Hey there,

I'm trying to build a procedure that can function as an adapted form of Prim's algorithm. The idea is that on a graph with just vertices, the procedure has a starting point, and from there will find out which vertex is the cheapest to connect to (currently expressed purely by the lowest distance). Once this is found, the connected vertex is removed from a list that has vertices that aren't connected yet, and added to a list of vertices that are in the minimal spanning tree.

My problem is that I get an error returned that says "invalid Boolean expression", and I'm not sure how to solve it. Can anybody here point me in the right direction?

The procedure is defined as follows:

Primmetje := proc (aantal, posities, begin)
local knopenover, knopeninmst, huidig, V, kaart, e, a;
knopenover := [seq(i, i = 1 .. aantal)];
knopeninmst := {};
huidig := [0, 0];
if begin <> {} then
  V := [begin];
knopeninmst := knopeninmst union {V}
end if;
remove(V, knopenover);
kaart := Graph(aantal);
SetVertexPositions(kaart, posities);
while nops(knopeninmst) < aantal
do for e in knopeninmst
   do for a in knopenover
     do if huidig = [0, 0] or Distance(posities[e], posities[a]) < Distance(posities[huidig[1]], posities[huidig[2]]) then
   huidig := [e, a];
knopeninmst := knopeninmst union {a};
remove(a, knopenover);
AddEdge(kaart, huidig)
end if
end do
end do
end do
end proc

When I try to execute it with some parameters the return is this:

vp := [2.5, 21], [6, 13.5], [8, 10], [11, 24.5], [14.3, 19.4], [16.8, 26], [22, 21.5], [22, 17], [22.2, 12.5], [26.8, 23], [28, 20.5], [30, 25.5], [32, 21], [29.5, 16];
Primmetje(14, vp, 1);
Error, (in Primmetje) invalid boolean expression: [[6, 13.5]]

I think it has something to do with the double brackes, but I'm not sure how to solve it.
 

restart;
M := -6*alpha*eta^2*mu*(lambda^2*mu+1)/`&vartheta;`+12*alpha*eta^2*lambda*mu^(3/2)*(sqrt(mu)*cot(A+sqrt(mu)*eta*(-t*(-alpha*eta^2*mu+f)+x))*lambda+1)/(`&vartheta;`*cot(A+sqrt(mu)*eta*(-t*(-alpha*eta^2*mu+f)+x)))-6*alpha*eta^2*mu*(sqrt(mu)*cot(A+sqrt(mu)*eta*(-t*(-alpha*eta^2*mu+f)+x))*lambda+1)^2/(`&vartheta;`*cot(A+sqrt(mu)*eta*(-t*(-alpha*eta^2*mu+f)+x))^2)+6*mu*eta^2*alpha*sqrt(sigma*(1+cot(A+sqrt(mu)*eta*(-t*(-alpha*eta^2*mu+f)+x))^2))/(sqrt(sigma)*theta*cot(A+sqrt(mu)*eta*(-t*(-alpha*eta^2*mu+f)+x))^2);
             2    /      2       \   
  6 alpha eta  mu \lambda  mu + 1/   
- -------------------------------- + 
             &vartheta;              

                                1                               
  ------------------------------------------------------------- 
                /      (1/2)     /   /          2       \    \\ 
  &vartheta; cot\A + mu      eta \-t \-alpha eta  mu + f/ + x// 

  /            2          (3/2) /  (1/2)    / 
  \12 alpha eta  lambda mu      \mu      cot\A

       (1/2)     /   /          2       \    \\           \\   
   + mu      eta \-t \-alpha eta  mu + f/ + x// lambda + 1// - 

                                1                                
  -------------------------------------------------------------- 
                                                               2 
                /      (1/2)     /   /          2       \    \\  
  &vartheta; cot\A + mu      eta \-t \-alpha eta  mu + f/ + x//  

  /           2    /  (1/2)    / 
  \6 alpha eta  mu \mu      cot\A

       (1/2)     /   /          2       \    \\           \  \   
   + mu      eta \-t \-alpha eta  mu + f/ + x// lambda + 1/^2/ + 

  /                /      / 
  |        2       |      | 
  \6 mu eta  alpha \sigma \1

                                                       2\\      \/
        /      (1/2)     /   /          2       \    \\ ||      | 
   + cot\A + mu      eta \-t \-alpha eta  mu + f/ + x// //^(1/2)/ 

  /                 
  |     (1/2)       
  \sigma      theta 

                                                    2\
     /      (1/2)     /   /          2       \    \\ |
  cot\A + mu      eta \-t \-alpha eta  mu + f/ + x// /
alpha := 2;
                               2
eta := 3;
                               3
mu := 1.5;
                              1.5
lambda := 2;
                               2
theta := 3;
                               3
sigma := .5;
                              0.5
b := .5;
                              0.5
f := 5;
                               5
y := 0;
                               0
plot3d([abs(M)], x = -3 .. 3, t = -3 .. 3);
 

How I can extract data about figure as matrix format.

I want to save data from figure as matrix for each curve in figure as i.e.,  x and y .

for example as this format.

fig10.mw

sol := Vector[column](4, [` 1001 x 4 `*Matrix, `Data Type: `*anything, `Storage: `*rectangular, `Order: `*Fortran_order])

hi everyone,

I am trying to find maxima for a multi variable function. As a first step I want to find out partial differentiation and equate each expression to zero.
 

Normally I would expect to solve these equations to get the variable value(s). But I came across a perticular function which cannot be solved as above explained.

Expression :=30*a*sin(theta)+80*cos(theta)*b
variable :=  {a, b, theta} # using indets(Expression,name)

PartialDifferentiationExpressions :={80*cos(theta), 30*sin(theta), 30*a*cos(theta)-80*sin(theta)*b}

How can I find the critical Points from these?

My approach is to find Hessian Matrix and ten find out maxima for the function(at which critical point it happens)

Note Edit: This is an example. I am writing a script which takes an equation as Input and solve for the maxima. Hence, I would like to know how to when this kind of situation arises. Thank You

Hi

How can i solve this system of equations:

N and P are variables and the others are constant numbers.

I used "solve" function but the results are:

piecewise(And(1 < c, 1 < d, 1 < k, 0 < -RootOf(exp(_Z)*c*d*k+exp(_Z)*_Z*d-exp(_Z)*c*k-c*d*k+c*k-_Z)*d/((d-1)*c), 0 < -RootOf(exp(_Z)*c*d*k+exp(_Z)*_Z*d-exp(_Z)*c*k-c*d*k+c*k-_Z)), [{N = d*k*(exp(RootOf(exp(_Z)*c*d*k+exp(_Z)*_Z*d-exp(_Z)*c*k-c*d*k+c*k-_Z))-1)/(d*exp(RootOf(exp(_Z)*c*d*k+exp(_Z)*_Z*d-exp(_Z)*c*k-c*d*k+c*k-_Z))-1), p = c*k*(d*exp(RootOf(exp(_Z)*c*d*k+exp(_Z)*_Z*d-exp(_Z)*c*k-c*d*k+c*k-_Z))-d-exp(RootOf(exp(_Z)*c*d*k+exp(_Z)*_Z*d-exp(_Z)*c*k-c*d*k+c*k-_Z))+1)/(d*exp(RootOf(exp(_Z)*c*d*k+exp(_Z)*_Z*d-exp(_Z)*c*k-c*d*k+c*k-_Z))-1)}], [])

Is it possible to export data from Maple to Excel where data are generated by Maple using Optimization:-LSSolve or DirectSearch:-SolveEquations function?

I have attached a file, where n is the primary variable. For n=1, 2, 3, .......... 20, I have got solution individually. But it is more time consuming event. Is it possible to use array or a vector for the variable n and solve equations at a time for all value of n and stored a excel file?
Any advice for me? @Preben Alsholm , @tomleslie or anybody
system_equations_(10.12.2008).mw
 

I've made calculation as array := function(array1,array2). Then summ of all array[i] all in one cycle. If I look in summ variable I see fu(0+fu() +.. agebraical representation. I tried evalf(summ). still algebraical. How to derive numerically?

I want to calculate the following summation for a,b;

when it is odd, 

vanishes.

MY TRY:

restart:
J:=proc(a,b,c) local u,v,p,w;
p:=(a+b+c):
if p=2*k then
u:=b+c-a: 
v:=c+a-b: 
w:=a+b-c:
return (-1)^p*sqrt(binomial(2*u,u)*binomial(2*v,v)*binomial(2*w,w)/((2*p+1)*binomial(2*p,p)))
else 
return 0
end if
end proc:
aa:=2:bb:=4:
sum(J(aa,bb,cc)*J(aa,bb,cc)*sqrt(2*(2*aa+1)*(2*bb+1)*(2*cc+1))*psi[1,cc],cc=abs(bb-aa)..(aa+bb));

Hi
I am confused by solve and SolveEqutions and their options!
Solving same equation results different solutions.
Which  solution is correct and why?
How i can be sure one of the solutions is correct about the other equations?
I attached my code.

Thanks.

Test2_MaplePrimes.mw

Its difficult for me to explain what I'm looking for so I will try with an example:

Example 1

Is there anyway any function that will say for example the sum of squares of integers for a given n say n=4, 1^2 + 2^2 + 3^2 + 4^2 = 30 that will give the equation ((n^3)/3) + ((n^2)/2) + (n/6) that gives the sum for a given n?

It would be determined from the series 1^2 + 2^2 + 3^2 + 4^2 + ... or the equation for the elements 1^2, 2^2, 3^2, 4^2 the equation that gives the sum; in this case the well known ((n^3)/3) + ((n^2)/2) + (n/6) .

So far I've got FindSequenceFunction[{1, 4, 9, 16}, n] but this only finds the individual elements not an equation for the sum of them. Of course I already know the equation for producing the individual elements.

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