The CauchyRiemann procedure (for older version  of Maple )doesn't work quite right in Maple 2024 .
Also ran the procedure through the AI for so-called code improvement and now it shows what the code stands for 
The output according to the original procedure would look like on the screenshot, but running original procedure does not give this output ? 
I also want to extend the procedure with a plot of the complex function. 
That differentiability of complex functions is not obvious even if the cauchy-riemann equation is satisfied ?

>  (1) >  >  CauchyRiemann:=proc(expr::algebraic) # original procedure   local x, y, u, v, u_x, u_y, v_x, v_y, flag1, flag2;   u:=evalc(Re(eval(expr, z=x+I*y)));   v:=evalc(Im(eval(expr, z=x+I*y)));   u_x:=diff(u,x);   u_y:=diff(u,y);   v_x:=diff(v,x);   v_y:=diff(v,y);   print('f(z)'=expr);   printf("\n");      print('u(x,y)'=u);   print('u[x](x,y)'=u_x);   print('u[y](x,y)'=u_y);   printf("\n");   print('v(x,y)'=v);   print('v[x](x,y)'=v_x);   print('v[y](x,y)'=v_y);   printf("\n");   if u_x=v_y then     print('u[x]=v[y]');     print(u_x=v_y);     flag1:=true;   else     print('u[x]<>v[y]');     print(u_x<>v_y);     flag1:=false;   end if;   if u_y=-v_x then     print('u[y]=-v[x]');     print(u_y=-v_x);     flag2:=true;   else     print('u[y]<>-v[x]');     print(u_y<>-v_x);     flag2:=false;   end if;    printf("\n"); if flag1=true and flag2=true then    print(`Fullfill the Cauchy-Riemann Equations`);    print(`The derivative is:`='u[x]+I*v[y]');    print('diff(f(z),z)'=u_x+I*v_y); else    print(`Cauchy-Riemann ?`); end if end proc: >  f(z):=1/(z+2): CauchyRiemann(f(z)) (2) >    Also ran the procedure through the AI for so-called code improvement and now it shows what the code stands for >  restart; # Improved and corrected version of the CauchyRiemann procedure :ASKED AI  CauchyRiemann := proc(expr::algebraic)     local x, y, u, v, u_x, u_y, v_x, v_y, CR1, CR2;     # Assign real and imaginary parts of the function     u := evalc(Re(eval(expr, z = x + I*y)));     v := evalc(Im(eval(expr, z = x + I*y)));     # Calculate partial derivatives     u_x := diff(u, x);     u_y := diff(u, y);     v_x := diff(v, x);     v_y := diff(v, y);     # Properly format and print function details     printf("f(z) = %a\n", expr);     printf("u(x, y) = %a, u_x = %a, u_y = %a\n", u, u_x, u_y);     printf("v(x, y) = %a, v_x = %a, v_y = %a\n", v, v_x, v_y);     # Evaluate and print Cauchy-Riemann equations     CR1 := u_x = v_y;     CR2 := u_y = -v_x;     printf("\nCauchy-Riemann Equations:\n");     printf("u_x = v_y: %a\n", CR1);     printf("u_y = -v_x: %a\n", CR2);     # Check both equations     if CR1 and CR2 then         printf("The function is analytic (holomorphic) at this point.\n");         printf("The derivative f'(z) is %a + I*%a\n", u_x, v_y);     else         printf("The function does not satisfy the Cauchy-Riemann equations and is not analytic.\n");     end if; end proc; # Test the procedure with a specific function f := z -> 1/(z + 2); CauchyRiemann(f(z)); f(z) = 1/(z+2) u(x, y) = (x+2)/(y^2+(x+2)^2), u_x = 1/(y^2+(x+2)^2)-(x+2)/(y^2+(x+2)^2)^2*(2*x+4), u_y = -2*(x+2)/(y^2+(x+2)^2)^2*y v(x, y) = -y/(y^2+(x+2)^2), v_x = y/(y^2+(x+2)^2)^2*(2*x+4), v_y = -1/(y^2+(x+2)^2)+2*y^2/(y^2+(x+2)^2)^2 Cauchy-Riemann Equations: u_x = v_y: 1/(y^2+(x+2)^2)-(x+2)/(y^2+(x+2)^2)^2*(2*x+4) = -1/(y^2+(x+2)^2)+2*y^2/(y^2+(x+2)^2)^2 u_y = -v_x: -2*(x+2)/(y^2+(x+2)^2)^2*y = -y/(y^2+(x+2)^2)^2*(2*x+4) The function does not satisfy the Cauchy-Riemann equations and is not analytic.

Download CAUCHY_RIEMANN_-FORUM_VRAAG.mw

I want to solve or try to solve this equation 

PDE := diff(G(a, H, phi, PI), a)(aH) + diff(G(a, H, phi, PI), H)(k/a^2 - kappa^2/2*PI^2/a^6) + diff(G(a, H, phi, PI), phi)(PI/a^3) = diff(G(a, H, phi, PI), PI)(a^3*diff(V(phi), phi))

with pdsolve(PDE, G)

and maple answer me the next

Error, (in pdsolve/info) first argument does not have a differentiated function with name G

I nw in maple, maybe I´m make a mistake, but I can't find what

How can I  get the result for the integration when m =n or m is not equal to n and How can I add assumption that m, or n can be even or odd?

Thanks in advance for your help.

Download test1.mw

I have  4 worksheets with derived equations. So I export the equations and  possibly some procedures (but they can be handled seperately if needed)  from each worksheet as a .mpl file. 

I want to combine the .mpl files  together without using copy/paste. Then I can open that single file in the VS code editor.
There may be other ways to achieve this so I am open to suggestions.

Good evening!

I am Athanasios Paraskevopoulos, a graduate student specializing in applied mathematics. Recently, I've started exploring Maple through its trial version and I'm considering making a purchase. My question for you all is: Am I restricted to buying only the graduate student package, or am I free to choose from any of the Maple packages available? Any guidance or personal experiences with different packages would be greatly appreciated!

Thank you in advance for your help!

I chased down a problem to factoring a square that has sqrt in the coefficients. All numbers are real,
The code is inside a procedure in a package. Iso I could do with something robust.

expand((sqrt(A+B)*x+sqrt(7-K)*y)^2)
     2      2            (1/2)          (1/2)        2      2
  A x  + B x  + 2 (A + B)      x (7 - K)      y - K y  + 7 y 

factor(%) 

Using a command in succession can coax maple into different outputs. 

eq:=(a*x+b)/(c*x+d)=1
                                
Getting rid of the denominator is probably the most deciding factor in how maple displays an output.

isolate(eq,b)
                      

isolate(%,x,1)
                      

Using the isolate command, you couldn't arrive at that output without using it twice. 

I was trying to solve a system of polynomial equations, which contains three equations and six variables $a_0,a_1,a_2,b_0,b_1,b_2$. However, as I swap the variable name, Maple solve function gives me a totally different solutions. Only the solutions before swapping the variables are useful for the problem I study. I have already attached the file. Could anyone tell me if the choice of variable name really matters, or if i just misuse this function?

Choice_of_name_infolevel.mw

Hello, everyone,

I am new to Maple and I am trying to get use of it.

I tried to plot the following linear systems in different ways. I realized that the Student Linear Algebra is not as flexible as Linear Algebra. My question is the following. Is there any other way to create a plot without defining the implicit plots?

Download linear_systems.mw

How can I get the roots of this equation: besselJ,  J0(xn)=0?

Multiplication is possible by a numerical constant

a < b;
-1*%;
                             a < b

                            -b < -a

However, using a name with assumptions

with(RealDomain):# just to make sure that a and b are real, probably not required
ineq := a/c < b/c;
assume(c < 0);
c*ineq;
expand(%);
eval(%);
                                 a   b
                         ineq := - < -
                                 c   c

                           /a   b   \
                          *|- < -, c|
                           \c   c   /

                           /a   b   \
                          *|- < -, c|
                           \c   c   /

                           /a   b   \
                          *|- < -, c|
                           \c   c   /

does not work. Have I missed an essential assumption/trick?

What else can I do?

Or: is there a mathematical reasons not to offer this possibilty.

Context: For manual case studies of parameters in inequalties I find it sometimes helpfull to remove rational expressions. Doing this manually with lhs/rhs and denom and flipping < or > is prone to errors

How can numbers be displayed inside the contour plot?

 restart;
with(plots);
contourplot(x*exp(-x^2 - y^2), x = -2 .. 2, y = -2 .. 2, axes = boxed);
like this

Dear All,

I solved the following partial differential equation numerically using pdsolve. As mentioned in the help pages, plotting the solution versus x at a given time is possible. 
Can anyone help me extract the plot of the solution versus time at a given space variable x? Also, how can I extract the data of the mentioned plot, for example, u_numeric (0.75, 0.5)?
pd_numeric:=(D[2,2])(u_numeric)(x,t)+(D[1,1,1,1])(u_numeric)(x,t)-0*h(x,t)=0;
bc_numeric[1]:=u_numeric(0,t)=0;
bc_numeric[2]:=(u_numeric)(1,t)=0; 
bc_numeric[3]:=D[1,1](u_numeric)(0,t)=0;
bc_numeric[4]:=D[1](u_numeric)(1,t)=0; 
ic_numeric[1]:=u_numeric(x,0)=0.1*x*(x-1)^2;
ic_numeric[2]:=D[2](u_numeric)(x,0)=0;

sol:=pdsolve(pd_numeric, {seq(bc_numeric[i], i=1..4), seq(ic_numeric[i],i=1..2)}, u_numeric(x,t), time=t, range=0..1, numeric, spacestep=1/2000, timestep=1/2000);
sol:-plot(t=1);
sol:-animate(t=1, frames=2000, title="time=%f");
Best wishes