Question

MATLAB

Given a table like this one: i xLeft xRightyLeftyRight xNew yNew 3 8.75 10.000 173.05 -200.0. 9.3750 -12.7441 4 8.75 9.3750 173.05 -12.74 9.0625 80.4260 On iteration 5 of Bisection, what is xLeft? xRight? Complete the Newton's Method Root Finding code below: function [out] = myNewton (p, x0, tol) % p is a function handle to a non-linear function of x % x0 is an initial guess of the root of p % tol is a tolerance around the root

Answer 1

part 1)

x_left = 9.0625

x_right = 9.3750

we are trying to minimize the interval for searching of solution by bisection method and simultaneously find and reduce the mid which is x_new to zero, so from the given table we can deduce the above x_left and x_right.

part2)

CODE

function [out] = myNewton(f,x0, tol)

%initial high error
err=1;
syms x
%calculating differentiation
df = eval(['@(x)' char(diff(f(x)))])

while err > tol
del= -f(x0)/df(x0);   
x0=x0 +del ; %using newton - raphsn method
err= abs(del/x0); %calulating error in each iteration
end

out = x0;

end

SCREENSHOTS

function [out] = myNewton (f,x0, tol) tinitial err=1; high error calculating differentiation df eval(' (x)char (diff (f (x)))) 10 while err >tol del--f (x0)/df (x0) 12 13 14 15 %using newton - raphsn method abs (del/x0); %calulating error in each iteration err= end out x0 17 end Command Window [out] myNewton (@(x) x^2 +2*x 1, 1, 1e-6) >> + = 2 (x) 2*x+2 out -1.0000

Please go through above code and explanation and if any doubts then comment.

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