MaplePrimes Questions

has anybody experienced issues with Maple connectiong to Solidworks?  There doen't seem to be much info ion trouble shooting a failed "OpenConnection" call.

 

thanks,

Bill

From a Maple Primes answer two years ago:

f(x,y) is the equation of a line through point [m,n]. The solve command finds values of a and b for which f(x,y) are lines through [m,n] and tangent to x^2 + y^2 = r^2.

f := proc (x, y) options operator, arrow; a*(x-m)+b*(y-n) end proc

solve([f(0, 0) = r, a^2+b^2 = 1], [a, b])

These commands are far from the conventional solution. Why do they provide the correct answers?

Dear all,

Thank you for helping me  to generate a table of values of f(x) starting with x=0 to 100 in steps of 1, that is for x=0,1,2,3,...,100.

 

I tried:

f:=x->2*sqrt(3)*a1*a2*(sum(pochhammer(1/3,k)*3^k*x^(3*k)/(3*k)! ,k=0..infinity)*sum(pochhammer(2/3,k)*3^k*x^(3*k+2)/(3*k+2)!  ,k=0..infinity)-sum(pochhammer(2/3,k)*3^k*x^(3*k+1)/(3*k+1)!  ,k=0..infinity)*sum(pochhammer(1/3,k)*3^k*x^(3*k+1)/(3*k+1)!  ,k=0..infinity));

tab_values:=[evalf(simplify(seq(Ni1(xx),xx=0..100)))];

But I the result is amazing.... I don't understand the problem.

Thanks

 

Hello,

I would like to plot gait diagrams (the lines you can see on the picture belowà from the solutions obtained with a NL oscillator (composed with 8 coupled odes). Here the result that I would like to obtain.

Initial plot:

 

Desired plot

 

 

I would like to obtain 4 lines corresponding to the 4 elliptic trajectories obtained with the NL oscillator. The four lines should be done like this. When the trajectory is above 0, the line should be colored in green. When the trajectory is below 0, the line should be colored in black. 

May you help me to define this kind of graph called gait diagrams from the solution of the NL oscillator ?

Here you can find my maple code:

K:=Matrix([<0, -1, 1, -1>,<-1, 0, -1, 1>,<-1, 1, 0,-1>,<1, -1, -1,0>]);

for i to 4
do
r[i]:=sqrt((u[i](t))^2+(v[i](t))^2):
omega[i]:=omega[sw]/(1+exp(b*v[i](t)))+omega[st]/(1+exp(-b*v[i](t))):
Equ[i]:=diff(u[i](t),t)=Au*(1-r[i]^2)*u[i](t)-omega[i]*v[i](t):
Eqv[i]:=diff(v[i](t),t)=Av*(1-r[i]^2)*v[i](t)+omega[i]*u[i](t)+MatrixVectorMultiply(K,<seq(v[i](t),i=1..4)>)[i]:
EqSys[i]:=[Equ[i],Eqv[i]]:
end do:

paramsCycle:=omega[st]=4*2*Pi,omega[sw]=2*Pi,Au=5,Av=50,b=100;
params:=paramsCycle;

Differential system 
sys:=map(op,eval([seq(EqSys[i],i=1..4)],[params]));
ic:=[u[1](0)=0, v[1](0)=0,u[2](0)=0, v[2](0)=-0.1,u[3](0)=0, v[3](0)=0.1,u[4](0)=0, v[4](0)=0.1];
Résolution1
res:=dsolve([sys[],ic[]],numeric):
Initial boundaries
tcalc:=4;
ic2:=[seq(u[i](0)=eval(u[i](t), res(tcalc)),i=1..4),seq(v[i](0)=eval(v[i](t), res(tcalc)),i=1..4)];
Résolution2
res:=dsolve([sys[],ic2[]],numeric):

tmax:= 40:
numpts:=100*tmax:
plots:-odeplot(res,[t,v[1](t)],0..tmax,thickness=2, view=[0..5, -1.5..1.5],numpoints = numpts);
plots:-odeplot(res,[t,v[2](t)],0..tmax,thickness=2, view=[0..5, -1.5..1.5],numpoints = numpts);
plots:-odeplot(res,[t,v[3](t)],0..tmax,thickness=2, view=[0..5, -1.5..1.5],numpoints = numpts);
plots:-odeplot(res,[t,v[4](t)],0..tmax,thickness=2, view=[0..5, -1.5..1.5],numpoints = numpts);
plots:-odeplot(res,[seq([t,v[i](t)+i*5],i=1..4)],0..tmax,thickness=2,view=[0..5,0..25], numpoints = numpts);

Thanks a lot for your help

Dear all;

Thank you for helping me to solve this  question.

I solve an ode, but I have an error when I would like to plot the solution.

uanble to achieve continuous solution with requested accuracy of 0.1e-5 with maximum 128 point mesh (was able to get 0.14e-5), consider increasing `maxmesh` or using larger `abserr`
Error, (in plots/odeplot) input is not a valid dsolve/numeric solution

I try to increase the point mesh or take a large abserr but always I have the same problem.

 

restart:
with(plots):

ode := diff(y(x), x, x) = x*y(x)+sqrt(x);

ics := y(0) = 0, y(1) = 1;
sol:=dsolve({ode,ics}, numeric):
odeplot( sol,[x, y(x)], x=0..1, maxmesh=1000);

I'm not really an expert in maple. I'm stuck with a problem that could be related to the precision of the computational engine...

I defined a function which is relatively complicated as follows:

m := proc (x, l, T) options operator, arrow; (exp(l*(x-T)/(1-exp(-l*T)))*exp(l*T/(1-exp(-l*T)))*(1-exp(-l*T))+(2-exp(-l*T)-exp(l*T))*exp(l*(x-T)/(1-exp(-l*T))))/(exp(l*T)-1)-(l*(x-T)*exp(l*(x-T)/(1-exp(-l*T)))-2*exp(-l*T)+4-2*exp(l*T))/(exp(l*T)-1) end proc

whenever i simplify the equation or factor some of the terms, I get completely different results.

As an example, you can define this function as follows

m := proc (x, l, T) options operator, arrow; (exp(x*l/(1-exp(-l*T)))*(1-exp(-l*T))+(2-exp(-l*T)-exp(l*T))*exp(l*(x-T)/(1-exp(-l*T))))/(exp(l*T)-1)-(l*(x-T)*exp(l*(x-T)/(1-exp(-l*T)))-2*exp(-l*T)+4-2*exp(l*T))/(exp(l*T)-1) end proc

 

By drawing the function, you can see how things get messy.

plot(m(450, (1/250)*x, 250), x = 0 .. 100)

 

I know the problem occurs as a results of computation round-off. How could I ameliorate that? I have already increase the precision in the option menu, but that didn't help. By the way, how can I be sure that the anwers maple gives me is correct, except using my intuition?

 

Hello,

I've tried to find the solution to my problem, but none of my attempts was succesful.

I have a function which is one-to-one in a particular domain which I am interested in. I would like to get the inverse function of it only in this domain. Here is my function and plot for xp=0..10000:

x := xp-> (-1)*720.5668720*sinh(0.2043018094e-3*xp-0.8532729286)+84952.59423+4.014460003*10^5*arcsinh(0.1219272144e-1*sinh(0.2043018094e-3*xp-0.8532729286)-0.2032498888)

I would appreciate any help,

Iza

I am tryng to write and solve within Maple the equations of movement for a single body subject to a gravitatinal field, F_=-GMm/r2

But I get an error message when trying to define the Angular Momentum which doesn't make sense for me.

Thank you for any help on this topíc.


restart; with(Physics[Vectors]); conventions, Setup(mathematicalnotation = true);
conventions, [mathematicalnotation = true]
r_ := rho*_rho;
r_ := rho _rho
_rho(t);
_rho(t)
rho(t);
rho(t)
v_ := diff(rho(t)*_rho(t), t);
/ d \ / d \
v_ := |--- rho(t)| _rho(t) + rho(t) |--- phi(t)| _phi(t)
\ dt / \ dt /
a_ := diff(%, t);
/ 2 \
| d | / d \ / d \
a_ := |---- rho(t)| _rho(t) + 2 |--- rho(t)| |--- phi(t)| _phi(t)
| 2 | \ dt / \ dt /
\ dt /

/ 2 \ 2
| d | / d \
+ rho(t) |---- phi(t)| _phi(t) - rho(t) |--- phi(t)| _rho(t)
| 2 | \ dt /
\ dt /

eq[1] := -G*M*_rho(t)/r^2-a_ = 0;
/ / 2 \ 2\
| G M | d | / d \ |
eq[1] := _rho(t) |- --- - |---- rho(t)| + rho(t) |--- phi(t)| |
| 2 | 2 | \ dt / |
\ r \ dt / /

/ / 2 \
| / d \ / d \ | d |
+ _phi(t) |-2 |--- rho(t)| |--- phi(t)| - rho(t) |---- phi(t)|
| \ dt / \ dt / | 2 |
\ \ dt /

\
|
| = 0
|
/
Eq[1, 2] := seq(Component(lhs(eq[1]), n) = 0, n = 1 .. 2);
/ 2 \ 2
G M | d | / d \
Eq[1, 2] := - --- - |---- rho(t)| + rho(t) |--- phi(t)| = 0,
2 | 2 | \ dt /
r \ dt /

/ 2 \
/ d \ / d \ | d |
-2 |--- rho(t)| |--- phi(t)| - rho(t) |---- phi(t)| = 0
\ dt / \ dt / | 2 |
\ dt /
NULL;

L_ := `&x`(r_, m*v_);
Error, (in Physics:-Vectors:-&x) found the unit vector _rho present also as a function _rho(t); either one form or the other - not both - can be present in an algebraic expression

 

HI, I am trying to solve two PDEs but in boundry conditions there is arising an error plz help.
Nazi.mw

Hello,

I have many tests to conduct on a MapleSim model. At each test simulations, I would like to change slightly some parameters.

As I have many tests to do (with different parameters), I would like to use a script so as to do like this :

Parameters test1
Launch MapleSim simulation
Store results of test 1

And so on ...

How can I do this kind of script ?

Is it possible to store the 3D animation results ?

Thank you for your help.

 

Hi,

I recently noticed that Maple 2015 has become irresponsive on mac (macbook pro) with the latest java release (Java 8 update 66). Did anywone experience the same problem ?

Hi,

I have a first order differential eq. for some variable say $r(x)$, where $x$ is the independent variable.

After solving this differential equation numerically, I want to use its solution in other expression for $r(x)$ and plot the expession with $x$.

Please let me know how to do it.

Thanks in advance.

 

 

I'm trying to use the CodeGeneration package to generate code for a series expansion. I'd like to wrap it in a function that specifies the arguments, so that the code generation package can generate a complete function definition along with definitions for all the temporary variables.

with(CodeGeneration):
f := proc(r): x->r end proc:
to_translate := f(convert(series(sin(x),x,20),polynom));

to_translate := proc (x) options operator, arrow; x-(1/6)*x^3+(1/120)*x^5-(1/5040)*x^7+(1/362880)*x^9-(1/39916800)*x^11+(1/6227020800)*x^13-(1/1307674368000)*x^15+(1/355687428096000)*x^17-(1/121645100408832000)*x^19 end proc


CodeGeneration['C']( to_translate );


Warning, procedure/module options ignored
double to_translate (double x)
{
  return(r);
}

Instead of using the value for 'r' passed in, CodeGenerate is producing a function which returns the bare symbol 'r' which is treated as a double. It shouldn't be an issue with lazy evalution because 'to_translate' is evaluated on the statement before the call to CodeGeneration, and to_translate has the full expression that I want to generate code for. How do I get CodeGeneration to produce the intented result?

psi = x^(4/5)*f(x, eta)/(1+x)^(1/20)

eta = y/(x^(1/5)*(1+x)^(1/20))

how can i (diff(psi, x))

how can i (diff(psi, y))

Has anyone solved this problem from an older Putnam paper?

An ellipse sitting in the first quadrant with its major axis parallel to the x axis is tangent to the positive x and y axes.

It slides clockwise within the first quadrant while maintaining tangency to both positive axes until its major axis is parallel to the y axis.

Prove that the locus of its centre is the arc of a circle.

I have crudely animated this motion by sliding the axes around the stationary ellipse. Is there a more elegant animation which slides the ellipse against stationary axes?

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