You may have forgotten to load the `plots` package before calling `display`. You can also call is as,

plots:-display( ... )

acer

Maple 18.00 is build id 922027.

That is the value shown on the main menubar via Help->About Maple. It is also given by issuing the commands,

  kernelopts(version);

  version();

acer

Using Maple 17.02 on 64bit Windows 7,

restart;
g:=0.88641:
e:=2.53128:
eq:=tan(g)= e*sin(f)/(1+e*cos(f)):

fsolve(eq,f);
                          1.197496352

fsolve(eq,f,avoid={f=%});
                          -2.566269006

acer

restart:

with(Physics,diff);
                             [diff]

diff(y(x)^2,y(x));
                             2 y(x)

restart:

use diff=Physics:-diff in
  diff(y(x)^2,y(x));
end use;
                             2 y(x)

restart:

Physics:-diff(y(x)^2,y(x));
                             2 y(x)

acer

Hopefully the Online Help will be updated soon, so that the new links similar to this (and in its side panel) will be available,

http://www.maplesoft.com/support/help/Maple/view.aspx?path=updates/Maple17/index

acer

restart:

F:=proc(x) local y;
     y:=sprintf("%a",op(1,x));
     [seq(parse(y[i]),i=1..length(y))];
end proc:

F(0.567);
                           [5, 6, 7]

x:=evalf[100000](5555/7):

CodeTools:-Usage( F(x) ):
memory used=210.69MiB, alloc change=0 bytes, cpu time=765.00ms, real time=762.00ms

For reasons I don't fully understand, above approximately 10^5 or 10^6 digits it seems that garbage collection dominates the computation and the timing grows worse than linearly w.r.t. digits. This seems to affect both Carl's original approach and mine. I'm not sure where the culprit is, but I suspect the very many calls to builtin `parse`. I got no better with sscanf.

acer

The procedure that produces the list from the set has to be told how high it should check, no?

restart:                                          

m:=LL->{seq(`if`(LL[i]=1,i,NULL),i=1..nops(LL))}: 

unm:=(SS,k)->[seq(`if`(member(i,SS),1,0),i=1..k)]:

L:=[1,0,1,1,0,0]:                                 

S:=m(L);                                          

                                S := {1, 3, 4}

unm(S,nops(L));                                   

                              [1, 0, 1, 1, 0, 0]

acer

I don't know if there is a way to truly clear the 'value' of an already inserted PlotComponent.

I usually just code something to make the component's 'value' be something effectively close to an empty plot, using the `plot` command and without having to build a PLOT structure manually.

Eg, if the identity of the PlotComponent is "Plot0",

  DocumentTools:-SetProperty("Plot0",':-value',plot([[undefined,undefined]],':-axes'=':-none'));

So try that as the Action code behind your "Clear" Button.

acer

It's not very difficult to find two real roots of the system using `fsolve` if eq[1] is first solved for `a` and then substituted into eq[2], or if some special range for variable x is supplied.

restart:
f := x*exp((1 - x)*a):
f[2] := subs(x=f,f):
eq[1] := f[2] - x:
eq[2] := diff(f[2],x) + 1:
eq[4] := a=solve(eq[1],a);
                               /  x - 2\
                             ln|- -----|
                               \    x  /
                       a = - -----------
                               -1 + x   
R3 := subs( eq[4], eq[2] ):

#plot(R3, x=0..2);

rsols := [ fsolve(R3, x=0..1), fsolve(R3, x=1..2) ];
                  [0.2777041405, 1.722295859]

eval( subs(eq[4],[eq[1],eq[2]]), x=rsols[1] );
                      [     -10        -9]
                      [-5 10   , 1.9 10  ]

[x=rsols[1], eval( eq[4], x=rsols[1])];
              [x = 0.2777041405, a = 2.526467726]

[x=rsols[2], eval( eq[4], x=rsols[2])];
               [x = 1.722295859, a = 2.526467725]

Digits:=100:

rsols := [ fsolve(R3, x=0..1), fsolve(R3, x=1..2) ]:

eval( subs(eq[4],[eq[1],eq[2]]), x=rsols[1] );
                      [    -100       -99]
                      [1 10    , -1 10   ]

It is interesting that `fsolve` has difficulty solving the system unless it gets help with at least one of the ranges.

fsolve( [eq[1],eq[2]], {a,x}, x=0..1 );
              {a = 2.526467726, x = 0.2777041405}

fsolve( [eq[1],eq[2]], {a,x}, x=1..2 );
               {a = 2.526467726, x = 1.722295859}

And what is this nonsense? (Happens in M17.02 with Digits=10 but not Digits>10 it seems.)

s:=fsolve( {eq[1],eq[2]}, {a,x}, x=-3..0 );
              {a = 6.564724757, x = -2.865536950}

eval( [eq[1],eq[2]], s );
   [               856928220820                856928220833]
   [-1.472743519 10            , 2.009082419 10            ]

acer

acer

One of the advantages of ArrayInterpolation is that is can be used to compute interpolatory results for each input value in an Array in just a single call. If you have many values at which to interpolate, and if the input values do not depend upon each other, then this can improve performance as the problem size increases.

In your example the values at which to interpolate are the first column of Matrix `A`, and they are all known in advance of the need to interpolate. So you can bring a vectorized call to ArrayInterpolation outside of your do-loop.

InterProc1 := proc(N)
uses CurveFitting;
local i, A, variable1, variable2;
     A := Matrix(N, 4);
     variable1 := evalf([0, 100]);
     variable2 := evalf([12, 20]);
     for i from 1 to N do
          A(i, 1) := evalf(3+i);
          A(i, 2) := 2*i+A(i, 1);
          A(i, 3) := A(i, 2)-1;
          A(i, 4) := ArrayInterpolation(variable1, variable2, A(i, 1));
     end do;
     A;
end proc:

ans1 := CodeTools:-Usage( InterProc1(1000) ):
memory used=17.20MiB, alloc change=24.00MiB, cpu time=220.00ms, real time=228.00ms

InterProc1b := proc(N)
uses CurveFitting;
local i, A, variable1, variable2;
     A := Matrix(N, 4);
     variable1 := evalf([0, 100]);
     variable2 := evalf([12, 20]);
     for i from 1 to N do
          A(i, 1) := evalf(3+i);
          A(i, 2) := 2*i+A(i, 1);
          A(i, 3) := A(i, 2)-1;
     end do;
     A[1..N, 4] := ArrayInterpolation(variable1, variable2, A(1..N, 1));
     A;
end proc:

ans1b := CodeTools:-Usage( InterProc1b(1000) ):
memory used=459.64KiB, alloc change=0 bytes, cpu time=0ns, real time=11.00ms

The two results are the same.

LinearAlgebra:-Norm(ans1b-ans1);
                               0.

Some additional refinements might squeeze a bit more speed out of it (which might only be apparent for much larger sizes such as N=10^4 here). And perhaps more still could be done, reusing Vector workspaces and using ArrayTools:-Alias. But the subsequent improvements seem relatively minor compared to the improvement of making just a single call to ArrayInterpolation.

InterProc2 := proc(N)
uses CurveFitting;
local V1, i, A, variable1, variable2;
     A := Matrix(N, 4, datatype=float[8]);
     variable1 := Vector([0, 100], datatype=float[8]);
     variable2 := Vector([12, 20], datatype=float[8]);
     V1:=Vector(N, i->3+i, datatype=float[8]);
     A[1..N, 1] := V1;
     A[1..N, 2] := Vector(N, i -> 2*i+A[i,1], datatype=float[8]);
     A[1..N, 3] := Vector(N, i -> A[i,2]-1, datatype=float[8]);
     A[1..N, 4] := ArrayInterpolation(variable1, variable2, V1);
     A;
end proc:

ans2 := CodeTools:-Usage( InterProc2(1000) ):
memory used=171.74KiB, alloc change=0 bytes, cpu time=0ns, real time=7.00ms

LinearAlgebra:-Norm(ans2-ans1);
                               0.

acer

Student:-Calculus1:-Roots(cosh(C+cosh(C))/cosh(C)-2, C = -3 .. 3, numeric);

                  [-2.383208606, 0.3230741020]

acer

CurveFitting/ArrayInterpolation

acer

Your call to `animate` is passing A(tranz) which evaluates before `tranz` gets any numeric value. So `plot3d` is seeing just the valueless symbol `tranz`, and cannot produce a plot. That's what the error message is about.

This is a common usage mistake. There are several ways to correct for it. One easy way is to use so-called "uneval" quotes (single right-quotes) to delay the evaluation of the call A(tranz).

restart:

with(plots):
with(plottools):

A :=tranz-> plot3d(-2/sqrt(x^2+y^2)+5/sqrt((x-1.6)^2+y^2),
                   x = -5 .. 5, y = -5 .. 5,
                   view = [-5 .. 5, -5 .. 5, -5 .. 5],
                   transparency = tranz):
B := sphere([0, 0, 0], 2*(1/10), color = magenta, style = patchnogrid):
C := sphere([1.6, 0, 0], 5*(1/10), color = green, style = patchnogrid):
E := plot3d(0, x = -5 .. 5, y = -5 .. 5, view = [-5 .. 5, -5 .. 5, -5 .. 5],
            style = wireframe, shading = zgrayscale):

animate(display, ['A'(tranz), B, C, E, scaling = constrained,
                  view = [-5 .. 5, -5 .. 5, -5 .. 5],
                  axes = normal], tranz = 0.1 .. 0.9);

acer

rsolve( {x(n+2)-5*x(n+1)+6*x(n)=4*n+1, x(0)=1, x(1)=2}, x(n) );

                         n   7         3  n
                     -4 2  + - + 2 n + - 3 
                             2         2   

You can check that,

eval(sol,n=0);

                               1

eval(sol,n=1);

                               2

simplify( eval(sol,n=k+2)-5*eval(sol,n=k+1)+6*eval(sol,n=k) );

                            1 + 4 k

acer