I got a plot by following the code in Mpale 2022.
restart;
with(plots);
P2 := plot3d(-t^2 + x, t = 0 .. 20, x = 0 .. 400, labels = ['t', 'x', 'rho'], labelfont = [Times, 15], viewpoint = "circleleft", colorscheme = ["ygradient", ["Green", "Purple", "Blue", "Red", "BlueViolet"]]);
plotsetup(ps, plotoutput = "P2");
print(P2);

Then I tried the following code for the overleaf. However, I did not get the desired result.

\documentclass[11pt]{article}
\usepackage{graphicx}
\usepackage{epsfig}

\begin{document}
Plot obtained from Maple 2022. \\

\includegraphics[width=0.9\textwidth]{P2.eps}
\end{document}

plotting_file.pdf

Any help is highly appreciated.
Thanks
Deb

I am wondering why Maple simplifies (x^(1/3))^3 to x ,  but not (x^3)^(1/3) .
I even tried the surd function. I believe the surd function is for real number arguments, so it should simplify to x.

Download inverse1.mw

I asked maple to solve a basic log inequality.
;

This is what happened.

Here is a link to the document to replicate this behavior.

log_inequality.mw

I know there is a solution , if you look at the [graph](https://www.desmos.com/calculator/3n7uzwrak4).

I also tried fsolve, but you have to narrow down the solution interval to look for a solution, and use an equality instead of an inequality.

For t not equal to nT,   

dS/dt = delta- mu*S+ omega*V; 

 dV/dt = -(omega+mu)*V

For t=nT, 

 S(nT⁺)=(1-gamma) S(nT);

V(nT⁺)=V(nT)+ gamma* S(nT),

with the initial conditions  S(0⁺)=s₀

V(0⁺)=v₀

    how to plot the graph with this system of equations,impulsive points and initial conditions  

i am facing problem while solving differential equation in loops where conditons are given also in loops. the particular problem occurs while solving Differential equation involving He's Homotopy purturbation method

How  to make the integration 

into the following form:

Thanks in advance,Q1.mw

IsFrobeniusGroup(SmallGroup(20, 3)) will get true, but IsFrobeniusPermGroup(SmallGroup(20, 3)) will get false. What happen? As the documentation, it will get same result:

The two definitions are equivalent in the following sense.  If G is a Frobenius permutation group, then G is Frobenius as an abstract group

I want a maple code to solve the caputo fabrizio differential equations using Runge Kutta method with implicit functions and impulsive conditions in maple. Is there any code structure for that. 

restart;
with(Student[NumericalAnalysis]);
with(plots);
with(DEtools);
f := proc(u, r) local res; res := 1/25*r^2 + (sin(u(r)) + sin(diff(u(r), [r $ 1/5])))/(r^2 + 47); return res; end proc;

RK4 := proc(f, u0, r0, h, n) local u, r, i, k1, k2, k3, k4; u := Vector(n + 1); r := Vector(n + 1); u[1] := u0; r[1] := r0; for i to n do k1 := f(u[i], t[i]); k2 := f(u[i] + 1/2*h*k1, r[i] + 1/2*h); k3 := f(u[i] + 1/2*h*k2, r[i] + 1/2*h); k4 := f(u[i] + h*k3, r[i] + h); u[i + 1] := u[i] + 1/6*h*(k1 + 2*k2 + 2*k3 + k4); r[i + 1] := r[i] + h; end do; return [u, r]; end proc;
RK4 := proc (f, u0, r0, h, n) local u, r, i, k1, k2, k3, k4; u 

   := Vector(n+1); r := Vector(n+1); u[1] := u0; r[1] := r0; 

   for i to n do k1 := f(u[i], t[i]); k2 := f(u[i]+(1/2)*h*k1, 

   r[i]+(1/2)*h); k3 := f(u[i]+(1/2)*h*k2, r[i]+(1/2)*h); k4 := 

   f(u[i]+h*k3, r[i]+h); u[i+1] := u[i]+(1/6)*h*(k1+2*k2+2*k3+k4\

  ); r[i+1] := r[i]+h end do; return [u, r] end proc

u0 := cos(abs(0.9))/15;
                      u0 := 0.04144066455

r0 := 0;
                            r0 := 0

h := 0.1;
                            h := 0.1

n := 100;
                            n := 100

solution := RK4(f, u0, r0, h, n)

u := solution[1];
r := solution[2];
plot(u, r, style = line, color = blue, labels = ["Time (r)", "Solution (u)"]);
 is this correct to solve the implicit fractional differential equations using 4th order Runge-Kutta Method. will fsolve command  solve the fractional differential equations ?

How to convert this set of PDEs into ODEs? PDEs.mw

I have a list as follows:
lst := [`A=70`, `B=17`, `C=27`, `D=37`, `E=74`, `F=57`, `G=67`, `H=08`, `I=81`, `J=28`, `K=38`, `L=48`, `M=58`, `N=68`, `O=90`, `P=19`, `Q=29`, `R=39`, `S=49`, `T=59`, `U=96`, `V=010`, `W=110`, `X=210`, `Y=310`, `Z=410`, "SPACE=105", "DOT=106"]

however using tt:=convert(lst, table) gives

tt:=table([1 = `A=70`, 2 = `B=17`, 3 = `C=27`, 4 = `D=37`, 5 = `E=74`, 6 = `F=57`, 7 = `G=67`, 9 = `I=81`, 8 = `H=08`, 11 = `K=38`, 10 = `J=28`, 13 = `M=58`, 12 = `L=48`, 15 = `O=90`, 14 = `N=68`, 18 = `R=39`, 19 = `S=49`, 16 = `P=19`, 17 = `Q=29`, 22 = `V=010`, 23 = `W=110`, 20 = `T=59`, 21 = `U=96`, 27 = "SPACE=105", 26 = `Z=410`, 25 = `Y=310`, 24 = `X=210`, 28 = "DOT=106"])

I want this to be such that I can retrieve the numerical value assigned to each alphabet e.g.
tt[A] should produce 70.

Are the results consistent either we use simplify(expression) or simplify(expression,size)? It seems (2) and (3) are not consistent.

Download simplisize.mw

I am trying to define a function f(x1+tau,x2)= (x1+tau)*a+x2*b. I need this to be able to take derivations w.r.t. the sum (x1+tau) and to only tau as well. Is there a way to do this?

I want to approximate a positive function that is decreasing in Gamma, say f(Gamma), that is very complicated yet very smooth. I need this in order to obtain a tractable and compact version of its derivative, which enters in the partial derivative of another (very simple) function.

Along the way, three related questions emerge: Derivatives_and_Approximations.mw

Thanks a lot!

I don't why the limit is not providing me the answer; What did I do wrong?