You, probably, use one of earlier Maple versions. It works in Maple 10. Both typechecking constructions are equivalent in Maple 10 (inside if), with :: being slightly faster,

t:=RandomTools:-Generate(list(integer,1000000)):
f1:=proc() local n; if args::nonnegint then n:=1 else n:=0 fi end:
f2:=proc() local n; if type(args,nonnegint) then n:=1 else n:=0 fi end:
gc();
time(map(f2,t));

                                15.078
gc();
time(map(f1,t));

                                13.015

I did more time testing and the results are consistent. That's why and because it is also shorter, I prefer using :: __________ Alec Mihailovs http://mihailovs.com/Alec/

Check ?plot,option and links near the bottom of it. tickmarks is the option for doing that. title is another option. It should look like title="some string", with title in lower case - Maple is case sensitive, and with right hand side being a string, i.e. being included in double quotes. Also, values of y from -10 to 10 won't work great in this case, because your function values are near 100. __________ Alec Mihailovs http://mihailovs.com/Alec/

Check ?plot,option and links near the bottom of it. tickmarks is the option for doing that. title is another option. It should look like title="some string", with title in lower case - Maple is case sensitive, and with right hand side being a string, i.e. being included in double quotes. Also, values of y from -10 to 10 won't work great in this case, because your function values are near 100. __________ Alec Mihailovs http://mihailovs.com/Alec/

It is somewhat possible in Classic Maple using textplot, but "}" is too bold, so that doesn't look nice,

plots[display](
  plots[textplot]([[0,3,"x + 2 y = 0"], 
    [0,2,"x + 4y + 3z = 5"], [0,1,"x + y + z = 0"]],
    font=[TIMES,BOLDITALIC,12]),
  plots[textplot]([2.4,2.4,"}"],font=[TIMES,ROMAN,70]),
axes=none,view=[-6..9,-6..9]);

It doesn't work in Standard Maple because view option seems to be broken there. __________ Alec Mihailovs http://mihailovs.com/Alec/

It is somewhat possible in Classic Maple using textplot, but "}" is too bold, so that doesn't look nice,

plots[display](
  plots[textplot]([[0,3,"x + 2 y = 0"], 
    [0,2,"x + 4y + 3z = 5"], [0,1,"x + y + z = 0"]],
    font=[TIMES,BOLDITALIC,12]),
  plots[textplot]([2.4,2.4,"}"],font=[TIMES,ROMAN,70]),
axes=none,view=[-6..9,-6..9]);

It doesn't work in Standard Maple because view option seems to be broken there. __________ Alec Mihailovs http://mihailovs.com/Alec/

Yes, it works in Maple 10, but it doesn't work in earlier versions. I don't have Maple 9 on this computer, but I checked that the following works in Maple 9.5,

Calc:=proc()
  local t;
  global m,Timer;
   use Maplets, Maplets:-Elements in
    Timer:=proc() 
      t:=iolib(25);
      while iolib(25)-t<3 do od end;
    m:=Maplet(
      'onstartup'=Action(RunWindow('Splash'),
        Evaluate(function="Timer"), CloseWindow('Splash'), 
        RunWindow('Main')),      
      Window['Splash']('title'="Welcome",
        [Label(Image("E:\\image.jpg"))]),
      Window['Main']('title'="Calculator",
        'width'=640, 'height'=480, []));
    Display(m) end use end:

Calc();

One should replace E:\\image.jpg with a link to the image. __________ Alec Mihailovs http://mihailovs.com/Alec/

Yes, it works in Maple 10, but it doesn't work in earlier versions. I don't have Maple 9 on this computer, but I checked that the following works in Maple 9.5,

Calc:=proc()
  local t;
  global m,Timer;
   use Maplets, Maplets:-Elements in
    Timer:=proc() 
      t:=iolib(25);
      while iolib(25)-t<3 do od end;
    m:=Maplet(
      'onstartup'=Action(RunWindow('Splash'),
        Evaluate(function="Timer"), CloseWindow('Splash'), 
        RunWindow('Main')),      
      Window['Splash']('title'="Welcome",
        [Label(Image("E:\\image.jpg"))]),
      Window['Main']('title'="Calculator",
        'width'=640, 'height'=480, []));
    Display(m) end use end:

Calc();

One should replace E:\\image.jpg with a link to the image. __________ Alec Mihailovs http://mihailovs.com/Alec/

Georgios, What you get, can be done just as

x1+y1+z1;

                 [14, 15, 16, 13, 12, 11, 19, 18, 17]

I don't see how that related to his problem. Perhaps, you meant

Vector(x1).Vector(y1)+Vector(y1).Vector(z1)+Vector(z1).Vector(x1)+a;

                               615 + a

or

add(x1[i]*y1[i]+y1[i]*z1[i]+z1[i]*x1[i]+a,i=1..9);

                              615 + 9 a

__________ Alec Mihailovs http://mihailovs.com/Alec/

Georgios, What you get, can be done just as

x1+y1+z1;

                 [14, 15, 16, 13, 12, 11, 19, 18, 17]

I don't see how that related to his problem. Perhaps, you meant

Vector(x1).Vector(y1)+Vector(y1).Vector(z1)+Vector(z1).Vector(x1)+a;

                               615 + a

or

add(x1[i]*y1[i]+y1[i]*z1[i]+z1[i]*x1[i]+a,i=1..9);

                              615 + 9 a

__________ Alec Mihailovs http://mihailovs.com/Alec/

If func1 and func2 in fsolve are not equations (as in this example), then fsolve solves the system of equations {func1=0, func2=0). {x=a..b,y=c..d} tells that the solution should satisfy these conditions - x between a and b, and y between c and d. After some experimentations it should become clear that for the second value of k (close to 7.4) the tangent line to the curve is horisontal - that means that the first derivative is 0 at that point. Also, it is clear that it is an inflection point - that means that the second derivative is 0. For the first value of k (close to 4.5) the situation is slightly different - in that case the first, the second, and the third derivatives are 0. Since there are 2 unknowns, x, and k, the system good for solving should have 2 equations. I chose the second and the third derivatives because they change more rapidly at that point than the first derivative does, so it is easier for fsolve to find solutions. __________ Alec Mihailovs http://mihailovs.com/Alec/

If func1 and func2 in fsolve are not equations (as in this example), then fsolve solves the system of equations {func1=0, func2=0). {x=a..b,y=c..d} tells that the solution should satisfy these conditions - x between a and b, and y between c and d. After some experimentations it should become clear that for the second value of k (close to 7.4) the tangent line to the curve is horisontal - that means that the first derivative is 0 at that point. Also, it is clear that it is an inflection point - that means that the second derivative is 0. For the first value of k (close to 4.5) the situation is slightly different - in that case the first, the second, and the third derivatives are 0. Since there are 2 unknowns, x, and k, the system good for solving should have 2 equations. I chose the second and the third derivatives because they change more rapidly at that point than the first derivative does, so it is easier for fsolve to find solutions. __________ Alec Mihailovs http://mihailovs.com/Alec/

Another possibility is to use an option in a module. It doesn't provide such unique type checking, but it is easy to use,

demo1 := module()
  export New;
  local ModuleLoad;
    ModuleLoad := proc()
      TypeTools:-AddType('abcd', 
        x->type(eval(x,2),'And'('`module`','attributed(abcd)'))) end;
    New := proc()
      module() option abcd; end end;
    ModuleLoad();
end:

m2 := demo1:-New():
type( m2, abcd );
                                 true

__________ Alec Mihailovs http://mihailovs.com/Alec/

Well, I'll post here values with 100 digits, and you can easily get more :)

Digits:=100:

fsolve({diff(h3d(x,k),x),diff(h3d(x,k),x,x)},{x=0..40,k=3..9});

  {k = 7.400440848315302765167038232957419577297476432902563662056\
        490588253571383516334981890957252682040895, 
   x = 32.7118226\ 
        3298704917256466385867897726084231272381453699470095350128\
        590543972593666800386193671386922}

fsolve({diff(h3d(x,k),x,x),diff(h3d(x,k),x,x,x)},{x=0..40,k=3..9});

  {k = 4.486645549752612902368306021318062533435298382327812806814\
        665820760072631094645733641558565790506992, 
   x = 28.4423157\  
        2096965142579905499529202880216451551339048949143819162063\
        806982876568090045137372799412865}

__________ Alec Mihailovs http://mihailovs.com/Alec/

Well, I'll post here values with 100 digits, and you can easily get more :)

Digits:=100:

fsolve({diff(h3d(x,k),x),diff(h3d(x,k),x,x)},{x=0..40,k=3..9});

  {k = 7.400440848315302765167038232957419577297476432902563662056\
        490588253571383516334981890957252682040895, 
   x = 32.7118226\ 
        3298704917256466385867897726084231272381453699470095350128\
        590543972593666800386193671386922}

fsolve({diff(h3d(x,k),x,x),diff(h3d(x,k),x,x,x)},{x=0..40,k=3..9});

  {k = 4.486645549752612902368306021318062533435298382327812806814\
        665820760072631094645733641558565790506992, 
   x = 28.4423157\  
        2096965142579905499529202880216451551339048949143819162063\
        806982876568090045137372799412865}

__________ Alec Mihailovs http://mihailovs.com/Alec/

What variables? Post the result that you would like to get and I'll show how to get it. __________ Alec Mihailovs http://mihailovs.com/Alec/