Error using ==> inline.subsref at 14 Not enough inputs to inline function.

hi guys

Please if anyone can help me with my code. i face an error that says

**Error using ==> inline.subsref at 14
Not enough inputs to inline function.

x1=-4; x2=5;
F=inline('4*(sqrt(x1^2 + (10-x2)^2 )-10)^2  +  0.5*(sqrt(x1^2 + (10+x2)^2 )-10)^2 -5*(x1+x2)','x1','x2'); 
for j=1:200  
s1=-dF1;
 s2=-dF2;% dF1,dF2 can be found
x3=x1+s1*h;
 x4=x2+s2*h;
%the problem is i want to use inline function as function of (h).
g=4*(sqrt(x1^3 + (10-x4)^2 )-10)^2  +  0.5*(sqrt(x3^2 + (10+x4)^2 )-10)^2 -5*(x3+x4).
f=inline('g','h');

% later in my loop i will use f(a) where a is known but i always got

*** Error using ==> inline.subsref at 14
Not enough inputs to inline function.

What I need is f as function in h so i can work with.

My email; [e-mail]

Answer 1

[Walter Roberson]

Anonymous functions are easier.

F = @(x1, x2) 4*(sqrt(x1^2 + (10-x2)^2 )-10)^2  +  0.5*(sqrt(x1^2 + (10+x2)^2 )-10)^2 -5*(x1+x2);  %you do not appear to use F!

g = @(x1, x2, h) 4*(sqrt(x1^3 + (10-(x2+s2*h))^2 )-10)^2  +  0.5*(sqrt((x1+s1*h)^2 + (10+(x2+s2*h))^2 )-10)^2 -5*((x1+s1*h)+(x2+s2*h));

x1 = -4;  %are these really constants??
x2 = 5;   %are these really constants??

for j = 1 : 200
  s1 = -dF1;  %is this differential? You cannot differentiate an inline function or an anonymous function
  s2 = -dF2;  %differentiating what?
  f = @(h) g(x1, x2, h);
end

[Abdulaziz]

I appreciate your time.

x1,x2 are changing each loop.

dF1: I just differentiate F separately, outside the program.

I will send you all the program. do not wary about all the steps. the problem occurred because of f=inline('g','h'). I just want to make f as a function of h which i need to find the minimum of f(h), but i can not write h separately I should substitute x3 and x4 which are function of x1,h and x2,h. respectively.

The code is********************

x1=-4; x2=5;

F=inline('4*(sqrt(x1^2 + (10-x2)^2 )-10)^2 + 0.5*(sqrt(x1^2 + (10+x2)^2 )-10)^2 -5*(x1+x2)','x1','x2');

E_n=100; E_t=10^-3;Tol=10^-10;

for j=1:200

dF1=(8*(sqrt(x1^2 + (10-x2)^2) - 10) * x1 )/sqrt(x1^2 + (10-x2)^2) +((1*(sqrt(x1^2 + (10+x2)^2) - 10) * x1 )/sqrt(x1^2 + (10+x2)^2)) - 5 ;

dF2=((8*(sqrt(x1^2 + (10-x2)^2) - 10)*(x2-10)) / (sqrt(x1^2 + (10-x2)^2))) + (((sqrt(x1^2 + (10+x2)^2) - 10)*(10+x2))/(sqrt(x1^2 + (10+x2)^2))) - 5 ;

s1=-dF1;

s2=-dF2;

x3=x1+s1*h;

x4=x2+s2*h;

% now i want to set g as function of h

g=4*(sqrt(x3^2 + (10-x4)^2 )-10)^2 + 0.5*(sqrt(x3^2 + (10+x4)^2)-10)^2 -5*(x3+x2);

f=inline('g','h');

% Finding the bounds on the minimum of the function

%values of the initial bounds [a,b]=[0.0,0.1]

a(1)=0.0;

b(1)=0.1;

r=0.61803;

GSR=1.61803; %Golden Section Ratio = (r/r-1)

c(1)=r*a(1)+(1-r)*b(1); % c is a point between [a,b]

for i=1:100

if if f(a(i))>f(c(i)) && f(b(i))>f(c(i))

% This means there is a minimum value of the function f in the interval [a,b]

break;

else

%since the slope is negative we will shift the interval to the right by

%using the previous values of 'c' and 'b' and the Golden Section Ratio GSR

a(i+1)=c(i); c(i+1)=b(i);

%b can be determined from {(b-c/c-a)=(r/r-1)=GSR, then

b(i+1)=c(i+1)*(1+GSR)-(GSR*a(i+1));

end

end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Reduction of original interval using the golden section algorithm.

a(1)=a(i); b(1)=b(i);

c(1)=r*a(1)+(1-r)*b(1);

d(1)=(1-r)*a(1)+r*b(1);% c and d are points between [a,b]

for n=1:100

if f(c(n))<=f(d(n))

a(n+1)=a(n);

b(n+1)=d(n);

d(n+1)=c(n);

c(n+1)=r*a(n+1)+(1-r)*b(n+1);

else

if f(c(n))>f(d(n))

a(n+1)=c(n);

b(n+1)=b(n);

c(n+1)=d(n);

d(n+1)=(1-r)*a(n+1)+r*b(n+1);

end

end

E_n=abs((b(n+1)-a(n+1))/(b(1)-a(1)));

if E_n < E_t;

break

end

end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%cubic plunomyal fit to the points obtained at the last iteration

X1=a(n+1);X2=c(n+1);X3=d(n+1);X4=b(n+1);

q1=X3^3*(X2-X1)-X2^3*(X3-X1)+X1^3*(X3-X2);

q2=X4^3*(X2-X1)-X2^3*(X4-X1)+X1^3*(X4-X2);

q3=(X3-X2)*(X2-X1)*(X3-X1);

q4=(X4-X2)*(X2-X1)*(X4-X1);

q5=f(X3)*(X2-X1)-f(X2)*(X3-X1)+f(X1)*(X3-X2);

q6=f(X4)*(X2-X1)-f(X2)*(X4-X1)+f(X1)*(X4-X2);

a3=(q3*q6-q4*q5)/(q2*q3-q1*q4);

a2=(q5-a3*q1)/q3;

a1=((f(X2)-f(X1))/(X2-X1))-(a3*((X2^3-X1^3)/(X2-X1)))-a2*(X1+X2);

a0=f(X1)-a1*X1-a2*X1^2-a3*X1^3;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% finding the minimum

delta=a2^2-(3*a1*a3);

XI=(sqrt(delta)-a2)/(3*a3); XII=(-a2-sqrt(delta))/(3*a3);

if f(XI)<= f(XII)

h=XI;

else

h=XII;

end

x3=x1+s1*h;

x4=x2+s2*h;% we already used this in lines 11,12 for g=f(h)

if abs(F(x3,x4)-F(x1,x2)) <= 10^-10

break;

end

if j==5

break;

end

x1=x3;

x2=x4;

end

disp('f(x1,x2) x1 x2 iteration');

m=[F(x1,x2),x1,x2,j];

disp(m);

[Abdulaziz]

HI I really appreciate your help it incredibly works.

I hope I ask in the first day because I worked all two days long to figure it out finally i decided to write to your website.

Thaaaaaaaaaank yoooooo