Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

Hello

I have such problem:

I calculated rather complicated function, and i want to obtain the array or list of data with definite step.

For this purpose I used next expression:

Out:=[seq(t,F(t),t=0..tmax,step]

I need minimum 1000 points, but calculation for the 100 points requires about 300 second, for 1000 - it will be about 3000 seconds. It's too long for me.

 

How could I accelerate this process?

 

P.S.

Plotting of my function requires about 90 seconds for the entire range. So, this is very strange, thad converting to data list requres so much time.

Hi everybody,

Ever since i updated to maple 2015, the output of calculations display a disproportionate amount of zeros. For instance:

If i write:

2.3*10^(-4)

i get:

0.000230000000000

Is there anyway to make maple NOT display the last 10 zeros? I've tried Tools ->Options ->Precision -> Round screen display to X. But then it rounds all results to that X amount of digits, wich is also anoying. Like, if i set X to be 10, it'll write 3.000000000 instead of just 3.

Hope someone can help me :-)


Anton

the binomial coefficient  n k  can be defined recursively as follows for all nonnegative integers n, k:

(n)  = {0,      k>0

(k)  = {1       k=0, k=n

         {(n-1)+(n-1), otherwise.

          (k-1)   (k)

I need to complete a deinition of binom so that m so that binom(n,k) returns  n k  for all n greater than 0, and k greater than or equal to 0 using the definition of the binomial above..Any help appreciated..

binom:=
proc(n::TYPE1,k::TYPE2)
description "Compute a binomial coefficient";
option remember;
---MORE STUFF HERE---
end proc; # binom

 

I'm trying to find lypunov exponent for this  system of ODEs. I know I need to take the Jacobian but not sure if it's possible the way it's currently defined. If anyone could provide insight it would be much appreciated.

Eqns:= diff(omega(t),t)=-(G*MSat*beta^(2)*(xH(t)*sin(theta(t))-yH(t)* cos(theta(t)))*(xH(t)*cos(theta(t))+yH(t)*sin(theta(t))))/((xH(t)^(2)+yH(t)^(2))^(2.5)),diff(theta(t),t)=omega(t), diff(xH(t),t)=vxH(t),diff(vxH(t),t)=-(G*M*xH(t))/((xH(t)^(2)+yH(t)^(2))^(1.5)),diff(yH(t),t)=vyH(t),diff(vyH(t),t)=-(G*M*yH(t))/((xH(t)^(2)+yH(t)^(2))^(1.5)): ;

ICs := omega(0) = omega0, theta(0) = theta0, xH(0) = a*(1+e), yH(0) = 0, vxH(0) = 0, vyH(0) = sqrt(G*M*(1-e)/(a*(1+e)));

I want the exponent for omega, I  procedure that takes some initial conditions, changing just w0, and computes the long term value of omega. This plots a sort of bifurcation diagram. I'd like an estimate of the exponent to compare what I see. 

thanks for the help

ft

Variable exists but is not shown when using save/read with extension .m

My example maple code is

 

a:=b^2:

save a, "test1.m"

restart:

read "test1.m"

a;

 

after the read command, I can access the variable, but it is not shown under "Variables".

This lead to some confusion when debugging the worksheet. Can I change this somehow?

Using the input type file format is not a solution, since then reading takes forever for complicated expressions.

Further, in the read command documentation it says "This functionality is not intended for end users" for saving the file as .m. What does that mean?

I am a Mathematica user trying to make the switch to Maple, so first of all I apologise if I am making a stupid mistake here. The complete script is attached below.

I have the following line in my script:

    b := Trace(map(diff, HermitianTranspose(W), coords[1]) . map(diff, W, coords[1]))

W is a matrix defined earlier in the script. I can provide the context if it would help but I'm not sure if it's relevant to the issue I'm having. The problem is that

    simplify(b) assuming real, r>0

gives a vastly different result to

    simplify(Trace(map(diff, HermitianTranspose(W), coords[1]) . map(diff, W, coords[1]))) assuming real, r>0

How can this be? What can I do to make Maple simplify the expression correctly?

conifold_metric.mw

hello guys,

 

i have a system of autonomous equations which i want to plot its 3D phase space with directional field,

i have some problem with it :dy.mw , and i dont know how to command for add some directional field for 3D phase space .

 

thank you guys

 

I do not know! I am a novice in maple as well as in graph theory. I attach a doc in which i attempted to learn what is network. It seems alright with maple as it gives as what it understands as horizontal and vertical network. I do not expect a horizontal line as horizontal network and vertical line as vertical network, in which case a horizontal line with nodes 1 and 2 will be vertical network if  draw it vertically. The maple, in my opinion ( i am not a maple man) gives the answer in a vertical pattern, rightly so for clarity and space constraint.

I am sorry if i am confusing.  The document in my attempt to know more may give you also more insight. Thanks.

with(GraphTheory):

with(RandomGraphs):

N := GraphTheory:-RandomGraphs:-RandomNetwork(4, 0):

IsNetwork(N):

DrawNetwork(N, vertical)

 

with(GraphTheory):

with(RandomGraphs):

O1 := GraphTheory:-RandomGraphs:-RandomNetwork(4, 1):

GraphTheory:-IsNetwork(O1):

GraphTheory:-DrawNetwork(O1, vertical)

 

``

with(GraphTheory):

with(RandomGraphs):

P := GraphTheory:-RandomGraphs:-RandomNetwork(4, 0):

GraphTheory:-IsNetwork(P):

GraphTheory:-DrawNetwork(P, horizontal)

 

with(GraphTheory):

with(RandomGraphs):

Q := GraphTheory:-RandomGraphs:-RandomNetwork(4, 1):

GraphTheory:-IsNetwork(Q):

GraphTheory:-DrawNetwork(Q, horizontal)

 

``

 

Download MyAttempt.mwMyAttempt.mw

When I use the option assume = nonnegint or integervariables = {...} in Optimization[Minimize], or Optimization[LPSolve] I've got the message "kernel connection lost'. 

 

Thie even happens when I use the simple example from the Maple help:

with(Optimization):
LPSolve(2*x+5*y,{3*x-y=1,x-y<=5},assume={nonnegative,integer});

 

It says; mserver.exe has stopped working. 

What's wrong? 

How can I compute F from G according to the following text? (I implemented this but I need a more efficient implementation.)

 

Given a set G of polynomials which are a subset of k[U, X] and a monomial order with U << X, we want to comput set F from G s.t.


1. F is subset of G and for any two distinct f1, f2 in F , neither lpp (f1) is a multiple of lpp (f2) nor lpp (f2) is a multiple of lpp (f1).


2. for every polynomial g in G, there is some polynomial f in F such that lpp (g) is a multiple of
lpp (f ), i.e. ⟨lpp (F )⟩ = ⟨lpp (G)⟩,

--------------------------------------------------------------------------------------

It is worth nothing that F is not unique.

Example:  Let us consider G = {ax^2 − y, ay^2 − 1, ax − 1, (a + 1)x − y, (a + 1)y − a} ⊂ Q[a, x, y], with the lexicographic order on terms with a < y < x.

Then F = {ax − 1, (a + 1)y − a} and F ′ = {(a + 1)x − y, (a + 1)y − a} are both considered set.

please not that K[U,X] is a parametric polynomial ring (U is e sequence of parameters and X is a sequence of variables).

lpp(f) is leading monomial of f w.r.t. variables X. For example lpp(a*x^2+b*y)= x^2.

In Maple 2015.1 we have

restart;

solve([sin(2*x)/cos(x+3*Pi/2)=1,  x>-4*Pi, x<-5*Pi/2], x, allsolutions, explicit);

solve([sin(2*x)/cos(x+3*Pi/2)=1, x>0, x<2*Pi], x, allsolutions, explicit);

 

 

In the first example, the error message is not clear (actually there exists a unique root  x=-11*Pi/3), in the second example, one root  (x=5*Pi/3) is lost.

 

I've got this huge chunk of code which leads to an optimiazation at the very last line (Bestangles:=minimize(maximize()-minimize))). This minization is taking a very long time (havent solved it yet) and I would very much like to reduce that time. As I've understood maple does optimization by differentiating and then finding all extremes and comparing. Would this mean that since I minimize and optimize within a minimization command, it differentiates a ton of times? And if this is the case, can I somehow do the differentiation beforehand, since it is the same function being differentiate all the time? Or is there some other way I can improve the code? 
Thanks alot!

Heres the full code:

 

Consider a taper steel plate of uniform thickness t := 25mm as shown in the Fig. In addition to its self weight, the plate is subjected to a point load P := 100N at its mid point. Find the global force vector [F] , global stiffness matrix [K] , displacement in each element (1 and 2) , stresses in each element  (1 and 2) and reaction force at the support.Take E := 2*10^5N/mm2; rho := 8.2*10^(-5)kg/m3;

restart

t__1 := 150:

t__3 := 75:

w := 25:

l := 600:

t__2 := (t__1-t__3)/l*((1/2)*l)+t__3 = 225/2

A__1 := t__1*w = 3750``

A__2 := t__2*w = 5625/2``

A__3 := t__3*w = 1875``

Revised areas:

A__1e := (A__1+A__2)*(1/2) = 13125/4``

A__2e := (A__2+A__3)*(1/2) = 9375/4``

  E := 2*10^11:m2; F__1 := R__1:is support reaction N; F__2 := 100:N;``

rho__1 := 82*10^(-6) = 41/500000  N/mm2

rho__2 := 82*10^(-6) = 41/500000 N/mm2

l := 600:``

Number of elements,

n__e := 2:

l__e := 300 = 300````

q__0 := 100:N/m ; l := 1: m; n__e := 4:  elementsl  l__e := l/n__e: m;

We shall consider a two element system as shown in the Fig.
For element 1 Stiffness matrix K is

                                           Vector[row](2, {(1) = 1, (2) = 2})
K__1 := A__1e*E/l__e.(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1})) = Matrix([[2625000000000000, -2625000000000000], [-2625000000000000, 2625000000000000]])  Vector(2, {(1) = 1, (2) = 2})

For element 2 Stiffness matrix K is

                                         Vector[row](2, {(1) = 2, (2) = 3})
K__2 := A__2e*E/l__e.(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1})) = Matrix([[1875000000000000, -1875000000000000], [-1875000000000000, 1875000000000000]])  Vector(2, {(1) = 2, (2) = 3})

Global stiffness matrix obtained by adding all the elemental stiffness matrices and given b

           Vector[row](3, {(1) = 0, (2) = 0, (3) = 0})

K__g := Matrix(3, 3, {(1, 1) = K__1[1, 1], (1, 2) = K__1[1, 2], (1, 3) = 0, (2, 1) = K__1[2, 1], (2, 2) = K__1[1, 2]+K__2[1, 1], (2, 3) = K__2[1, 2], (3, 1) = 0, (3, 2) = K__2[2, 1], (3, 3) = K__2[2, 2]}) = Matrix([[K__1[1, 1], K__1[1, 2], 0], [K__1[2, 1], K__1[1, 2]+K__2[1, 1], K__2[1, 2]], [0, K__2[2, 1], K__2[2, 2]]])  Vector(3, {(1) = 0, (2) = 0, (3) = 0})

For element 1 Load matrix F is

  F__1e := (1/2)*`&rho;__1`*A__1e*l__e*(Vector(2, {(1) = 1, (2) = 1})) = Vector[column]([[861/25600], [861/25600]]) Vector(2, {(1) = 1, (2) = 2})

``

For element 2 Load matrix F isNULL

F__2e := (1/2)*A__2e*l__e*`&rho;__2`*(Vector(2, {(1) = 1, (2) = 1})) = Vector[column]([[123/5120], [123/5120]]) 

``

 

Download wrong_answers.mwwrong_answers.mwwrong_answers.mw

Ramakrishnan V

rukmini_ramki@hotmail.com

hello everyone

can any one tell me what is this anti reduction method. In the paper of serdal palmuk,the link is given bellow

http://www.hindawi.com/journals/mpe/2009/202307/

in this paper question #4 is first solved by anti reduction method for  exact solution.

but i dont understand this method,

if anybody know this then please also tell me how to solve this,

and in the next  (6 & 7 ) examples "in the pourus media equation" they first find its particular exact solution.i also dont understand this,so please tell me

actually i know how to solve ODE to find its exact solution but  i dont know how we find exact solutions of partial differtial equations,

so please help me to solve this problem

thanks

 

How to find the integral
,

assuming k and n  integer?
It is known (McCrea W. H., Whipple F. J. W.Random paths in two and three dimensions, Proc. Roy. Soc. Edinburgh. 1940. V. 60. P. 281–298) that

G(n,n)=2/Pi*sum(1/(2*k-1),k=1..n).

The general case is reduced to the case k=n.
This is not a creature of pure reason: the one appears in electric circuits
(see M. Skopenkov, A. Paharev, A. Ustinov, Through resistor net, Mat. pros. Issue 18 (2014), 33-65, in Russian, http://www.mccme.ru/free-books/matpros/pdf/mp-18.pdf).
I found G(8,8) = 182144/(45045*Pi) in 657.797 s and G(9,9) = 3186538/(765765*Pi) in 4157.687 s on my comp by

restart; s := time():(1/2)*VectorCalculus:-int((1-cos(9*Pi*x)*cos(9*Pi*y))/(sin((1/2)*Pi*x)^2+sin((1/2)*Pi*y)^2), [x, y] = Rectangle(0 .. 1, 0 .. 1)); time()-s;
Mathematica 10.3.0 does G(9,9) in 250.391 s on my comp.

 

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