can we load part of a package not the package whole !? for example from physics package i only need diff tool not all of the package tools,can i do sth for that !? 

restart:

Eq1:=diff(psi(y),y$4)-diff(psi(y),y$2)=0;

dsolve({Eq1});

bcs:=psi(h1)=F,D(psi)(h1)=-1,psi(h2)=-F,D(psi)(h2)=-1;

dsolve({Eq1,bcs});

Error, (in dsolve) found differentiated functions with same name but depending on different arguments in the given DE system: [{psi(h1), psi(h2), psi(y)}, {psi(h2), psi(y)}]

Any suggestion?

Cheers!

Dear all;

Please see only the last lines of this procedure,

1)I ask if this procedure give an output epsilon(x0,h). Really, I need your help. Thanks.

2) Can we plot epsilon(x0,h), versus h

> heun := proc (x0, h)

local x, y, i, N, k, f, ode, k1, k2, x1, x2, y1, y2;

f :=(x,y)-> 1/(1+cos(y)) ;

ode := diff(y(x), x) = f(x, y);

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

y2 := Array(0 .. N);

x2 := Array(0 .. N);

y1 := Array(0 .. 2*N);

x1 := Array(0 .. 2*N);

x2[0] := 0; y2[0] := (1/4)*Pi;

for i from 0 to N-1 do

x1[2*i+2] := (2*i+2)*h;  k1[1] := f(x1[2*i], y1[2*i]); 

k1[2] := f(x1[2*i]+h, y1[2*i]+h*k1[1]);

y1[2*i+2] := y1[2*i]+(1/2)*h*(k1[1]+k1[2]);

x2[i+1] := (2*i+2)*h;

k2[1] := f(x2[i], y2[i]);

k2[2] := f(x2[i]+2*h, y2[i]+2*h*k2[1]);

y2[i+1] := y2[i]+h*(k2[1]+k2[2])

end do;

return firstresulat = evalf([seq([x2[i], y2[i]], i = 0 .. N)]);

return secondresulat =evalf([seq([x1[2*i], y1[2*i]], i = 0 .. N)]);

epsilon:=(x0,h)->add((firsttest[i][2]-secondtest[i][2])^2 , i=1..round(x0/(2*h))+1)

end proc;

Dear all

Please, I need you help, to make a simple procedure with two output.  I try this simple code :

test:= proc(a,b)

local variable .....;

for i from 1 to 10 do

expression

end do

retur  fff;

return ggg;

end proc

Then I do  : test(a,b) but I get only ggg and not fff

Any help please

how maple calculate exp(x) with e.g. 100000 decimal numbers

a divsion of the series x^k/k! with e.g. 1/25000!/25001 lasts longer than the exp(1.xx) calculation

is there a faster way to calculate exp(x) than with the x^k/k! series

thanks

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?