tomleslie

It depends...

Answered: tomleslie 13876

May 23 2018

0 0

on exactly how nt) and m(t) were obtained.

Suggest you upload your complete worksheet including these calculations using the big green up-arrow in the Mapleprimes toolbar

Separate and simultaneous fits...

Answered: tomleslie 13876

May 22 2018

0 0

The attached worksheet

fits u() and v() independently, thus producing two (different) sets of fitting parameters a_1, a_2, b_1, b_2
fits u()^2+v()^2, producing a single set of fitting parameters a_1, a_2, b_1, b_2, which then apply to both of the constituent functions u() and v()

>

restart;

>	with(Statistics);

(1)

>

#
# Define the functions u() and v()
#
  u := proc(x, t)
              return (sin(x)+sin(x)*a_1*x*t-sin(x)*t+cos(x)*sin(x)*b_1*t^2*x+(1/2)*sin(x)*t^2-sin(x)*((1/2)*cos(x)*a_1*x*t^2+(1/2)*sin(x)*a_1*t^2-(1/2)*cos(x)*t^2)-cos(x)*((1/2)*sin(x)*b_1*x*t^2-(1/2)*sin(x)*t^2)+cos(x)*((1/2)*sin(x)*a_1*x*t^2-(1/2)*sin(x)*t^2)-sin(x)*((1/2)*cos(x)*b_1*x*t^2+(1/2)*sin(x)*b_1*t^2-(1/2)*cos(x)*t^2)+a_1*x*((1/2)*sin(x)*a_1*x*t^2-(1/2)*sin(x)*t^2)-cos(x)*sin(x)*t^2+sin(x)*a_2*t^2*x^2+((1/2)*sin(x)*a_1*x-(1/2)*sin(x))*a_1*t^2*x-sin(x)*a_1^2*x^2*t^2+(1/2)*sin(x)^2*a_1*t^2+(1/2)*sin(x)^2*b_1*t^2)/(1+a_1*x*t+a_2*x^2*t^2);
     end proc:
  v := proc(x, t)
            return (sin(x)+sin(x)*b_1*x*t-sin(x)*t+cos(x)*sin(x)*a_1*t^2*x+sin(x)*b_2*t^2*x^2+((1/2)*sin(x)*b_1*x-(1/2)*sin(x))*b_1*t^2*x-sin(x)*b_1^2*x^2*t^2+(1/2)*sin(x)*t^2-sin(x)*((1/2)*cos(x)*a_1*x*t^2+(1/2)*sin(x)*a_1*t^2-(1/2)*cos(x)*t^2)+cos(x)*((1/2)*sin(x)*b_1*x*t^2-(1/2)*sin(x)*t^2)-cos(x)*((1/2)*sin(x)*a_1*x*t^2-(1/2)*sin(x)*t^2)-sin(x)*((1/2)*cos(x)*b_1*x*t^2+(1/2)*sin(x)*b_1*t^2-(1/2)*cos(x)*t^2)-cos(x)*sin(x)*t^2+(1/2)*sin(x)^2*a_1*t^2+(1/2)*sin(x)^2*b_1*t^2+b_1*x*((1/2)*sin(x)*b_1*x*t^2-(1/2)*sin(x)*t^2))/(t^2*x^2*b_2+t*x*b_1+1);
        end proc:

>	# # Set some values for the parametes a_1, a_2, b_1 and # b_2. Then plot the exact functions u() and v() over the # range x=0..20, t=0..20 # a_1 := 1: a_2 := 5: b_1:= 6: b_2 := 3: plot3d(u, 0 .. 20, 0 .. 20, color = red); plot3d(v, 0 .. 20, 0 .. 20, color = red);

>

#
# Now generate a grid of values for u(), v() and
# a notional combined function u()^2+v()^2, for the
# set of parameters defined above.
#
  uVals:= Matrix( [ seq( seq( [i, j, evalf(u(i,j))], i=0..20), j=0..20)]):
  vVals:= Matrix( [ seq( seq( [k, l, evalf(v(k,l))], k=0..20), l=0..20)]):
  uvVals:= Matrix( [ seq( seq( [i, j, evalf(u(i,j))^2+evalf(v(i,j))^2], i=0..20), j=0..20)]):

>	a_1 := 'a_1': a_2 := 'a_2': b_1:= 'b_1': b_2 := 'b_2':

>

#
# Undefine the parameters (because these are to be
# evaluated from the NonlinearFit() commands. Do this
# separately for u() and v(), which means that the
# obtained parameter values for the two fits are
# independent.
#
# Then fit the combined notional function u()^2+v()^2,
# which will produce a single set of parameters which
# for both the functions u() and v()
#
  sol1:= NonlinearFit(u(x,t), uVals, [x, t], output = parametervalues);
  sol2:= NonlinearFit(v(x, t), vVals, [x, t], output = parametervalues);
  sol3:= NonlinearFit(u(x,t)^2+v(x,t)^2, uvVals, [x, t], output = parametervalues);
#
# Plot the functions u() and v() using the parameters
# obtained by fitting these two functions separately
#
  plot3d( 'eval'(u(x, t), sol1), x=0..20, t=0..20, color=blue);
  plot3d( 'eval'(v(x, t), sol2), x=0..20, t=0..20, color = blue);
#
# Plot the functions u() and v() using the parameters
# obtained by fitting the notional function u()^2+v()^2
# In other words u() and v() are fitted simultaneously, and
# therefore use the same parameters
#
  plot3d( 'eval'(u(x,t),sol3), x=0..20, t=0..20, color=blue);
  plot3d( 'eval'(v(x,t),sol3), x=0..20, t=0..20, color=blue);

>

Download fitProb6.mw

I still have...

Answered: tomleslie 13876

May 22 2018

0 0

little idea why you want to do the calculaton in a specifc (and rather complicated) way. The attached takes your worksheet pde_nonlinearfit.mw and "fixes" it so that at least everything works. Although I am not sure qhat it ultimately proves

>	restart; with(plots): with(Statistics): with(numapprox):

>

#
# Set up equations and boundary/initial conditions
#
  equ1:= [ diff(U(x,t),t)-diff(U(x,t),x$2)-2*U(x,t)*diff(U(x,t),x)+diff(U(x,t),x)*V(x,t)+U(x,t)*diff(V(x,t),x),
           diff(V(x,t),t)-diff(V(x,t),x$2)-2*V(x,t)*diff(V(x,t),x)+diff(U(x,t),x)*V(x,t)+U(x,t)*diff(V(x,t),x)
         ]:
  ics:=[U(x, 0) = sin(x), V(x, 0) = sin(x)]:
  Bcs:=[U(0, t) = 0, U(1, t) = 0, V(0, t) = 0, V(1, t) = 0]:

>

#
# Attempt an analytic solution of this PDE system.
# This return NULL (ie no solution), very quickly!
#
# Tried removing the ics/Bcs and supplying various
# hints for separable solution, TWS etc, but didn't
# get anything (at least in the time I was prepared
# to wait
#
sol1:=pdsolve( {equ1[], ics[], Bcs[]});
# sol1:=pdsolve( equ1, HINT=`*`);

(1)

>

#
# Solve this PDE system numerically, and plot the
# the functions U(x,t), V(x,t). Not a lot
# happening except at t=0
#
  sol2:=pdsolve( equ1,
                 { ics[], Bcs[] },
                 numeric
               ):
  sol2:-plot3d( U(x,t),
                x = 0..1,
                t = 0..1,
                title = U(x,t),
                titlefont = [times, bold, 20]
              );
  sol2:-plot3d( V(x,t),
                x = 0..1,
                t = 0..1,
                title = V(x,t),
                titlefont = [times, bold, 20]
              );

>

#
# Having already obtained a numerical solution for U(x,t)
# and V(x,t), the OP *seems* to want to obtain (another)
# numerical solution, but calculated in a rather complicated
# way. No idea why, or even if the proposed solution process
# is meaningful
#
  M := 5: N := 4:
#
# Define functions U(x,t), V(x,t) in terms of sums of other
# functions u[k](x,t) and v[k](x,t), as well as some parameters
# alpha[k] and beta[k]
#
  U := (x,t) -> (sum(u[j](x,t)*p^j, j = 0 .. M))/(1+sum(alpha[j]*t^j*p^j*x^j, j = 1 .. N));
  V := (x,t) -> (sum(v[j](x,t)*p^j, j = 0 .. M))/(1+sum(beta[j]*t^j*p^j*x^j, j = 1 .. N));
#
# Now set a couple of PDEs, using U and V, which, by the above
# can/will be expressed in terms of these new functions, u[k](x,t)
# and v[k](x,t), (and the parameters alpha[k], beta[k]. Also a
# further parameter p, is introduced, although OP subsequently
# sets this to zero - No idea why
#
  HU:= (U, V, p) -> (1-p)*diff(U, t)+p*(diff(U, t)-(diff(U, x, x))-2*U*diff(U, x)+diff(U, x)*V+U*diff(V, x)):
  HV:= (U, V, p) -> (1-p)*diff(V, t)+p*(diff(V, t)-(diff(V, x, x))-2*V*diff(V, x)+diff(U, x)*V+U*diff(V, x)):
#
# From the above generate specific PDEs, with p=0 (Why?)
# Notice that for the specific case p=0, the resulting
# equations are really trivial
#
  eqs := eval(HU(U(x,t), V(x,t), p), p = 0) = 0,
         eval(HV(U(x,t), V(x,t), p), p = 0) = 0;
  ics := u[0](x,0) = sin(x), v[0](x,0) = sin(x):

proc (x, t) options operator, arrow; (sum(u[j](x, t)*p^j, j = 0 .. M))/(1+sum(alpha[j]*t^j*p^j*x^j, j = 1 .. N)) end proc

proc (x, t) options operator, arrow; (sum(v[j](x, t)*p^j, j = 0 .. M))/(1+sum(beta[j]*t^j*p^j*x^j, j = 1 .. N)) end proc

(2)

>

#
# Simple solution shows that both u[0](x,t) and
# v[0](x,t) are independent of 't' and can be
# arbitrary functions of 'x'
#
  sol:= pdsolve({eqs});
#
# Evaluate this solution at t=0, substitute in the
# boundary conditions
#
  isolate~(subs( ics, eval(sol, t=0)),[_F1(x), _F2(x)]);
#
# Substitute these in the original solution
#
  sol:=subs(%, sol);
#
# Double check the above by using pdetest to verify
# the solution - returns all zeros, showing that
# solution is valid for both equations and boundary/
# initial conditions
#
  pdetest( sol, [eqs, ics]);

${u[0](x, t) = _F1(x), v[0](x, t) = _F2(x)}$

(3)

>

Download fitProb5.mw

I'm missing the objective...

Answered: tomleslie 13876

May 21 2018

0 0

I can fix the code in your worksheet so that

it executes with no errors
using numeric values for t, x[1](t), values for alpha[1]..alpha[4] can be obtianed so that U(t) and x[1](t) pretty much coincide
using numeric values for t, x[2](t), values for beta[1]..beta[4] can be obtained so that V(t) and x[2](t) pretty much coincide

See the attached.

BUT I have no idea what your are going to do with this information in order to solve the partial differential system given by PDESYS in your latest response. These seem to be two entirely unrelated problems. I will however make the simple observation that PDESYS will need six boundary/initial conditions ofr a complete solution. Since you only appear to have two, I can guarantee that you aren't going to get very far!

>

  restart;
  with(plots):
  with(Statistics):
  with(numapprox):
#
# Why do the following two commands exist?
# Neither x1E nor x2E are used for anything,
# anywhere. Are these commands only here to
# confuse???
#
  x1E := t -> (95/47)*exp(-2*t)-(48/47)*exp(-96*t):
  x2E := t -> (48/47)*exp(-96*t)-(1/47)*exp(-2*t):

  HU := (U, V, p) -> (1-p)*(diff(U, t)+U)+p*(diff(U, t)+U-95*V):
  HV := (U, V, p) -> (1-p)*(diff(V, t)+97*V)+p*(diff(V, t)+U+97*V):
  M := 5: N := 4:
  U := t -> (sum(u[j](t)*p^j, j = 0 .. M))/(1+sum(alpha[j]*t^j*p^j, j = 1 .. N)):
  V := t -> (sum(v[j](t)*p^j, j = 0 .. M))/(1+sum(beta[j]*t^j*p^j, j = 1 .. N)):

>

  eqs := eval(HU(U(t), V(t), p), p = 0) = 0,
         eval(HV(U(t), V(t), p), p = 0) = 0:
  ics := u[0](0) = 1, v[0](0) = 1:
  dsolve({eqs, ics}):
  assign(%);
  for i to M do
      eqs := eval(diff(HU(U(t), V(t), p), [p$i]), p = 0) = 0,
             eval(diff(HV(U(t), V(t), p), [p$i]), p = 0) = 0;
      ics := u[i](0) = 0, v[i](0) = 0;
      dsolve({eqs, ics});
      convert(%, int);
      assign(%);
  end do:
#
# What has been achieved at this point - well one now
# has explicit expressions for U(t) and V(t), in terms
# parameters alpha[1]..alpha[4] and beta[1]..beta[4]
# respectively. The following just shows the two
# relevant expressions
#
  eval(U(t), p=1);
  eval(V(t), p=1);

(1)

>

#
# Now define a couple of ODEs which *seem* entirely
# unrelated to anything which has gone before. OP had
# an overly complicated method of obtaining explicit
# numerical results for 't', 'x[1](t)', and 'x[2](t)'.
# Definitely easier just to specify the option for
# dsolve() which returns an output array with values
# for t, x[1](t) and x[2](t) over the required range
# and with the required time spacing
#
  ode := diff(x[1](t), t) = -x[1](t)+95*x[2](t),
         diff(x[2](t), t) = -x[1](t)-97*x[2](t):
  ics := x[1](0) = 1,
         x[2](0) = 1:
  bb:=10: points:=8000:
  dsol := dsolve( {ics, ode},
                  numeric,
                  range = 0 .. 10,
                  output=Array( [seq( j, j=0..bb, bb/points)])
                ):

>

#
# Now have explicit definitions for U(t) and V(t), as well
# as numerical values value t, x[1](t) and x[2](t). OP seems
# to want to use the NonlinearFit() command to
#
# 1) use values of t and x[1](t) to fit the function U(t),
#     thereby producing values for the parameters alpha[1]..
#     alpha[4] in the latter
# 2) use values of t and x[2](t) to fit the function V(t),
#     thereby producing values for the parameters beta[1]..
#     beta[4] in the latter
#
  fitU:= NonlinearFit( eval(U(t), p=1),
                       dsol[2][1][..,1],
                       dsol[2][1][..,2],
                       t,
                       output=parametervalues
                     );
  fitV:= NonlinearFit( eval(V(t), p=1),
                       dsol[2][1][..,1],
                       dsol[2][1][..,3],
                       t,
                       output=parametervalues
                     );

(2)

>

#
# Plot the (numeric) values of x[1](t) and the
# function U(t)( with the parameter values of
# alpha[1]..alpha[4] obtained from the NonlinearFit
# command. The two are pretty much indistinguishable
#
  p1:= pointplot( dsol[2][1][..,1],
                  dsol[2][1][..,2],
                  connect,
                  color=red
                ):
  p2:= plot( eval
             ( U(t),
               [ p=1, fitU[] ]
             ),
             t=0..10,
             color=blue
           ):
  display( [p1, p2]);

>

#
# Plot the (numeric) values of x[2](t) and the
# function V(t)( with the parameter values of
# beta[1]..beta[4] obtained from the NonlinearFit
# command. Again, the two are pretty much
# indistinguishable
#
  p3:= pointplot( dsol[2][1][..,1],
                  dsol[2][1][..,3],
                  connect,
                  color=red
                ):
  p4:= plot( eval
             ( V(t),
               [ p=1, fitV[] ]
             ),
             t=0..10,
             color=blue
           ):
  display( [p3, p4]);

>

Download fitProb4.mw

u and v are not dependent...

Answered: tomleslie 13876

May 21 2018

0 0

They are two completely separate functions, which happen to depend on the same variables (x,t) and the same parameters (a_1, a_2, b_1, b_2)

The nonLinearFit() function only accepts a single model function - so you can't simultaneously minimise u(x,t) and v(x,t). The only way I can think of to achieve this is to minimise some definitely positive combination of these two functions. An obvious choice would be to minimise the single function u(x,t)^2 + v(x,t)^2. This *ought* to work, but would require you to select the same 'grid' of points for both u() and v(). In you latest post the two 'grids' are different

In the definition of v,...

Answered: tomleslie 13876

May 21 2018

0 0

the same confusion between the parameter names a[1], a[2], b[1], b[2] and a_1, a_2, b_1, b_2 whihc I mentioned in my first response. I changed all references to use the latter, and worksheet now executes with no errors

>

u := proc (x, t)
return (sin(x)+sin(x)*a_1*x*t-sin(x)*t+cos(x)*sin(x)*b_1*t^2*x+(1/2)*sin(x)*t^2-sin(x)*((1/2)*cos(x)*a_1*x*t^2+(1/2)*sin(x)*a_1*t^2-(1/2)*cos(x)*t^2)-cos(x)*((1/2)*sin(x)*b_1*x*t^2-(1/2)*sin(x)*t^2)+cos(x)*((1/2)*sin(x)*a_1*x*t^2-(1/2)*sin(x)*t^2)-sin(x)*((1/2)*cos(x)*b_1*x*t^2+(1/2)*sin(x)*b_1*t^2-(1/2)*cos(x)*t^2)+a_1*x*((1/2)*sin(x)*a_1*x*t^2-(1/2)*sin(x)*t^2)-cos(x)*sin(x)*t^2+sin(x)*a_2*t^2*x^2+((1/2)*sin(x)*a_1*x-(1/2)*sin(x))*a_1*t^2*x-sin(x)*a_1^2*x^2*t^2+(1/2)*sin(x)^2*a_1*t^2+(1/2)*sin(x)^2*b_1*t^2)/(1+a_1*x*t+a_2*x^2*t^2);
end proc:

>

v := proc (x, t)
return (sin(x)+sin(x)*b_1*x*t-sin(x)*t+cos(x)*sin(x)*a_1*t^2*x+sin(x)*b_2*t^2*x^2+((1/2)*sin(x)*b_1*x-(1/2)*sin(x))*b_1*t^2*x-sin(x)*b_1^2*x^2*t^2+(1/2)*sin(x)*t^2-sin(x)*((1/2)*cos(x)*a_1*x*t^2+(1/2)*sin(x)*a_1*t^2-(1/2)*cos(x)*t^2)+cos(x)*((1/2)*sin(x)*b_1*x*t^2-(1/2)*sin(x)*t^2)-cos(x)*((1/2)*sin(x)*a_1*x*t^2-(1/2)*sin(x)*t^2)-sin(x)*((1/2)*cos(x)*b_1*x*t^2+(1/2)*sin(x)*b_1*t^2-(1/2)*cos(x)*t^2)-cos(x)*sin(x)*t^2+(1/2)*sin(x)^2*a_1*t^2+(1/2)*sin(x)^2*b_1*t^2+b_1*x*((1/2)*sin(x)*b_1*x*t^2-(1/2)*sin(x)*t^2))/(t^2*x^2*b_2+t*x*b_1+1);
end proc:

>

>	a_1 := 1: a_2 := 5: b_1:= 6: b_2 := 3:

(1)

>	uVals := Matrix([seq( seq( [i, j, evalf(u(i,j))], i=0..40), j=0..20)]);

(2)

>	vVals := Matrix([seq( seq( [k, l, evalf(v(k,l))], k=0..20), l=0..10)]);

(3)

>	ut1 := plot3d(u, 0 .. 40, 0 .. 20, color = red)

>	vt1 := plot3d(v, 0 .. 20, 0 .. 10, color = red)

>

>	a_1 := 'a_1': a_2 := 'a_2': b_1:= 'b_1': b_2 := 'b_2':

>	sol1:=NonlinearFit(u(x,t), uVals, [x, t], output = parametervalues);

(4)

>

>	sol2:=NonlinearFit(v(x,t), vVals, [x, t], output = parametervalues);

(5)

>	ut2:=plot3d( 'eval'(u(x,t),sol1), x=0..20, t=0..10, color=blue);

>	vt2 := plot3d(('eval')(v(x, t), sol2), x = 0 .. 20, t = 0 .. 10, color = blue);

(6)

>	evalf(eval(u(5,5), [a_1=1.0, a_2=5.0, b_1=6.0]));

(7)

>	evalf(eval(u(5,5), sol1));

(8)

>	evalf(eval(u(10,3), [a_1=1.0, a_2=5.0, b_1=6.0])); #assumed evalf(eval(u(10,3), sol1));

(9)

>	evalf(eval(u(7,2), [a_1=1.0, a_2=5.0, b_1=6.0])); #assumed evalf(eval(u(7,2), sol1))

(10)

>

expr:=(sin(x)+sin(x)*a_1*x*t-sin(x)*t+cos(x)*sin(x)*b_1*t^2*x+(1/2)*sin(x)*t^2-sin(x)*((1/2)*cos(x)*a_1*x*t^2+(1/2)*sin(x)*a_1*t^2-(1/2)*cos(x)*t^2)-cos(x)*((1/2)*sin(x)*b_1*x*t^2-(1/2)*sin(x)*t^2)+cos(x)*((1/2)*sin(x)*a_1*x*t^2-(1/2)*sin(x)*t^2)-sin(x)*((1/2)*cos(x)*b_1*x*t^2+(1/2)*sin(x)*b_1*t^2-(1/2)*cos(x)*t^2)+a_1*x*((1/2)*sin(x)*a_1*x*t^2-(1/2)*sin(x)*t^2)-cos(x)*sin(x)*t^2+sin(x)*a_2*t^2*x^2+((1/2)*sin(x)*a_1*x-(1/2)*sin(x))*a_1*t^2*x-sin(x)*a_1^2*x^2*t^2+(1/2)*sin(x)^2*a_1*t^2+(1/2)*sin(x)^2*b_1*t^2)/(1+a_1*x*t+a_2*x^2*t^2);

(11)

>

Download fitProb3.mw

Observation...

Answered: tomleslie 13876

May 20 2018

0 0

kernelopts(cputime) will just return the total time since the start of your Maple session. AFAIK it will never stop anything. If you want to stop your worksheet running after a certain time then you need kernelopts(cpulimit-someValue), where someValue is a time in seconds.

To resolve your original problem you need to be clear about which of two cases occur

the fsolve() command within your loop executes, but returns no values. This is not uncommon: it just means that fsolve() cannot find a solution. This situation can be detected and 'handled'. For example, ou could stop the loop at this point or carry on to subsequent loop index values (if required)
the fsolve() command never terminates - unusual, but I guess possible: in this case I can't think of anything other than the 'global' kernerlopts(cpulimit=someValue) mentioned above

What makes you think...

Answered: tomleslie 13876

May 19 2018

0 0

that settings are being kept?

Off the top of my head, the only other location normally used by Maple for configurations-type files is

C:\Users/yourUserName\AppData\Roaming\Maple\MapleVersion\Maple.ini

You will have to change the emboldened parts of the above to something appropriate for your machine

As an a general rule...

Answered: tomleslie 13876

May 19 2018

0 0

the best(?) way o produce a file in latex format is to do it from the menus, as in

File->Export As

and in the popup, change the 'Files of Type' selector, to LaTeX

Clarify...

Answered: tomleslie 13876

May 18 2018

0 0

Your code executes correctly - so exactly what is the problem?

Hmmm Maple 13...

Answered: tomleslie 13876

May 18 2018

0 0

was superseded in 2009, so nearly 10 years old. I have no access to this version - or even its help

I can tell that the syntax I used for the NonLinearFit() command only became available in Maple 15. Unfortunately I only have Maple versions from 18 onwards loaded on my machine, ie everything from the last five years. So I'm having to guess (a little)

It would be rare for Maplesoft to remove command alternatives which used to work, so *one* of the syntaxes which is currently available *ought* to work in Maple 13!

The only such alternative is that rather than supply the points to be fitted as an nX3 matrix where rows are

x-value, t-value, p1(x-value, t-value)

one has to supply two matrices. The first matrix is nx2 and contains the coordinates of the independent variables, ie rows are

x-value, t-value

The second matrix is nx1 and contains the function values, ie

p1(x-value, t-value)

So the command I used previously, ie

sol:=NonlinearFit(p1(x,t), p1Vals, [x, t], output = parametervalues);

has to be replaced with

sol:=NonlinearFit(p1(x,t), p1Vals[..,1..2], p1Vals[..,3], [x, t], output = parametervalues);

I have used this in the attached. It works in recent Maple versions, but I can't absolutely guarantee that it will work in Maple 13, because I can't test it!!

>

  restart;
  with(Statistics):
#
# Define function
#
  p1 := proc (x, t)
              return (sin(x)+sin(x)*a_1*x*t-sin(x)*t+cos(x)*sin(x)*b_1*t^2*x+(1/2)*sin(x)*t^2-sin(x)*((1/2)*cos(x)*a_1*x*t^2+(1/2)*sin(x)*a_1*t^2-(1/2)*cos(x)*t^2)-cos(x)*((1/2)*sin(x)*b_1*x*t^2-(1/2)*sin(x)*t^2)+cos(x)*((1/2)*sin(x)*a_1*x*t^2-(1/2)*sin(x)*t^2)-sin(x)*((1/2)*cos(x)*b_1*x*t^2+(1/2)*sin(x)*b_1*t^2-(1/2)*cos(x)*t^2)+a_1*x*((1/2)*sin(x)*a_1*x*t^2-(1/2)*sin(x)*t^2)-cos(x)*sin(x)*t^2+sin(x)*a_2*t^2*x^2+((1/2)*sin(x)*a_1*x-(1/2)*sin(x))*a_1*t^2*x-sin(x)*a_1^2*x^2*t^2+(1/2)*sin(x)^2*a_1*t^2+(1/2)*sin(x)^2*b_1*t^2)/(1+a_1*x*t+a_2*x^2*t^2);
        end proc:
#
# Set some values for the 'test' parameters.
# These will become the 'unknowns' which the
# nonLinearFit() command will try to determine
#
# NB p1() does not depend on b_2
#
  a_1 := 1: a_2 := 5: b_1:= 6: b_2 := 3:
#
# Generate a matrix of points which will be used
# for the fit. This matrix is organised with three
# columns
#
# column1 = x_value
# column2 = t_value
# column3 = p1( x_value, t_value)
#
  p1Vals := Matrix([seq( seq( [i, j, evalf(p1(i,j))], i=0..20), j=0..10)]);
#
# Produce a plot of the function with the assumed values
# of the parameters, for comparison with the yet-to-be-
# obtained plot of the function with the fitted parameters
#
  plt1:=plot3d( p1, 0..20, 0..10, color=red);
#
# Reset the parameters to be unknown
#
  a_1 := 'a_1': a_2 := 'a_2': b_1:= 'b_1': b_2 := 'b_2':
#
# Generate the values for these parameters from the
# NonLinearFit() command. Note that b_2 will not be
# obtained because it does not exist in p1()
#
#
# sol:=NonlinearFit(p1(x,t), p1Vals, [x, t], output = parametervalues);
#
# Above command replaced by a slight variation which
# *might* work in Maple 13 - although I can't be sure
#
  sol:=NonlinearFit(p1(x,t), p1Vals[..,1..2], p1Vals[..,3], [x, t], output = parametervalues);
#
# Note that a_1 and a_2 are very close to the 'assumed'
# values, but b_1 seems to be off by a mile!
#
# Produce a plot of the target function with these fitted parameters
# Looks pretty similar to the original plot - even with a grossly
# different value for b__1
#
  plt2:=plot3d( 'eval'(p1(x,t),sol), x=0..20, t=0..10, color=blue);
#
# Generate a few values of p1(xt) with [a_1=1, a_2=5, b_1=6],
# ie the original assumed values, and then with [a_1=1, a_2=5,
# b_1=-154326.052324557] ie the fitted values: just to see
# if the difference is significant - and it doesn't seem to be.
# I guess the original function just isn't that dependent on
# b_1!
#
  evalf(eval(p1(5,5), [a_1=1.0, a_2=5.0, b_1=6.0])); #assumed
  evalf(eval(p1(5,5), sol)); #fitted

  evalf(eval(p1(10,3), [a_1=1.0, a_2=5.0, b_1=6.0])); #assumed
  evalf(eval(p1(10,3), sol)); #fitted

  evalf(eval(p1(7,2), [a_1=1.0, a_2=5.0, b_1=6.0])); #assumed
  evalf(eval(p1(7,2), sol)); #fitted
#
# The reason why there is little/no dependence on the b_1
# parameter can be seen if one just applies a simplify()
# command to the expression with the procedure p1(). The
# simplified expression does not contain b_1!!!
#
  expr:=(sin(x)+sin(x)*a_1*x*t-sin(x)*t+cos(x)*sin(x)*b_1*t^2*x+(1/2)*sin(x)*t^2-sin(x)*((1/2)*cos(x)*a_1*x*t^2+(1/2)*sin(x)*a_1*t^2-(1/2)*cos(x)*t^2)-cos(x)*((1/2)*sin(x)*b_1*x*t^2-(1/2)*sin(x)*t^2)+cos(x)*((1/2)*sin(x)*a_1*x*t^2-(1/2)*sin(x)*t^2)-sin(x)*((1/2)*cos(x)*b_1*x*t^2+(1/2)*sin(x)*b_1*t^2-(1/2)*cos(x)*t^2)+a_1*x*((1/2)*sin(x)*a_1*x*t^2-(1/2)*sin(x)*t^2)-cos(x)*sin(x)*t^2+sin(x)*a_2*t^2*x^2+((1/2)*sin(x)*a_1*x-(1/2)*sin(x))*a_1*t^2*x-sin(x)*a_1^2*x^2*t^2+(1/2)*sin(x)^2*a_1*t^2+(1/2)*sin(x)^2*b_1*t^2)/(1+a_1*x*t+a_2*x^2*t^2);
  simplify(expr);

(2+(2*a_2*x^2-2*a_1*x+1)*t^2+(2*a_1*x-2)*t)*sin(x)/(2*a_2*t^2*x^2+2*a_1*t*x+2)

(1)

>

Download fitProb2.mw

In my first response to this question...

Answered: tomleslie 13876

May 17 2018

0 0

I demonstrated the multiplication of matrices with complex entries which does work. Which part of this don't you understand?

Your life would be so much easier if you would just post a worksheet, demonstrating your problem, using the big green up arrow in the Mapleprimes toolbar. Someone here would then fix your worksheet and repost it. How hard is that to grasp?

Hmmmmmm...

Answered: tomleslie 13876

May 17 2018

0 0

The worksheet you post works for me with no errors, so I have to assume that this is a Maple Version issue - or possibly OS.

Which Maple version are you running o0n what OS?

All of the options...

Answered: tomleslie 13876

May 17 2018

0 0

in the attached worksheet will rovide the correct answer

>

restart;

>	(1/2+I)(3+2I);# 1-D input

(1)

>

(2)

>

(3)

>	restart; # # redefine the imaginary unit # interface(imaginaryunit=j):

>	(1/2+j)(3+2j);# 1-D input

(4)

>

(5)

>

(6)

>

Download plexInput.mw

Observations...

Answered: tomleslie 13876

May 17 2018

0 0

As Preben has said, the square root of -1 in Maple is represented by capital I: you can change this to (pretty much) anything you want, for example the character 'j', by setting interface(imagiunaryunit=j) at the top of your worksheet.
When entering compex numbers, ensure that you have a multiplication sign between the character 'I' and its associated coefficient, so for 1+2I, you would actually type 1+2*I.
If you are using 2-D input (as in your original worksheet), then you can use 'implied multiplication', which requires you just to type a space beween the 'I' and its associated coefficient. This will work 99.9%(!?) of the time, but my advice would be to avoid ithis practice, and type the explicit multiplication symbol as in (2) above

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It depends...

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I'm missing the objective...

u and v are not dependent...

In the definition of v,...

Observation...

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Hmmm Maple 13...

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