Hi,

I have been looking at some new models of Casio Scientific Calculators and came across with "Fx-115es Plus" Model which seem to have a some sort of simple CAS(Computer Algebra System) built into it.

Two new features which i really liked were

(i) Ability to make any part of the expression inert and simplying the rest.

(ii) Fully Integrated Repeated decimal display for fractions.

I want to ask if there is any builtin commands that can achieve these two effects in maple.

I will give some example for each of these

(i) simplifying say 2^3*2^4 in maple gives 32.

but forexample if i want to make 2 in the bases inert then simplifying the result should give 2^7

if i make 3 inert then the result is 16*2^3

if i make 4 inert then the result is 8*2^4

another example say (2^3)^4 in maple gives 4096

but if i make 2 inert then the result should be 2^12

if i make 3 inert then the result is 16^3

if i make 4 inert then the result is 8^4

In this way it is possible to keep any interesting part of large complex expression unevaluated and simplifying the rest across it to maintain focus on the interesting part.

I know i can try to achieve this effect by using unevaluation quotes but they get messy and harder to track in large nested forms.

Another approach might be to replace the inert parts by explicit undeclared symbols with required assumptions and simplifying, but this is not it.

I know in Maple 18 they have introduced some package called InertForm or something, can it achieve this effect and also mark inert parts of the expression as grey like it is possible for some operators.

(ii) the example for the second is quite obvious, say given the fraction 237/14, evalf of this gives 16.92857143 but a result like 16.9Overscript[285714, _] is more closer to differentiation it from a irrational expansion. Sorry i donot know how to pretty print this here.

Another advantage is when i want to give some large repeating decimal expansion and have maple convert it to fractional form. Currently i have no idea how many times to repeat the decimals explicitly to make maple understand that it is a repeating decimal expansion.

I use "dsolve([ode_1])" command to solve an ODE, and the solution contains lots of " I * ln(cos(m)+I*sin(m)) " expression. As "m" is real, I think this expression is equal to "-m", but the maple command "simplify" do nothing for this expression. Any one who can simplify the expression by using maple ? or ever puzzled by similar problem ?

Hi,

Please I need you to add in the output of my code the order of error defined in the procedure.

Thanks for helping me.

Here, the code.

QuestionNumber2.mw

Hi, I would like to solve this nonlinear problem :

with :

I use the NLPSolve command and i get this error : Error, (in Optimization:-NLPSolve) integer variables are not accepted.

Can you help me ?

Hi users:

i will solve the system of equations with laplace method in maple,every body can help me?

EQ1:= -1.383117237*10^(-9)*q2(T)+4.345190953*10^(-9)*q1(T)+3.643814673*10^(-17)*(diff(q1(T), T, T)) = 0
EQ2 := 3.522088316*10^(-10)*q2(T)-1.106493791*10^(-9)*q1(T)-6.094634321*10^(-16)*q4(T)+6.094634321*10^(-16)*q5(T)+2.429209782*10^(-24)*(diff(q2(T), T, T)) = 0
EQ3 := 1.378642186*10^(-8)*q3(T)+4.570975739*10^(-12)*q4(T)+4.570975739*10^(-12)*q5(T)+3.643814673*10^(-17)*(diff(q3(T), T, T)) = 0
EQ4 := .6366197720-.5208074999*(diff(q5(T), T))-0.5140163994e-5*q4(T)-4.092794921*q5(T)+.7896341498*(diff(q3(T), T))-.5208074999*(diff(q4(T), T)) = 0
EQ5 := .3183098860-1.041615857*q4(T)+3.088010748*q5(T)-0.6580284584e-4*(diff(q2(T), T))-0.8680125003e-1*(diff(q4(T), T))+0.8680125003e-1*(diff(q5(T), T)) = 0

with initial conditions:

q1(0) = 0, q2(0) = 0, q3(0) = 0, q4(0) = 0, q5(0) = 0, (D(q1))(0) = 0, (D(q2))(0) = 0, (D(q3))(0) = 0

Hi,

I solve laplace equation in a square. All the lines of my code is okay.

Please just look to the last part of my code titled procedure:

When I run my code without (last funciton f #f := (x,y) -> 0;) see last lines to find "f". It's runing, there is no problem. But when I put add f, there is an error. Many think  for any help.

Procedure
Using the previous suty in section stencil we can write the procedure to solve the Laplace equation in [0,1]*[0,1] with the boundary condition Neumann conditions on the vertical boundary and Dirichlet boundary condition on the horizontal baoundary. In our study we will use the same stepsize h in x and y direction.

PoissonSolve:=proc(N,_f)
local Z,i,h,y,x,sys,w,f,sol,j,u,Data;
# define basic grid parameters
Z := i -> (1/(N+1))*i;
x[0] = Z(0),x[N+1] = Z(N+1),y[0] = Z(0),y[N+1] = Z(N+1);
 h := evalf(Z(1)-Z(0));
# Fix the boundary data and the source matrix
for i from 0 to N+1 do:
    # Neumann boundary condition
     u[N+1,i] :=  u[N,i] ;    
     u[0,i] := u[1,i];
     # Dirichlet boundary condition
     u[i,0] := 0;
     u[i,N+1] := 0;
   od:
   f := Array(0..N+1,0..N+1,[seq([seq(evalf(_f(Z(i),Z(j))),i=0..N+1)],j=0..N+1)],datatype=float);
# Write down the system of equations to solve and solve them
     sys := [seq(seq(Stencil(h,i,j,u,f),i=1..N),j=1..N)];
  w := [seq(seq(u[i,j],i=1..N),j=1..N)];
  sol := LinearSolve(GenerateMatrix(sys,w));
   # parse the solution vector sol back into "matrix" form
   for i from 1 to N do:
     for j from 1 to N do:
        u[i,j] := sol[(j-1)*N+i]:
     od:
   od:
# generate a 3D plot of the solution using the surfdata command
   Data := [seq([seq([Z(i),Z(j),u[i,j]],i=0..N+1)],j=0..N+1)]:
surfdata(Data,axes=boxed,labels=[`x`,`y`,`u(x,y)`],shading=zhue,style=patchcontour);
end proc:



Here is an example of the output when the source function is set to zero
                                 f(x, y) = 0
; i.e., when  reduces down to Laplace's equation:
#f := (x,y) -> 0;
#PoissonSolve(10,f);

Question8.mw

Hi

I need you to help me in writting procedure with input "r" ( order of derivative) and some coefficients seq(alpha[i],i=1..N).  My code work very well, need only put all the element in procedure with output The Taylor series obtained in last line of my code and the order of error.  I want the procedure return the coefficients beta used in the series and the order of Error  and the coefficients beta[i]

May thinks.

Second_Question.mw

I will not modify my previous question: only I add this remark. I tried to write these lines of procedure.
Add_display.mw

I need only to add( if the procedure is true) these lines in my procedure.  These lines gives the degree of error computed in the code, but when I put these lines in the codes, there is an error. Thanks for your help.

Degree := degree(Error,stepsize);
    if (showorder) then:
       print(cat(`This stencil is of order `,Degree));
    fi:
    if (showerror) then:
       print(cat(`This leading order term in the error is `,Error));
    fi:
    convert(D[r$n](f)(vars) = stencil,diff);

Hello my dear friends:

i will obtain value f(4) from below equation:(f is funciton of t)

f[0](t)=a*cos(omega*t)
f[n+1](t)=f[0](t)+(1/omega)*int((omega^2*f[n](tau)-f[n](tau)^3)*sin(omega*(t-tau)),tau=0..t)

THANKS...

Hi:

how obtain value a4 from below equation in maple:

a[0]=1

a[n+2]=a[0]+a[n+1]

I calculate eigenvalues and  eigenvectors of a floating-point square matrix M with the command Eigenvectors(M, output = 'list').  How can I estimate errors of my results?

I have two 6x1 Matrices which are the results of a calculation process in Maple. One with a set of equations and the other one with a set of variables: 

A := [0, f(x6), f(x6), 0, 0, f(x6)];

b := [x1, x2, x3, x4, x5, x6];

I'd like to solve the following system:

for i from 1 to 6 do

eq[i] := A[i] = b[i]:

od;

which is

eq[1] := 0 = x1;

eq[2] := f(x6) = x2;

eq[3] := f(x6) = x3;

...

If I type in the eqations manually, and execute "s := solve({eq[1],..,eq[6]},{x1,..,x6})" everything solves fine.

If I use the "for i from..." - structure, and execute "s := solve({eq[1],..,eq[6]},{x1,..,x6})" I get an empty space as solution.

I've tried to convert both matrices into lists, but it doesn't work.

Could it be that Maple doesnt know that x6 has to be the x6 in the function f(x6) ?

Can anyone tell me how to solve this please?

Hi,

I need your help to fix the error in this code.  many thinks
restart:
pde:=diff(u(t,x),t$2)=diff(u(t,x),x$2)-sin(u(t,x)):
f:=x->x^2:
IBC := {u(0,x)=f(x),u(t,-50)=0,D[2](u)(t, 50)=0,D[1](u)(0, x)=-diff(f(x),x)}:

pds := pdsolve(pde, IBC, numeric):

p1 := pds:-plot(t = 0);
p2 := pds:-plot(t = 1/10);
p3 := pds:-plot(t = 1/2);
p4 := pds:-plot(t = 1);
p5 := pds:-plot(t = 2);
plots[display]({p1, p2, p3, p4, p5}, title = `Sine Gordaon at t=0,0.1,0.5,1,2`);

Hi Maple-Prime-ers!

I have a system of equations, containing 18 variables and 13 equations, making this a 5 degree of freedom (DOF) system.  I would like to analytically solve each of the equations in terms of each of these DOFs.  Normally I would use solve(system, dof_variables) to accomplish this, but it doesn't return anything.  Not even [].

I can solve this system by hand.  I've included a hand-solution involving isolate() and subs() in the attached worksheet.  I'm looking to incorporate this in an optimization algorithm with varying system, so I would like an automated way of doing this.

Does anybody have any suggestions to get solve to work as intended?

3driversys_FD_BRAKE_ICE_GEN.mw

Here is the system I am talking about:

The free variables are:  {FD_T, FD_W, ICE_T, EM2_T, BRAKE_T}

I'm looking for a solution in this form:

For context, I'm designing a work sheet based around quantum tunneling. Currently I'm looking at the boundary conditions.

What I want to be able to do is to set the expression psi[1] equal to psi[2], but only for the value x = 0. Is this possible? I've tried using if statements, and I've considered converting these expressions into functions for this purpose, but I'm not having much luck. 

Thanks

Blanky

Dear all,

Please, I need help to remplace in the code attached,  addadd(beta[i]*u(x+[i]*h),i=0..N)+Error; 
by add(beta[i]*u(x+alpha[i]*h),i=0..N)+Error;  I see, that we must add a function, alpha, but how.

I think I must modify this rows: but how:

stencil := add(a[ii[i]]*subsop(r=op(r,phi)+ii[i]*stepsize,phi),i=1..N);

Question2.mw

Many thinks