Unanswered Questions

This page lists MaplePrimes questions that have not yet received an answer

Hi,

thank for any  help.

I need you help to make a movies ( animation)  according to the index k=1..10.  k is used in Vect_T[k], vector of time to make an animation of the true solution, and it's also used in Matrix(M[..,..,k]), h

True solution:ow can I make a seq of plots for different value of k. 

 

plot3d(True_solution(x,y,Vect_T[k]),x=0..a,y=0..b,axes=boxed,grid= [11,11]);

Analytic solution

plots:-matrixplot(Matrix(M[..,..,k]),axes=boxed,tickmarks= [[1="0", 2=" ", 3=" ", 4=" ", 5=".4", 6=" ", 7=" 0.6", 8=" ", 9=".8", 10=" ", 11="1"],[1="0", 2=" ", 3=" ", 4=" ", 5=".4", 6=" ", 7=" 0.6", 8=" ", 9=".8", 10=" ", 11="1"],default],labels=[x,y,u],axes = boxed);

I tried, this:

S:=seq(plots:-matrixplot(Matrix(M[..,..,i]),labels=[x,y,u],axes = boxed),i=1..10):
plots:-display(S);
 But I get only one figure.

 also,

I tried

BoxTrue:=Null;
for k from 1 to 10 do
BoxTrue:=BoxTrue, plot3d(True_solution(x,y,Vector_time[k]),x=0..a,y=0..b,axes=boxed,grid= [11,11]): od:
display(BoxTrue, insequence=true);
also, no display

 

want to write in maple code

to generate all commutative diagram 

with adjacency matrix

 

however, i only know a -> b, b-> c , a->d , d-> c

google no information about all commutative diagram, 

 

another problem is

would like to enrich theory , however, do not know how to connect property such as equations with diagram

Hi all,

I will use the following dummy example.

with function,f

f:=(xid,yid)->sum(x[i],i=1..xid)*sum(y[i],i=1..yid);

 

and a complicated term, myterm


myterm:=(f(3,4)+f(2,2))*f(1,1):
myterm:=expand(myterm);

 

'if' i have some previous knowledge, or know a bit of the term, i can find the structure by doing this


repar:=[f(1,1),f(2,2),f(3,4)];  # Or with more f(xid,yid) terms

tmp:=seq(repar[i]=ff[i],i=1..3);
simplify(myterm,{tmp});          # This is fine, gives me what i want

 

But, can we go further, and more 'obvious'

 

Given the fucntion f, same as before, and the same 'myterm'

can I have this
restart:
iwant:=(f(3,4)+f(2,2))*f(1,1);  # as a result, straightforward

so I dont have to go back to 'repar' and find that the terms exactly are.

 

Thanks,

 

Dear all;

 

I need a help in this question.

u(x,y,t) my solution of PDE. x,y space, and t=time.

In the case without t. .i.e. u(x,y). Here is a visualization of the lattice u(x_i,y_j). i=1..3, anf j=1..3.  Please try this example, it's working. 

my question, if i would like to add a third variable t, i.e.  and get u(x_i,y_j,t_k)  on each points. I would like to show the lattice in (x,y) plan for each t_k. I Think I will get many parallel lattice. 

Please can some one, modify this code to get the parallel lattice. Thanks.


with(LinearAlgebra):
with(plots):
with(PDEtools):
interface(rtablesize=20):

Here,


L := 'L':
N := 'N':
g := 'g':
Z := i -> -L+2*L/(N+1)*i;
x[0] = Z(0),x[N+1] = Z(N+1),y[0] = Z(0),y[N+1] = Z(N+1);

 

N := 4;
L := 1;
r := L/(N+1)/4;
ngon := (n,x,y,r,phi) -> [seq([x+r*cos(2*Pi*i/n+phi), y+r*sin(2*Pi*i/n+phi)], i = 1 .. n)]:
p[1] := display([seq(polygonplot(ngon(4,Z(0),Z(j),r,Pi/2),color=magenta),j=0..N+1),

seq(polygonplot(ngon(4,Z(N+1),Z(j),r,Pi/2),color=magenta),j=0..N+1),

seq(polygonplot(ngon(4,Z(i),Z(0),r,0),color=magenta),i=1..N),

seq(polygonplot(ngon(4,Z(i),Z(N+1),r,0),color=magenta),i=1..N),

seq(seq(polygonplot(ngon(20,Z(i),Z(j),r,0),color=white),i=1..N),j=1..N),

textplot([seq(seq([Z(i+0.1),Z(j),typeset(u[i,j])],i=0..N+1),j=0..N+1)],align={above,right})],view=[Z(-1)..Z(N+2),Z(-1)..Z(N+2)],tickmarks=[[seq(Z(i)=typeset(x[i]=evalf[2](Z(i))),i=0..N+1)],[seq(Z(i)=typeset(y[i]=evalf[2](Z(i))),i=0..N+1)]],

axes=boxed,scaling=constrained,labels=[``,``]):
p[1];

 

 

 

 

 

 

What is the weak solution integral equation for 

du(x)/dx2 -(1+x2 ) u(x)-1=0

okay so we have to, In the Notation of this question http://www.mapleprimes.com/questions/201278-Fix-A-Syntax-Error-In-My-Simple-Function-please-Help , let f0 be contained in V be any given function and de ne
a sequence (fn) with n contained in (N subscript 0) of functions fn contained in V by

(f subscript 0) := f0  and   fn+1=Afn  for all n contained in (N subscript 0).


prove that this sequence converges pointwise, i.e. that for all   i,j contained in [N]X[N],
(f subscript infinity)(i, j) := (lim n-> infinity  fn(i, j)) exists and that "triangle"(f subscript infintiy) =0
and that
f1 = 0 :

any help would be much appreciated guys <3  

I have a plot of a sinusoidal function(cos_phi) which changes with incrementing values of variable k. When this function goes above 1 or below -1 I would like to have the range of k for which it occurs outputed somehow. Would anybody know how to do this?

I am having trouble printing out a limit cylce on maple 16.  I have the attached file and if anybody could look at it and perhaps help me out it would be greatly appreciated.  The first limit cycle is supposed to look somewhat like the second one.  I'v tried many different things but nothing seems to be working.  an explenation would also be nice too.  if the file does not open correctly also let me know. thank you very much.  

 Math_4710_Hilbert_16.mw

Hi , i need to write some maple functions for the equations below and i am not sure if this should be arrays or not. please see the question below;

Consider a factory which manufactures only one product. Raw material is bought from external supplier and stored until required.

Finished items are held in a warehouse. The operation of factory and its warehouse

can be modelled as a set of equations as folllows.

 Let us define at time t :

R(t) = Raw material stored (units)

F(t) = finished goods stock (units)

B(t) = order backlog (units)

T(t) = target stock level for finished goods (units)

 

All variables defined above give quantities at the start of week t.

 

X( t, t+1 )= weekly orders received from customers

M( t, t+1 )= raw material supplied per week.

P( t, t+1 )= production per week.

D(t, t+1 )= amount dispatched to customers per week.

 

All variables defined above give quantities over week t to t+1

(i.e over the week t). The operation of the factory and its warehouse can be expressed

as a set of equations given as follows:

 

Backlog and Stock Position

(1) B(t+1) = B(t)+X (t,t+1)–D(t,t+1)

(2) T(t+1) =(m +1)/m(X(t,t+1)+X(t–1,t)+…+X (t–m+1,t–m+2))

 

(assuming that the company wishes to maintain m (suppose m=5)) weeks stock of

finish items and hence the target level is m times the average of the last m–1 weeks)

(3) R (t+1)=R(t)+M(t,t+1)–P(t,t+1)

(4) F (t+1)=F (t) +P(t,t+1)–D(t,t+1)

Rates

(5) D (t,t+1) = B(t) if B(t)<F(t)

F (t) otherwise

(6) M (t,t+1) = P(t–1,t)

(7) P (t,t+1) = T(t) – F(t) + D(t,t+1)

= R(t) if result exceeds R(t)

= 0 if the result is negative

 

Given the initial values for the variables, it is possible to simulate this system to study

how the system will respond to the order rate. Suppose that all is calm, and the factory

has operated as follows for the last five weeks.

Target warehouse stock = 250

Finished goods stock = 250

Raw material stock = 150

Production rate = 50/week

Material supply rate = 50/week

Order rate = 50/week

Order backlog = 50

 

Suppose the behaviour continues for the first week of the simulation but that during

next week orders double due to the sales promotion. During the third week orders

drops to zero as all demand returns of the previous week was satisfied. For the fourth

week and the succeeding weeks, demand returns to an order rate of 50/week. What

happens elsewhere in the system? A deterministic simulation will provide the answer

to the above mentioned question.

 

 For this compute the following.

i) The values of the equations (1)–(4) at the start of week t.

ii) The values of the equations (5)–(7) i.e. the new values of the rates during

the following week.

iii) Move simulation time to the start of the next week.

Next simulation should be presented in tabular form and plot production and demand

rate to examine the performance of the system.

 

All help will be much appreciated.

Best Regards,

 

I do not have any code to produce yet.  I want to have Maple animate a bouncing ball on some 3d surface.

and if the A are an inversable matrix

Hi

worksheet output doesnt display the eigenvalues of a 19X19 matrix. Just gives in the form Root of(...). when datatype is indicated as anything other than complex(sflot) it gives error "unable to store". solution is required in compact mathematical expression form (may involve square roots etc). Please help.

Thanks

When I try "touch and drag" in Maple documents, the documentation for example, I end up selecting text instead of scrolling. Unfortunately the alternatives for scrolling are a) the scrollbar, b) page-up/down or c) moving the cursor. Each of those alternatives is about 2.5311 orders of magnitude more inconvenient than touch&drag.

Does anybody know if its possible to get Maple (18) to scroll instead of select text when I touch the screen?

 

Hi everyone.

I'm trying to visualize some some field lines for a 3D magnetic field I have simulated, and I have read that the Student[VectorCalculus] package is the best package for this type of visualization. I tried implementing it as described online and in the manual, and I got an error:

Error, (in plots/fieldplot3d) no non-zero vectors found

 

Does anyone have any ideas as to what could be causing this issue?

Attached is the Maple document I am using. The section entitled "Problem Area" is where the issue is.

 

turtle-rec-bend.mw

Hello,

• Is there a simple way to find the domain for the real solutions of f(x)?

• And is there a way to let maple get the part of f(x) with the sqrt?
   (not by typing it by hand as I dit below)

• Is there a way to write the summary of the found domains in one line?

Thanks for your help. 





restart:
# How to find the Domain for real solutions for x?
f(x):=(x-1+sqrt(x^2-3*x+2))/(x-1);
discont_for_x=discont(f(x),x);
# x<>+1 (because the de denom=0 is not allowed)
denom(f(x))=0;
x={solve(denom(f(x))=0,x)};
# x<=1 union  2<=x (because the part under the sqrt must be >=0 to give Real solutions)
sqrt(x^2-3*x+2);
0<=x^2-3*x+2;
x=solve(0<=x^2-3*x+2,x);




First 264 265 266 267 268 269 270 Last Page 266 of 361