Here is the question:Prior to this question I was given f(z)=z^2+1, N(z)=(z^2-1)/(2z), T(z)=z-I/z+I such that T(N^k (z))=(T(z))^2^k. And L is a set of number on the real axis. Now the question is that given we have two regions of the complext plane as follow:

R₊ = {z : N^k{z) -> i as k -> ∞}; R_- = {z : N^k(z) -> -i as k -> ∞}.

Draw a diagram to illustrate these regions, the line L and the roots i and -i. We call R₊ the basin of attractionfor the root +i, and similarly R_-is the basin of attraction for the root -i.

 Show that if z is on the set L (the common boundary of the two regions R₊ and R_{_}, then N^k(z) stays on L for all values of k. (This is easy once you identify what L is.) So in this case iteration does not produce a root at all.

So basically my problem is that the fact I'm not very familar with the commands to draw such diagram, and I don't know much about Newton's method to compute complex roots. It would be appreciated if anyone can help me how to get start with the question. Thanks.

I have a diff equation in cartesian coordinates I need to transform to a certain cylindrical system. The de looks like this:

I define my new system with addcoords like this:

addcoords('AccCylinder',[r,theta,y],[r*cos(theta),y,r*sin(theta)]); # Note: y is the longitudinal axis here!

and also do

VectorCalculus:-AddCoordinates('AccCylinder'[r,theta,y],[r*cos(theta),y,r*sin(theta)],overwrite) assuming r >= 0;

and note that I had to overwrite as my system was already known, so maybe addcoords is reduncant(?)

I then do the transformation ("(7)" is the label of my above de):

changecoords((7),[x,y,z],AccCylinder,[r,theta,y]);

and get

This may be correct, but it has the expressions in the differentials, which diff does not know how to handle. I need to convert things like diff(xpr,r*cos(theta)) = diff(xpr,x)*diff(x,r) where x would be r*cos(theta). I can do this "by hand", but that seems overly tedious and error-prone. Somehow I'd expect the coordinate transforms to be able to do this but I can't figure out how.

Any idea?

Thanks,

Mac Dude,

Given the following system of first order ODE,

dx/dt=0.2x(1-0.5 x)-(1.5 xy)/(1+0.116 x),

dy/dt=(1.3 xy)/(1+0.1x)-0.8y.

Draw a DEplot (for t from 0 to 50) and indicate the particular

trajectory with the initial conditions x(0)=1,y(0)=2. If I

switched to forward Euler method,what would the DE plot look

like then? Is it possible to make the plot made by the

forward Euler method look close to the one which used the

default method?

Hello !

I installed Maple 14 on my windows8 computer (with compatibility windows7). My program worked the 2 first times, and suddently it bugged on the third, though I changed NOTHING.

It puts me an "Error: recursive assignment". The concerned line is: 

if a < 2 then broSet := broSet union {[param[1, j], param[2, i]]} fi:

There should be no problem.

Does anyone have any idea ?

по_альтернативной_ок.mw

Hi!

i have problems with surfdata command. I want to see beautifull surface, but see only few black lines.

and i dont know why, because i do all like in example.

Thank you vwry much for answers)

Hi everybody,

When doing calculations, I often run in the following problem.  I have the final solution wich I simplify symbolically so many terms are cancelling.  But I get this:

While I would like to regroup all the terms into the square root.  But look, even in this forum, Maple get sqrt(2) automatically out of the square root.

I know that it is the simple form.  But in some instances, I need the square root to stay together so I can show a property.  But is there a way to be able, sometimes, to tell Maple to leave all the terms under the square root?

Thank you in advance for your help.

--------------------------------------
Mario Lemelin

Maple 17.01 Ubuntu 13.10 - 64 bits
Maple 17 Win 7 -  64 bits
messagerie : mario.lemelin@cgocable.ca
téléphone :  (819) 376-0987

when a user liked a particular photo for the first launch of a mobile app, and the time different where by the event started and stoped was k seconds which is at constant (does not change). At this constant thus triggered the like application operation or function that was connected to the application server. Then later the user unlike the photo and the same event and operation was carried out at of (k + 2) seconds. The time variation that was carried out during the sequential or running loop event carried out by 2 seconds.

check if this derived equation is correct first f(X_n) = e_n(X_nⁿ  + X_n⁰ + 1) . if its correct do the second question.

what is the rate at which the operation will be finish given an equation f(X_n) = e_n(X_nⁿ  + X_n⁰ + 1) 

Hi

I have a simple Maple application with two TextArea components (TextArea0 and TextArea1) inside.

The start up code is following:

#++++++++++++++++++
Actions := module()
export DoNothing;
   
    DoNothing := proc()
    end proc;
    
end module; #Actions
#++++++++++++++++++

the action when cotents change for both TextArea components is

#++++++++++++++++++
use Actions in
DoNothing();
end use;
#++++++++++++++++++

The problem:
When I change the focus from one TextArea component to another and press Restart maple server then an error appears:

Error in Component TextArea1 with caption "TextArea": `Actions` is not a module or member.
Haw can it be fixed? Thank you.

problem_with_modul.mw

The dot product of vector v and u= ||v|| ||u||fosters also vector u*v=XvXu+Yv=A, v=<2,5>, u=<4,9>.

As it says in the title, I would like to solve the following ODE numerically using forward Euler method, without using the Student Package.

(dy(t))/(dt)=t(1-0.3t)-(ty)(1+0.6t)

with initial condition y(0)=1. I want to solve it for up to t=1, and then plot both the solution by Euler's method and the solution by "dsolve" on the same graph so I can compare them.

Also, can I make a separate DEplot with t extending to 5?

Thanks in advance.

When I open the MapleCloud palette, I see a small list of user documents under the rubric "Popular". Is there a significantly longer list I can access to browse?

Assume that |T(z)| < 1. Why does it follow that (T(z))^r → 0 as r → ∞?

(Hint |(T(z))^r| = |T(z)|^r.) Deduce from the above that, as k→∞, we have T(N^k(z)) → 0. Why does it follow that N^k(z)→i ?

What is the corresponding result when |T(z)| > 1 ?

(Hint: Let l/T(z) = U(z)so that |U(z)| < 1, then apply the same argument (check U(N(z)) = (U(z))², noting that U^-1(0) = -i).

Now consider the two regions of the plane:

R₊ = {z : N^k{z) → i as k → ∞}; R_- = {z : N^k(z) → -i as k → ∞}.

Draw a diagram to illustrate these regions, the line L and the roots i and -i. We call R₊ the basin of attraction for the root +i, and similarly R_-is the basin of attraction for the root -i.

Show that if z is on the set L (the common boundary of the two regions R₊ and R_{_}, then N^k(z) stays on L for all values of k. (This is easy once you identify what L is). So in this case iteration does not produce a root at all.

It would be appreciated if any one could help me with clear up my confusion with questions.

Thanks a lot.

Dear users

A friend of of mine has a problem with an integral and since it's for his thesis, it's pretty important. 

That's why I ask it here cause I don't know where to ask it elsewhere, so if it's wrong posted, completely my bad.

l:=(y*o)/(v);

R:=(Phi*o)/v;

A:=5*(a*ln(R)+b);

P:=sqrt(1+4*k^2*l^2*(1-exp^(-l/a)));

M:=int((2)/(1+P),o);

With other words, I want to integrate the wole thing to the variable o, who appears in the variables l and R.

Somehow, when I put this in Maple, it won't solve it. Probably it's just a stupid fault or i just forget something, but i don't find it. Does anyone knows how to solve it?

Already a lot of thanks!

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