Undefined operator './' for input arguments of type 'function_handle'.

Hello,

The part of the code I developed is exhibited below:

deriv1 = @(x) (sum(log(ti(x))-log(x(2))-(x(3)./Ti)-(x(4)./Vi)-log(tj(x)./(log(x(2))+(x(3)./Tj)+(x(4)./Vj))).*(tj./(log(x(2))+(x(3)./Tj)+x(4)./Vj)).^x(1))+sum(- log(ti(x)./(log(x(2))+x(3)./Vi + x(4)./Vi)).*(ti(x)./(log(x(2))+ x(3)./Vi + x(4)./Vi)).^x(1) + 1./x(1)));
% Derivative dL/dA
%deriv2 = @(x) -(x(1).*(sum((Vj.^x(3).*tj(x).*x(2)).^x(1))+sum((Vi.^x(3).*ti(x).*x(2)).^x(1))-1))./x(2);
deriv2 = @(x) (sum((x(1).*tj(x).*(tj(x)./(log(x(2)) + x(3)./Tj + x(4)./Vj)).^(x(1) - 1))./(x(2).*(log(x(2)) + x(3)./Tj + x(4)./Vj).^2) - x(1)./x(2)) + sum((x(1).*ti(x).*(ti(x)./(log(x(2)) + x(3)./Vi + x(4)./Vi)).^(x(1) - 1))./(x(2).*(log(x(2)) + x(3)./Vi + x(4)./Vi).^2)));  
% Derivative dL/dx(3)
deriv3 = @(x) (sum((x(1).*tj(x).*(tj(x)./(log(x(2)) + x(3)./Tj + x(4)./Vj)).^(x(1) - 1))./(Tj.*(log(x(2)) + x(3)./Tj + x(4)./Vj).^2)) + sum (- x(1)./Ti + (x(1).*ti(x).*(ti(x)./(log(x(2)) + x(3)./Vi + b./Vi)).^(x(1) - 1))./(Vi*(log(x(2)) + x(3)./Vi + x(4)./Vi).^2)));
% Derivative dL/b
deriv4 = @(x) (sum((x(1).*tj(x).*(tj(x)./(log(x(2)) + x(3)./Tj + x(4)./Vj)).^(x(1) - 1))./(Vj.*(log(x(2)) + x(3)./Tj + x(4)./Vj).^2)) + sum( - x(1)./Vi + (x(1).*ti(x).*(ti(x)./(log(x(2)) + x(3)./Vi + x(4)./Vi)).^(x(1) - 1))./(Vi.*(log(x(2)) + x(3)./Vi + x(4)./Vi).^2)));

I keep receiving the error message:

Undefined operator './' for input arguments of type 'function_handle'.

Error in
Example>@(x)(sum(log(ti(x))-log(x(2))-(x(3)./Ti)-(x(4)./Vi)-log(tj(x)./(log(x(2))+(x(3)./Tj)+(x(4)./Vj))).*(tj./(log(x(2))+(x(3)./Tj)+x(4)./Vj)).^x(1))+sum(-log(ti(x)./(log(x(2))+x(3)./Vi+x(4)./Vi)).*(ti(x)./(log(x(2))+x(3)./Vi+x(4)./Vi)).^x(1)+1./x(1)))

I was wondering where I was making the mistake. I have checked this part several times. Any help will be appreciated.

[the cyclist]

The definitions of those functions works fine, so that is not helpful to us.

Can you give us a full example of code that we can run, that has the error?

[Walter Roberson]

My guess is that Ti is a function handle and that where you refer to Ti you should have Ti(x)

[guest10011]

Here are the full codes.

clc
clear
% ========================================================================
% Initialize data and constants
% ========================================================================
% Definitions of different variables used in this code
% x(1): beta parameter
% x(2): A parameter
% x(3): Fi parameter
% x(4): b parameter

% Stress Levels 1 through 4 in Kelvin
T = [65 75 85 95] + 273.15;
V = [0.7 0.8 0.85 0.95];

% The end time of each of the Stress Levels 1 through 3 (in hours)
tend = [7 18 22 24 26];

% Times to failure
TTF = [15 18 22 24 28 30 33]';

% ========================================================================
% The next step is to define the the failure and survival times as well as
% failure and survival stress levels for each of the six step stress
% levels.  
% ========================================================================
% STEP 1:
% Failure Times in hours 
TTF1=0;
% Survival times in hours 
TTS1 = tend(1).*ones(15,1);
% Failure Stress (V)
VF1 = 0;
MF1 = 0;
% Survival Stress (V)
VS1 = T(1).*ones(15,1);
MS1 = V(1).*ones(15,1);

% STEP 2: 
% Failure Times in hours 
TTF2 = TTF(1:2);
% Survival times in hours 
TTS2 = tend(2).*ones(13,1);
% Failure Stress (Kelvin)
VF2 = T(2).*ones(2,1);
MF2 = V(2).*ones(2,1);
% Survival Stress (Kelvin)
VS2 = T(2).*ones(13,1);
MS2 = T(2).*ones(13,1);

% STEP 3:
% Failure Times in hours 
TTF3 = TTF(3:5);
% Survival times in hours 
TTS3 = tend(3).*ones(10,1);
% Failure Stress (Kelvin)
VF3 = T(3).*ones(3,1);
MF3 = V(3).*ones(3,1);
% Survival Stress (Kelvin)
VS3 = T(3).*ones(10,1);
MS3= V(3).*ones(10,1);

% STEP 4: 
% Failure Times in hours 
TTF4 = TTF(5:7);
% Survival times in hours 
TTS4 = tend(4).*ones(7,1);
% Failure Stress (Kelvin)
VF4 = T(4).*ones(3,1);
MF4 = V(4).*ones(3,1);
% Survival Stress (Kelvin)
VS4 = T(4).*ones(7,1);
MS4 = V(4).*ones(7,1);

% ========================================================================
% SETUP FOR TAU (tau1 through tau5 for stress levels 2 through 6
% This establishes the equivalent time (tau_(i-1)) of step i-1 if the item
% is operated at step stress level i.  
% tau for step 2
tau1 = @(x) (tend(1)).*exp(x(3).*(1./T(2) - 1./T(1)) + x(4).*(1./V(2) - 1./V(1)));
% tau for step 3
tau2 = @(x) (tend(2)-tend(1)+tau1(x)).*exp(x(3).*(1./T(3) - 1./T(2)) + x(4).*(1./V(3) - 1./V(2)));
% tau for step 4
tau3 = @(x) (tend(3)-tend(2)+tau2(x)).*exp(x(3).*(1./T(4) - 1./T(3)) + x(4).*(1./V(4) - 1./V(3)));
% ========================================================================
% SETUP FOR ADJUSTED TIME VECTORS
tvecti = [0;tend(1).*ones(2,1);tend(2).*ones(3,1);tend(3).*ones(3,1)];
tvectj = [zeros(15,1);tend(1).*ones(13,1);tend(2).*ones(10,1);tend(3).*ones(7,1)];

taui = @(x) [tau1(x);tau1(x);tau2(x);tau2(x);tau2(x);tau2(x);tau3(x);tau3(x);tau3(x)];
tauj = @(x) [zeros(15,1);tau1(x);tau1(x);tau1(x);tau1(x);tau1(x);tau1(x);tau1(x);tau1(x);tau1(x);tau1(x);tau1(x);tau1(x);tau1(x);tau2(x);tau2(x);tau2(x);tau2(x);tau2(x);tau2(x);tau2(x);tau2(x);tau2(x);tau2(x);tau3(x);tau3(x);tau3(x);...
    tau3(x);tau3(x);tau3(x);tau3(x)];

ti = @(x) [TTF1;TTF2;TTF3;TTF4] - tvecti + taui(x);
tj = @(x) [TTS1;TTS2;TTS3;TTS4] - tvectj + tauj(x);
% The failure and survival stress levels are assigned a vector of the same
% sizes as the adjusted time vectors for the MLE operation.
Ti = [VF1;VF2;VF3;VF4];
Tj = [VS1;VS2;VS3;VS4];

Vi = [MF1;MF2;MF3;MF4];
Vj = [MS1;MS2;MS3;MS4];
% ========================================================================
% SETUP FOR DERIVATIVES
% Derivative dL/dbeta
% deriv1 = @(x)
% -((sum((x(2).*Vj.^x(3).*tj(x)).^x(1).*log(x(2).*Vj.^x(3).*tj(x)))+sum((x(2).*Vi.^x(3).*ti(x)).^x(1).*log(x(2).*Vi.^x(3).*ti(x))-log(x(2).*Vi.^x(3).*ti(x)))).*x(1)-1)./x(1);
deriv1 = @(x) (sum(log(ti(x))-log(x(2))-(x(3)./Ti)-(x(4)./Vi)-log(tj(x)./(log(x(2))+(x(3)./Tj)+(x(4)./Vj))).*(tj./(log(x(2))+(x(3)./Tj)+x(4)./Vj)).^x(1))+sum(- log(ti(x)./(log(x(2))+x(3)./Vi + x(4)./Vi)).*(ti(x)./(log(x(2))+ x(3)./Vi + x(4)./Vi)).^x(1) + 1./x(1)));
% Derivative dL/dA
%deriv2 = @(x) -(x(1).*(sum((Vj.^x(3).*tj(x).*x(2)).^x(1))+sum((Vi.^x(3).*ti(x).*x(2)).^x(1))-1))./x(2);
deriv2 = @(x) (sum((x(1).*tj(x).*(tj(x)./(log(x(2)) + x(3)./Tj + x(4)./Vj)).^(x(1) - 1))./(x(2).*(log(x(2)) + x(3)./Tj + x(4)./Vj).^2) - x(1)./x(2)) + sum((x(1).*ti(x).*(ti(x)./(log(x(2)) + x(3)./Vi + x(4)./Vi)).^(x(1) - 1))./(x(2).*(log(x(2)) + x(3)./Vi + x(4)./Vi).^2)));  
% Derivative dL/dx(3)
deriv3 = @(x) (sum((x(1).*tj(x).*(tj(x)./(log(x(2)) + x(3)./Tj + x(4)./Vj)).^(x(1) - 1))./(Tj.*(log(x(2)) + x(3)./Tj + x(4)./Vj).^2)) + sum (- x(1)./Ti + (x(1).*ti(x).*(ti(x)./(log(x(2)) + x(3)./Vi + b./Vi)).^(x(1) - 1))./(Vi*(log(x(2)) + x(3)./Vi + x(4)./Vi).^2)));
% Derivative dL/b
deriv4 = @(x) (sum((x(1).*tj(x).*(tj(x)./(log(x(2)) + x(3)./Tj + x(4)./Vj)).^(x(1) - 1))./(Vj.*(log(x(2)) + x(3)./Tj + x(4)./Vj).^2)) + sum( - x(1)./Vi + (x(1).*ti(x).*(ti(x)./(log(x(2)) + x(3)./Vi + x(4)./Vi)).^(x(1) - 1))./(Vi.*(log(x(2)) + x(3)./Vi + x(4)./Vi).^2)));

% Derivative vector [dL/dbeta;dL/dK;dL/dn]
F1 = @(x) [deriv1(x);deriv2(x);deriv3(x);deriv4(x)];

% Initial estimate for parameters
x0 = [2 1e-14 10 10];

% initialize PARAMHAT estimates
paramhatv = x0;

for i = 1:100
    if i == 1
        paramhat = fsolve(F1,x0);
    else
        paramhat = fsolve(F1,paramhat);
    end
    paramhatv(i+1,:) = paramhat;
end
% Display the derivative vector based on the PARAMHAT estimate
F1(paramhat)

Answer 1

[Walter Roberson]

(tj./(log(x(2))+(x(3)./Tj)+x(4)./Vj))

Has function handle tj divided by something

[guest10011]

Yes. Did I apply it in the wrong way? What would be the correct way?

[guest10011]

Thank you! I understand now.

[Walter Roberson]

tj(x) instead of tj in that context.