Download Approximating the Square Root of Two using Newton's Method and Secant Method and more Assignments Cryptography and System Security in PDF only on Docsity!
Math 151A Homework #2 – due Wednesday 10/18, in class
Show all your work!
- The square root of two, revisited. Use Newton’s method and the secant method to approximate the value of
2 to within 10−^4. As usual, submit your code and a list of intermediate approximations.
a. Use Newton’s method with an initial guess p 0 = 1. How does this compare with the bisection method (from last week’s homework), in terms of number of iterations required? Answer: Create a function file called f.m containing: function f = f(x) f = x.^2 - 2; end and another called f1.m containing the first derivative: function f1 = f1(x) f1 = 2*x; end and a script called newton.m containing, for example: %---------------------------------------------------------- % Matlab/Octave program to find the root of a function % using Newton’s method. % - the function must be defined in f.m, and its % derivative in f1.m %----------------------------------------------------------
% set initial guess p0, and the desired precision p0 = 1; tol = 1.0e-4;
% prevent infinite loops, by setting a maximum % number of iterations maxits = 1000;
for i = 1:maxits
% compute the subsequent approximation p1 = p0 - f(p0)/f1(p0);
% nicely formatted printing of each approximation printf("%.2d: p = %.8f\n",i,p1);
% check to see if we are at the root if f(p1)==
break endif
% check to see if we’ve reached the desired precision if( abs(p1-p0) < tol) break endif
% new approximation becomes old aproximation p0 = p1;
end Then run the script: octave:1> newton 01: p = 1. 02: p = 1. 03: p = 1. 04: p = 1.
Notice that we converge much faster than with the bisection method (from last week’s homework).
b. Use Newton’s method with an initial guess p 0 = 0.0001. What happens to the number of iterations required, and why? Answer: Modify the newton.m code to start from initial condition p0 = 0.0001 and run it: octave:2> newton 01: p = 10000. 02: p = 5000. 03: p = 2500. 04: p = 1250. 05: p = 625. 06: p = 312. 07: p = 156. 08: p = 78. 09: p = 39. 10: p = 19. 11: p = 9. 12: p = 5. 13: p = 2. 14: p = 1. 15: p = 1. 16: p = 1. 17: p = 1. 18: p = 1.
octave:3> secant 01: p = 2. 02: p = 1. 03: p = 1. 04: p = 1. 05: p = 1. 06: p = 1. The number of iterations required is comparable to Newton’s method, although slightly more. This increase in number of iterations may be offset by not having to compute the derivative.
- Linear convergence. (problem 6b of section 2.4) Show that the sequence defined by pn = 1/n^2 (for n ≥ 1) converges linearly to p = 0. How large must n be before we have |pn − p| ≤ 5 × 10 −^2?
Answer: Obviously pn converges to 0, since limn→∞ 1 /n^2 = 0. Now consider the ratio:
pn+1 − p pn − p
1 /(n + 1)^2 1 /n^2
n^2 (n + 1)^2
lim n→∞
( (^) n 2
(n + 1)^2
= lim n→∞
n^2 n^2 + 2n + 1
= lim n→∞
2 n 2 n + 2
= lim n→∞
Which means that
lim n→∞
|pn+1 − p| |pn − p|
so we have convergence of order α = 1 with asymptotic error constant λ = 1. Note that this is not linear convergence, since linear convergence requires that α = 1 and λ < 1.
- Quadratic convergence. (problem 8, section 2.4) Show that the sequence defined by pn = 10−^2
n converges quadratically to 0. Answer: Since limn→∞ 2 n^ = ∞, then limn→∞ 10 −^2 n = 0. Now consider the ratio:
pn+1 − p (pn − p)^2
10 −^2
n+ ( 10 −^2 n^
) 2 =^
10 −^2
n· 2 ( 10 −^2 n^
10 −^2
n ) 2 ( 10 −^2 n^
So then
lim n→∞
|pn+1 − p| |pn − p|^2
and we have the order of convergence to be α = 2.
- Dealing with multiple zeros. Consider the function f (x) = x^3 + 4x^2 + 8x.
a. Show that p = 0 is a simple zero of f (x). Answer: We know that 0 is a zero of f , since f (0) = 0^3 + 4 · 02 + 8 · 0 = 0. It is a simple zero, since the first derivative f ′(0) = 3 · 02 + 8 · 0 + 8 6 = 0 is non-zero. b. Use Newton’s method to find the root to within 10−^6 , with initial guess p 0 = 1. Answer: We can use the same Newton’s method code from the first problem, with the function files f.m: function f = f(x) f = x.^3 + 4x.^2 + 8x; end and its derivative f1.m: function f1 = f1(x) f1 = 3x.^2 + 8x + 8; end which gives the result: octave:1> newton 01: p = 0. 02: p = 0. 03: p = 0. 04: p = 0. 05: p = 0.
Now consider the similar function g(x) = x^3 + 4x^2.
c. Show that p = 0 is a zero of multiplicity 2 of g(x). Answer: We know that 0 is a zero of g since g(0) = 0^3 + 4 · 02 = 0. It is not a simple zero, because the first derivtive is also zero at 0: g′(0) = 3 · 02 + 8 · 0 = 0. However, the second derivative is non-zero (g(2)(0) = 6 · 0 + 8 6 = 0), so 0 is a zero of g of multiplicity two. d. Use Newton’s method to find the root to within 10−^6 , with initial guess p 0 = 1. Answer:
octave:2> newton 01: p = 0. 02: p = 0. 03: p = 0. 04: p = 0. 05: p = 0. 06: p = 0. 07: p = 0. 08: p = 0. 09: p = 0.
