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One approach using *polyhedraplot*...

gulliet 266
July 02 2009
0 0
Let assume that the entries are a list of lists that denotes the height z, i.e. [[3, 2, 1, 1], [2, 2, 1], [1, 1]]. Then, we transform it (manually in the example below) in a list of x-, y-, z-coordinates (for a given point, the x-coordinate is equal to its row index, and, similarly, the y-coordinate is equal to its column index). Finally, we plot the diagram thanks to the function polyhedraplot. > with(plots); > polyhedraplot([[1, 1, 3], [1, 2, 2], [1, 3, 1], [1, 4, 1], [2, 1, 2], [2, 2, 2], [2, 3, 1], [3, 1, 1], [3, 2, 1]], polyscale = .5, polytype = hexahedron, scaling = CONSTRAINED, orientation = [20, 70]); 97_FerrersDiagram.jpg HTH, -- Jean-Marc

One possible way, which uses...

gulliet 266
March 10 2009
0 0
One possible way, which uses functional programming, is as follows: 
restart;

myInt := proc(fun, var, n)
  foldl((f, v) -> int(f, v), fun, seq(var, i = 1..n));
end proc:

myInt(sin(x)*exp(x), x, 2);

                                1              
                              - - cos(x) exp(x)
                                2              

seq(myInt(sin(x)*exp(x), x, i), i = 1..4);

            1                 1                  1                
          - - cos(x) exp(x) + - sin(x) exp(x), - - cos(x) exp(x), 
            2                 2                  2                

              1                 1                  1              
            - - cos(x) exp(x) - - sin(x) exp(x), - - sin(x) exp(x)
              4                 4                  4
Regards, -- Jean-Marc

piecewise() rather than abs() and discon...

gulliet 266
March 10 2009
0 0
Also, you may want to convert the abs() into piecewise() and set discont = true in the graphs. 
convert(abs(x^2-6*x+5), piecewise);

                         /  2                       
                         | x  - 6 x + 5       x <= 1
                         |                          
                         |   2                      
                        <  -x  + 6 x - 5      x < 5 
                         |                          
                         |  2                       
                         | x  - 6 x + 5       5 <= x
                         \                          

f := unapply(%, x);
Df := unapply(diff(f(x), x), x);
DDf := unapply(diff(Df(x), x), x);

                     /         2                     2                    
  f := x -> piecewise\x <= 1, x  - 6 x + 5, x < 5, -x  + 6 x - 5, 5 <= x, 

     2          \
    x  - 6 x + 5/
Df := x -> piecewise(x < 1, 2 x - 6, x = 1, undefined, x < 5, -2 x + 6, x = 5, 

  undefined, 5 < x, 2 x - 6)
DDf := x -> piecewise(x < 1, 2, x = 1, undefined, x < 5, -2, x = 5, undefined, 

  5 < x, 2)

a := plot(f(x), x = -2 .. 7, y = -10 .. 10):
b := plot(Df(x), x = -2 .. 7, y = -10 .. 10, 
    discont = true, linestyle = 2):
c := plot(DDf(x), x = -2 .. 7, y = -10 .. 10, 
    discont = true, linestyle = 3):
display([a, b, c], title = 
    `function in solid, 1st deriv in dotted, 2nd deriv dashed`);
97_multipleplotfirstsecondderivdiscont.jpeg Regards, -- Jean-Marc

"z" is not "_Z1"...

gulliet 266
March 10 2009
0 0 
restart; with(plots):
f := proc (x) options operator, arrow; sin(x^2-7*x+3) end proc; 
Df := unapply(diff(f(x), x), x); 
_EnvAllSolutions := true; 
critpts := solve(Df(x) = 0, x); 
indets(critpts[1]);
                                      / 2          \
                         f := x -> sin\x  - 7 x + 3/
                                 / 2          \          
                   Df := x -> cos\x  - 7 x + 3/ (2 x - 7)
                          _EnvAllSolutions := true
                          7   1                       (1/2)  
               critpts := - + - (37 + 4 Pi _Z1 + 2 Pi)     , 
                          2   2                              

                 7   1                       (1/2)  7
                 - - - (37 + 4 Pi _Z1 + 2 Pi)     , -
                 2   2                              2
                     /                           (1/2)\ 
                    { _Z1, (37 + 4 Pi _Z1 + 2 Pi)      }
                     \                                / 

a := plot(f(x), x = 0 .. 6); 
b := plot(Df(x), x = 0 .. 6, linestyle = 2); 
c1 := plot([subs({_Z1 = -3}, critpts[1]), t, t = -6 .. 6], color = green); 
c2 := plot([subs({_Z1 = -3}, critpts[2]), t, t = -6 .. 6], color = green); 
c3 := plot([critpts[3], t, t = -6 .. 6], color = green); 
c4 := plot([subs({_Z1 = -2}, critpts[2]), t, t = -6 .. 6], color = green); 
c5 := plot([subs({_Z1 = -1}, critpts[2]), t, t = -6 .. 6], color = green); 
c6 := plot([subs({_Z1 = 0}, critpts[2]), t, t = -6 .. 6], color = green); 
c7 := plot([subs({_Z1 = -2}, critpts[1]), t, t = -6 .. 6], color = green); 
display([a, b, c1, c2, c3, c4, c5, c6, c7], 
title = `function in solid, derivative in dotted`);
97_multipleplotfirstsecondderiv.jpeg Regards, -- Jean-Marc

What is "nt"?...

gulliet 266
March 09 2009
0 0
Before going any further, could you check that the expressions below are actually what you intend to work with. (Note that I have fixed several syntax errors and assumed that "nt" stands for "int".) By the way, what are "c" and "t"? Also, my understanding is that you do not have any equation but an integral that must be computed and then plotted, perhaps thanks to the function plot3d(), am I right? 
restart;
g := wp*sqrt((w^2-wp^2)/wh^2-w^2)/c; 
varphi := (1/2)*w*t; 
Phi0 := sin((1/2)*w*t)/((1/2)*w*t);

                                               (1/2)
                                / 2     2     \     
                                |w  - wp     2|     
                             wp |-------- - w |     
                                |    2        |     
                                \  wh         /     
                        g := -----------------------
                                        c           
                                         1    
                               varphi := - w t
                                         2    
                                         /1    \
                                    2 sin|- w t|
                                         \2    /
                            Phi0 := ------------
                                        w t     

A := int((exp(-(1/4)*g^2*l^2)/g*((w^2+c^2*g^2)/(wh^2-
w^2)))*Phi0*(sin(w*y-x+varphi)+sin(w*y+x+varphi)), w = wp .. wh);
        /                                       /     /  
        |                                       |     |  
        |                                       |     |  
        |                                       |     |  
        |                  1                    |     |  
A := int|-------------------------------------- |2 exp|- 
        |                  (1/2)                |     |  
        |   / 2     2     \                     \     \  
        |   |w  - wp     2|      /  2    2\              
        |wp |-------- - w |      \wh  - w / w t          
        |   |    2        |                              
        \   \  wh         /                              

      / 2     2     \   \                                                     
    2 |w  - wp     2|  2|                                                     
  wp  |-------- - w | l |                                                     
      |    2        |   |   /         / 2     2     \\                        
      \  wh         /   |   | 2     2 |w  - wp     2||    /1    \ /   /       
  ----------------------| c |w  + wp  |-------- - w || sin|- w t| |sin|w y - x
              2         |   |         |    2        ||    \2    / \   \       
           4 c          /   \         \  wh         //                        

                                   \              \
                                   |              |
                                   |              |
                                   |              |
     1    \      /          1    \\|              |
   + - w t| + sin|w y + x + - w t|||, w = wp .. wh|
     2    /      \          2    //|              |
                                   /              |
                                                  |
                                                  |
                                                  |
                                                  /
Regards, -- Jean-Marc

The constant Pi is written...

gulliet 266
March 08 2009
0 0
The constant Pi is written Pi, whereas pi is just the greek letter pi without any numeric value associated to it. 
> evalf(Pi);

                                 3.141592654
> evalf(pi);

                                     pi
Regards, -- Jean-Marc

Use Pi (with a capital P) instead of pi ...

gulliet 266
March 08 2009
0 0
Use Pi (with a capital P) instead of pi (which is just the greek letter without any specific meaning). 
> evalf(Pi);

                                 3.141592654
> evalf(pi);

                                     pi
Regards, -- Jean-Marc

msolve - solve equations in Z mod m...

gulliet 266
March 08 2009
0 0
You should have a look at the function msolve(). See help("msolve") for more info and examples. [EDITED] One possible way: 
> sols := msolve(x^i = 4, 12);

  sols := {i = 1, x = 4}, {i = 2, x = 2}, {i = 2, x = 4}, {i = 2, x = 8}, 

    {i = 2, x = 10}, {i = 3, x = 4}, {i = 3, x = 10}, {i = 4, x = 2}, 

    {i = 4, x = 4}, {i = 4, x = 8}, {i = 4, x = 10}, {i = 5, x = 4}, 

    {i = 5, x = 10}, {i = 6, x = 2}, {i = 6, x = 4}, {i = 6, x = 8}, 

    {i = 6, x = 10}, {i = 7, x = 4}, {i = 7, x = 10}, {i = 8, x = 2}, 

    {i = 8, x = 4}, {i = 8, x = 8}, {i = 8, x = 10}, {i = 9, x = 4}, 

    {i = 9, x = 10}, {i = 10, x = 2}, {i = 10, x = 4}, {i = 10, x = 8}, 

    {i = 10, x = 10}, {i = 11, x = 4}, {i = 11, x = 10}

> map(proc (s) options operator, arrow; subs(s, x^i) end proc, {sols});

{4, 16, 64, 100, 256, 1000, 1024, 4096, 10000, 16384, 65536, 100000, 262144, 

  1000000, 1048576, 4194304, 10000000, 16777216, 100000000, 1000000000, 

  1073741824, 10000000000, 100000000000}
Regards, -- Jean-Marc

Assuming you want you to use...

gulliet 266
March 08 2009
0 0
Assuming you want you to use the piecewise command (see help("piecewise") for more info), defining a multivariate function can be done as follows: 
f := (x, y) -> piecewise(x <> 0 and y > 0, sin(x*y)/x, 0);
For instance, 97_definingmultivariatepiecewisefun.jpg Regards, -- Jean-Marc

Perhaps "method = sqp" ?...

gulliet 266
March 04 2009
0 1 
> restart; with(Optimization); 
datax := [10, 15, 20, 25, 30, 45, 60, 90, 120, 150, 180]; 
datay := [428, 334.1, 280.7, 262.7, 242.4, 212.9, 182.2, 151, 130.5, 116.9, 109.5]; 
tau := 1; 
model := N1*exp(lambda1*(t-tau))+N2*exp(lambda2*(t-tau)); 
residuals := [sum((datay[i]-(eval(model, t = datax[i])))^2, i = 1 .. nops(datax))]; 
NLPSolve((428-N1*exp(9*lambda1)-N2*exp(9*lambda2))^2+
(334.1-N1*exp(14*lambda1)-N2*exp(14*lambda2))^2+
(280.7-N1*exp(19*lambda1)-N2*exp(19*lambda2))^2+
(262.7-N1*exp(24*lambda1)-N2*exp(24*lambda2))^2+
(242.4-N1*exp(29*lambda1)-N2*exp(29*lambda2))^2+
(212.9-N1*exp(44*lambda1)-N2*exp(44*lambda2))^2+
(182.2-N1*exp(59*lambda1)-N2*exp(59*lambda2))^2+
(151-N1*exp(89*lambda1)-N2*exp(89*lambda2))^2+
(130.5-N1*exp(119*lambda1)-N2*exp(119*lambda2))^2+
(116.9-N1*exp(149*lambda1)-N2*exp(149*lambda2))^2+
(109.5-N1*exp(179*lambda1)-N2*exp(179*lambda2))^2, 
initialpoint = {N1 = 1.122*10^3, N2 = 1.2830*10^3, 
lambda1 = -0.39e-1, lambda2 = -.326}, method = sqp);


 [439.485735349365996, 
    [N1 = 257.198668654980509, 
    N2 = 531.115292604514593, 
    lambda1 = -0.00532918849969983306, 
    lambda2 = -0.119870376119667749]]
Regards, -- Jean-Marc

convert(theta1(2),sum)...

gulliet 266
March 04 2009
0 0 
restart;
pol:=(-1)^n/(2^n*n!)*(1 - x)^(-a) * (1 + x)^(-b) *diff((1 - x)^(a+n)* 
(1 + x)^(b+n),x$n):
Lx:=unapply(pol,n,a,b):
theta1(2) := Lx(2,0,0):
convert(theta1(2),sum);

              1        2   1                        1        2
              - (1 - x)  + - (-2 + 2 x) (2 x + 2) + - (1 + x) 
              4            4                        4
Regards, -- Jean-Marc

If I have made no mistake in...

gulliet 266
March 04 2009
0 0
If I have made no mistake in translating from Maple to Mathematica, here is what is returned by Mathematica 6.0.3: 97_doubleintefrommathemarica603.jpg Regards, -- Jean-Marc

side assumptions...

gulliet 266
March 04 2009
0 0
Not exactly the form you are looking for but here is a possible method. You can pass side assumptions to simplify(), for instance, {alpha+beta = 1} and use assuming for the rest of the constraints. 
simplify(simplify((beta*p-beta-p)*alpha/(p*(beta-1)),{alpha+beta = 1}))
assuming 0 < alpha, alpha < 1, 0 < beta, beta < 1;

                             1 + alpha p - alpha
                             -------------------
                                      p
HTH, -- Jean-Marc

D((diff...) rather than D*(diff...)...

gulliet 266
March 04 2009
0 1
You are missing a parenthesis in the definition of PDE: write D((diff(u(t, z), `$`(z, 2*z)))*z) rather than D*(diff(u(t, z), `$`(z, 2*z)))*z). 
PDE := diff(u(t, z), t)-v*(diff(u(t, z), z)) = 
D((diff(u(t, z), `$`(z, 2*z)))*z)-k*u(t, z)
Regards, -- Jean-Marc

mtaylor - multivariate Taylor series exp...

gulliet 266
March 03 2009
0 0 
> restart;
> f := proc (x, d) options operator, arrow; sin(x^2+d^2) end proc;

                                           / 2    2\
                         f := (x, d) -> sin\x  + d /

> mtaylor(f(x, d), [x = 0, d = 0], 4);

                                    2    2
                                   x  + d 
> help("mtaylor");
Regards, -- Jean-Marc
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