Obtain an estimate to the solution of this equation (Use Radian) X-0.8 = 0.2 sin

Question

Obtain an estimate to the solution of this equation (Use Radian)

X-0.8 = 0.2 sin (X)

(e) Using Newton-Raphson method. Start with x=0.9 as an initial guess and do four iterations. Give an estimate of the absolute error.

(f) Using Simple fixed point iteration. Start with x=0.9 as an initial guess and do four iterations. Give an estimate of the absolute error.

(g) using Secant method. Use the following an initial points x0 = 0.8; x1 = 0.9 and do four iterations. Give an estimate of the relative error.

Explanation / Answer

e)

i = 1;
p0 = 0.9;            %initial conditions
N = 4;                %maximum number of iterations
error = 0.01;         %precision required

syms 'x'
f(x) = x -0.8 - 0.2*sin(x);      %function we are solving
df = diff(f);           %differential of f(x)

while i <= N
    p = p0 - (f(p0)/df(p0));          %Newton-Raphson method
  
    if (abs(p - p0)/abs(p)) < error                     %stopping criterion when difference between iterations is below tolerance
        fprintf('Solution is %f ', double(p))
        return
    end

    i = i + 1;
    p0 = p;             %update p0
end

fprintf('Solution after %d iterations is %f and absolute error is %f ', N, eval(p), abs(eval(p)-0.9531))     %error for non-convergence within N iterations

f)

i = 1;       
p0 = 0.9;           %our guess is p = 1.5.
error = 0.0001; %the precision we want to find our solution in.
N = 4;            %max number of iterations.

syms 'x'
g(x) = 0.8 + 0.2*sin(x); %i.e. what we are trying to solve

while i <= N
    p = g(p0);                  %the next p value
    if abs((p - p0)) < error    %stopping criterion using the required precision
        fprintf('Solution is %f ', double(p))     %the output p is symbolic so need ot force it into a decimal
        return
    end
  
i = i + 1;
p0 = p;         %update the value of i and p to continue the loop with

end
fprintf('Solution is %f after %d iterations and absolute error is %f ', eval(p), N, abs(eval(p)-0.9531))

g)

error = 0.0001;          %predefine some tolerance.
N = 4;                   %number of steps
i = 2;

p0 = 0.8;                  %the two guesses
p1 = 0.9;

syms 'x'
f(x) = x-0.8-0.2*sin(x);          %what we are solving for f(x) = 0

q0 = f(p0);              %two coressponding values of the function evaluated at po and p1
q1 = f(p1);

while i <= N
    p = p1 - (q1*((p1 - p0)/(q1 - q0)));               %actual Secant Method (finds x axis intercept between two guesses)
  
    if abs(p - p1) < error                            %stopping criterion
        fprintf('Solution is %f ', double(p))
        return
    end

    i = i + 1;
    p0 = p1;             %update all the new parameters, po, p1, q0, q1
    p1 = p;
    q0 = q1;
    q1 = f(p);
  
end

fprintf('Solution is %f after %d iterations and error is %f ',eval(p), N, abs(eval(p)-0.9531))

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