How to calculate c.d.f from probability mass function. Suppose that the pmf of a discrete random variable is given : f(x)=(2*x+1)/25, x=0,1,2,3,4

Hi -

    It is often useful useful to generate two procedures --- one to evaluate a function and one to evaluate its gradient.  The procedure codegen[GRADIENT] does not treat functions of array variables.  Why doesn't GRADIENT support array variables?  Would it be possible to replace the array variables by variables, apply GRADIENT, and then replace the array variables by variables again?

Best wishes,

David

I want to find  the volume contribution x^2+y^2=z and x^2+y^2=2x over xy with Maple.

Hello, I was trying to control color of a plot3d. 
I find this answer : http://www.mapleprimes.com/questions/148397-Plot3d-Color-Range
And this post of @Carl Love : 
"

Here's how to do it with a continuous transformation to your existing color function, which is presumed to return a value between 0 and 1 (the HUE color scale). Keeping it continuous is very very nice when you want colors to represent  numeric values. Let's say your existing color function is C, and your coordinate functions for a parametrized surface are Fx, Fy, Fz.

Gamma:= 1.15:
plot3d(
     [Fx, Fy, Fz],  a..b, c..d, 
     color=  [
          (x,y)-> (1-C(x,y))^Gamma/3, #Hue
          (x,y)-> 1-C(x,y)/4,         #Saturation
          (x,y)-> 1-C(x,y)/7,         #Value
          colortype= HSV
     ],
     lightmodel= NONE,
     style= patchnogrid     
);

There are several parameters that can be adjusted; I've chosen some of them by my personal taste for color .

• Gamma controls the evenness of the distribution between red and green. I gave this one a name because this is a well-known concept (see the Wikipedia article "Gamma correction").
• The 3 in the Hue selects the fraction (1/3 in this case) of the full color spectrum that you want. If you want green to red, it will need to be pretty close to 3.
• The Hue value is subtracted from 1 to make the scale go green to red rather than red to green.
• The 4 in the Saturation controls (to some extent) how "light" the light-green is.
• The 7 in the Value controls (to some extent) how dark the dark-red is (lower values will make it darker).
• lightmodel= NONE is used so that the colors will not change due to shadows when the plot is rotated. "

I made some test to see the impact of the Gamma parameter. 
And with Gamma = 1, it's odd. 

It looks like with gamma = 1, plot3d makes an automatic scaling of the colors.
But I don't understand why.
Does anyone know ?

Download oddity.mw

I'm writing a simple Maple program to test the Generalized Finite Element Method: main_screened_Poisso.mw

When trying to define the Neumann boundary conditions, I have to define a directional derivative dudn=dudx*n. However, I can't seem to define a unit vector normal to Gamma, which is defined by a LineSegments objects.

Other than that, the row reduction is very slow, even though I'm using floating point arithmatic and not exact arithmatic, I believe.

How can I solve these problems? Thanks in advance!

The following limit does not return a value. Then the evalf gives a wrong answer.

The answer should be "undefined" or -infinity .. infinity.

limit(exp(n)/(-1)^n, n = infinity) assuming n::posint; evalf(%);

                       /exp(n)              \
                  limit|------, n = infinity|
                       |    n               |
                       \(-1)                /

                               0.

The same happens if you delete the assumption.

A similar problem occurs with

limit(sin(Pi/2+2*Pi*n), n = infinity) assuming n::posint;
                            -1 .. 1
without the assumption this would be appropriate.

Hi!

I still have a problem and im looking forward to any suggestions

this is the previous code that i have wrote

P := array([[8, 4], [8, 3], [8, 2], [7, 1], [6, 0], [5, 0], [4, 0], [2, 1], [1, 1], [1, 4]]);

> for j from 2 to 5 do k[j] := j+1;

x[j] := add(P[j, 1], j = j-1 .. j+2);

X[j] := add(P[j, 1]^2, j = j-1 .. j+2);

y[j] := add(P[j, 2], j = j-1 .. j+2);

Y[j] := add(P[j, 2]^2, j = j-1 .. j+2);

xy[j] := add(P[j, 1]*P[j, 2], j = j-1 .. j+2);

cx[j] := evalf(x[j]/k[j]);

cy[j] := evalf(y[j]/k[j]);

c11[j] := evalf(X[j]/k[j]-cx[j]^2);

c22[j] := evalf(Y[j]/k[j]-cy[j]^2);

c12[j] := evalf(xy[j]/k[j]-cx[j]*cy[j]);

C[j] := evalf(Matrix(2, 2, [[c11[j], c12[j]], [c12[j], c22[j]]]));

E[j] := simplify(fnormal(LinearAlgebra[Eigenvalues](C[j])));

if E[j][1] > E[j][2] then a[j] := E[j][2]/(E[j][1]+E[j][2]) else b[j] := E[j][1]/(E[j][1]+E[j][2])

 end if;

a[j];b[j]

 end do;

now, my question is how to put the output from the above looping in a  matrix form.

in my matrix, i need to call a[j],b[j], E[j][1], E[j][2] and the coordinate points. so my matrix dimension 4 x 5

I guess its a simple task but i tried hard and didnt get it worked.

thank you.

https://drive.google.com/file/d/0B2D69u2pweEvU3NpWWQwS3U1XzQ/edit?usp=sharing
https://drive.google.com/file/d/0B2D69u2pweEvMnFabkdiX1hpYVk/edit?usp=sharing

a1 := Diff(x1(s,t),s$2) = a*x1(s,t)+b*x2(s,t)+c*x3(s,t)+d*u(t);
a2 := Diff(x1(s,t),t)=x1(s,t);
b1 := Diff(x2(s,t),s$2) = e*x1(s,t)+f*x2(s,t)+g*x3(s,t)+h*u(t);
b2 := Diff(x2(s,t),t)=x2(s,t);
c1 := Diff(x3(s,t),s$2) = i*x1(s,t)+j*x2(s,t)+k*x3(s,t)+l*u(t);
c2 := Diff(x3(s,t),t)=x3(s,t);
sys := [a1, a2, b1, b2, c1, c2];
sol := pdsolve(sys);

length exceed limit

hi, i have a problem with maple. i took the codes from a book, it should be true but even i copied and past exactly the same, again the maple gave me error which i couldnt solve. could you please help me? how can i correct it ? incorrect codes are below, but if you need all codes i can write here. i need a solution immediately :(

>relativefrequencies := proc(text, language)
evalf(frequencies(text, language)/StringTools:-Length(text))
end proc:
VigenereKeyFind := proc(ciphertext, max:=floor(StringTools:-Length(ciphertext)/15),
{language:=en})
uses StringTools;
local freqs;
freqs := map(x -> relativefrequencies(x, language),
map(Implode, partit(Explode(ciphertext),
keylength(ciphertext, max, ’:-language’=language))));
Implode(map(li -> frequencyanalysis(li, ’:-language’=language), freqs))
end proc;

>VigenereKeyFind(c);
Error, (in VigenereKeyFind) ``’`` does not evaluate to a module

a1 := Diff(x1(s,t),s$2) = a*x1(s,t)+b*x2(s,t)+c*x3(s,t)+d*u(t);
b1 := Diff(x2(s,t),s$2) = e*x1(s,t)+f*x2(s,t)+g*x3(s,t)+h*u(t);
c1 := Diff(x3(s,t),s$2) = i*x1(s,t)+j*x2(s,t)+k*x3(s,t)+l*u(t);
sys := [a1, b1, c1];
sol := pdsolve(sys);

sol := pdsolve(sys);
[Length of output exceeds limit of 1000000]

Hi all, 

I am quite new to maple. Thanks for your help in advance. 

Could I define a general function f of a single variable x, f(x), without specifying its explicit form? I know it is simple to define functions of specific forms, e.g., f(x)=2*x (In maple: f:=x->2*x). I try to define a function without knowing its form, e.g., f:=x->f(x), expecting maple return f(y) if I type >>f(y). But this does not work (Maple says:

Error, (in temp) too many levels of recursion). 

Is it possible to define a general function f(x) without specific form in maple? Thank you again. 

Hi,

I get the error in the following code

restart:

gama1:=0.01:

zet:=0;
#phi0:=0.00789:
Phiavg:=0.02;
lambda:=0.01;
Ha:=1;


                               0
                              0.02
                              0.01
                               1
rhocu:=2/(1-zet^2)*int((1-eta)*rho(eta)*c(eta)*u(eta),eta=0..1-zet):

eq1:=diff(u(eta),eta,eta)+1/(mu(eta)/mu1[w])*(1-Ha^2*u(eta))+((1/(eta)+1/mu(eta)*(mu_phi*diff(phi(eta),eta)))*diff(u(eta),eta));
eq2:=diff(T(eta),eta,eta)+1/(k(eta)/k1[w])*(-2/(1-zet^2)*rho(eta)*c(eta)*u(eta)/(p2*10000)+( (a[k1]+2*b[k1]*phi(eta))/(1+a[k1]*phi1[w]+b[k1]*phi1[w]^2)*diff(phi(eta),eta)+k(eta)/k1[w]/(eta)*diff(T(eta),eta) ));
eq3:=diff(phi(eta),eta)+phi(eta)/(N[bt]*(1+gama1*T(eta))^2)*diff(T(eta),eta);
      /  d   /  d         \\   mu1[w] (1 - u(eta))
      |----- |----- u(eta)|| + -------------------
      \ deta \ deta       //         mu(eta)      

           /             /  d           \\               
           |      mu_phi |----- phi(eta)||               
           | 1           \ deta         /| /  d         \
         + |--- + -----------------------| |----- u(eta)|
           \eta           mu(eta)        / \ deta       /
                                /      /                        
                                |      |                        
/  d   /  d         \\     1    |      |  rho(eta) c(eta) u(eta)
|----- |----- T(eta)|| + ------ |k1[w] |- ----------------------
\ deta \ deta       //   k(eta) |      |         5000 p2        
                                \      \                        

                                /  d           \
     (a[k1] + 2 b[k1] phi(eta)) |----- phi(eta)|
                                \ deta         /
   + -------------------------------------------
                                          2     
         1 + a[k1] phi1[w] + b[k1] phi1[w]      

            /  d         \\\
     k(eta) |----- T(eta)|||
            \ deta       /||
   + ---------------------||
           k1[w] eta      ||
                          //
                                      /  d         \
                             phi(eta) |----- T(eta)|
          /  d           \            \ deta       /
          |----- phi(eta)| + ------------------------
          \ deta         /                          2
                             N[bt] (1 + 0.01 T(eta))
mu:=unapply(mu1[bf]*(1+a[mu1]*phi(eta)+b[mu1]*phi(eta)^2),eta):
k:=unapply(k1[bf]*(1+a[k1]*phi(eta)+b[k1]*phi(eta)^2),eta):
rhop:=3880:
rhobf:=998.2:
cp:=773:
cbf:=4182:
rho:=unapply(  phi(eta)*rhop+(1-phi(eta))*rhobf ,eta):
c:=unapply(  (phi(eta)*rhop*cp+(1-phi(eta))*rhobf*cbf )/rho(eta) ,eta):
mu_phi:=mu1[bf]*(a[mu1]+2*b[mu1]*phi(eta)):

a[mu1]:=39.11:
b[mu1]:=533.9:
mu1[bf]:=9.93/10000:
a[k1]:=7.47:
b[k1]:=0:
k1[bf]:=0.597:
zet:=0.5:
#phi(0):=1:
#u(0):=0:
phi1[w]:=phi0:
N[bt]:=0.2:
mu1[w]:=mu(0):
k1[w]:=k(0):

eq1:=subs(phi(0)=phi0,eq1):
eq2:=subs(phi(0)=phi0,eq2):
eq3:=subs(phi(0)=phi0,eq3):

#A somewhat speedier version uses the fact that you really need only compute 2 integrals not 3, since one of the integrals can be written as a linear combination of the other 2:
Q:=proc(pp2,fi0) local res,F0,F1,F2,a,INT0,INT10,B;
global Q1,Q2;
print(pp2,fi0);
if not type([pp2,fi0],list(numeric)) then return 'procname(_passed)' end if:
res := dsolve(subs(p2=pp2,phi0=fi0,{eq1=0,eq2=0,eq3=0,u(1)=lambda/(phi(1)*rhop/rhobf+(1-phi(1)))*D(u)(1),D(u)(0)=0,phi(1)=phi0,T(1)=0,D(T)(1)=1}), numeric,output=listprocedure):
F0,F1,F2:=op(subs(res,[u(eta),phi(eta),T(eta)])):
INT0:=evalf(Int((1-eta)*F0(eta),eta=0..1-zet));
INT10:=evalf(Int((1-eta)*F0(eta)*F1(eta),eta=0..1-zet));
B:=(-cbf*rhobf+cp*rhop)*INT10+ rhobf*cbf*INT0;
a[1]:=2/(1-zet^2)*B-10000*pp2;
a[2]:=INT10/INT0-Phiavg;
Q1(_passed):=a[1];
Q2(_passed):=a[2];
if type(procname,indexed) then a[op(procname)] else a[1],a[2] end if
end proc;
#The result agrees very well with the fsolve result.
#Now I did use a better initial point. But if I start with the same as in fsolve I get the same result in just about 2 minutes, i.e. more than 20 times as fast as fsolve:

Q1:=proc(pp2,fi0) Q[1](_passed) end proc;
Q2:=proc(pp2,fi0) Q[2](_passed) end proc;
Optimization:-LSSolve([Q1,Q2],initialpoint=[6.5,exp(-1/N[bt])]);


proc(pp2, fi0)  ...  end;
proc(pp2, fi0)  ...  end;
proc(pp2, fi0)  ...  end;
              HFloat(6.5), HFloat(0.006737946999)

the error is :

Error, (in Optimization:-LSSolve) system is singular at left endpoint, use midpoint method instead

how can I fix it.

Thanks

Amir

How to draw an object in 3d which is like a spring

3 layers of circle already enough

Who was the creator of Maplesoft?