Can I please have help on problem 1? Matlab is needed for this problem. Thank you!
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Problem 1 [6 pts]: For an iterative method, we define true relative error, Et, and approximate relative error, Ca, at iteration k (k = 0, 1, 2, ... ) as [current approximation - true solution) Xk+1 - x Et true solution x* current approximation - previous approximation Xk+1 - Xk Eg = [current approximation Xk+1 Consider function f (x) = xex - cosx. (1) Apply the bisection method to find the root of f (x) = 0 within interval (0, 1). Use (0, 1) as the initial bracketing interval. Iterate until &q is less than 10-. Show &t at each iteration - that is, plot et vs. iteration number. (To get the "true" solution, you may run the bisection method until Eq is very small, say, less than 10-7.) (2) Repeat (1) using the false-position (linear interpolation) method. (3) For the same equation, f(x) = 0, apply Newton's method with an initial guess of Xo = 1, and the same stopping criterion Ea < 10-4. Again, plot &t vs. iteration number. Then, try initial guess xo = -0.4 and xo = -0.5, and show what you get. (4) Solve the same equation using the secant method. Try initial guess xo = 0.0 and x1 = 1.0. Use the same stopping criterion Ea < 10-4. Problem 2 [2 pts]: Prove that using the bisection method, the true error, Ixk+1 - x*|, is always less than or equal to the approximate error, xx+1 - Xkl, at any iteration. Note: In practice, it is usually impossible to calculate the true error, because the true solution x* is unknown. Therefore, what you have proven is a very nice feature of the bisection method. Problem 3 [2 pts]: Prove the following theorem. Theorem: Let p be a fixed point of g(x), i.e. g(p) = p. Suppose that g' (p) = 0, g"(p) # 0. If the method of fixed-point iteration, Xk+1 = g(Xk), converges to x = p, then it converges quadratically.
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