I need to evaluate a variable inside the function definition in the moment it is defined and not every time it is called. Is there a way around?

a:=3;
f := x-> a*x;
f(t);

a:=4;
f(t);

I want the function "f" to be 3*x even if "a" is modified after the declaration. In Mathematica I am used to the so called immediate assignmet(=) and delayed assignment(:=) for which I cannot find an equivalent command in Maple. 

Any help is appreciated!

im trying to write differential quadrature method i have successfully  generated system of equations using code but now im stuck with rk4 method to solve these equation in a loop kindly give me suggestions or modify the code

Download dq_code.mw

I have a Prime Version Abo. On my iPhone works everyrhing fine. But on IPad there is a Limit of 5 step by step solutions. Same AppleID. What's the Problem?

Can the display of an object o in Maple, i.e.
> o;
                                                      display of the object
be controlled in a standard Maple worksheet, e.g. with Typesetting or something similar?

Hello, 

I have come across a problem in Maple, when I try plotting af slope field from my differential equation. 

I have tried different ways of defining the ODE, but when I use it in the command of "linjeelementer" (Danish for line elements - It's a command that's part of a package called "With (Gym)" in Maple, it gives me the error: Error, (in DEtools/DEplot/CheckDE) - Derivatives must be given explicitly. 

I don't understand what derivatives I must "give explicitly"?

Translation of the relevant sentences if anyone needs it when looking at the screenshot:

"I need to draw a slope field for the differential equation along with the solution curve for the population growth in India after 1960.

I define the differential equation:
ODE := -0.000032229*y^2 + 0.065843*y - 15.103"

"I am drawing the slope field for the differential equation along with the solution curve for the population growth in India after 1960, i.e., the solution curve that passes through the point (0,449):I am drawing a slope field for the differential equation along with the solution curve for the population growth in India after 1960, i.e., the solution curve that passes through the point (0,449):
linjeelementer(ODE, y(t), t = 0 .. 350, y = 0 .. 2000)"

Is there a way to change the font on the help pages?               

restart;
with(geometry);
with(plots);
_EnvHorizomtalName = 'x';
_EnvVerticalName = 'y';
_local(D);
line(delta, y = 1/3*x - 2, [x, y]);
line(deltap, y = (-1)/4*x + 1, [x, y]);
line(D, y = 3*x - 5, [x, y]);
point(S, 3, 0);
omega := Pi/3;
intersection(P, delta, D);
intersection(Pp, deltap, D);
projection(H, S, delta);
projection(K, S, deltap);
projection(M, S, D);
circle(c1, [H, K, M], 'centername' = O1);
display*[textplot*([[coordinates(S)[], "S"], [coordinates(P)[], "P"], [coordinates(H)[], "H"], [coordinates(K)[], "K"], [coordinates(M)[], "M"]], font = [times, bold, 16], align = [above, right]), draw*([delta(color = blue), deltap(color = blue), D(color = red), c1(color = black), S(color = black, symbol = solidcircle, symbolsize = 16), P(color = black, symbol = solidcircle, symbolsize = 16)], scaling = constrained, axes = none, view = [-15 .. 15, -15 .. 15])];
               [                /[             [9  -13     ]  
plots:-display [plots:-textplot |[[3, 0, "S"], [-, ---, "P"], 
               [                \[             [8   8      ]  

  [33  -9     ]  [52  4      ]  [9  2     ]]  
  [--, --, "H"], [--, --, "K"], [-, -, "M"]], 
  [10  10     ]  [17  17     ]  [5  5     ]]  

                                                  \              
  font = [times, bold, 16], align = [above, right]|, geometry:-d\
                                                  /              

  raw ([delta(color = blue), deltap(color = blue), 

  D(color = red), c1(color = black), 

  S(color = black, symbol = solidcircle, symbolsize = 16), 

  P(color = black, symbol = solidcircle, symbolsize = 16)], 

  scaling = constrained, axes = none, view = [-15 .. 15, -15 .. 15])];
  display*[textplot*([[3, 0, "S"], [9/8, -13/8, "P"], [33/10, -9/10, "H"], [52/17, 4/17, "K"], [9/5, 2/5, "M"]], font = [times, bold, 16], align = [above, right]), draw*([delta(color = blue), deltap(color = blue), D(color = red), c1(color = black), S(color = black, symbol = solidcircle, symbolsize = 16), P(color = black, symbol = solidcircle, symbolsize = 16)], scaling = constrained, axes = none, view = [-15 .. 15, -15 .. 15])]
Why the program will not run; Thank tou.

Hello, 

This morning when i tried to open my Maple file, it didn't work and instead wrote "There were problems during the loading process. Your worksheet may be incomplete". Is there anyone that can help me save the document? 

I read in here that i may be due to something in the file but i do not know how to remove it? 

I caught aome serious errors with Maple 22 calculating the Ricci tensor not only wrong for known metrics, but also producing garbage entries in the Ricci tensor e.g. bare thetas where only trigonometric functions of angles are expected. Completely wrong.

That aside,

I need to replace the built in Ricci tensor calculation in Maple and therefore need maple to contract the Christoffel Symbols.

d_[alpha](Christoffel[~alpha, mu, nu]) - d_[nu](Christoffel[~alpha, mu, alpha]) + Christoffel[~beta, mu, nu]*Christoffel[~alpha, beta, alpha] - Christoffel[~beta, mu, alpha]*Christoffel[~alpha, nu, beta]

How do I tell Maple to contract thae alpha and beta indices in above expression, without crudely having to set up Summing.?

Hi,

I'm trying out the 2024 version of Maple and I'm getting the following warning message:

Warning, not a built-in function (`rtable_alias`)

which I didn´t get for the 2023 version. I have no clue where it is coming from since it happens even when I start a new worksheet:

I've also attached print outs of the same worksheets (from Maple help examples and from Maple Portal), one using Maple 2023 version and the other one using Maple 2024 version so youcould see the warning and some other problems.

I really appreciate if someone would have an idea of what is going on here. Thanks very much in advance.

interpolation_2023.pdf

interpolation_2024.pdf

optimization_2023.pdf

optimization_2024.pdf

I want like this plots here two phase are there boundary is -2 to0 and 0 to 2

f(±2)=0,g(±2)=0,f(0)=1,h(±2)=1,H(±2)=1

This is perhaps a stupid question. It is quite easy to add a vector with minutes to 60 minutes. However, when I try to do this stating 1 hour instead of 60 minutes it does no longer do the calculation. Why is this? Thank you for any explanation.

QuestionMP.mw

In the example below, the second call to simplify is ineffective

eq := (a*x + b)/(c*x + d) = 1;

`simplify/nodenom`:= x -> (numer@(lhs - rhs) = 0)(x);                       
                      
`simplify/nodenom`(eq);
simplify(eq, nodenom);

Why is that?
Convert, for example, works

`convert/nodenom`:= x -> (numer@(lhs - rhs) = 0)(x);

`convert/nodenom`(eq);
               
convert(eq, nodenom);

Let M a matrix of zeros and ones for instance.

N := 10^5:  # This is the order of magnitude I have to consider (10^4 may be enough if L is "small")
L := 5:     # L is between 3 and 8 in practice (see context at the end of this post)
M := LinearAlgebra:-RandomMatrix(N, L, generator=0..1):

Let C a set of colums of M and |C| the number of elements of C.
Let P a list of length |C| of  zeros and ones.

I want to count al the rows in M such that 

`and`( seq( M[n, C[k]] = P[k], k = 1.. |C| ) )

For instance

pattern := n -> `if`(evalb(`and`(seq(S[n, C[k]]=P[k], k=1..3))), 1, 0):
t0 := time():
add(pattern(n), n=1..N);
time()-t0
                             12351
                             0.431  # seconds

This code appears to be quite slow when one consider that I have to run it for all the possible choices of C and P (the contect is detailed below).

Do you have any some ideas to improve the efficiency (in terms of computational time) of this counting operation?
Thanks in advance

The context:
Let A = {1, 2, .., L} and P(A) the power set of A.
I set Q(A) = P(A) \ ( { } union A}.
Let C and C'  two disjoint subsets of Q(A).
Example

# With L=3

Q(A) = { {1},  {2},  {3},  {1, 2},  {1, 3},  {2, 3} }

# if A chose C = 
C = {2}

# Then C' is member of 
C' in { {1}, {3}, {1, 3}

For any couple (C, C')  I want to do this

1. Let |C| (|C'|) cardinal of C (resp C').
    
2. Let us consider that members of C (resp C') "point to" the corresponding columns of M.
   for instance C = {2, 3} refers to columns 2 and 3 of M.
   As C and C' are ordered iI will write them indistinctely as lists (for instance C = [2, 3] in example above) when necessary.
    
3. To a given |C| (|C'|) one may associate 2^|C| (resp 2^|C'|) sequences of 0 and 1.
   For instance,  to C = {2, 3} are associated sequences [0, 0], [0, 1], [1, 0], [1, 1].
   Let s(C) (s(C')) the list, or set, of 0-1 sequences associated to C (resp C')..
    
4. For any couple (C, C') and any of the 2^|C|+|C'| combinations (p, p') where p belongs to s(C) and p' belongs to s(C') I want to count the number of occurrences defined by: 

   seq( M[n, C[k]] = p[k], k=1..|C|) and seq( M[n, C'[k]] = p'[k], k=1..|C'|) 

   For instance, if C={2} and C={1, 3} I want to count the number M rows M such that 

   p = [0], p'=[1, 0]
   M[n, 2]=0 and M[n, 1]=1 and M[n, 3]=0;
   
   p = [1], p'=[1, 1] 
   M[n, 2]=1 and M[n, 1]=1 and M[n, 3]=1
   
   ... and so on

The computational time may ne quite large because of the number of possible couples (C, C') when L is large ( there exists  Stirling2(L+1, 3) such couples: already 90 couples for L=5  ... and 28501 for L=10).

Hi,

I hope everyone is fine here. I want to write the following nested loop in my code

restart; TOL := 10^(-3); v[i] := 10^(-4); u[i] := 10^(-4);
if abs(v[i]-1)<=TOL then    if abs(u[i])<=TOL then break    else t[i+1]:=t[i]+alpha; else; fi; fi;

it is not working. If the first condition (v[i]-1)<=TOL is fulfilled, then we have to check the second condition, u[i]<=TOL. That’s why I don’t use “and” here. If both conditions are fulfilled, then stop the loop and if second the condition not fulfilled (but first fulfill) then t[j+1]=t[j]+alpha. 

I shall be waiting for your positive response.