Write a series of tests for Newton\'s Method (attached below) that verifies that

Question

Write a series of tests for Newton's Method (attached below) that verifies that the function produces the correct value with the indicated inputs:

input: (@(x) x.^2 -2,@(x) 2*x,1)

output: sqrt(2)

input: (@(x) x.^2 -2,@(x) 2*x,-1)

output: -sqrt(2)

input: (@(x) x.^2 - sin(x), 2*x - cos(x),1)

output: 0.8767262153950624

You may have to set a tolerance for the error instead of just using verifyEqual

Also: add a special error message to Newton's Method that triggers when the derivative is too close to zero (say 1e-8), where the program stops.

Add a test that checks for this error.

Explanation / Answer

function dummy=Newtonm()
clc;
clear all;

f=@(x)x^2-2;%function

f1=@(x)2*x; %derivative of function

x0=1;%initial guess

y1=Newton(f,f1,x0)
x01=-1;
y1=Newton(f,f1,x01)
f2=@(x)x^2-sin(x);
f21=@(x)2*x-cos(x)
x02=1;
y1=Newton(f2,f21,x02)
function y1=Newton(f,f1,x0)
      n=1;
erorr=0.1;
del=1e-10;
x(1)=x0;
while (abs(erorr>del))
    
y1=x0-(f(x0)/f1(x0));% Newton method

erorr(n+1)=abs((y1-x0)); %erorr

if abs(erorr<1e-3)
break
end

x0=y1; % update xold
x(n+1)=x0;
n=1+n;
end
if (n<100)
disp('Root is')
x(end)
disp('num_iter          x_value                erorr')
disp('_______________________________________________________________________________')
for i=1:n-1
fprintf('%d %20f %20f ',i ,x(i),erorr(i))

end
else
      fprintf('The required accuracy is not reached in 100 iteration')
end
end
end

%%%% Solution %%%%%

Root is

ans =

   1.414213562373095

num_iter          x_value                erorr
_______________________________________________________________________________
1                  1.000000                  0.100000
2                  1.500000                  0.500000
3                  1.416667                  0.083333
4                  1.414216                  0.002451
5                  1.414214                  0.000002

y1 =

   1.414213562373095

Root is

ans =

-1.414213562373095

num_iter          x_value                erorr
_______________________________________________________________________________
1                 -1.000000                  0.100000
2                 -1.500000                  0.500000
3                 -1.416667                  0.083333
4                 -1.414216                  0.002451
5                 -1.414214                  0.000002

y1 =

-1.414213562373095

f21 =

    @(x)2*x-cos(x)

Root is

ans =

   0.876726215395063

num_iter          x_value                erorr
_______________________________________________________________________________
1                  1.000000                  0.100000
2                  0.891396                  0.108604
3                  0.876985                  0.014411
4                  0.876726                  0.000259
5                  0.876726                  0.000000

y1 =

   0.876726215395063

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