Hello dear!

I want to find the vector field of a vector field my approch is below:

with(VectorCalculus);
SetCoordinates(cartesian[x, y, z]);
V := VectorField(<(u(x, y, z), v(x, y, z), w(x, y, z))>);
Gradient(V);

I want answer as bellow:

Matrix(3, 3, {(1, 1) = Diff(u, x), (1, 2) = Diff(u, y), (1, 3) = Diff(u, z), (2, 1) = Diff(v, x), (2, 2) = Diff(v, y), (2, 3) = Diff(v, z), (3, 1) = Diff(w, x), (3, 2) = Diff(w, y), (3, 3) = Diff(w, z)});

Please help me to fix this problem. I am waiting the quick answer.

THANKS in advance

Hello people in mapleprimes

I asked a question for a similar thing before.
https://www.mapleprimes.com/questions/222751-PDEtoolsdeclare

The question I had then is not solved yet in my mind, so I think  I will ask you
again. It is about differentiation of a composite function.

with(PDEtools):
declare(f(x,y,z),g(t,y));
a:=f(x,y,z);
b:=x=g(t,y);
c:=eval(a,b);
d:=diff(c,y);
OFF;
e:=d;

h:=convert(e,diff);

For both e and d, the notation of D__1 and D__2 appear.
And, even if, with convert, I changed them to another form,
what I can get is more complicated form of h.

Is it inevitable to use D__1 and D__2 brabra in the differentiation of
composite function?

I am writing this question thinking that if there is a way of go around it,
 I want to know.

And, though this is the second question, even if I use latex( ) command,
natually, codes of Tex I can get is that of the output shown on the screen, including D__1 and D__2, even though I want it to be shown with f__x and f__y or that including prime. Then, I have to edit that tex file, changing D__1 to f__x etc. which I think is very complicated modification of the text file.
I wish there is a way to modify it in Maple so as it to be direct TeX codes which  do not require me to modify.
Is it impossible to do such a thing?

Thanks in advance.

diff_2.mw

How to find the range of poset?

i would like to search the range of range is equal to the range of domain after solve system.

it may not exactly equal but around 80% of elements in range belong to the range of domain

if it is not factorisation domain,

how to generate ideal with function in maple?

On the help page, the description says the remove function does the opposite of select.
when I run:

remove(has, D[1](xx[0]), {xx[0]})

it gives: D[1]()

while: select(has, D[1](xx[0]), {xx[0]})

produces: D[1](xx[0])

I find this during programming. 

Is there a way to remove the whole expression "D[1](xx[0])"

Many thanks!

1. Make a plot wich shows CD(R) as a function of R for R=1..10000. V0=abs(V), V=[vx,vy] (d,rho,mu are constants)

2. Also make a plot wich shows log10(CD(R) as a function of log10(R) in same range R=1..10000
Use y=log10(R)

I appreciate any help .

Hi everyone.    How i can introduce an acximetric tolerance?. For example +10/-5. Thanks!

f:= z-> MeijerG([[1/2], []], [[], [1]], z);

simplify(f(z));
                               0
convert(f(z), hypergeom);
                               0
f(.2);
Error, (in evalf/MeijerG/limit) numeric exception: division by zero

f(2.2);
                         -0.4721129136

It seems that when MeijerG is not analytic on the unit circle, simplify and convert give a result that's valid only inside the unit circle.

This solve command produces both unconditional and conditional solutions.

Is there a way to have it only output unconditional solutions, or, failing that, is there a way to select the unconditional solutions from its output?

solve({2*x+y+z = 3, x^2+y^2+z^2 = 3}, [x, y, z], real, parametric)

Hello,

When i try to open my 583 KB maplefile in maple 2017, a box shows up, asking: "How do you wanna open this text file?"  with 3 options. No matter what option i pick, a white empty page shows up. I have windows 10 and office 2016 installed on my PC.

This is the second time this problem happens to me, and both times have i lost hours of work. I have red multiple chats of people with the exsact same problem trying to get support from MaplePrime, but unfortunately, you have never been able to repair the files or find the source of the problem. 

Is this problem related to Maple 2017 or why does this happen? how can I trust Maple not to "loose" my work a third time?

My file: Download Hjemmeopgavesæt_2.mw

Kind regards Anna

Is there a way in Maple to numerically integrate (with some symbolic variables retained), involving vectors?

Alternatively, is there a way to simply numerically integrate the following expression involving vectors, without any remaining symbolic variables remaining?

Integrate over vectors p1, p3
(with vector x remaining as free variable after integration)
( e^{-i p1 . x} (p1² + p3²) )  / ( (p3 - p1)² (p3²*a² +1)² )

If absolutely required for numerical integration, i.e. no other way to get maple to perform a semi-numerical integration, then vector x, and the scalar variable "a" can also be specified a value, but x should remain a vector. 

Of course, if the integration above can be done analytically, or even partly analytically (e.g. if the vectors are expressed in spherical polar co-ordinates, and some of the variables like Sin \theta etc. can be integrated over), that would be very useful as well. 

 I wand to plot the following expression involving an imaginary number. Can I get some help??

((1/10)*exp((2/135)*sqrt(-11)*sqrt(225)*t-(2/3*I)*x+1/15)/((3/10)*exp((2/135)*sqrt(-11)*sqrt(225)*t-(2/3*I)*x+1/5)

Hi,

I am trying to complete this assignment but I am having trouble on this question. It is asking me to use a graphical approach to estimate a 𝛿 > 0 such that for all x...

0 < | x - c | < 𝛿 implies that | f(x) - L | < ε

It says I can do this by plotting my f(x), y₁, and y₂ over the interval [c - 𝛿, c + 𝛿] with a y-range of [L - ε, L + ε]. y₁, and y₂ are banding lines defined by ( L - ε ) and ( L + ε ), respectively. The value of ε is shown below. I dont know where to start but this is what I have so far...

f(x) := ( x ( 1 - cos(x) ) ) / ( x - sin(x) );

The limit ( L ) of this function as x approaches 0 ( c = 0 ) is...

L := ( limit ( f(x), x = 0) );

The limit is 3 ( L = 3 ).

 ε = r = Student Id / 5000000   = 0621748 / 5000000   = 155337 / 1250000

 ε := 155337 / 1250000;

y₁ := ( L - ε );

y₂ := ( L + ε );

Originally, it had asked me to graph the function with the banding lines (y₁, y₂) together within the interval [-0.2, 0.2], which I have done. I just do not know how to find a value for 𝛿 in this case. Please help.

Thanks.

abs(-((-(1/20)*sqrt(5)-1/20)*cos((1/5)*Pi)-(1/20)*sqrt(2)*sqrt(5-sqrt(5))*sin((1/5)*Pi))*24^(1/5)*5^(3/5)-I*((1/20)*sqrt(2)*sqrt(5-sqrt(5))*cos((1/5)*Pi)+(-(1/20)*sqrt(5)-1/20)*sin((1/5)*Pi))*24^(1/5)*5^(3/5))

I am trying to compute the modulus the above complex number. However, the "abs" command does not affect the complex expression. Any reason for this?

Many thanks!

Hello,

I'm writting a Modelica External component which use a C function, like this :

model SAMS.BlockUdpReceiver
    parameter Integer relativeSockId = 0 "Relative socket id (0..10)";
    parameter Integer stateUdp = 2 "state of receiver 1=receive data from udp ; 2=state from recorder config file";
    parameter Real step_time = 0.05 "sample time in second";
    parameter Integer nbDouble = 3 "number of double value to send";

    function udpReceiveDataMapleSim
        input Integer relativeSockId;
        input Integer stateUdp;
        input Real step_time;
        input Real currentTime;
        input Integer nbDouble;
        output Real dataToReceived[nbDouble];
    external "C"udpReceiveDataMapleSim(SAMS.BlockUdpReceiver.udpReceiveDataMapleSim.relativeSockId, SAMS.BlockUdpReceiver.udpReceiveDataMapleSim.stateUdp, SAMS.BlockUdpReceiver.udpReceiveDataMapleSim.step_time, SAMS.BlockUdpReceiver.udpReceiveDataMapleSim.currentTime, SAMS.BlockUdpReceiver.udpReceiveDataMapleSim.nbDouble, SAMS.BlockUdpReceiver.udpReceiveDataMapleSim.dataToReceived)
            annotation (
                Library = "F:/SAMSSVN/workspaceMaple/SamsLibrary/x64/Debug/SamsDll.dll",
                __Maplesoft_callconv = "stdcall");
    end udpReceiveDataMapleSim;

    Modelica.Blocks.Interfaces.RealOutput dataReceived[nbDouble] annotation (Placement(
        visible = true,
        transformation(
            origin = {120, 50},
            extent = {
                {-20, -20},
                {20, 20}},
            rotation = 0),
        iconTransformation(
            origin = {110, 50},
            extent = {
                {-10, -10},
                {10, 10}},
            rotation = 0)));
equation
    when {initial(), sample(0, step_time)} then
        dataReceived = udpReceiveDataMapleSim(relativeSockId, stateUdp, step_time, time, nbDouble);
    end when;

    annotation (
        Diagram(coordinateSystem(extent = {
            {-100, -100},
            {100, 100}})),
        Icon,
        experiment(__Maplesoft_engine = 2),
        __Maplesoft_none);
end SAMS.BlockUdpReceiver;

I want MapleSim to call C function at initial() time and with a sampled period of step_time.

But, the function is called at each simulation step (here at 0.4 ms)

for demonstrate it, I print a log file each time the function is called.
this is a part of the result :

id Receiver =11 ; CurrentTime = 0.151200 while t_step=0.002000
id Receiver =11 ; CurrentTime = 0.151200 while t_step=0.002000
id Receiver =11 ; CurrentTime = 0.151200 while t_step=0.002000

id Receiver =11 ; CurrentTime = 0.151600 while t_step=0.002000
id Receiver =11 ; CurrentTime = 0.151600 while t_step=0.002000
id Receiver =11 ; CurrentTime = 0.151600 while t_step=0.002000

id Receiver =11 ; CurrentTime = 0.152000 while t_step=0.002000
id Receiver =11 ; CurrentTime = 0.152000 while t_step=0.002000
id Receiver =11 ; CurrentTime = 0.152000 while t_step=0.002000

Why C function is called so many time (3 times in a step), and why is called for each simulation step (0.4ms) instead of each 2 ms ?

Maybe I make a bug in my Modelica code ? or not ?

thanks for any reply.

restart;
globale values, small water droplet in gravity field  , through air with velocity V=[vx,vy]  ,v=abs(V)  total dragforce kan be written FDD=abs(FD)

mu := 0.18e-4; rho := 1, 2; d := .2; Co := .4;
#abs(FD) divided into  FDx og FDy
%;

FDx := (3*Pi*mu*d+(1/8)*Pi*Co*rho*d^2*abs(v))*vx;




FDy := (3*Pi*mu*d+(1/8)*Pi*Co*rho*d^2*abs(v))*vy;

# Define v=[vx,vy] to abs(v) wich gives the length of v (scalar value)


v := proc (vx, vy) options operator, arrow; (vx^2+vy^2)^.5 end proc;

#Make FDx og FDy to  abs(FD) 

FD := sqrt(FDx^2+FDy^2);

#Define total dragforce

FDD := (1/8)*Cd*abs(v)*Pi*d^2*abs(v^2);

#Total dragforce can be writte like FDD=abs(FD), solve for Cd(v)

FD1 := solve(FD = FDD, Cd);

#Make a function which returns Cd(v) 

Cd := proc (v) options operator, arrow; 63.66197724*FD1/abs(v)^3 end proc;

HOWTO GET MY Cd(v) to return values ?????

Ive got some problem understand the use of = and := , and how to get my Cd function to mangage the v expression. A copy (in norwegian) from my assignment to see the formulas is attached. Im thankfull for any help.

inl1d.pdf