Hi,

    This is not an answer. I can't ask question, since I can't add any 'tags'! This problem appeared long time ago, but I hadn't solved it until now. So I ask here.

    I have a question about the function coeffs(). I don't know why it cann't get coefficients for me.

question.mw

Thanks.

Variable disappear after read file and then pass parameter to create thread

I have a question about animating 3d space curves (vector-valued functions). I have a curve that has a component x(t), component y(t), component z(t), and a time, t. The problem says to create an animation of a point moving around the curve. How do I do this?

Thank you.

first I define some constants (note I may change the constants later based on the context of the application)

p_l := 10^(-15);     
epsilon := 1.09*10^(-10);
p_B := 1.09*10^(-8);

n_A := 10^7;         
k_A := 0;
n_B := 10^8; 
k_B := 0;

then I define a function l(x,y):

l := (x, y) -> x^k_A*(1 - x)^(n_A - k_A)*y^k_B*(1 - y)^(n_B - k_B)

Now I use both with(GlobalOptimization) and with(Optimization) to maximize l(x,y) give some constraints and I get:

GlobalSolve(l(x_1, y_1), x_1 = p_l .. epsilon, y_1 = p_B .. 1, maximize, initialpoint = [x_1 = 0, y_1 = 0]);
  [-0., [x_1 = 1.09000000000000 10^(-10)   , y_1 = 0.633548870211381]]

Maximize(l(x_1, y_1), x_1 = p_l .. epsilon, y_1 = p_B .. 1, initialpoint = [x_1 = 0, y_1 = 0]);
 [0.33621648834727435318,  [x_1 = 1.00000000000000 10^(-15)   , y_1 = 1.0900000000000000000 10^(-8)  ] ]

Clearly the second answer is right and the first is wrong... I am not sure why the ``global optimazation'' is doing worse than the normal and free ``optimazation''.. Am I using the GlobalSolve in a wrong way??

A warm greeting for everyone

I have a function in one variable r defined in spherical coordinates f(r)=r^2+1.
I want to getting field plot or gradient plot for this function in spherical coordinates: 0<r<1,  0<theta<Pi/2 and 0<phi<2Pi.
is there any suggestion

Amr

Hi, for each region shown in the figure, there is a single point. I want to define the location constraints of the point in each region

e.g. for region 1 and 5, it is easy to define:

R1 := {p_B <= y_1 <= 1, p_l <= x_1 <= epsilon}

R5:={y_5 <= x_5, p_l <= x_5<= epsilon, p_l <= y_5 <= epsilon}

but what about region 2? it seems wrong if I define like this in maple:

R2 := {epsilon <= x_2 <= p_B, p_B <= y_2 <= 1} union {x_2 <= y_2, p_B <= x_2  <= 1, p_B <= y_2 <= 1}

Or it is actually right....

Hello people in mapleprimes,
I have a question.

I use maple2019 with mac os 10.14.6.

$ sw_vers
ProductName:	Mac OS X
ProductVersion:	10.14.6
BuildVersion:	18G95

With maple2019, errors appears.

> kernelopts(version);
        Maple 2019.1, APPLE UNIVERSAL OSX, Jun 6 2019, Build ID 1403154

> assume(a>-1,b>0);
> additionally(a<=1);
> about(a);
Originally a, renamed a~:
  is assumed to be: FAIL

> assume(tau<1,tau>0,s<1,s>0):
> a_e1:=tau*s*(1+tau)<tau*s+tau+s-1:
> b_e2:=expand(lhs(a_e1)-rhs(a_e1))<0:
>  b_e3:=collect(b_e2,s,factor):
> solve(b_e2,s) assuming tau<1;
Error, (in assuming) when calling 'property/ConvertProperty'. Received: 'FAIL
is an invalid property'

On the other hand, with maple2018, they do not.

> kernelopts(version);
       Maple 2018.2, APPLE UNIVERSAL OSX, Nov 16 2018, Build ID 1362973

> assume(a>-1,b>0);
> additionally(a<=1);
> about(a);
Originally a, renamed a~:
  is assumed to be: RealRange(Open(-1),1)

> assume(tau<1,tau>0,s<1,s>0):
> a_e1:=tau*s*(1+tau)<tau*s+tau+s-1:
> b_e2:=expand(lhs(a_e1)-rhs(a_e1))<0:
> b_e3:=collect(b_e2,s,factor):
> solve(b_e2,s) assuming tau<1;
                                    1
                               [{-------- < s~}]
                                 1 + tau~

Are errors due to some bugs in maple2019 when being used with mac os 10.14.6?

Take care.

taro

Can anyone know how to speed up Directsearch package 2 optimization command

model1.mw

When I run the program it shows an error about one of the dependant variables.

Hello

I want to find abs(psi)^2 for below psi

psi=int(int(exp(-a*(mu-b)^2)*exp(-c*(V-d)^2)*exp(I*(Q*u-(5/3*Q)*v^(mu/V)-3/16*(v^(4/V)/Q)))*v^((1/2)*(1-V)/V), mu = 0 .. 2), V = -15 .. 15) when mu=0..2 and V=-15..15 for Q:=3, a=10, c=5, b=3 and d=0.

unfortunatelly, I write above psi in maple 18 but it does not work, please tell me how to calculate (abs(psi))^2 for above wave equation (psi).

with the best regard.

Hello,

I'm working in MapleSim to model a mechanism called Schatz which has a six-Revolut, known whit TURBULA name.
I modeled the mechanism in CREO to have a CAD part to use in maplesim.
everything is okay when I run the simulation without a speed constant. as shown in the picture above.
when I add a speed source, MapleSim finds an error (Internal error occurred during the simulation).
Can someone help me please?

Is there or will there be a german version or is there a language pack available somewhere?

I was wondering if you can index arrays in maple with a loop. In MATLAB, you can index all rows and columns in the Matrix, A, with A(:,:), or a portion of the elements such as A(2:10,5:20) to index, or "loop" through these specific elements. In Maple, I am trying to index the following elements, as done in MATLAB with the command 

dIdQ(nmid-(imax-1)/2:nmid+(imax-1)/2,1) = -(omega_t(1:imax,1) - omega_1(nmid-imax*jmax-(imax-1)/2:nmid-imax*jmax+(imax-1)/2,1));

Is there a way to convert this matlab code to maple code?

Any help is very appreciated. Thank you!

r=a+b sin(x), this is the function I want to plot the graph in polar co-ordinates but how to manage those arbitrary constants "a and b" ?

Hi everyone, I need someone’s expertise with the Maple Soft Application, I am currently into my 4^th day of my Maple Soft experience and also returning to School after many years. Can someone help me with the following codes to solve step by step Differential Equation 1 below;

Please help me with following codes below

1. Boundary Condition 1
2. Boundary Condition 2
3. General Solution
4. The integration Part of the Equation
5. Solving the Differential Equation  

Equation 1.

Determine the equilibrium temperature for a one dimensional rod with a constant thermal properties with the following source and boundary conditions.

1. Q = 0, U(0) = 10, u(L) = 20
2. Q/K0 = x, u(0) = 0, u(L) = 10

I understand how to obtain the solution to the equation above,

(a) Equilibrium satisfies

                                                U’’(x) = 0,

Whose general solution is u = c1 + c2x.

The boundary condition u(0) = 0 implies c1 = 0 and u(L) = T implies c2 = T/L so that u = T x/L.

(f) In equilibrium, u satisfies

U’’(x) = −Q/K0 = −x^2,

Whose general solution (by integrating twice) is

u = −x ^4 /12 + c1 + c2x.

The boundary condition u(0) = T yields c1 = T, while u’(L) = 0 yields c2 = L^3/3.

Thus u = −x ^4 /12 + L^3x/3 + T.