MaplePrimes Questions

restart:

sys:={-diff(v(x,t),t)+0.5*p*diff(u(x,t),x,x)+q*u(x,t)*(u(x,t)^2+v(x,t)^2)=0,diff(u(x,t),t)+0.5*p*diff(v(x,t),x,x)+q*v(x,t)*(u(x,t)^2+v(x,t)^2)=0};
        /                      /  2         \
        |/ d         \         | d          |
sys := < |--- u(x, t)| + 0.5 p |---- v(x, t)|
        |\ dt        /         |   2        |
        \                      \ dx         /

               /       2          2\       / d         \
   + q v(x, t) \u(x, t)  + v(x, t) / = 0, -|--- v(x, t)|
                                           \ dt        /

           /  2         \                                      \ 
           | d          |             /       2          2\    | 
   + 0.5 p |---- u(x, t)| + q u(x, t) \u(x, t)  + v(x, t) / = 0 >
           |   2        |                                      | 
           \ dx         /                                      / 
eq1 := diff(u(x,t),t) = u__t(x,t):
eq2 := diff(v(x,t),t) = v__t(x,t):

sys_tmp := subs(eq1, eq2, sys):

sys_new := sys_tmp union {eq1, eq2}:

Boundary conditions:
bc :=
    u(0,t) = 2,
    v(0,t) = 0;
                 bc := u(0, t) = 2, v(0, t) = 0
Initial conditions:
ic :=
    u(x,0) = tanh(2*Pi),
    v(x,0) = tanh(2*Pi),
    u__t(x,0) = 0,
    v__t(x,0) = 0;
 
ic := u(x, 0) = tanh(2 Pi), v(x, 0) = tanh(2 Pi), u__t(x, 0) = 0, 

  v__t(x, 0) = 0
Solve the system:
pdsol := pdsolve(subs(p=1, q=0.5, sys_new), {ic, bc}, numeric);

I want to change my variables. The coding below shows the influence of the Nb and Nt on the local Nusselt number (-theta') against the convection Bi.

But now I want it to be  the influence of the Bi on the local Nusselt number (-theta') against the Nb and Nt.

Can anyone help me?

NULL

with(plots)

DE1 := diff(f(eta), `$`(eta, 3))+f(eta)*(diff(f(eta), `$`(eta, 2)))-(diff(f(eta), eta))^2 = 0

diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2 = 0

(1)

DE2 := diff(theta(eta), `$`(eta, 2))+Pr*f(eta)*(diff(theta(eta), eta))+Pr*Nb*(diff(theta(eta), eta))*(diff(phi(eta), eta))+Pr*Nt*(diff(theta(eta), eta))^2 = 0

diff(diff(theta(eta), eta), eta)+Pr*f(eta)*(diff(theta(eta), eta))+Pr*Nb*(diff(theta(eta), eta))*(diff(phi(eta), eta))+Pr*Nt*(diff(theta(eta), eta))^2 = 0

(2)

DE3 := diff(phi(eta), `$`(eta, 2))+Le*(diff(phi(eta), eta))+Nt*(diff(theta(eta), `$`(eta, 2)))/Nb

diff(diff(phi(eta), eta), eta)+Le*(diff(phi(eta), eta))+Nt*(diff(diff(theta(eta), eta), eta))/Nb

(3)

BC1 := f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0

f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0

(4)

BC2 := theta(10) = 0, (D(theta))(0) = -Bi*(1-theta(0))

theta(10) = 0, (D(theta))(0) = -Bi*(1-theta(0))

(5)

BC3 := phi(0) = 1, phi(10) = 0

phi(0) = 1, phi(10) = 0

(6)

``

  getRes:= proc(a, x);
                local sol;
                if   type(x, numeric)
                then sol:= dsolve
                           ( eval
                             ( {BC1, BC2, BC3, DE1, DE2, DE3},
                               [Nt = a, Nb = a, Pr = 2, Le = 5, Bi = x]
                             ),
                             numeric,
                             output = listprocedure,
                             abserr = 0.0001,
                             maxmesh = 1024,
                             initmesh = 512
                           );
                else return 'procname(a,x)';
                fi;
                return -eval(diff(theta(eta), eta), sol)(0)
           end proc:

  cl:=[red, green, blue, yellow, black]:
  L:=[0.2, 0.4, 0.6,0.8,1.0]:
  display
  ( [ seq
      ( plot
        ( getRes( L[k], x),
          x=0..5,color = colorList[k], legend = ["Nb=Nt" = L[k]], legendstyle = [location = top],
          color=cl[k]
        ),
        k=1..5
      )
    ]
  )
               

 

 

 

Download odeProb2.mw

Contour integration notation

" (&int;)[+infinity]^(+infinity)((-x)^(z))/((e)^(x)-1). (&DifferentialD;x)/(x)"

 

The limits of integration are intented to indicate a path of integration which begins at + ∞, moves to th e left down the positive real axis, circles the orign once in positive ( counterclockwise) direction, and returns up to the positive real axis to  +∞

-How does this contour look like  in a  graph ?
- the "(&DifferentialD;x)/(x)" notation  ?
- calculating this complexe contour integral?

Seems that the concept of the contour integration is similar wit a line integral in real calculus ?

NULL

Download contourintegraal_vraag1.mw

I have just come across this curious but really boring problem.
In the code snippet below, expr1 and expr2 are identical.

restart:
data := [a=1, b=2]:
convert(
  piecewise(And(x(t) > a, x(t) < b), 1, 0),
  Heaviside
):
expr1 := eval(eval(%, data),  x(t)=z):
plot(expr1, z=0..3);


convert(
  piecewise(And(z > a, z< b), 1, 0),
  Heaviside
):
expr2 := eval(%, data),:
plot(expr2, z=0..3);


But if I change the parameterization of the problem, expr2 is still correct but expr1 is not

restart:
data := [d=1.5, a=0.5]:
convert(
  piecewise(And(x(t) > d-a, x(t) < d+a), 1, 0),
  Heaviside
):
expr1 := eval(eval(%, data),  x(t)=z):
plot(expr1, z=0..3);


convert(
  piecewise(And(z > d-a, z< d+a), 1, 0),
  Heaviside
):
expr2 := eval(%, data),:
plot(expr2, z=0..3);

Where does this come from?

PS: I'm sorry not to be able loading the mw file

Hello, 

I would like to write to a file the output of the command CodeGeneration[Matlab]. I couldn't find a way to write the output. 

What I am trying to do is to generate complex symbolic equations and expressions and write them as either a function or variables in matlab. 

As an example: 

exp := [sin(theta(1)), cos(theta(1))]; 

func := unapply(exp,[theta]); 

CodeGeneration:-Matlab(func)

If i can write the output of the above command in a matlab file, it will generate a matlab function file. 

I was using the command 'writeto' but this one basically outputs every line a command written to a file instead of the terminal, which is not what I want (it will output even the warnings, I know I can silent them!). 

can you please help.. thanks. 

Hello guys,

I want to plot two functions such as x(t) and y(t) in a unique diagram as a function of each other. In a routine way, one needs to solve one of these functions as t and then input its results in others. for example, solving x(t) to find t and then input this t(x) into y(t) to have y(x). but here problem is that I cannot solve x(t) to find t and so this routine solution is not accelssible.

x(t):=1 + (3*n*(v - 1)^2*A*(t^v)^2*(1 + alpha*(v - 1)^2*A^2*(t^v)^2*ln(m^2*t^2/((v - 1)^2*A^2*(t^v)^2))/(3*n^2*m^2*t^2) + A^2*beta*(v - 1)^2*(t^v)^2/(3*n^2*m^2*t^2))*((4*A^2*alpha*t^v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + 6*t^(-v + 2)*m^2*n^2 + 4*t^v*(beta - alpha/2)*A^2*(v - 1)^2)*sqrt(n^2*(v - 1)^2*A^2*t^(2*v)/v^2) - 3*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2))*(1 + ((4*A^2*alpha*t^v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + 6*t^(-v + 2)*m^2*n^2 + 4*t^v*(beta - alpha/2)*A^2*(v - 1)^2)*sqrt(n^2*(v - 1)^2*A^2*t^(2*v)/v^2) - 3*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2))/(3*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2))))/(2*v^2*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2)) - (3*n^2*(v - 1)^2*A^2*(t^v)^2*(1 + alpha*(v - 1)^2*A^2*(t^v)^2*ln(m^2*t^2/((v - 1)^2*A^2*(t^v)^2))/(3*n^2*m^2*t^2) + A^2*beta*(v - 1)^2*(t^v)^2/(3*n^2*m^2*t^2))*(((4*A^2*alpha*t^v*v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2))/t + 4*A^2*alpha*t^v*(v - 1)^2*(2 - 2*v)/t + 6*t^(-v + 2)*(-v + 2)*m^2*n^2/t + 4*t^v*v*(beta - alpha/2)*A^2*(v - 1)^2/t)*sqrt(n^2*(v - 1)^2*A^2*t^(2*v)/v^2) + (4*A^2*alpha*t^v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + 6*t^(-v + 2)*m^2*n^2 + 4*t^v*(beta - alpha/2)*A^2*(v - 1)^2)*n^2*(v - 1)^2*A^2*t^(2*v)/(sqrt(n^2*(v - 1)^2*A^2*t^(2*v)/v^2)*v*t) - 3*n*A*(2*A^2*alpha*t^(2*v)*v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2))/t + A^2*alpha*t^(2*v)*(v - 1)^2*(2 - 2*v)/t + 2*A^2*beta*(v - 1)^2*t^(2*v)*v/t + 6*n^2*m^2*t))/(3*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2)) - (((4*A^2*alpha*t^v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + 6*t^(-v + 2)*m^2*n^2 + 4*t^v*(beta - alpha/2)*A^2*(v - 1)^2)*sqrt(n^2*(v - 1)^2*A^2*t^(2*v)/v^2) - 3*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2))*(2*A^2*alpha*t^(2*v)*v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2))/t + A^2*alpha*t^(2*v)*(v - 1)^2*(2 - 2*v)/t + 2*A^2*beta*(v - 1)^2*t^(2*v)*v/t + 6*n^2*m^2*t))/(3*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2)^2)))/(2*v^2):

y(t):=1 + ((4*A^2*alpha*t^v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + 6*t^(-v + 2)*m^2*n^2 + 4*t^v*(beta - alpha/2)*A^2*(v - 1)^2)*sqrt(n^2*(v - 1)^2*A^2*t^(2*v)/v^2) - 3*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2))/(3*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2)) - (((4*A^2*alpha*t^v*v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2))/t + 4*A^2*alpha*t^v*(v - 1)^2*(2 - 2*v)/t + 6*t^(-v + 2)*(-v + 2)*m^2*n^2/t + 4*t^v*v*(beta - alpha/2)*A^2*(v - 1)^2/t)*sqrt(n^2*(v - 1)^2*A^2*t^(2*v)/v^2) + (4*A^2*alpha*t^v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + 6*t^(-v + 2)*m^2*n^2 + 4*t^v*(beta - alpha/2)*A^2*(v - 1)^2)*n^2*(v - 1)^2*A^2*t^(2*v)/(sqrt(n^2*(v - 1)^2*A^2*t^(2*v)/v^2)*v*t) - 3*n*A*(2*A^2*alpha*t^(2*v)*v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2))/t + A^2*alpha*t^(2*v)*(v - 1)^2*(2 - 2*v)/t + 2*A^2*beta*(v - 1)^2*t^(2*v)*v/t + 6*n^2*m^2*t))/(3*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2)) - (((4*A^2*alpha*t^v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + 6*t^(-v + 2)*m^2*n^2 + 4*t^v*(beta - alpha/2)*A^2*(v - 1)^2)*sqrt(n^2*(v - 1)^2*A^2*t^(2*v)/v^2) - 3*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2))*(2*A^2*alpha*t^(2*v)*v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2))/t + A^2*alpha*t^(2*v)*(v - 1)^2*(2 - 2*v)/t + 2*A^2*beta*(v - 1)^2*t^(2*v)*v/t + 6*n^2*m^2*t))/(3*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2)^2))*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2)/((4*A^2*alpha*t^v*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + 6*t^(-v + 2)*m^2*n^2 + 4*t^v*(beta - alpha/2)*A^2*(v - 1)^2)*sqrt(n^2*(v - 1)^2*A^2*t^(2*v)/v^2) - 3*n*A*(A^2*alpha*t^(2*v)*(v - 1)^2*ln(m^2*t^(2 - 2*v)/((v - 1)^2*A^2)) + A^2*beta*(v - 1)^2*t^(2*v) + 3*n^2*m^2*t^2)):

there are x(t) and y(t). I want to plot y as x, not t. so please help me.

with best

Hello people in Mapleprimes,
I haven't written in this site for a long time.

I have a question in the below program, which is to write ribbons.
For the implementation of this program, 
I wrote this.
ribbonplot5([cos, sin, cos + sin], -Pi .. Pi,numpoints=20);

In the part of pattern matching, as -Pi .. Pi above is a range, so it is OK.
But, when I changed this part to x=-Pi..Pi, an error message appears.

ribbonplot5([cos, sin, cos + sin],x=-Pi .. Pi,numpoints=20);

brings error messages:
"Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct"

In the code of ribbonplot5,  x=-Pi..Pi which is the type of name=range shoud satisfy the pattern matching, as I wrote in the first part of the program as 
ribbonplot5 := proc(Flist, r1::{range,name=range})

I cannot know why the part of the program:
 
" else
       newFlist := map(unapply, Flist, lhs(r1));
     opts := ['labels'=[lhs(r1), " "," "],
                 args[3 .. nargs]];
    ribbonplot5(newFlist,rhs(r1),op(opts)):
"

wouldn't work well.

I wish I could get an answer to this. 

Thanks in advance.

This is a program in maple9 Advanced Programming Gude p. 253

graphic_ribbonplot.mw

restart;
extend := proc(f)
     local x, y;
     unapply(f(x), x, y);
end proc:
p:=x->cos(2*x):
q:=extend(p);


ribbonplot5 := proc(Flist, r1::{range,name=range})
     local i, m, p, n, opts,newFlist;
     opts := [args[3 .. nargs]];
     if type(r1, range) then
         if not hasoption(opts, 'numpoints', 'n', 'opts')
         then n := 25
         end if; 
         m := nops(Flist);
         p := seq(plot3d(extend(Flist[i]), r1, (i-1) .. i,
                                  grid=[n, 2], op(opts)),
                       i = 1 .. m):
         plots[display](p):
     else
       newFlist := map(unapply, Flist, lhs(r1));
     opts := ['labels'=[lhs(r1), " "," "],
                 args[3 .. nargs]];
    ribbonplot5(newFlist,rhs(r1),op(opts)):
   end if:
end proc:
ribbonplot5([cos, sin, cos + sin], -Pi .. Pi,numpoints=20);

I'd like to get the simplest possible expression when using simplify. The problem is that this is all done in a program, without having the benifit of looking at the expression on the screen and trying things. So I need a method that works for large set of input all the time.

I noticed in this example that simplify did not do what I think it should have.

Given the expression -2*(x^3 + 2)/(x*(x + 1)*(sqrt(3)*I + 2*x - 1)*(sqrt(3)*I - 2*x + 1)) it can be simplified to (x^3 + 2)/(2*x^4 + 2*x) but  I had to go into many tries in order to get this final result.

Is there a better way to do this? I end up now   doing simplify(expand(numer(expr))/expand(denom(expr))) in the hope to get better simplification.  Here is an example


 

interface(version);

`Standard Worksheet Interface, Maple 2021.2, Windows 10, November 23 2021 Build ID 1576349`

restart;

w:=-2*(x^3+2)/x/(x+1)/(I*3^(1/2)+2*x-1)/(I*3^(1/2)-2*x+1);

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

simplify(w)

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

simplify(evalc(w))

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

simplify(w) assuming x::real

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

simplify(w,size)

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

simplify(w,symbolic)

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

simplify(w,sqrt)

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

simplify(expand(numer(w))/expand(denom(w)))

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


It would be nice if Maple simplify would just do it directly as follows in Mathematica. May be there is a an  option I am overlooking

 

Download simplify.mw

I drew a methane molecula and display three times because I wanted to get two-fold and three-fold symmetry axis. ,I would is that I would like to get a more exact output and doing it without clicking, it's say, using code. Can someone help me?tareaclase6.mw
 

University of Costa Rica 

Course: SP-1493

Student: Lic. Marcus Vinicio Mora Salas

Instructor: Dr. Erick Castellón Elizondo

 

Assigment5

Description

 

In this assigment exercises e2.113,  e2.114, e2.115 and e2.117 was solved.

NULL

***e2.113

   

 

 

 

NULL


 

Download tareaclase6.mw

 

Suppose I have a real function which is defined on these intervals [x0,x1],...[x1,xn], and on each interval it's defined differently, for example f(x):=sqrt(x) if x>=0; f(x):=1/x if x<0.

For the last example I tried my last resort to use the assuming operator the following way:

f := ((x -> sqrt(x)) assuming (0 <= x)) or ((x -> 1/x) assuming (x < 0)) plot(f, x)

But it doesn't seem to be working, how do you propose to define this function? perhaps proc?

I tried looking for an example in maple's help, but didn't find any.

Your help is appreciated.

I read that abs(v) should find the magnitude of complex number. But I find cases where it does not.

Should one then use sqrt( Re(v)^2 + Im(v)^2 ) instead or is there different command to use?  Here is an example

interface(version)

`Standard Worksheet Interface, Maple 2021.2, Windows 10, November 23 2021 Build ID 1576349`

restart;

v:=1/16*(I*5^(1/2)+I-4*sin(1/5*Pi))*2^(3/10)*(5-5^(1/2))^(1/2)-1/8*2^(4/5)*((I*5^(1/2)+I)*sin(1/5*Pi)+1/2*5^(1/2)+3/2)

(1/16)*(I*5^(1/2)+I-4*sin((1/5)*Pi))*2^(3/10)*(5-5^(1/2))^(1/2)-(1/8)*2^(4/5)*((I*5^(1/2)+I)*sin((1/5)*Pi)+(1/2)*5^(1/2)+3/2)

abs(v)

-(1/16)*(I*5^(1/2)+I-4*sin((1/5)*Pi))*2^(3/10)*(5-5^(1/2))^(1/2)+(1/8)*2^(4/5)*((I*5^(1/2)+I)*sin((1/5)*Pi)+(1/2)*5^(1/2)+3/2)

sqrt( Re(v)^2 + Im(v)^2 );

((-(1/4)*2^(3/10)*(5-5^(1/2))^(1/2)*sin((1/5)*Pi)-(1/8)*2^(4/5)*((1/2)*5^(1/2)+3/2))^2+((1/16)*2^(3/10)*(5-5^(1/2))^(1/2)*(5^(1/2)+1)-(1/8)*2^(4/5)*(5^(1/2)+1)*sin((1/5)*Pi))^2)^(1/2)

 

Download complex.mw

Here is the same thing in Mathematica:

I searched help for modulus but could not find one.

Maple 2021.2

Can someone confirm, that autosave is not implemented for workbooks?

The only thing I can restore form a workbook session are specific worksheets of the workbook that I was working on.

None of the Maple codes are saved for example.

This is a major issue, folks. If you haven't implemented it, you need to tell us about it in the help document.

IN CAPITAL LETTERS!

Hi all,

Do you know how one can ask Maple to show expressions involving special functions using their"classical symbols" instead of the associated Maple's command ?

Thanks !

Kevin

I following a example of products multiplication like this one

u:=n->Product(2*k-1,k=1..n)/Product(3*k-1,k=1..n)*x^n;

Calculating with  this with maple 1d input is correct, but when i convert a maple 1d input  to 2D input ( i did somewhere) and use this then there is difference with the maple 1d calculation

Seems to be not a advisable to use converted maple 1d to 2 D input for calculation : ( for a mixed calculation(maple input/2D input)  or solely 2d input) , but only for purpose of seeing what the expression in maple input is standing for.   

Note: i did the calculation again with mixed input and now the correct sequenze of answers shows up ?

Just wanted to post that I had some data loss because I opened two different workbooks with the same name from different locations.

This could lead to loss of data.

I had code attachments from the "old" workbook implemented in the "new" workbook, while the new code was gone.

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