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Newtons method for finding minimum of a function.

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Sarah Smith
Sarah Smith on 10 Feb 2020
Answered: Jim Riggs on 10 Feb 2020
I want to mark the solution point (x,f(x)) obtained by the Newton's algorithm on the curve of the function f(x)=(e^(2*sinx))-x to see if it is a local minimum or something else, but I am stuck. I used myfunction in order to obtain the function value and its derivatives Can someone help me with this? I will show you what I have so far.
% Clears all data and screen
clear;
close all;
clc;
f1 = @(x) exp(2*sin(x))-x; %function
x0 = 3; %initial guess
maxIter = 1e6; %max iteration of the function
epsilon = 10e-6; %tolerance value
% Calling the Newton's Method function
[xk,i,error,errorVect] = NewtonsMethod(x0,maxIter,epsilon);
% Printing Newton's Method Results in the console
fprintf('\nThe solution of the function exp(2*sin(x))-x = 0 is %4.5f in %u iteration with %e error.\n', xk,i,error)
xk;
ymin = f1(xk);
errorVect;
% Function for Newton's Method function
%Graphing the error vector
x_axis = 1:1:length(errorVect); %x-axis values
semilogy(x_axis,errorVect,'-mo'); %graph of error vector in logarithmic y-axis
grid on %logarithmic grid (y-axis only)
function[xk,i,error,errorVect] = NewtonsMethod(xk,maxIter,epsilon)
xold = xk;
for i = 1:maxIter
[f1x,df1x,ddf1x]=myfunction(xk);
xk = xk - (df1x/ddf1x);
error = abs(xk-xold);
xold = xk;
errorVect(i) = error;
if (error<epsilon)
break;
end
end
end
% Function for calling the value and the derivative of the function
function [f,g,h] = myfunction(x)
f1 = @(z) exp(2*sin(z))-z; %function
f = f1(x);
syms z
if nargout > 1
g1(z) = diff(f1(z));
g = double(g1(x));
h1(z) = diff(g1(z));
h = double(h1(x));
end
end
  2 Comments
Jim Riggs
Jim Riggs on 10 Feb 2020
for your myFunction, try the following;
function [f,g,h] = myfunction(x)
f1 = @(z) exp(2*sin(z))-z; %function
ff = f1([x-1, x, x+1]); % this is a 3-element vector
f = ff(2); % the middle value is f1(x)
if nargout > 1 % (not sure why you use this)
gg = diff(ff); % this is a 2-element vector from diff(ff)
g = mean(gg);
h= diff(gg); % single value from diff(gg)
end
end

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Answers (1)

Jim Riggs
Jim Riggs on 10 Feb 2020
Here is another variation;
function [f,g,h] = myfunction(x)
f1 = @(z) exp(2*sin(z))-z; %function
dx = 0.1;
ff = f1([x-dx, x, x+dx]); % this is a 3-element vector
f = ff(2); % the middle value is f1(x)
if nargout > 1 % (not sure why you use this)
gg = diff(ff)./dx; % this is a 2-element vector from diff(ff)
g = gg(1);
h= diff(gg)./dx; % single value from diff(gg)
end
end

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