Maple 2024 Questions and Posts

These are Posts and Questions associated with the product, Maple 2024

In this kind of contour plot i have two line but when i change time variable t just contour of one line wil move the other is not do any movement and is stop how i can  make the second plot one second line move too? also there is any way for ploting this kind any other option?

line-2-done.mw

i did for two of them base on the information but one of them is not make my odetest to be zero? where is problem

test.mw

It is possible to perform the simplest QFT calculations with second quantization, in Maple? Bosons in a box. See attached example. bosons_in_a_box.mw

Sure any general purpose programming language is capable of performing this task with enough effort. What I am interested in is if the physics tools has a standard way of dealing with these calculations. The general impedement when attempting the calculation is that integrations are perfomed by replacements with delta functions or kronecker delta functions, and its not clear how to force the Maple Physics package to recognize this or if that's possible. Part of the problem is that integrations in maple are defined in one dimension at a time where as in QFT the integration element is almost always atleast three dimensional, d^3x or dxdydzy, the later of which can get extremely cumbersome with even a small number of fields under consideration. I don't find much of what I am refering to mentioned in the help pages and I doubt these types of QFT calculations are possible to perform in Maple without addressing these issues.

bosons_in_a_box.mw

i don't  know  why my graph make a problem and what is issue i did plot  but this time make issue for me which i don't know where is problem there is anyone which can help and even modified the plot?

explore-chaotic.mw

i did substitution but my result is so different from the author i think he just take the linear term of theta but i didn't do that so how take just linear term of that function and find unknwon , and how afeter replacing eq(12) inside eq(11) we can remove thus exponential and find w? also i think author did a mistake which the equation 12 is theta(x,t) not Q(x,t)

restart

with(PDEtools)

undeclare(prime, quiet)

declare(u(x, t), quiet); declare(U(xi), quiet); declare(V(xi), quiet); declare(theta(x, t), quiet)

pde := diff(u(x, t), `$`(t, 2))-s^2*(diff(u(x, t), `$`(x, 2)))+(2*I)*(diff(u(x, t)*U^2, t))-(2*I)*alpha*s*(diff(u(x, t)*U^2, t))+I*(diff(u(x, t), `$`(x, 2), t))-I*beta*s*(diff(u(x, t), `$`(x, 3)))

diff(diff(u(x, t), t), t)-s^2*(diff(diff(u(x, t), x), x))+(2*I)*(diff(u(x, t), t))*U^2-(2*I)*alpha*s*(diff(u(x, t), t))*U^2+I*(diff(diff(diff(u(x, t), t), x), x))-I*beta*s*(diff(diff(diff(u(x, t), x), x), x))

(1)

T := u(x, t) = (sqrt(Q)+theta(x, t))*exp(I*(Q^2*epsilon*gamma+Q*q)*t); T1 := U = sqrt(Q)+theta(x, t)

u(x, t) = (Q^(1/2)+theta(x, t))*exp(I*(Q^2*epsilon*gamma+Q*q)*t)

 

U = Q^(1/2)+theta(x, t)

(2)

P := collect(eval(subs({T, T1}, pde)), exp)/exp(I*(Q^2*gamma*`ε`+Q*q)*t)

diff(diff(theta(x, t), t), t)+(2*I)*(diff(theta(x, t), t))*(Q^2*epsilon*gamma+Q*q)-(Q^(1/2)+theta(x, t))*(Q^2*epsilon*gamma+Q*q)^2-s^2*(diff(diff(theta(x, t), x), x))+(2*I)*(diff(theta(x, t), t)+I*(Q^(1/2)+theta(x, t))*(Q^2*epsilon*gamma+Q*q))*(Q^(1/2)+theta(x, t))^2-(2*I)*alpha*s*(diff(theta(x, t), t)+I*(Q^(1/2)+theta(x, t))*(Q^2*epsilon*gamma+Q*q))*(Q^(1/2)+theta(x, t))^2+I*(diff(diff(diff(theta(x, t), t), x), x)+I*(diff(diff(theta(x, t), x), x))*(Q^2*epsilon*gamma+Q*q))-I*beta*s*(diff(diff(diff(theta(x, t), x), x), x))

(3)

 

TT := Q = alpha[1]*exp(I*(k*x-t*w))+alpha[2]*exp(-I*(k*x-t*w))

Q = alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w))

(4)

S := eval(subs(TT, P))

diff(diff(theta(x, t), t), t)+(2*I)*(diff(theta(x, t), t))*(gamma*epsilon*(alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))^2+(alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))*q)-((alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))^(1/2)+theta(x, t))*(gamma*epsilon*(alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))^2+(alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))*q)^2-s^2*(diff(diff(theta(x, t), x), x))+(2*I)*(diff(theta(x, t), t)+I*((alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))^(1/2)+theta(x, t))*(gamma*epsilon*(alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))^2+(alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))*q))*((alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))^(1/2)+theta(x, t))^2-(2*I)*alpha*s*(diff(theta(x, t), t)+I*((alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))^(1/2)+theta(x, t))*(gamma*epsilon*(alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))^2+(alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))*q))*((alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))^(1/2)+theta(x, t))^2+I*(diff(diff(diff(theta(x, t), t), x), x)+I*(diff(diff(theta(x, t), x), x))*(gamma*epsilon*(alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))^2+(alpha[1]*exp((k*x-t*w)*I)+alpha[2]*exp(-I*(k*x-t*w)))*q))-I*beta*s*(diff(diff(diff(theta(x, t), x), x), x))

(5)

Download steps.mw

or for this equation 

steps-2.mw

I test a lot of them but some of them make a problem i  don't know i am do it in wrong way or the author did wrong i need verifying thus solution of odes specially in case 4 when we have not equal sign how use that?

and case 5 is Weierstrass elliptic function which i don't know how set up and use i think is a on kinf of odes but why they use that sign for this function?

ode-17.mw

Hi,

I’m having fun animating a beautiful geometric shape starting from a few trigonometric functions. I’m wondering if there’s a way to link each curve in the animation to its name.

restart

plots:-animatecurve([sin(x), sin(x)^2, sin(x)^3, sin(x)^4, sin(x)^5, sin(x)^6, surd(sin(x), 2), surd(sin(x), 3), surd(sin(x), 4), surd(sin(x), 5), surd(sin(x), 6)], x = 0 .. Pi, thickness = 2.5, background = "AliceBlue", labels = ["", ""], size = [800, 800])

 

plot([sin(x), sin(x)^2, sin(x)^3, sin(x)^4, sin(x)^5, sin(x)^6, surd(sin(x), 2), surd(sin(x), 3), surd(sin(x), 4), surd(sin(x), 5), surd(sin(x), 6)], x = 0 .. Pi, thickness = 2.5)

 

NULL

Download Animation_Trigo.mw

I recently upgraded from Maple 23 to Maple 24. While many display issues have been resolved, I’ve encountered a new real problem: when entering operations or a factorial in a denominator or exponent, the cursor unexpectedly jumps to the inline position. This forces me to manually reposition the cursor using backspace plus left-arrow keys, and forgetting to do so can lead to errors. I have a perpetual Maple license through my university and haven’t purchased maintenance this time. Is there any way to fix or work around this cursor-jumping issue in Maple 24 without purchasing a new license?

Already  by Help of my favorite Dr.David he did find the thus three step for non schrodinger equation but in here i got some issue of coding which is so different from before, is about transformation of pdes to two parts od real and imaginary part and then substitution our function the functions is clear but combining them and findinf leading exponent and resonance point and finding function in step 3 in different and jsut the function is different with eperate the real and imaginary part for finding step one ...

note:=q=u*exp(#) then |q|=u

schrodinger-test.mw

paper-1

paper-2

In the attached file, the trigonometric term (2, term) is transformed into a term (3, term1) consisting of radicals. Is there a Maple procedure that can be used to reverse this process? Given an algebraic term (e.g., consisting of radicals, powers, etc.), under what conditions can it be transformed into a trigonometric form (not a Fourier series) in the sense of (3) according to (2)?test.mw

 interface(version);

`Standard Worksheet Interface, Maple 2024.2, Windows 11, October 29 2024 Build ID 1872373`

(1)

restart

term := 2*cos(5*arcsin((1/2)*x))

2*cos(5*arcsin((1/2)*x))

(2)

term1 := expand(term)

-3*(-x^2+4)^(1/2)*x^2+(-x^2+4)^(1/2)+(-x^2+4)^(1/2)*x^4

(3)

convert(term1, trig)

-3*(-x^2+4)^(1/2)*x^2+(-x^2+4)^(1/2)+(-x^2+4)^(1/2)*x^4

(4)

simplify(term1, trig)

(-x^2+4)^(1/2)*(x^4-3*x^2+1)

(5)

solve(term1 = sqrt(2), x)

(1/2)*(8-2*(10+2*5^(1/2))^(1/2))^(1/2), (1/2)*(8+2*(10-2*5^(1/2))^(1/2))^(1/2), (1/2)*(8+2*(10+2*5^(1/2))^(1/2))^(1/2), -(1/2)*(8-2*(10+2*5^(1/2))^(1/2))^(1/2), -(1/2)*(8+2*(10-2*5^(1/2))^(1/2))^(1/2), -(1/2)*(8+2*(10+2*5^(1/2))^(1/2))^(1/2)

(6)

evalf(solve(term1 = sqrt(2), x))

.3128689302, 1.782013048, 1.975376681, -.3128689302, -1.782013048, -1.975376681

(7)

plot(term, x = -2.5 .. 2.5)

 

plot(term1, x = -2.5 .. 2.5)

 

NULL

Download test.mw

is been a while i work on a test still i am study and there is a lot paper remain and is so important in PDEs, a lot paper explain in 2003 untill know and there is other way to find it too but i choose a easy one but is 2025 paper  which is explanation is so beeter than other paper, also some people write a package for take out this test with a second but maybe is not work for all i  search for that  but i didn't find it i will ask the question how we can find thus as shown in graph i did my train but need a little help while i am collect more information and style of solving 

Download p1.mw

In thus manuscript i got some reviewer comment which is asked to simplify this expresion and there is a lot of them maybe if i do by hand i  made a mistake becuase a lot of variable so how i can fix this issue and make thus square root are very simple as they demand

restart

B[2] := 0

0

(1)

K := sqrt(-(1/2)*sqrt(2)*sqrt(lambda*a[5]/a[4])+sqrt(-a[5]/(2*a[4]))*(B[1]*sqrt(-lambda)*sinh(xi*sqrt(-lambda))+B[2]*sqrt(-lambda)*cosh(xi*sqrt(-lambda)))/(B[1]*cosh(xi*sqrt(-lambda))+B[2]*sinh(xi*sqrt(-lambda))+mu/lambda)+sqrt(-(lambda^2*B[1]^2*a[5]-lambda^2*B[2]^2*a[5]-mu^2*a[5])/(2*lambda*a[4]))/(B[1]*cosh(xi*sqrt(-lambda))+B[2]*sinh(xi*sqrt(-lambda))+mu/lambda))*exp(I*(k*(xi+v*tau^alpha/alpha)+w*tau^alpha/alpha+gamma))

(1/2)*(-2*2^(1/2)*(lambda*a[5]/a[4])^(1/2)+2*(-2*a[5]/a[4])^(1/2)*B[1]*(-lambda)^(1/2)*sinh(xi*(-lambda)^(1/2))/(B[1]*cosh(xi*(-lambda)^(1/2))+mu/lambda)+2*(-2*(lambda^2*B[1]^2*a[5]-mu^2*a[5])/(lambda*a[4]))^(1/2)/(B[1]*cosh(xi*(-lambda)^(1/2))+mu/lambda))^(1/2)*exp(I*(k*(xi+v*tau^alpha/alpha)+w*tau^alpha/alpha+gamma))

(2)

simplify(K)

(1/2)*exp(I*((k*v+w)*tau^alpha+alpha*(k*xi+gamma))/alpha)*2^(3/4)*((lambda*(a[5]*(-lambda^2*B[1]^2+mu^2)/(lambda*a[4]))^(1/2)+(-B[1]*cosh(xi*(-lambda)^(1/2))*lambda-mu)*(lambda*a[5]/a[4])^(1/2)+sinh(xi*(-lambda)^(1/2))*lambda*(-a[5]/a[4])^(1/2)*(-lambda)^(1/2)*B[1])/(B[1]*cosh(xi*(-lambda)^(1/2))*lambda+mu))^(1/2)

(3)

subsindets(K, `&*`(rational, anything^(1/2)), proc (u) options operator, arrow; (u^2)^(1/2) end proc)

(1/2)*(-2*2^(1/2)*(lambda*a[5]/a[4])^(1/2)+2*(-2*a[5]/a[4])^(1/2)*B[1]*(-lambda)^(1/2)*sinh(xi*(-lambda)^(1/2))/(B[1]*cosh(xi*(-lambda)^(1/2))+mu/lambda)+2*(-2*(lambda^2*B[1]^2*a[5]-mu^2*a[5])/(lambda*a[4]))^(1/2)/(B[1]*cosh(xi*(-lambda)^(1/2))+mu/lambda))^(1/2)*exp(I*(k*(xi+v*tau^alpha/alpha)+w*tau^alpha/alpha+gamma))

(4)

latex(%)

\frac{\sqrt{-2 \sqrt{2}\, \sqrt{\frac{\lambda  a_{5}}{a_{4}}}+\frac{2 \sqrt{-\frac{2 a_{5}}{a_{4}}}\, B_{1} \sqrt{-\lambda}\, \sinh \left(\xi  \sqrt{-\lambda}\right)}{B_{1} \cosh \left(\xi  \sqrt{-\lambda}\right)+\frac{\mu}{\lambda}}+\frac{2 \sqrt{-\frac{2 \left(\lambda^{2} B_{1}^{2} a_{5}-\mu^{2} a_{5}\right)}{\lambda  a_{4}}}}{B_{1} \cosh \left(\xi  \sqrt{-\lambda}\right)+\frac{\mu}{\lambda}}}\, {\mathrm e}^{\mathrm{I} \left(k \left(\xi +\frac{v \,\tau^{\alpha}}{\alpha}\right)+\frac{w \,\tau^{\alpha}}{\alpha}+\gamma \right)}}{2}

 

KK := sqrt(-(1/2)*sqrt(2)*sqrt(lambda*a[5]/a[4])+sqrt(-a[5]/(2*a[4]))*(B[1]*sqrt(-lambda)*sinh(xi*sqrt(-lambda))+B[2]*sqrt(-lambda)*cosh(xi*sqrt(-lambda)))/(B[1]*cosh(xi*sqrt(-lambda))+B[2]*sinh(xi*sqrt(-lambda))+mu/lambda)+sqrt(-(lambda^2*B[1]^2*a[5]-lambda^2*B[2]^2*a[5]-mu^2*a[5])/(2*lambda*a[4]))/(B[1]*cosh(xi*sqrt(-lambda))+B[2]*sinh(xi*sqrt(-lambda))+mu/lambda))*exp(I*(k*(xi+v*tau^alpha/alpha)+w*tau^alpha/alpha+gamma))

(1/2)*(-2*2^(1/2)*(lambda*a[5]/a[4])^(1/2)+2*(-2*a[5]/a[4])^(1/2)*B[1]*(-lambda)^(1/2)*sinh(xi*(-lambda)^(1/2))/(B[1]*cosh(xi*(-lambda)^(1/2))+mu/lambda)+2*(-2*(lambda^2*B[1]^2*a[5]-mu^2*a[5])/(lambda*a[4]))^(1/2)/(B[1]*cosh(xi*(-lambda)^(1/2))+mu/lambda))^(1/2)*exp((k*(xi+v*tau^alpha/alpha)+w*tau^alpha/alpha+gamma)*I)

(5)

latex(KK)

\frac{\sqrt{-2 \sqrt{2}\, \sqrt{\frac{\lambda  a_{5}}{a_{4}}}+\frac{2 \sqrt{-\frac{2 a_{5}}{a_{4}}}\, B_{1} \sqrt{-\lambda}\, \sinh \left(\xi  \sqrt{-\lambda}\right)}{B_{1} \cosh \left(\xi  \sqrt{-\lambda}\right)+\frac{\mu}{\lambda}}+\frac{2 \sqrt{-\frac{2 \left(\lambda^{2} B_{1}^{2} a_{5}-\mu^{2} a_{5}\right)}{\lambda  a_{4}}}}{B_{1} \cosh \left(\xi  \sqrt{-\lambda}\right)+\frac{\mu}{\lambda}}}\, {\mathrm e}^{\mathrm{I} \left(k \left(\xi +\frac{v \,\tau^{\alpha}}{\alpha}\right)+\frac{w \,\tau^{\alpha}}{\alpha}+\gamma \right)}}{2}

 

NULL

Download simplify.mw

i  can determine the pdes by one variable which is work so good but in some of the pdes i have two function i can separate by hand but how i can do by maple?

Download linear.mw

i did a lot of trail to avoid for find my parameter in the last step of that i get this `[Length of output exceeds limit of 1000000]` and i don't know how to fix it i need to find that parameter but when i do substitution  is said this there is any way for hundle this situation 

help-parameter.mw

My goal is to export images of curves where the output image has no background or equivalently, a fully transparent background. Is there a standard way to perform this task? I know that I can do this with the help of external software but I want to do it at the export stage in maple and avoid increasing the number of tools needed to perform the job. Eventually, I may want to generate many thousdands of curves/images with transparent backgrounds. 

Below I show how to make a plot while choosing a background with the color blue. I have found no way to select a fully transparent background. I expect this background transparency will be applied at the export image stage. 

how_do_I_plot_with_no_background.mw

To give an explicit example of what a plot with no background looks like, I make a plot in an unrelated software of a curve with no background. It's impossible to differentiate a white background from a transparent background in this environment, so we also show the same image embedded in an unrelated colored background also generated in an unrelated software.

In the image below, we can see how a image of a curve with a transparent background can be embedded over top of any other image without the blue square shown the the first example.

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