Question

clearvars
close all
clc

tol = 0.0001; % this is the tolerance for root identification
xold = 0.5; % this is the initial guess

test = 1; % this simply ensures we have a test value to enter the loop below.
%If we don't preallocate this, MATLAB will error when it trys to start the
%while loop below

k = 1; %this is the iteration counter. Similar to "test" we need to preallocate it
%to allow the while loop to begin

%while loops will execute as long as the condition (in this case "test > tol && k < 1000")
%evaluates true. We have two conditions. "test" is greater than our
%tolerance (tol) and our counter is less than 1000. The purpose of the
%counter is to enure if we screw something up, the while loop will
%terminate eventually, even if we don't get convergence below tol
while test > tol && k < 1000 % this loop will continue until the tolerance is reached or 1000 iterations
    disp(['Iteration Number: ' num2str(k)]) %display current iteration number
    
    %insert code for Newton's Method here
    
    %the below will output information about the interations
    disp(['   Test value: ' num2str(fnew)])
    disp(['   Estimated root: ' num2str(xnew)])
    
    xold = xnew; %updates for the next iteration. By overwriting xold with the new value,
    %we will be able to use the current value when the loop repeats
    k = k+1; %this counts the iterations, and increases k by 1 each iteration
end

R = xnew; %this ensures you output the value R, so that it can be automatically checked

the code above is provided, pls fill in the code to solve the problem below
2) Newton's method (also known as Newton-Raphson) is another root-finding algorithm. Starting from an initial guess, we can improve our estimate using the following formula: f(xold) Xnew = Xold f'(xold) Where f denotes the derivative. This process can be iterated to achieve progressively better approximations of the root. Using MATLAB (HW_5_P2_partial.m) and Newton's Method, identify where In(x) intersects sin(x) from an initial guess of 0.5. Recall that we can express this as In(x)-sin(x) = 0. Recall that the derivative of In(x)-sin(x) is 1/x- cos(x). Does this work for an initial guess of 4? How about 5? For full credit, find the root of the function within the tolerance of 0.0001, starting from an initial guess of 0.5.

Show transcribed image text2) Newton's method (also known as Newton-Raphson) is another root-finding algorithm. Starting from an initial guess, we can improve our estimate using the following formula: f(xold) Xnew = Xold f'(xold) Where f denotes the derivative. This process can be iterated to achieve progressively better approximations of the root. Using MATLAB (HW_5_P2_partial.m) and Newton's Method, identify where In(x) intersects sin(x) from an initial guess of 0.5. Recall that we can express this as In(x)-sin(x) = 0. Recall that the derivative of In(x)-sin(x) is 1/x- cos(x). Does this work for an initial guess of 4? How about 5? For full credit, find the root of the function within the tolerance of 0.0001, starting from an initial guess of 0.5.

Answer 1

clearvars
close all
clc

% create symbolic variables
syms x;
syms f(x);
syms df(x);

% function
f(x)=log(x)-sin(x);

% differentiation of function f
df(x)=diff(f);

tol = 0.0001; % this is the tolerance for root identification
xold = 0.5; % this is the initial guess

test = 1; % this simply ensures we have a test value to enter the loop below.
%If we don't preallocate this, MATLAB will error when it trys to start the
%while loop below

k = 1; %this is the iteration counter. Similar to "test" we need to preallocate it
%to allow the while loop to begin

%while loops will execute as long as the condition (in this case "test > tol && k < 1000")
%evaluates true. We have two conditions. "test" is greater than our
%tolerance (tol) and our counter is less than 1000. The purpose of the
%counter is to enure if we screw something up, the while loop will
%terminate eventually, even if we don't get convergence below tol
while test > tol && k < 1000 % this loop will continue until the tolerance is reached or 1000 iterations
disp(['Iteration Number: ' num2str(k)]) %display current iteration number
  
% calculate new functional value
fnew=eval(f(xold));
  
% calculate differentiation value
f_new=eval(df(xold));
  
% calculate new value for x
xnew=xold-fnew/f_new;
  
% calculate difference between two value of x
test=abs(xold-xnew);
  
%the below will output information about the interations
disp([' Test value: ' num2str(fnew)])
disp([' Estimated root: ' num2str(xnew)])
  
xold = xnew; %updates for the next iteration. By overwriting xold with the new value,
%we will be able to use the current value when the loop repeats
k = k+1; %this counts the iterations, and increases k by 1 each iteration
end

R = xnew; %this ensures you output the value R, so that it can be automatically checked

Iteration Number: 1 Test value: -1.1726 Estimated root: 1.5447 Iteration Number: 2 Test value: -0.56484 Estimated root: 2.4538 Iteration Number: 3 Test value: 0.26286 Estimated root: 2.2311 Iteration Number: 4 Test value: 0.012713 Estimated root: 2.2191 Iteration Number: 5 Test value: 4.235e-05 Estimated root: 2.2191

for initial guess x = 4 , output is

Iteration Number: 1 Test value: 2.1431 Estimated root: 1.6284 Iteration Number: 2 Test value: -0.51076 Estimated root: 2.3888 Iteration Number: 3 Test value: 0.18713 Estimated root: 2.2259 Iteration Number: 4 Test value: 0.007144 Estimated root: 2.2191 Iteration Number: 5 Test value: 1.3486e-05 Estimated root: 2.2191

root is same .

again for x=5 ,

Test value: -0.67097-0.080931 Estimated root: -5.7224-1.90041 Iteration Number: 146 Test value: -0.021696-0.05223i Estimated root: -5.7351-1.9105i Iteration Number: 147 Test value: 0.00040763+0.00014772i Estimated root: -5.735-1.9105i Iteration Number: 148 Test value: 6.7796e-09-2.4192e-08i Estimated root: -5.735-1.9105i

root is imaginary, so for x=5 , this method for this function does not work ,