I've been poking around with convolutions on Maple, and some weird behavior came up---if I let it compute the convolution of a piecewise function, then take the convolution of that, it comes out differently than if I enter a function from scratch as the middle step---file attached (PiecewiseProblem.mw).  I'm not really a Maple pro, so am I'm doing something crazy here?

Thanks!

hello, 

so i've been having trouble with this one for a while. I think i'm just missing something simple.. maybe yous could help.

all we have to do is to write a maple procedure that takes an integer N and a boolean function F as in14 as arguments, returns nothing, and plots a square N N lattice of points, coloring the points (i; j) with F(i; j) true in red and the other ones blue.

thaaaanks.

why mapleprimes.com does not send me email notifications ?! could anyone help ? i do not receive email notification of related posts.

Dear all,

Thank you for your Help.

h: stepsize;

x in [0,x0];

I give all the step of my code, but I think there is a mistake. I wait for your Help.

I would like to compute the error between  Method Huen with step size h and step size 2h using the definition of epsilon given below:

 ## The error written epsilon(x0,h)= sqrt(1/(N+1) * sum_i=0^N  (y_i^{2h}-y_(2i)^h)^2 ), where y_i^(2h) is the approximation of y(i*2*h).

 ## We want : loglog epsilon versus h.

  epsilon:=(x0,h)->sqrt( 1/(N+1)*add( (Heun1(f,x0,i)-Heun2(f,x0,i))^2,i=0..N ) );

  f:=(x,y)=1/(1+cos(y)); 

  ode:=diff(y(x),x)=f(x,y);

ic:=y(0)=1;  h:=x0/(2*N);

## Method Heun with step size 2h

> Heun1 := proc (f, x0,)

local x, y, i, h, k;

y := Array(0 .. N);

x := Array(0 .. N);

h := evalf((1/2)*x0/N);

x[0] := 0;

y[0] := 1;

for i from 0 to N do

x[i+1] := (2*i+2)*h;

k[1] := f(x[i], y[i]);

k[2] := f(x[i]+h, y[i]+h*k[1]);

y[i+1] := y[i]+h*((1/2)*k[1]+(1/2)*k[2]);

end do;

[seq([x[i], y[i]], i = 0 .. N)];

end proc;

### Now Heun with step size h  ( the same h)

> Heun2 := proc (f, x0,)

local x, y, i, h, k;

y := Array(0 .. N);

x := Array(0 .. N);

h := evalf((1/2)*x0/N);

x[0] := 0;

y[0] := 1;

for i from 0 to N do

x[i+1] := (i+1)*h;

k[1] := f(x[i], y[i]);

k[2] := f(x[i]+h, y[i]+h*k[1]);

y[i+1] := y[i]+h*((1/2)*k[1]+(1/2)*k[2]);

end do;

[seq([x[2*i], y[2*i]], i = 0 .. N)];

end proc;

Thanks you for your help.

Hi

Is MapleSim 6.1 fully Modelica 3.1 compliant? If not, which features does it miss for the moment?

Lookup tables? (modelica ones, not maplesim's)

Thank you very much

> with(DETools);
> eq1 := diff(y(t), t) = v(t);
d
--- y(t) = v(t)
dt
> eq2 := diff(v(t), t) = -3*v(t)+10*y(t);
d
--- v(t) = -3 v(t) + 10 y(t)
dt
> phaseportrait([eq1, eq2], [y, v], t = 0 .. 10, [[y(0) = 0, v(0) = 1]], y = -10 .. 10, v = -10 .. 10, linecolor = blue);

why my phase portrait not working

Dear all

I want to display the value of k   for which the stop test is verified. Thanks for your help. 

epsilon:=(y0,N)->evalf(abs(

 testerror:=array(1..75);
for k from 2 by 2 to 50 do
testerror[k]:=data[RK3][k][2];
if testerror[k]<=0.0001 then break; end if; end do;

i_before_e$.mw

My Homies

I'm trying to find words which contradict the rule "i before e except after c". according to BBC program "QI" there are 923 exceptions, but below i get less than that. i guess its only as good as the dictionary, the 'inbuilt' has only 25k words but 'ospd3' (with 80k words) seems to stop before letter Q (with my sledgehammer code) when there are more words out there like "weird", "veil" and "reciept".

with(StringTools); with(PatternDictionary); bid := Create(ospd3);
words :=([seq])(LowerCase(Get(bid, i)), i = 1 .. Size(bid)-1);
tot := 0;
for n to nops(words) do
C := searchtext(cie, words[n]); E := searchtext(ei, words[n]);
if`and`(C = 0, E = 0) then next else tot := tot+1;
print([tot, words[n]]) end if
end do

Dear collegues

I wrote the following code

restart:
Digits := 15;
a[k]:=0;
b[k]:=7.47;
a[mu]:=39.11;
b[mu]:=533.9;
mu[bf]:=9.93/10000;
k[bf]:=0.597;
ro[p]:=3880 ;
ro[bf]:= 998.2;
c[p]:= 773;
c[bf]:= 4182;
#mu[bf]:=1;
Gr[phi]:=0; Gr[T]:=0;
#dp:=0.1;
Ree:=1;
Pr:=1;
Nbt:=cc*NBTT+(1-cc^2)*6;

#######################
slip:=0.1;         ####
NBTT:=2;           ####
lambda:=0.1;       ####
phi_avg:=0.02;    ####
#######################


eq1:=diff( (1+a[mu]*phi(eta)+b[mu]*phi(eta)^2)*diff(u(eta),eta),eta)+dp/mu[bf]+Gr[T]*T(eta)-Gr[phi]*phi(eta);
eq2:=diff((1+a[k]*phi(eta)+b[k]*phi(eta)^2)*diff(T(eta),eta),eta)+lambda*T(eta)/k[bf];
eq3:=diff(phi(eta),eta)+1/Nbt*diff(T(eta),eta);
Q:=proc(pp2,fi0) local res,F0,F1,F2,a,INT0,INT10;
global Q1,Q2;
print(pp2,fi0);
if not type([pp2,fi0],list(numeric)) then return 'procname(_passed)' end if:
res := dsolve({subs(dp=pp2,eq1)=0,eq2=0,eq3=0,u(0)=slip*D(u)(0),u(1)=-slip*D(u)(1),D(T)(0)=0,D(T)(1)=1,phi(0)=fi0}, numeric,output=listprocedure,continuation=cc);
F0,F1,F2:=op(subs(res,[u(eta),phi(eta),T(eta)])):
INT0:=evalf(Int(F0(eta),eta=0..1));
INT10:=evalf(Int(F0(eta)*F1(eta),eta=0..1));
a[1]:=evalf(Int(F0(eta),eta=0..1))-Ree*Pr;;
a[2]:=INT10/INT0-phi_avg;
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;
Q1:=proc(pp2,fi0) Q[1](_passed) end proc;
Q2:=proc(pp2,fi0) Q[2](_passed) end proc;
Optimization:-LSSolve([Q1,Q2],initialpoint=[0.3,0.0007]);




se:=%[2];
res2 := dsolve({subs(dp=se[1],eq1)=0,eq2=0,eq3=0,u(0)=slip*D(u)(0),u(1)=-slip*D(u)(1),D(T)(0)=0,D(T)(1)=1,phi(0)=se[2]}, numeric,output=listprocedure,continuation=cc);
G0,G1,G2:=op(subs(res2,[u(eta),phi(eta),T(eta)])):
TTb:=evalf(Int(G0(eta)*G2(eta)*(G1(eta)*ro[p]*c[p]+(1-G1(eta))*ro[bf]*c[bf] ),eta=0..1))/evalf(Int(G0(eta)*(G1(eta)*ro[p]*c[p]+(1-G1(eta))*ro[bf]*c[bf] ),eta=0..1));
with(plots):
odeplot(res2,[[eta,phi(eta)/phi_avg]],0..1);
odeplot(res2,[[eta,T(eta)/TTb]],0..1);
odeplot(res2,[[eta,u(eta)/(Ree*Pr)]],0..1);

res2(1);
Nuu:=(1/TTb);
1/((1+a[k]*G1(1)+b[k]*G1(1)^2)/(1+a[k]*phi_avg+b[k]*phi_avg^2));
(1/TTb)*(((1+a[k]*G1(1)+b[k]*G1(1)^2)/(1+a[k]*phi_avg+b[k]*phi_avg^2)));
>

I want to run the code for the value of NBTT in the range of 0.2 to 10. this code gave the results in the range of 4-10 easily. So, I used the continuation which improve the range of the results between 2-10. However, I coudnt gave the results when 0.2<NBTT<2. Would you please help me in this situation.

Also, It is to be said that the values of phi should be positive. in some ranges, I can see that phi(1) is negative. Can I place a condition in which the values phi restricted to be positive.

Thanks for your attentions in advance

Amir

I want to define the budget set of a consumer, which consists of an horizontal, slant and vertical line segments. How can I define and plot the budget set?

I find this a little frustrating as I learn Maple, so I think there is a better way to handle this.

I find myself having to keep wrapping expressions with evalf() in order to compare them, since when I use a constant such as Pi in these expressions and then compare them,  Maple complains.

In a large program, one does not know if an expression contains Pi or not beforehand, so is one really supposed to convert every expression to float just in case they might need to compare 2 expressions? 

Let me explain with simple example:

x:=1.2;  #it does not matter if this was 12/10 or 1.2, same error will result.
y:=Pi/3;
if x<y then
   print("x<y");
else
   print("x>=y");
fi;

The above gives the error "Error, cannot determine if this expression is true or false: 1.2<(1/3)*Pi"

So I changed the y assignment above to y:=evalf(Pi/3); or evalf(Pi)/3; and now Maple is happy.

But this for me looks awkward. In Mathematica, I can simply write the same, using symbolic Pi, and it works as is:

x = 1.2;  #even if this is symbolic 12/10 it will also work
y = Pi/3;
If[x < y, Print["x<y"], Print["x>=y"]]

I did not have to write  y=N[Pi/3]  where N[] is the equivalent function to Maple's evalf() which converts its argument to real.

So, now in Maple, I find myself writing evalf() around so many things, since I have to anticipate I might need to compare them in some logic later on and I can't keep track which one has some symbolic constant such as Pi in them or not. While in Mathematica I never had to worry about this.

Is there a way to reduce the need to having to use evalf() so much?
It seems to me, Maple should be able to decide if  1<Pi without me having to write 1<evalf(Pi) ?

I have two equations that are valid under the substitution sin <-> cos, so a simple way to generate the second equation is to replace all occurrences of sin with cos. But Maple gets the wrong answer when I do this, because of its built-in simplification. Here is an example. Z1 and Y1 shows the problem; Z2 and Y2 shows that my attempt to overcome the problem doesn't work.  Z3 doesn't work either, proving that the problem is internally generated by Maple because Maple insists on ordering variables in its own way, no matter how I write them.

________simplified example from Maple 15

restart;
Z1:=sin(-a*x+b);
Z2:='sin(-a*x+b)';
Z3:=sin(b-a*x);



                         -sin(a x - b)
                         sin(-a x + b)
                         -sin(a x - b)
Y1:=subs(sin=cos,Z1);
Y2:=subs(sin=cos,Z2);


                         -cos(a x - b)
                         -cos(a x - b)
 correct answer should be cos(-a*x+b) but the calculated results are off by a minus sign.

Ans1:=evalf(subs(a=1,b=2,x=3,[Y1,Y2,cos(-a*x+b)]));
          [-0.5403023059, -0.5403023059, 0.5403023059]

Question: How do I override Maple's desire to stick the "-" sign outside the sin function?

Dear all;

Than you for help.

how  many steps are required to achieve a error of 1.e-3 in the numerical value of y(1).

Here The 3 -step procedure  Range Kutta Method.

## Exact  solution  

### We will modifty N ( number of steps to get error =10^(-3). )

## Procedure Range Kutta

> RK3 := proc (f, a, b, y0, N)

local x, y, n, h, k, vectRK3;

y := Array(0 .. N);

x := Array(0 .. N);

h := evalf(b-a)/N;

x[0] := a; y[0] := 1;

for n from 0 to N-1 do

x[n+1] := a+(n+1)*h;

k[1] := f(x[n], y[n]);

k[2] := f(x[n]+(1/2)*h, y[n]+(1/2)*h*k[1]);

k[3] := f(x[n]+h, y[n]+h*(-k[1]+2*k[2]));

y[n+1] := y[n]+(1/6)*h*(k[1]+4*k[2]+k[3])

end do;

[seq([x[n], y[n]], n = 0 .. N)]; y[1];

end proc;

## Now  we compute the error between y(1) and exact  solution for different value of  N

### I have a problem in this part

 errorRk3 := array(1 .. 29);
 for N from  2 to 30 do

errorrRk3[N] := abs(eval(rhs(res), x = 1)-RK3((x,y)->-y,0,1,N));

if errorrRk3[N] =10^{-3} end ;
end  do ;

Hi all,

Say I have some list like this,

tmp:=[[0, 0, 1], [0, 1, 0], [0, 1, 1], [0, 1, 2], [1, 0, 0], [1, 0, 1], [1, 0, 2], [1, 1, 0], [1, 1, 1], [1, 1, 2], [1, 2, 0]];

And I have worked out some probabilities for each of them, a,b,c,d, ect.

I want to print them like this

Pr( 001 ) = 1

Pr( 010 ) = 1-phi[2]+phi[2]*(1-p[3])*(1-phi[3])

Pr( 011 ) = phi[2]*p[3]*(1-phi[3])

and so on.

I there a way to do that?

The probabilities can be extracted from a Vector. I have no problem to print them.

I dont know how to convert the 0,1,2 into the desired format as shown above.

This is the best I can do.

Also, is it possible to convert all the subscripte [] to _ when printing the output?

and get ride of all * as well.

Thanks,

casperyc

ali := (1/12)*(12*c[4]*ln(x+1)*x^4+48*c[4]*ln(x+1)*x^3+72*c[4]*ln(x+1)*x^2+48*c[4]*ln(x+1)*x+12*c[1]*x+30*c[2]*x^2+12*c[2]*x+36*c[3]*x^3+36*c[3]*x^2+12*c[3]*x+12*c[5]*x^5+48*c[5]*x^4+72*c[5]*x^3+48*c[5]*x^2+3*c[0]*x^4+12*c[0]*x^3+18*c[0]*x^2+12*c[0]*x+4*c[1]*x^4+16*c[1]*x^3+24*c[1]*x^2+6*c[2]*x^4+24*c[2]*x^3+12*c[3]*x^4+12*c[4]*ln(x+1)+12*c[5]*x)/(x+1)^4+.1

but simplify(ali-simplify(ali)) is not equal to 0!