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hi all.

I have wrore the following program for optimization with bernstein and block pulse hybrid functions.

the program have some errors which i can't understand.

Bernestien1.mws

restart:

alias(C=binomial):
with(LinearAlgebra):
macro(LA= LinearAlgebra):


HybrFunc:=proc(N, M,  tj)               # N=Number of subintervals,  M=Number of functions in subintervals
 
local B, n, m;

global b;

for n from 1 to N do
for m from 0 to M-1 do

B := (i,m,t) -> C(m,i)*(1-t)^(m-i)*t^i:

b[n,m]:=unapply(piecewise(t>=(n-1)*tj/N and t<n*tj/N, B(m,2,N*t-(n-1)*tj), 0), t):
 od:od:


Array(1..N, 0..M-1, (n,m)->b[n,m](t)):

#convert(%,vector);
end proc:

HybrFunc(3, 3, 1);




                                       # End Of Definition
 
g2(t):=t;            #*exp(t-1):                      # Any other function can be replaced here
    

g1(t):=add(add(c[n,m]*b[n,m](t), m=0..2), n=1..3);
Optimization[Minimize](sqrt(int((g2(t)-g1(t))^2, t=0.. 1)));
assign(op(%[2]));
plot([g2(t),g1(t)], t=0..1, 0..5, color=[blue,red],thickness=[1,3],discont, scaling=constrained);

Array(1 .. 3, 0 .. 2, {(1, 0) = piecewise(0 <= t and t < 1/3, (1-3*t)^2, 0), (1, 1) = piecewise(0 <= t and t < 1/3, (6*(1-3*t))*t, 0), (1, 2) = piecewise(0 <= t and t < 1/3, 9*t^2, 0), (2, 0) = piecewise(1/3 <= t and t < 2/3, (2-3*t)^2, 0), (2, 1) = piecewise(1/3 <= t and t < 2/3, (2*(2-3*t))*(3*t-1), 0), (2, 2) = piecewise(1/3 <= t and t < 2/3, (3*t-1)^2, 0), (3, 0) = piecewise(2/3 <= t and t < 1, (3-3*t)^2, 0), (3, 1) = piecewise(2/3 <= t and t < 1, (2*(3-3*t))*(3*t-2), 0), (3, 2) = piecewise(2/3 <= t and t < 1, (3*t-2)^2, 0)}, datatype = anything, storage = rectangular, order = Fortran_order)

g2(t) := t

"g1(t):=c[1,0] ({[[(1-3 t)^2,0<=t and t<1/3],[0,otherwise]])+c[1,1] ({[[6 (1-3 t) t,0<=t and t<1/3],[0,otherwise]])+c[1,2] ({[[9 t^2,0<=t and t<1/3],[0,otherwise]])+c[2,0] ({[[(2-3 t)^2,1/3<=t and t<2/3],[0,otherwise]])+c[2,1] ({[[2 (2-3 t) (3 t-1),1/3<=t and t<2/3],[0,otherwise]])+c[2,2] ({[[(3 t-1)^2,1/3<=t and t<2/3],[0,otherwise]])+c[3,0] ({[[(3-3 t)^2,2/3<=t and t<1],[0,otherwise]])+c[3,1] ({[[2 (3-3 t) (3 t-2),2/3<=t and t<1],[0,otherwise]])+c[3,2] ({[[(3 t-2)^2,2/3<=t and t<1],[0,otherwise]])"

Error, (in Optimization:-NLPSolve) complex value encountered

Error, invalid left hand side in assignment

(1)



Download Bernestien1.mws

 I'll be so grateful if any one can help me.

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

in maple 15

https://drive.google.com/file/d/0B8F2D27rfQWgVXE1alN0V3JWU1U/edit?usp=sharing

there are 3 equation to be minimized

and i limit x between x + 5 and x - 5 as constraints

 

though f1 got a error in first line of command,

later i type a correct command for f1 in later part of script

f1 := -2*x1 - x2;
f2 := -x1 - 4*x2;
g1 := 2*x1 + 3*x2 - 6;
g2 := -x1;
g3 := -x2;
penalty := lambda1*Max(f1-M,0)+lambda2*Max(f2-M,0)+(M^2)*(max(g1,0) + max(g2,0) + max(g3,0));
M := 1;
k := 1;
s := 1;
lambda1 := 0.5;
lambda2 := 0.5;
with(Optimization):
Minimize(subs(lambda2=0.5, subs(lambda1=0.5, subs(M=1, penalty))));

correct answer for these parameters should be below, however, Minimize got error

when lambda1 = 0.5, lambda2 = 0.5
(x1, x2) = 1.551123, 0.965918
f(x1, x2) = (–4.068164, –5.414795);

what is the wrong with Pi set ::: in this function ::: Warning, no iterations performed as initial point satisfies first-order conditions

Optimization[Minimize](x^2 + y^2 + 25*(sin(x)^2+sin(y)^2), x=-2*Pi .. 2*Pi , y= -2*Pi .. 2*Pi);

Warning, no iterations performed as initial point satisfies first-order conditions
[0., [x = HFloat(0.0), y = HFloat(0.0)]]


Optimization[Maximize](x^2 + y^2 + 25*(sin(x)^2+sin(y)^2), x=-2*Pi .. 2*Pi , y= -2*Pi .. 2*Pi);

Warning, no iterations performed as initial point satisfies first-order conditions
[-0., [x = HFloat(0.0), y = HFloat(0.0)]]

--------------------------------

I got my good result when I apply it with this function :


f:= (x,y)->cos(x)*sin(y) -(x/(y^2+1));


Optimization[Maximize](f(x,y), x = -1 .. 2, y = -1 .. 1);


[0.994945017202501170,[x = HFloat(-0.6362676080636113), y = HFloat(1.0)]]

Optimization[Minimize](f(x,y), x = -1 .. 2, y = -1 .. 1);


[-2.02180678335978703,[x = HFloat(2.0), y = HFloat(0.10578346945175972)]]

Hi all 

I have the following segment of maple program which belongs to time delay systems dynamic. here C=X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P, is a matrix(vector) which comes from reordering the system terms and my goal is to minimizing J:=X.E.Transpose(X)+U.E.Transpose(U), subject to constraint C=0, but i don't know how to do so.

I will be so grateful if anyone can guide me

best wishes

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department


restart:
with(Optimization):
with(LinearAlgebra):
macro(LA= LinearAlgebra):
L:=1:  r:=2:  tau:= 1:
interface(rtablesize= 2*r+1):

Z:= Matrix(
     2*r+1, 2*r+1,
     [tau,
      seq(evalf((L/(2*(iz-1)*Pi))*sin(2*(iz-1)*Pi*tau/L)), iz= 2..r+1),
      seq(evalf((L/(2*(iz-1-r)*Pi))*(1-cos(2*(iz-1-r)*Pi*tau/L))), iz= r+2..2*r+1)
      ],
     scan= columns,
     datatype= float[8]
);
                        
Dtau00:= < 1 >:
Dtau01:= Vector[row](r):
Dtau02:= Vector[row](r):
Dtau10:= Vector(r):
Dtau20:= Vector(r):

Dtau1:= LA:-DiagonalMatrix([seq(evalf(cos(2*i*Pi*tau/L)), i= 1..r)]):
Dtau2:= LA:-DiagonalMatrix([seq(evalf(sin(2*i*Pi*tau/L)), i= 1..r)]):
Dtau3:= -Dtau2:
Dtau4:= copy(Dtau1):

Dtau:= < < Dtau00 | Dtau01 | Dtau02 >,
         < Dtau10 | Dtau1  | Dtau2  >,
         < Dtau20 | Dtau3  | Dtau4  > >;
 
P00:= < L/2 >:
P01:= Vector[row](r):
P02:= Vector[row](r, j-> evalf(-L/j/Pi), datatype= float[8]):
P10:= Vector(r):
P20:= Vector(r, i-> evalf(L/2/i/Pi)):
P1:= Matrix(r,r):
P2:= LA:-DiagonalMatrix(P20):
P3:= LA:-DiagonalMatrix(-P20):
P4:= Matrix(r,r):

P:= < < P00 | P01 | P02 >,
      < P10 | P1  | P2  >,
      < P20 | P3  | P4  > >;

interface(rtablesize=2*r+1):    # optionally
J:=Vector([L, L/2 $ 2*r]):      # Matrix([[...]]) would also work here

E:=DiagonalMatrix(J);

X:=  Vector[row](2*r+1,symbol=a);
U:=Vector[row](2*r+1,symbol=b);

X0:= Vector[row](2*r+1,[1]);
G:=Vector[row](2*r+1,[1]);
C:=simplify(X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P);

Z := Matrix(5, 5, {(1, 1) = 1., (1, 2) = 0., (1, 3) = 0., (1, 4) = 0., (1, 5) = 0., (2, 1) = 0., (2, 2) = 0., (2, 3) = 0., (2, 4) = 0., (2, 5) = 0., (3, 1) = 0., (3, 2) = 0., (3, 3) = 0., (3, 4) = 0., (3, 5) = 0., (4, 1) = 0., (4, 2) = 0., (4, 3) = 0., (4, 4) = 0., (4, 5) = 0., (5, 1) = 0., (5, 2) = 0., (5, 3) = 0., (5, 4) = 0., (5, 5) = 0.})

Dtau := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1., (2, 3) = 0, (2, 4) = 0., (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1., (3, 4) = 0, (3, 5) = 0., (4, 1) = 0, (4, 2) = -0., (4, 3) = -0., (4, 4) = 1., (4, 5) = 0, (5, 1) = 0, (5, 2) = -0., (5, 3) = -0., (5, 4) = 0, (5, 5) = 1.})

P := Matrix(5, 5, {(1, 1) = 1/2, (1, 2) = 0, (1, 3) = 0, (1, 4) = -.318309886100000, (1, 5) = -.159154943000000, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = .1591549430, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (3, 5) = 0.7957747152e-1, (4, 1) = .1591549430, (4, 2) = -.159154943000000, (4, 3) = 0, (4, 4) = 0, (4, 5) = 0, (5, 1) = 0.7957747152e-1, (5, 2) = 0, (5, 3) = -0.795774715200000e-1, (5, 4) = 0, (5, 5) = 0})

E := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1/2, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1/2, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 1/2, (4, 5) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 0, (5, 5) = 1/2})

X := Vector[row](5, {(1) = a[1], (2) = a[2], (3) = a[3], (4) = a[4], (5) = a[5]})

U := Vector[row](5, {(1) = b[1], (2) = b[2], (3) = b[3], (4) = b[4], (5) = b[5]})

X0 := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

G := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

C := Vector[row](5, {(1) = 1.500000000*a[1]-2.-.1591549430*a[4]-0.7957747152e-1*a[5]-.5000000000*b[1]-.1591549430*b[4]-0.7957747152e-1*b[5], (2) = a[2]+.1591549430*a[4]+.1591549430*b[4], (3) = a[3]+0.7957747152e-1*a[5]+0.7957747152e-1*b[5], (4) = a[4]+.3183098861*a[1]-.1591549430*a[2]+.3183098861*b[1]-.1591549430*b[2], (5) = a[5]+.1591549430*a[1]-0.7957747152e-1*a[3]+.1591549430*b[1]-0.7957747152e-1*b[3]})

(1)

J:=X.E.Transpose(X)+U.E.Transpose(U);

J := a[1]^2+(1/2)*(a[2]^2)+(1/2)*(a[3]^2)+(1/2)*(a[4]^2)+(1/2)*(a[5]^2)+b[1]^2+(1/2)*(b[2]^2)+(1/2)*(b[3]^2)+(1/2)*(b[4]^2)+(1/2)*(b[5]^2)

(2)

Minimize(J,{C=0});






Error, (in Optimization:-NLPSolve) invalid arguments

 

#XP:=-.015+X[1]+add(X[l+1]*f1(l)+X[r+l+1]*f2(l), l= 1..r):
#plot([XP,T1], t= 0..1);#,legend= "Solution Of x(t) with r=50"):

 

 

 

 

 

 

Download work1.mwswork1.mws

Hi all

In matlab software we have a command namely fmincon which minimizes any linear/nonlinear algebric equations subject to linear/nonlinear constraints.

Now my question is that: what is the same command in maple?or how can we minimize linear/nonlinear function subject to linear/nonlinear constraints in maple?

thanks a lot

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

I received an unexpected error message when trying to minimize a function: evaluating

returns the error message

Error, (in @) too many levels of recursion

Why am I getting this message?  It's hard for me to see how minimizing a function involves recursion, unless Maple is trying to iteratively approximate a solution.

hey maple followers,

i need, please to find the method used in command "minimize".
i looked into help maple and i found "theorema mean values" as example
some help please
thanks!

P.s: minimize not Minimize

> sol := pdsolve({ICS, sys1, sys2, sys3, sys4, sys5, sys6, sys7}, numeric, method = rkf45, parameters = [a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p], output = listprocedure);

Error, (in pdsolve) invalid input: `pdsolve/numeric` expects its 2nd argument, IBCs, to be of type {set, list}, but received method = rkf45

 

restart;

x11:=[1.05657970467127, .369307407127487, .400969917393968, .368036162749865, .280389875142339, .280523489139136, .283220960827744, .373941285224253, .378034013792196, .384412762008662, .358678988563716, .350625923673556, .852039817522304, .362240519978640, 1.03197080591829, .343650441408896, .982510654490390, .404544012440991, .422063867224247, 1.20938803285209, .455708586000668, 1.22503869712995, .388259397947667, .472188904769827, 1.31108028794286, 1.19746589728366, .572669348193002];

y11:= [.813920951682113, 10.3546712426210, 2.54581301217449, 10.2617298458172, 3.82022939508992, 3.81119683373741, 3.90918914917183, 10.5831132713329, 10.8700088489538, 11.0218056177585, 10.5857571473115, 9.89034057997145, .271497107157453, 9.77706473740146, 2.23955104698355, 4.16872072216206, .806710906391666, 11.9148193656260, 12.0521411908477, 2.52812993540440, 12.6348841508094, 2.72197067934160, 5.10891266728297, 13.3609183272238, 3.03572692234234, 1.07326033849793, 15.4268962507711];

z11:= [8.93290500985527, 8.96632856524217, 15.8861149154785, 9.16576669760908, 3.20341865536950, 3.11740291181539, 3.22328961317946, 8.71094047480794, 8.60596466961827, 9.15440788281943, 10.2935566768586, 10.5765776143026, 16.3469510439066, 9.36885507010739, 2.20434678689869, 3.88816077008078, 17.9816287534802, 10.1414228793737, 10.7356141216242, 4.00703203725441, 12.0105837616461, 3.77028605914906, 5.01411979976607, 12.7529165152417, 3.66800269682059, 21.2178824031985, 13.9148746721034];

u11 := [5.19, 5.37, 5.56, 5.46, 5.21, 5.55, 5.56, 5.61, 5.91, 5.93, 5.98, 6.28, 6.24, 6.44, 6.58, 6.75, 6.78, 6.81, 7.59, 7.73, 7.75, 7.69, 7.73, 7.79, 7.91, 7.96, 8.05];

u11 := [seq(close3(t+t3), t3=0..26)];

sys1:=Diff(a1(s,t),s) = a*a1(s,t)+ b*a2(s,t)+ c*a3(s,t)+ d*u(t);

sys2:=Diff(a2(s,t),s) = e*a1(s,t)+ f*a2(s,t)+ g*a3(s,t)+ h*u(t);

sys3:=Diff(a3(s,t),s) = i*a1(s,t)+ j*a2(s,t)+ k*a3(s,t)+ l*u(t);

sys4:=Diff(y(t),t) = m*a1(s,t)+n*a2(s,t)+ o*a3(s,t)+ p*u(t);

sys5:= Diff(a1(s,t),t) = a1(s,t);

sys6:= Diff(a2(s,t),t) = a2(s,t);

sys7:= Diff(a3(s,t),t) = a3(s,t);

sol := pdsolve([sys1, sys2, sys3,sys4,sys5,sys6,sys7]);

t2 := [seq(i, i=1..27)];

xt1 := subs(_C1=1,sol[1]); # a1(t)

xt2 := subs(_C1=1,sol[2]); # a2(t)

xt3 := subs(_C1=1,sol[3]); # a3(t)

ut1 := subs(_C1=1,sol[4]); # u(t)

tim := [seq(n, n=1..27)];

N:=nops(tim):

ICS:=a1(1)=x11[1],a2(1)=y11[1],a3(1)=z11[1],u1(1)=u11[1];

sol:=pdsolve({sys1, sys2, sys3,sys4,sys5,sys6,sys7,ICS}, numeric, method=rkf45, parameters=[ a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p],output=listprocedure);

ans(.001,.002,.003,.001,.002,.003,.001,.002,.003,.003,.003,.003,.003,.003,.003,.003);

ans:=proc(a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p) sol(parameters=[ a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p]);

add((xt1(tim[i])-x11[i])^2,i=1..N)+add((xt2(tim[i])-y11[i])^2,i=1..N)+add((xt3(tim[i])-z11[i])^2,i=1..N)+add((ut1(tim[i])-u11[i])^2,i=1..N);

end proc;

result1 := Optimization:-Minimize(ans,initialpoint=[.001,.002,.003,.001,.002,.003,.001,.002,.003,.003,.003,.003,.003,.003,.003,.003]);

product.mw

Experts, I pose a question:

Separate the numbers 3,4,5,6,7,8,28,30,35 into three groups of three numbers each, so that the
product of the numbers in each group is equal.

The idea is to select numbers where the variance between the 3 groups is minimized.

my attempt doesn't get the anwer directly, there must be a better approach

Minimize doesn't work with dsolve porcedure?

experiment_real.mw

tr := proc (x, y)::integer; tr := x+y; result := x^(2+y) end proc

Warning, `result` is implicitly declared local to procedure `tr`

 

tr(5, 5)

78125

(1)

with(Optimization); Minimize(tr(x, y), x = 0 .. 1000, y = 1 .. 1, initialpoint = {x = 25, y = 1})

[0.117556065072605623e-15, [x = HFloat(4.898709434833346e-6), y = HFloat(1.0)]]

(2)

xxx := 97.39391293; yyy := -1.588898710

-1.588898710

(3)

xx := 100;

3

(4)

trool := proc (leng, alpha)::integer; global psi, zx, zy, xx, yy, xxx, yyy, sa, ca, ps, Vx, Vy, vx, vy, ode, ics, XX, YY, trool, G, str, start, ds; sa := evalf(sin(alpha)); ca := evalf(cos(alpha)); ps := evalf(evalc(Im(evalc(str*(x+I*y)-((1/2)*I)*G*ln(x+I*y-start)/Pi)))); psi := ps; xxx := evalf(xx+leng*ca); yyy := evalf(yy+leng*sa); Vx := diff(psi, y); Vy := -(diff(psi, x)); vx := Re(evalf(subs(x = xxx, y = yyy, subs(vvx = Vx, vvx)))); vy := Re(evalf(subs(x = xxx, y = yyy, subs(vvy = Vy, vvy)))); proc (X) options operator, arrow; X(t) end proc; proc (Y) options operator, arrow; Y(t) end proc; zx := proc (t) options operator, arrow; evalf(subs(x = X(t), y = Y(t), subs(vvx = Vx, vvx))) end proc; zy := proc (t) options operator, arrow; evalf(subs(x = X(t), y = Y(t), subs(vvy = Vy, vvy))) end proc; ode := diff(X(t), t) = zx(t), diff(Y(t), t) = zy(t); ics := X(0) = xxx, Y(0) = yyy; ds := dsolve([ode, ics], type = numeric, [X(t), Y(t)], method = rkf45, maxfun = 0, output = listprocedure, abserr = 0.1e-3, relerr = 0.1e-3, minstep = 0.1e-1); XX := rhs(ds[2]); YY := rhs(ds[3]); trool := XX(0.1e-3) end proc:

with(Optimization); Minimize(trool(alpha, leng), assume = nonnegative, alpha = 0 .. 2*Pi, leng = .2 .. 2, iterationlimit = 1000, initialpoint = {alpha = 1, leng = 1})

Error, (in XX) parameter 'alpha' must be assigned a numeric value before obtaining a solution

 

alpha = 0 .. 2*Pi, leng = .2 .. 2, output = solutionmodule

alpha := 1; leng := 1; XX(10)

HFloat(100.54666738117751)

(5)

``

trool(1, 11)

HFloat(100.00711298362239)

(6)

psi

3.*y-11.93662073*ln((x-100.)^2+y^2)

(7)

``

 

Download experiment_real.mw

with trool procedure minimize dosent work .... and its make me realy sad, couse i need to optimize alpha and leng in other (big one) porcedure with same dsolve.

get this errors:
"Warning, The use of global variables in numerical ODE problems is deprecated, and will be removed in a future release. Use the 'parameters' argument instead (see ?dsolve,numeric,parameters)"
"Error, (in XX) parameter 'alpha' must be assigned a numeric value before obtaining a solution"

Hello,

For a control systems project I'am working on I need to minimize the actuator effort required to control an input voltage to certain output voltage. As a first experiment I chose a sinusoidal input ug and a sinusoidal output ul. The function to minimize is the funtion Uint. We know that when both the input voltage and output voltage share the frequency, phase and amplitude, that the control effort is zero. Running the animation (see Maple input below) shows that if the input and output voltage frequencies approach eachother the function Uint goes to zero. However, if the frequencies exactly match and I evaluate the function Uint I get a devision by zero notification (this could already be seen from the function of Uint). How can I solve this?

restart

with(plots):

u[G] := sin(omega[G]*t)

u[L] := sin(omega[L]*t)

U[int] := int((u[G]-u[L])^2, t)

omega[G] := 9:

animate(plot, [[u[G], u[L], (1/10)*U[int]], t = 0 .. 5, legend = ["Public grid voltage", "Local grid voltage (control goal)", "Cum. Actuator voltage (scaled)"]], omega[L] = 8 .. 10, gridlines = true, labels = ["Time [s]","Voltage [V]"], labeldirections = ["horizontal", "vertical"], labelfont = ["ARIAL", "bold", 12])

omega[L] := 9; t := 10; evalf(U[int])

NULL

 

Download 20131109_Division_by.mw

 

Thanks

Let x, y, z be three nonnegative numbers and x^2 + y^2 + z^2 = 5. Find the minimum value of the expression

A = 1/2*x^2*y^2 + y^2*z^2 + z^2 *x^2 + 96/(x + y + z + 1).

How to find the minimum value of the expression?

Hi

i want to minimizez the "Eq1(h[1],h[2],...h[n])" for all the h[i] in the ranges of -1<h[i]<1.  "n" (number of unknown variables) can be determined during the code and was not know before. how can i do this?

It must be stated that when i have Eq1 and two unknown variables such as h[1] and h[2], i wrote the following code and it works good, but i dont know how can i write mentioned command for unknown number of variables.

NLPSolve(Eq1,h[1]=-1..1,h[2]=-1..1);

thanks for your attention in advance

hi,

i have one equation (like equ1) wich this equatioin is a function of x. i want to find x in such a way that the equ1 minimizes. unfortunetly the equ1 is large and when i want to use NLPsolve it takes several time. is there any quick method to find the minimum x?

thanks for your attention in advance

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