Has anyone tried animated ascii art, in this way? Say, a dancing stick-man? acer

It seems to be more and more common in the answers provided on mapleprimes that opinion appears before accuracy. The copy command has worked on tables and arrays for many releases. But of those two structures, it's only really needed for tables. But now consider the help-page for array. Where, up to and including release Maple 10, is the cross-reference to ?copy ? When was the array help-page ever updated to mention the copy() command? How about the vector or matrix help-pages? The lack of properly updated documentation is not so dramatically new. As Joe pointed out, Vector(V) produces a copy of Vector V. Things work similarly for arrays, Matrices, and Arrays. The command array(a) produces a copy of array a, and so on. So maybe it should be this functionality that should be well documented. Moreover, if the help-pages of Vector, Array, Matrix, and rtable are going to get a cross-reference to the copy help-page, then let them *also* mention that those constructors themselves can produce copies. And why not document this difference too, that copy preserves almost all rtable properties, while the constructors themselves may not. Eg, M:=Matrix(2,2,shape=symmetric): N:=copy(M): P:=Matrix(M): MatrixOptions(N); MatrixOptions(P); acer

The general form of your Matrix is not clear. You say that it is nxn, but the portion below the first row is itself nxn. Perhaps one of these below matches what you intend. with(LinearAlgebra): M := n -> Matrix(n,n,[[Vector[row](n,[seq(x||i,i=1..n)])], [Matrix(n - 1, n, Matrix(n-1,n-1,shape=identity))]]): Minv := n -> Matrix(n,n,[[<>], [Vector[row](n,[1/(x||n),seq(-(x||i)/(x||n),i=1..n-1)])]]): Norm(MatrixInverse(M(7))-Minv(7)); Norm(M(7).Minv(7)-IdentityMatrix(7)); MM := n -> Matrix(n,n,[[Vector[row](n,[seq(x||i,i=1..n)])], [<>]]): MMinv := n -> Matrix(n,n,[[Vector[row](n,[1/(x||1),seq(-(x||(i))/(x||1),i=2..n)])], [<>]]): Norm(MatrixInverse(MM(7))-MMinv(7)); Norm(MM(7).MMinv(7)-IdentityMatrix(7)); acer

Once you've created a procedure, you can add option remember to it. Eg, f := proc(x) 2*x; end proc; g := subsop(3=remember,eval(f)); acer

How about this, t:= seq(i*Unit(s),i=100..2000,100); acer

This is probably not near the most efficient way, but... dice := proc(n::posint) local i, N, p, new; new := convert(n, string); N := iquo(StringTools[Length](new) + 1, 3); p := Array(1 .. N); for i to N do p[i] := parse(StringTools[Take](new, 3)); new := StringTools[Drop](new, 3) end do; convert(p, list); end proc: dice(632096185); diceL := proc(l::list(posint)) local i, j, N, p, new, result; result:=Array(1..nops(l)); for j from 1 to nops(l) do new := convert(l[j], string); N := iquo(StringTools[Length](new) + 1, 3); p := Array(1 .. N); for i to N do p[i] := parse(StringTools[Take](new, 3)); new := StringTools[Drop](new, 3); end do; result[j]:=convert(p, list); end do; convert(result,list); end proc: diceL([632096185,123456789,214365879]); acer

p1:=nextprime(10^200); p2:=nextprime(p1); evalf(p1*p2); p1:=nextprime(10^399); p2:=11: evalf(p1*p2); acer

There are probably much easier and slicker ways to do the elimination, but four of the equations were so simple that it could be done "by hand". restart: C := 2: TY := .9: Y := Vector(1 .. C, TY/C): TM := 1: DL := 5: eq1 := X1 = (Y[1]+Z2)/TM: eq2 := X2 = (Y[2]-Z2)/M2: TX := X1+X2: eq3 := M2 = (1-X1)*TM+X1/(1/TM+W1): W0 := Y[1]/TM^2+Y[2]/M2^2: EF := W0/(1-TX): eq4 := W2 = EF+X1*min(W2, DL): eq5 := W1 = (TX*EF-X2*W2)/X1: eq6 := Z2 = Y[2]*max(0, W2-DL)/W2: # Four of the equations are so simple that we can eliminate # them by hand. list0 := subs(eq2,[eq1,eq3,eq4,eq5,eq6]): list1 := subs(eq3,list0): list2 := subs(list1[5],[list1[1],list1[3],list1[4]]): list3 := subs(list2[1],[list2[2],list2[3]]): # Solve the remaining two equations in two variables. sol1 := fsolve({op(list3)},{W1,W2}); # Now backsubstitute. z2 := eval(eq6,[op(sol1)]): x1 := eval(eq1,z2): m2 := eval(eq3,[x1,op(sol1)]): x2 := eval(eq2,[m2,z2]): # Check the original equations, by substition. eval([eq1,eq2,eq3,eq4,eq5,eq6],[z2,x1,m2,x2,op(sol1)]); # Here's a solution. z2,x1,m2,x2,op(sol1); # Find another solution, avoiding the first solution. # It helps fsolve to know a range, which may also tell # it that the ranges are purely real. sol2 := fsolve({op(list3)},{W1,W2},avoid={sol1}, W1=-1000..1000,W2=-1000..1000); z2 := eval(eq6,[op(sol2)]): x1 := eval(eq1,z2): m2 := eval(eq3,[x1,op(sol2)]): x2 := eval(eq2,[m2,z2]): # Check the original equations, by substition. eval([eq1,eq2,eq3,eq4,eq5,eq6],[z2,x1,m2,x2,op(sol2)]); # Here's a second solution. z2,x1,m2,x2,op(sol2); The two solutions I saw were, Z2 = -0., X1 = 0.4500000000, M2 = 0.6744105183, X2 = 0.6672493797, W1 = 2.617057514, W2 = -22.32043803 Z2 = 0.4462841010, X1 = 0.8962841010, M2 = 0.1052051185, X2 = 0.03532051532, W1 = 600.8482070, W2 = 605.5062301 Increasing Digits didn't radically alter the final answer, so the simplified two equations seem "stable". acer

The statement that "..in the above command, ALL numerical computations are being done with only two significant digits," doesn't seem quite right. > stopat(`evalf/sin`): > a:=sin(Pi/8): > evalf(a,2); The above indicates to me that computation is done something like, > xr := evalf(Pi/8,2): > evalf(evalhf(sin(xr)),2); That still produces .36, sure. But the sine computation is done in evalhf it seems. The input to sin is just .37, though, as you described. Is this next true? If one extra guard digits had been used in approximating Pi/8 then the final result would be accurate to 2 places. > restart: > xr := evalf(Pi/8,3): > evalf(evalhf(sin(xr)),2); 0.38 Is sin "atomic" enough to warrant guard digits for approximating the argument, assuming what I wrote above is true? acer

Does it work if you use (only) single-forward slashes? acer

Something like this? with(LinearAlgebra): U:=Matrix(2,2,symbol=u): V:=Matrix(2,2,symbol=v): complicated_expr:=Determinant(U) *CharacteristicPolynomial(U,x)/CharacteristicPolynomial(V,x): func:=unapply(complicated_expr, convert(map(rhs,rtable_elems(U) union rtable_elems(V)),list) ); acer

When you issue, Optimization[NLPSolve](proc2(a,b,c), a=50E-9..100E-9,b=100E-9..150E-9,c=50..100) the first argument proc2(a,b,c) gets evaluated right away. This is how Maple normally works. That is, the unassigned names a,b,c get passed to proc2 right away. And the result of that evaluated call is what actually gets used as the objective. There are several ways to work with this. The simplest was might be to put unevaluation quotes on the proc2 argument. Eg, Optimization[NLPSolve]('proc2'(a,b,c), a=50E-9..100E-9,b=100E-9..150E-9,c=50..100) Other ways might be, Optimization[NLPSolve](proc2, 50E-9..100E-9,100E-9..150E-9,50..100) Optimization[NLPSolve]((a,b,c)->proc2(a,b,c), 50E-9..100E-9,100E-9..150E-9,50..100) Or you might be able to edit proc2 so that when its arguments are not numeric it returns its own procname(args) unevaluated. For a simpler illustrative example, look at these, foo:=proc(a) if a > 1 then a else 0 end if; end proc: plot(foo(a),a=0..2); plot('foo'(a),a=0..2); plot(a->foo(a),0..2); plot(foo,0..2); foo:=proc(a) if type(a,numeric) then if a > 1 then a else 0 end if; else 'procname'(args); end if; end proc: plot(foo(a),a=0..2); acer

If I understand the problem correctly then, for a given Matrix M and Vector v of variables, you want that, M . v = v So, a logical progression of equivalent statements might be, M . v = IdentityMatrix(Dimension(v)) . v M . v - IdentityMatrix(Dimension(v)) . v = 0 ( M - IdentityMatrix(Dimension(v)) ) . v = 0 So, isn't this the same as saying that v must be in the Nullspace of ( M - IdentityMatrix(Dimension(v)) ) ? That is, with(LinearAlgebra): Nullspace( M - IdentityMatrix(Dimension(v)) ); acer

What are the variables? You've given three equations. Are the Greek letters supposed to be parameters, and {p,s,n} the variables? And what are the ranges for the variables? If the variables are {p,s,n} then you don't want just the trivial solution {0,0,0}, right? So, I'm guessing what you want here. But maybe this is close. ds := theta*p*(1-exp(-n))-rho*s; dp := xi*(1-exp(-s))*(lambda-p)-p; dn := mu*lambda*p*(1-exp(-n))-omega*n; Fs := proc(Lambda,Mu,Theta,Rho,Xi,Omega) local result; result:=fsolve(subs([mu=Mu,theta=Theta,rho=Rho,xi=Xi,lambda\ =Lambda,omega=Omega],{ds,dp,dn}),{s,p,n}); result := eval(s,result); if type(result,numeric) then return result; else return Float(undefined); end if; end proc; Fs(1,1,1,1,-1,1); plot('Fs'(L,1,1,1,-1,1),L=1..3); The above should plot s, as a function of lambda in the range from 1 to 3, holding all the other Greek letter parameters at constant values. If that's what you're after then editing it to plot p or n, or depend on another of the parameters, should be straightforward. acer

This is like counting, in base 3. Imagine that the symbol "3" stands for the digit 2, the symbol "2" for the digit 1, and the symbol "1" for the digit 0. > convert([2,2,2,2,2,2,2,2,2],base,3,10); [2, 8, 6, 9, 1] Reading the base-10 digits above from right to left, that's 19682. Add 1 to that, on account of the number zero, and you get 19683. This simple procedure might help illustrate this idea. It won't be nearly as fast as combinat[permute]. > makePAT := proc(i) local PAT, nPAT, j; > PAT := convert(i,base,3); > nPAT := nops(PAT); > PAT:=[seq(1,j=1..9-nPAT),seq(1+PAT[nPAT+1-j],j=1..nPAT)]; > end proc; > makePAT(0); [1, 1, 1, 1, 1, 1, 1, 1, 1] > makePAT(19682); [3, 3, 3, 3, 3, 3, 3, 3, 3] acer