MaplePrimes Questions

Why won't this procedure Compile? All I get is a cryptic and ungrammatical error message.

    step:= proc(
          n::integer[4],
          XYZ::Matrix(datatype= float[8]),
          E::Vector(datatype=integer[4]), F::Vector(datatype=integer[4]),
          W::Matrix(datatype= float[8]), #3x2 scratch matrix
          mu::integer[4]
     )
     option autocompile;
     local
          i::integer[4], j::integer[4],
          ed::float[8],
          fd::float[8],
          p::float[8], t::float[8]       
     ;
          to mu do
               for i to n do
                    ed:= 0;  fd:= 0;
                    for j to 3 do
                          p:= XYZ[i,j];
                          t:= XYZ[E[i],j] - p;
                          ed:= ed+t^2;
                          W[j,1]:= t;
                          t:= XYZ[F[i],j] - p;
                          fd:= fd+t^2;
                          W[j,2]:= t
                    od;
                    ed:= sqrt(ed) + .01;  fd:= sqrt(fd) + .01;
                    for j to 3 do
                         XYZ[i,j]:= 0.995*XYZ[i,j] - 0.01*W[j,1]/ed + 0.02*W[j,2]/fd
                    od
               od
          od
     end proc;
Compiler:-Compile(step);

Error, (in Compiler:-Compile1) In memory compilation failed

Dear all,

I have somme difficulties to calculate this integral.

int(I*sqrt((R*exp(I*theta)+1)/(R*exp(I*theta)-a)), theta = 0 .. Pi);

Thanks

Hello guys,

I was just playing around with differential equations, when I noticed that symbolic solution is  different from the numerical.What is the reason for this strange behavior?


ODE := (diff(y(x), x))*(ln(y(x))+x) = 1

sol := dsolve({ODE, y(1) = 1}, y(x))

a := plot(op(2, sol), x = .75 .. 2, color = "Red");
sol2 := dsolve([ODE, y(1) = 1], numeric, range = .75 .. 2);

with(plots);
b := odeplot(sol2, .75 .. 2, thickness = 4);
display({a, b});

 

 

Strange_issue.mw

Mariusz Iwaniuk

Hello,

I need to crate a function to be evaluated in a range of values, and this function i would to use in other expression, example:

cel1      "seq(i,i=0.001..2,0.001)"

cel2      "A:=&1";cel1

cel3      "f:=x->diff(KelvinBei(0,x),x)"

cel4      ""B:=map(x->f(x),[A])"

 

This is ok with a lot of function but with diff(KelvinBei(0,x),x) in cel4 show this error "Error,(in f) invalid input:.1e-2, which is not valid for its 2nd argument.

Why??? How can I do??

Hello guys,

I was just playing around with the Shanks transformation of a power series, when I noticed that polynomials aren't evaluated as I would expect.
I created this minimal working example; the function s should evaluate for z=0 to a[0], however it return simply 0.
Is there something I messed up?

restart

s := proc (n, z) options operator, arrow; sum(a[k]*z^k, k = 0 .. n) end proc;

proc (n, z) options operator, arrow; sum(a[k]*z^k, k = 0 .. n) end proc

(1)

series(s(n, z), z = 0)

series(a[0]+a[1]*z+a[2]*z^2+a[3]*z^3+a[4]*z^4+a[5]*z^5+O(z^6),z,6)

(2)

The value of s in z=0 should be a[0], however it returns 0:

s(n, 0)

0

(3)

s(1, 0)

0

(4)

Download evaluate_sum.mw

 

Thanks for your help,

Sören

 

Hi all,

 

It is believeble for me that the KummerU function is infinite or has a singular solution at r=0;

but without the command D(R) and dsolve ({er, D(R)(0)=0}, {R(r)}) that show the r=0 is a singular solution for the R(r), how could I prove it without using of these commands: for example is plotting of this function applicable??

while we have not the value of the  \lambda

````

restart

eq := diff(R(r), r, r)+(diff(R(r), r))/r+(-r^2+1)*lambda^2*R(r) = 0;

diff(diff(R(r), r), r)+(diff(R(r), r))/r+(-r^2+1)*lambda^2*R(r) = 0

(1)

``

dsolve(eq);

R(r) = _C1*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+_C2*exp(-(1/2)*lambda*r^2)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2)

(2)

``

dsolve({eq, (D(R))(0) = 0}, {R(r)})

R(r) = _C1*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)

(3)

``````

``

``

``

R := proc (r) options operator, arrow; C1*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+_C2*exp(-(1/2)*lambda*r^2)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2) end proc

proc (r) options operator, arrow; C1*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+_C2*exp(-(1/2)*lambda*r^2)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2) end proc

(4)

``

``

``

D(R)

proc (r) options operator, arrow; -C1*lambda*r*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+2*C1*exp(-(1/2)*lambda*r^2)*((lambda*r^2-1/2-(1/4)*lambda)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+(1/2+(1/4)*lambda)*KummerM(-1/2-(1/4)*lambda, 1, lambda*r^2))/r-_C2*lambda*r*exp(-(1/2)*lambda*r^2)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2)+2*_C2*exp(-(1/2)*lambda*r^2)*((lambda*r^2-1/2-(1/4)*lambda)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2)-KummerU(-1/2-(1/4)*lambda, 1, lambda*r^2))/r end proc

(5)

``

``


Download Kummer.mwKummer.mw

                         

         
               

Hi all,

 

I am generating a sphere with increasing radius that interacts at some point with a plan, just like this:

I would like to find a way to make the sphere "wave" (I agree this is not a wave...) reflected by the plan as in a mirror.

How could I do that ?

Thanks a lot for your help.

B

Im trying to draw a shpere but it always saying: 

Error, (in plot3d) unexpected option: z = -2 .. 2


this is the equation: x^2+y^2+z^2-4=0

i'm writing this way

plot3d(x^2+y^2+z^2-2^2, x = -2 .. 2, y = -2 .. 2, z = -2 .. 2)


what should I do? this is my first time with this software

 

best from Brazil,
Nina

Greetings to all.

With the following matter I am betting on there being a simple mistake on my part due to fatigue owing to a challenging session of intense computing. The following link at Math.Stackexchange.Com points to a computation involving complex residues. Consult the link for additional details.

I usually verify my computations with Maple, I did the same this time. Thereby I happened on a curious phenomenon which I have documented below. Please study the session data provided, I believe it speaks for itself.

user@host:~$ math
Mathematica 10.0 for Linux x86 (64-bit)
Copyright 1988-2014 Wolfram Research, Inc.

In[1]:= Residue[z^2/(z^4 + 2*z^2 + 2)^2, {z, 2^(1/4)*Exp[3*Pi*I/8]}]

            1/8      1/4
        (-1)    ((-1)    + Sqrt[2])
Out[1]= ----------------------------
            1/4      1/4           3
        16 2    ((-1)    - Sqrt[2])

In[2]:= N[%]

Out[2]= 0.117223 - 0.0083308 I

In[3]:=
user@host:~$ maple
    |\^/|     Maple 18 (X86 64 LINUX)
._|\|   |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2014
 \  MAPLE  /  All rights reserved. Maple is a trademark of
 <____ ____>  Waterloo Maple Inc.
      |       Type ? for help.
> residue(z^2/(z^4 + 2*z^2 + 2)^2, z=2^(1/4)*exp(3*Pi*I/8));
                                       0

> quit
memory used=0.9MB, alloc=8.3MB, time=0.07
user@host:~$

I am looking forward to learning what the correct syntax is to get the residue in this case and I hope I can assist other users who might have run into the same problem. I will cancel the question should it turn out to be trivial and of little potential use to the community.

Best regards,

Marko Riedel

Post Scriptum. Being a programmer myself I would be curious to learn more about the algorithms that are deployed here and how and why they did not succeed.

Please check this:

N:=3;

sum1 := lcm(N, 0)+lcm(N, 1)+lcm(N, 2)+lcm(N, 3);

sum2 := sum(lcm(N, k), k = 0 .. N);

 

Why is sum2 wrong?

 

Regards,

César Lozada

 

hi.

how i can solve two equation with respect to parameter sigma1

SOLL.mw

restart; pprime11 := -16395.36603*q1+5.811117425*q1*sigma1^2+3526.724044*p1-1.250000000*p1*sigma1^2+4.999870968*10^11*p1^3+4.999870970*10^11*p1*q1^2+7.967307034*10^14*p1^2*q1+4.999870966*10^12*sigma1*p1^2*q1-2.655769012*10^14*q1^3+4.999870968*10^12*sigma1*q1^3-17633.62022*q1*sigma1+6.250000000*q1*sigma1^3

qprime11 := 2.655769012*10^14*p1^3-7.967307034*10^14*p1*q1^2+4.999870970*10^11*p1^2*q1-4.999870968*10^12*sigma1*p1^3-4.999870966*10^12*p1*sigma1*q1^2+3526.724044*q1-1.250000000*q1*sigma1^2+16395.36603*p1-5.811117425*p1*sigma1^2+4.999870968*10^11*q1^3+17633.62022*p1*sigma1-6.250000000*p1*sigma1^3:

-50 < sigma1 and sigma1 < 50:

sigma1 <> 53.11665685, -53.11665685:

SOLL := solve({pprime11, qprime11}, real)

Warning,  computation interrupted

 

``

 

Download SOLL.mw

Hi all, 

Is it possible, and how, to disable the automatic completion in the "Advanced Search / keyword" field ?

Underlying this question is my looking for informations about cmaple (beyond what the help page displays : if I can't find any I will ask some specific ones later). But every time I type "cmaple" in the keyword textfield, it is replaced by "campée" (which is a french word ... I guess english people will obtain something different)

Hello,

does anyone know a way to combine two plots in one where one is created with ScatterPlot3D (Package:Statistics) and the other one with plot3D (Package:plots)?

 

Normally you would write something like this:

but that only works if the plots are from the same package...

The first example is very slow compared to the second.  The difference is in the definition of f().

restart;
y := int(1/(-0.4016e-1*m^(2/3)-0.211e-3*m^(5/3)), m);
f:= unapply(abs(y), m):
n := 500: ## sample size
M := <seq(2*idx/n,idx=1..n)>; ## m
Y := f~(M)+~Statistics:-Sample(Normal(0,3), n)^+; ## signal + noise

restart;
y := int(1/(-0.4016e-1*m^(2/3)-0.211e-3*m^(5/3)), m);
f:= (x) -> abs(subs(m=x,y));
n := 500: ## sample size
M := <seq(2*idx/n,idx=1..n)>; ## m
Y := f~(M)+~Statistics:-Sample(Normal(0,3), n)^+; ## signal + noise


find x852 given that

x0=0.250

x1=2.00

xn=(1/9)*sqrt(7+xn-1) /xn-2

 

I don't know how to solve this problem.

Anyone please help me with it!

Thanks a lot

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