Hi everyone:

How can I convert the following phrase (eq) into the phrase in the photo?

eq:= -w*z^2/(2*E*MI*alpha^2)+w*H*sinh(alpha*H)*cosh(alpha*z)/(alpha^3*E*MI*cosh(alpha*H))-w*H*sinh(alpha*H)/(alpha^3*E*MI*cosh(alpha*H))+w*cosh(alpha*z)/(alpha^4*E*MI*cosh(alpha*H))-w/(alpha^4*E*MI*cosh(alpha*H))+w*H*z/(E*MI*alpha^2)-w*H*sinh(alpha*z)/(E*MI*alpha^3):

the phrase in the photo:

tnx...

I want to write a proc, say f, that takes an single argument Z, as in
f := proc(Z::?) ...
where the only acceptable Z values are pairs [x,y], where x and y are selected from the set {a,b,c,d}. The entries a,b,c,d of that set are prescribed symbolic constants. 

Thus, the following are legal calls to f:
f([a,a]), f([a,b]), f([d,c])
but these are illegal:
f([a]), f([a,b,c]), f([a,x])

I don't know what to put for the "::?" in the proc's definition in order to enforce those constraints.  Any ideas?

Extra: In fact, I would prefer to ban the f([a,a]) option as well (no repeated symbols) but that's of secondary importance since it is easy to inspect and handle that within the proc's body.

Hello. I am a student who has been given access to Maple through my school. Normally, the program works just fine and without any problems. But today I tried to do a linear regression which did not work as it used to do. Though, the exponential (ExpReg) and the potential (PowReg) regression still works normally.

Here is a picture of what the error looks like:

As shown on the picture, I use a special extension called the 'Gym-pakke' which we use in my high school. As far as I know, this extension is created for Danish high schools?

Another thing that I had like to emphasize, is that the linear regression still works whenever I copy-paste the command from a previous document - a document from before the issue began. But I am tired of having to copy-paste every time I have to do a linear regression.

Please help me solve this problem, thanks! :)

Hi there!

I am trying to save the coefficients of a polynomial in a list to work with them in a rather complicated procedure. It is about representing a polynomial via a set of orthogonal polynomials phi_n which change depending on the input. For example, phi_s*phi_0=1*phi_s, so the coefficient of phi_s is 1 and the rest is 0. I save this as a[s][0][s]:=1. In this procedure, however, the coefficient of phi_{s+1} or phi_{s+10} might come up, and I have not declared them as 0, so the procedure stops whenever something like a[s][0][s+2] or a[s][0][s+10] appears. I could work with polynomials I guess, saving x^s for the result of

I did two attempts with different ideas
 

Download betounes_ex_set_2_task_5.mw

restart;
a := 10;
ff := proc()

       ##local a,b;

        b := a + 10:  #### implicit local

         return b:

        end proc:
ff();

output: 20

restart;

a := 10:
ff := proc()

       ##local a,b;

        b := a + 10:   #### implicit local

        a:=b:

        return b:

        end proc:
ff();

output: a+10

expected: 20

Please explain the reason

How can I create an even function, g, in Maple? I want Maple to give g(x) - g(-x) as 0.

Good day sirs, I write a system of DAE but giving me this code "(The use of global variables in numerical ODE problems is deprecated, and will be removed in a future release. Use the 'parameters' argument instead (see ?dsolve,numeric,parameters)". The code is attached below.

Thanking you in anticipating for your help.

Help!!!!.mw

Hello, I was given the problem "Set N:=100. (i) Form four lists L[1], L[3], L[5], L[7] where L[r] contains all primes 2< p < N such that p mod 8 = r ." and came up with the following code:

N := 100;
List1 := [];
List3 := [];
List5 := [];
List7 := [];
for p from 2 to N do
    if isprime(p) and p mod 8 = 1 then List1 := [op(List1), p]; end if;
    if isprime(p) and p mod 8 = 3 then
        List3 := [op(List3), p];
    end if;
    if isprime(p) and p mod 8 = 5 then
        List5 := [op(List5), p];
    end if;
    if isprime(p) and p mod 8 = 7 then
        List7 := [op(List7), p];
    end if;
end do;
List1;
List3;
List5;
List7;
 

this gave me the answer I wanted, however using the above code I have to answer this second question: 

In the notation of Problem 1, make a procedure with input = arbitrary  positive integer N and output = the list [nops(L[1]), nops(L[3]), nops(L[5]), nops(L[7])]. Do some experiments to see for which (if any) r, L[r] is largest. 

I am unsure of how to create a procedure out of the code I already have. I created this: 
 

restart;
F := proc(n)

local N, p, List1, List3, List5, List7;

List1 := [];

List3 := [];

List5 := [];

List7 := [];

for p from N do

if isprime(p) and p mod 8 = 1 then List1 := [op(List1), p]; end if;

if isprime(p) and p mod 8 = 3 then List3 := [op(List3), p]; end if;

if isprime(p) and p mod 8 = 5 then List5 := [op(List5), p]; end if;

if isprime(p) and p mod 8 = 7 then List7 := [op(List7), p]; end if;

return [nops(List1), nops(List3), nops(List5), nops(List7)];

end do;

end proc;

however, when I input F(100) or any other value of N, I am receiving the error message "Error, (in F) initial value in for loop must be numeric or character"

Any ideas on how to improve my program to get the output I desire?

thank you 

Download 23May(1).mw

Seems to make no difference with or without RETURN() for the procedure ?

vraag_op_forum_gesteld_over_boek_vb_binomium.mw

For to know what each graph is standing for in the plot legenda

Could not yet get it  in y , y' , y '' , to 5th derative 

Better is perhaps to have vertical table with the names and function expression together , but that could be difficult betounes_ex_set_task_4def.mw  

f := x -> exp(-x)*sin(x); intvx:= 0..3;
f := proc (x) options operator, arrow; exp(-x)*sin(x) end proc
intvx := 0 .. 3
 

And this one below ( i prefer this one ) , but got in worksheet now the one above
Probably a option issue ?

2 determinants equal, solve for the unknowns in one determinant/Linear Algebra

Hi guys, thank you for reading my question.

A is one determinant "Matrix(2, 3, [[1, 2, 3], [3, 1, 2]])".B is" Matrix(2, 3, [[1, x, 3], [y, 1, z]])"

I wrote 2 matrice in the maple

As I try to solve the unknowns,I wrote:

A:= B, solve(x, y, z) It doesn't work

I switch another phrase,

A = B;solve(B:x,y,z)    It doesn't work as well

Could you help me with it ?

Thank you

Dear All.

Please kindly help to correct the attached code on discretization of fourth order PDE using method of line.
Thank you and kind regards.

Download Discretization_of_PDE_Order_4.mw