Hi, I have the feeling I ran into this before, but can't find the answer, so here goes:

I want to create an animated plot of a histogram, changing the range of data to be histogrammed. So I type:

plots:-animate(Statistics:-Histogram,[dR*~WindowN(RP0,Ci,CangleWidth+Ci),ignore=true],Ci=-0.5 .. 0.2);

The error I am getting is

Error, (in MathPad:-WindowN) cannot determine if this expression is true or false: Ci < HFloat(0.39995) and HFloat(0.0) <= -HFloat(0.34995000000000004)+Ci

What happens here is that dR is a Vector of data with RP0 being another Vector equally long. I have a library function called WindowN that returns a Vector of the same length as its argument, with 1. for the elements within the window, undefined for the elements outside. The windowing algorithm is just a straightforward if, in a loop over the elements. This all works and I get a correct Histogram plot if I set the window numerically and call Histogram. CangleWidth is preset to 0.05.

Here the window to use is supposed to be set by the animate command. I expect Ci to be set to numbers in the range given (25 points default) and prepare the plots. But it looks like the WindowN routine gets an unevaluated Ci. I tried evalf(Ci) or eval(Ci) in the arglist, to no avail. Ci is not used or set before this call to animate.

I can see how I can workaround this but I think this should work as is. plots:-animate works for me in other contexts.

Any ideas?

Mac Dude

Following previous question at

http://www.mapleprimes.com/questions/149581-Improve-Algorithm-Dsolve

and also

http://www.mapleprimes.com/questions/149243-BVP-With-Constraining-Integrals

I wrote the following code

***********************

restart:

gama1:=0:


phi0:=0.00789:


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/(eta-1)+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]/(1-eta)*diff(T(eta),eta) )):
eq3:=diff(phi(eta),eta)-phi(eta)/(N[bt]*(1-gama1*T(eta))^2)*diff(T(eta),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,u(0)=0,eq1):
eq2:=subs(phi(0)=phi0,u(0)=0,eq2):
eq3:=subs(phi(0)=phi0,u(0)=0,eq3):

p:=proc(pp2) global res,F0,F1,F2:
if not type([pp2],list(numeric)) then return 'procname(_passed)' end if:
res := dsolve({eq1=0,subs(p2=pp2,eq2)=0,eq3=0,u(0)=0,u(1-zet)=0,phi(0)=phi0,T(0)=0,D(T)(0)=1}, numeric,output=listprocedure):
F0,F1,F2:=op(subs(res,[u(eta),phi(eta),T(eta)])):
evalf(2/(1-zet^2)*Int((1-eta)*(F1(eta)*rhop+(1-F1(eta))*rhobf)*( F1(eta)*rhop*cp+(1-F1(eta))*rhobf*cbf )/(F1(eta)*rhop+(1-F1(eta))*rhobf)*F0(eta),eta=0..1-zet))-pp2*10000:
end proc:


s1:=Student:-NumericalAnalysis:-Secant(p(pp2),pp2=[6,7],tolerance=1e-6);

                   HFloat(6.600456858832996)

p2:=%:



ruu:=evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta),eta=0..1-zet))):
phb:=evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta)*F1(eta),eta=0..1-zet))) / evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta),eta=0..1-zet))) :
TTb:=evalf(2/(1-zet^2)*(Int((1-eta)*F2(eta),eta=0..1-zet))):
rhouu:=evalf(2/(1-zet^2)*(Int((1-eta)*(F1(eta)*rhop+(1-F1(eta))*rhobf)*F0(eta),eta=0..1-zet))):
with(plots):
res(parameters=[R0,R1]):
odeplot(res,[[eta,u(eta)/ruu],[eta,phi(eta)/phb],[eta,T(eta)/TTb]],0..zet);

*************************************

as you can see at the second line of the code, the value of phi0:=0.00789. however, I want to modify the code in a way that phi0 is calculated with the following addition constraint

evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta)*F1(eta),eta=0..1-zet))) / evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta),eta=0..1-zet)))-0.02=0

I would be most grateful if you could help me in this problem.

Thanks for your attention in advance

Amir

I need to create a function from one list to another, but I don't want to define an operation.  I only want to assign a 1-1 mapping from a list of 64 elements to another list of 64 elements, i.e. the 1st element in list a is assigned to the 1st element in list b, etc.  I see plenty of examples of how to do this with an operation, but I don't want the mapping to perform any operation, just the mapping.  How is this done?

[> restart;
[> OPTIMAL:=proc(f,Condition::set)
local vars,L,S,SS,k,m,K,Fmax,Fmin,sol_min,sol_MIN,sol_max,sol_MAX;
vars:=indets(f,name) union indets(Condition,name);
L:=map(convert,Condition,equality);
K:=combinat[choose](L);
S:=NULL:
for k to nops(k) do if solve(K[k])<>NULL then
extrema(f,K[k],vars,'s||k'):S:=S,a||values(s||k):fi od:
SS:={}
for k to nops([s]) do
if type(s[k],set) then SS:=SS union evalc(s[k]) fi od;
SS:=remove(has,SS,l);
S:=NULL:
for k to nops(ss) do if type(SS[k],set(equation)) and
map(evalb,evalf(simplify(subs(SS[k],Condition))))={true}
then S:=S,vatue(ss[k]) fi od;
SS =[s];
Fmax:=-infinity:
for k to nops(SS) do
if evalf(simplify(subs(ss[k],f)))>evalf(Fmax)
then Fmax:=simplify(value(subs(SS[k],f)));sol_max:=SS[k]; fi;od;
sol_MAX:=sol_max:
for k to nops(SS) do
if Fmax=simplify(value(subs(SS[k],f)))
and (SS[k] minus sol_max)<>{} then
sol_MAX:=sol_MAX,Ss[k] fi od;
‘Fmax'=simplify(expand(Fmax)),sol_MAX:
Fmin:=infinity:
for k to nops(SS) do
if evalf(simplify((subs(SS[k],f))))<evalf(Fmin)
then Fmin:=simplify((subs(SS[k],f)));sol_min:=SS[k]; fi;od;
sol_MIN:=sol_min:
for k to nops(SS) do
if Fmin=simplify(value(subs(SS[k],f)))
and (SS[k] minus sol_min)<>{} then
sol_MIN:=sol_MlN,SS[k] fi od;
RETURN('F[min]'=simplify(expand(Fmin)),sol_MlN,
'F[max]'=simplify(expand(Fmax)),sol_max);
end: ↵

" Error, unable to match delimiters"
  Please, help me.
I don't understand. Thank you, thank you.

C(n,k)=((2*(k-2-n))/(k*(k-1)))*C(n,k-2)
if n in {even} then C(n,0)=((-1)^(n/2))*(factorial(n)/factorial(n/2)) and C(n,1)=0
if n in {odd} then C(n,0)=0 and C(n,1)=(2*(-1)^((n-1)/2))*(factorial(n)/factorial((n-1)/2))

i've been trying for hours to get a procedure involving this to work :(

Thanks

I want to make a plot of this function sine and imaginary cosine below,

psi(x):=sin(k*x)+I*k*l[2]*cos(k*x);

where

k:=n*pi/L;
 
L=1,n=1,l[2]=1;

plot(psi(x),x=0..5);

Plotting error, empty plot.

i always recieve this Error. Can someone help me with this. You can take any range of x including Pi if you wish , and as well provide me with refrences Text book on ploting functions if He/She knows. Looking forward for feedbacks. Thank you very much .
      
                                           

In geom3d. I want to find the vertices A(x1,y1,z1), B(x2,y2,z2), where x1, y1, z1, x2, y2, z2 are integer numbers so that the triangle OAB  (O is origin) and perimeter and area are integer numbers. I tried

> resrart:

N:=5:

L:=[]:

for x1 from -N to N do

for y1 from x1 to N do

for z1 from y1 to N do

for x2 from -N to N do

for y2 from -N to N do

for z2 from -N to N do

a:=sqrt(x1^2+y1^2+z1^2):b:=sqrt(x2^2+y2^2+z2^2):c:=sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2):

p:=(a+b+c)/2:

S:=sqrt(p*(p-a)*(p-b)*(p-c)):

if type(2*p, integer) and type(S, posint)

then L:=[op(L), [[0, 0, 0], [x1, y1, z1], [x2, y2, z2]]]: fi:

od: od: od: od: od: od:

nops(L);

But my computer runs too long. I can not receive the result. How to get the answer?

If I the length of the side are 6, 25, 29. I tried 

DirectSearch:-SolveEquations([(x2-x1)^2+(y2-y1)^2+(z2-z1)^2 = 6^2, (x3-x2)^2+(y3-y2)^2+(z3-z2)^2 = 25^2,  (x3-x1)^2+(y3-y1)^2+(z3-z1)^2 = 29^2], {abs(x1) <= 30, abs(x2) <= 20, abs(x3) <= 20, abs(y1) <= 20, abs(y2) <= 20, abs(y3) <= 20, abs(z1) <= 20,abs(z2) <= 20, abs(z3) <= 20}, assume = integer, AllSolutions, solutions = 1);

Does someone know how to ask Maple to compute intersection and union of ranges (for that matter Realrange)?

Does anyone know how Maple dsolve numeric integrators with adaptive stepsize like rkf45 and dverk78 select their stepsizes? Is there anyway to find out the stepsize the procedure chooses at each step? Currently, I can only obtain the value of my solution at a particular value but I do not know how many steps Maple took and why it selected these step sizes.

The DirectSearch package is a powerful Maple  tool. However, every soft has its advantages and disadvantages. In particular, the DS has problems in the case of a thin feasible set in higher dimensions. Recently a serious bug in the DS was detected by me. Solving an optimization problem, the DirectSearch produces the error communication

Warning, initial point [x1 = 1., x2 = 1., x4 = 2., y1 = 2., y2 = 3., y4 = 2.] does not satisfy the inequality constraints; trying to find a feasible initial point
Error, (in DirectSearch:-Search) cannot find feasible initial point; specify a new one
 while that initial point satisfies the constraints.

Download opti.mw

if I find fourier intergra solution if 1/pi∫ （form infintiy to 0）(a*cos(u*x)+u*sin(u*x))/a^2+u^2 du. can you polt the function to the integral converges? and can you plot the fourier intergral on a=1 and -1<=x<=3 with 0<=u<=20?

if  i use a fourier transform to find an integral representation for the solution is  U(x,t)=1-1/2*erf((x+a)/(2*square rootkt))+1/2*erf((x-a)/(2*square rootkt)), can you plot the sloution on the interval -5<=x<=5 with k=10^-6 and a=1 for t=10^5 and t=10^6 , give me two graph? erf mean is  Error function !

I used implitplot to plot solutions to some (tricky) equations in 2 variables, of the form implicitplot([f(x,y)=0]).  Now I have a (tricky) change of parameterization G:R^2->R^2 of the form G(x,y)=(g(x,y),h(x,y)). I'd like to plot the image of the solution set of f(x,y)=0 under the map G. Of course, if I could invert G I'd implicitplot           G^(-1)(f(x,y))=0, but the functions I have dont lend themselves to this.

Presumably MAPLE stores the points it plots somewhere, and I should be able to apply G to this set of points. But I don't know how to approach this. Anybody know? 

Hi

Long story short I had a detailed question and then the session timed out and killed it!

Quickly then, if we calculate something recursively Maple acts differently to other languages.

For example, in Python:

__________________________________________

>>>t=1;

>>>t+=1;

>>>print t;

___________________________________________

Is interpreted as:

__________________________________________

>>>t=1;

>>>t=t+1=2

>>>print t

2

___________________________________________

In Maple:

___________________________________________

>t:=1

>t:=t+1

>print(t)

___________________________________________

Is Interpreted as:

___________________________________________

>t=1

>t=t+1=1+1=2

>print(t); t=t+1=1+1=2

2

____________________________________________
And there in lies my problem. Logically, I wish to use a iterative algorithm to work out an expression of the nth derivative of a function from the (n-1) derivative. However, doing this is in maple brings up "error (in Test) too many levels of recursion".

For context, Minimum Working Example:

____________________________________________

>TestFunction:=(x)->cos(exp(-1/x^2))*F(x):

>limit(TestFunction(x),x=0)

F(0)

>TestDerivative:=(x)->eval(diff(TestFunction(y),y),y=x):

>limit(TestDerivative(x),x=0)

D(F)(0)

>for i from 1 to 50 do
print(D^(i)(TestFunction)(0)=limit(TestDerivative(x),x=0));

TestDerivative:=(x)->eval(diff(TestDerivative(y),y),y=x):

od:

D(TestFunction)(0)=D(F)(0)

Error, (in TestDerivative) too many levels of recursion

_______________________________________________

Ideally, this code would output the limit of the second derivative at zero by differentiating the first derivative and then the limit of the third derivative at zero by differentiating the second derivative etc. But what Maple is trying to do is to find the limit of the second derivative by differentiating the function then differentiating the result of that, then to find the limit of the third derivative it will first derivative by differentiate the function, then the second by differentiating the result, then third derivative by differentiating the result of that. If I have the analytic expression for the 5th derivative and I wanted the expression for the 6th derivative, I do not want to work out the 1st, 2nd, 3rd, 4th and then 5th derivative when I've already an of the expression of the 5th derivative!

I will note, it is possible to avoid the problem by using different names at each step but that does not solve the iterative problem.  Is there anyway to force maple to overwrite a function name? Is there a seperate solution? Or is maple just that daft in this case?

Thanks for the help,

Hamzaan

3rd_order.mw This 3rd order nonlinear ode is not giving the desire result,how do I resolve it?

Thanks