^2/omega0, omega = 1..1.5,
legend = [sprintf("%s = %.1f",`ω`[0],omega0)]);

# The character I want to be plotted inside the legend:

`ω`[0];

I suppose the problem is that " `ω[0]` is not considered a string and, therefore, I can not call it in the legend with a "%s". I do not know how to make it work, though. Does anyone know how to do so?

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