Hi

Can we sketch a Simulink model using Maple.

Hi

Is there any idea using Maple  to compute the leg lengths  in Stewart-Gough platform ( see the following figure)

I would like to make a code using cross-product and the unit Normal N 

·       P is of length 13 and displaced in the Y direction by 10 degrees from the vertical (Z axis)

·       N is displaced in the X direction by 18 degrees from the vertical (Z axis)

·       L_B is position [7 5] from the bottom plate centre in the XY plane

·       L_T is in position [3.5 4.2] from the top-plate centre in the AB plane

Many thanks for any help

Dear all;

I open this good discussion, and hope can get a nice and strong idea in this domain of approximation of Hankel funciton and order truncation of infinite series. Thanks for all idea, can improve the discussion. 

Using the asymptotics of Hankel function for large argument and large orders ( both together) and
find   an order of truncation N of the obove series so that we can ensure an error bound  of epsilon( epsilon very small given).  abs(sum ( c[m]*HankelH1(m,x)*exp(I*m*theta), m=-infinity..infinity)-sum ( c[m]*HankelH1(m,x)*exp(I*m*theta), m=-N..N))<epsilon.

A first idea come in mind: the series converge, so that the general terms of this series converge to zero, and in particularity,  abs(c[m]* HankelH1(m,x))<1:
then abs(c[m])<1/ abs(HankelH1(m,x)).
So we can ensure an error bound of epsilon on the coefficient c[m] by imposing  (HankelH1(m,x))<epsilon  this lead to abs(c[m])<epsilon.
I consider the case where m and x are very large, I can suppose for example m=x*(1+zeta), with 0<zeta<1. zeta parameter. So that our truncation N depend on zeta.
And then how can I find and approximation of the inverse of Hankel function for large argument and large order. using m=x*(1+zeta). I think this give us N the truncation order.
I hope get a good discussion in this subjet.
Of course maybe there are other strong idea to compute the truncation series.
I get the following error in the code:

Error, (in MultiSeries:-multiseries) unable to expand with respect to parameter

###### Code### and error
restart:

with(MultiSeries):
assume(0 <= x);

assume(0 <= zeta<1);

HankelH1(v,x):

sum(c[m]*HankelH1(m, x)*exp(I*m*theta),m=-infinity..infinity);

1/HankelH1(x*(1+zeta),x);

MultiSeries:-asympt(%,x, 4);

eval(%, O=0);

convert(%,exp);

simplify(%);

truncated_series.mw

Hello;

I need some help to compute the series approximation of the modulus and argument of hankel function for large x. The code display

 Error, (in asympt) unable to compute series

Thanks  for helping me.

#We define the hankel function as
#HankelH1(v,x) = BesselJ(v,x) + I*BesselY(v,x), where BesselJ and BesselY are bessel function.
#In this question the parameter "v " is  fixed. "

# Code

restart:
with(MTM):
HankelH1(v, x);
# The modulus of Hankel function
Mn:=x->abs(HankelH1(v, x));
thetan:=x->argument(HankelH1(v, x));
phin:=x->argument(diff(HankelH1(v, x),x));
dervthetan:=x->diff(thetan(x),x);
dervphin:=x->diff(phin(x),x);
# Compute series
series(thetan(x),x=infinity,7);
series(phin(x),x=infinity,7);
series(Nn(x),x=infinity, 7);
series(Mn(x),x=infinity,7);
# I define the following function
f:=x->(Nm(x)/Mn(x))*exp(I*(phin(x)-thetan(x)));

# Series approximation
series(f(x),x=infinity, 7);

Dear all;

Hello everybody, I need your help to dispaly some values obtained using my function f. When I run the code there is no results obtained. Many thanks.

restart:

# The vectors e(i) satify the folowing conditions
e(0)*e(1)=e(n-1) assuming  1<n;
e(0)*e(0)=e(2):
e(1)*e(1)=e(n-1) assuming  1<n: :
e(2)*e(1)=e(n) assuming  1<n:
#
for i from 1  to n-1 do
e(i)*e(0)=e(i+1);
end do:

# We define the function f
f:=e(0)->e(0)+(n-3)*e(1);
f:=e(1)->(n-2)*e(1);
for i from 2  to 3 do
f:=e(i)->(n+i-3)*e(i)+(i-1)*(n-3)*e(n-3+i);
end do:

for i from 4 to n do
f:=e(i)->(n+i-3)*e(i)
end do:

# We define the two vectors
x:=sum(alpha(k)*e(k),k=0..n);
y:=sum(beta(k)*e(k),k=0..n);

#Question : I would like to compute the following  but there is no display of the solution. 
(x*y);
f(x*y);
f(x);
f(y);
x*f(y);
f(x)*y;
f(x*y)- f(x)*y-x*f(y);