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CHE/COSC/MATH 4340-01 Numerical Analysis
Solution of Writing Assignment 03 Due Date: Tuesday, 02/12/
Problem 1.
To compute reciprocals without division, we can solve x = 1/R by finding a zero of the function f (x) = x^1 − R. Derive a short algorithm to find 1/R by applying Newton’s method applied to f (x). Do not use division or exponentiation in your algorithm. For positive R, what initial iterates are suitable?
Answer: Notice that f ′(x) = −x−^2. Newton’s method for this function reads
xn+ 1 = xn − x
− n 1 − R −x− n^2 = xn + x^2 n(x− n 1 − R) = 2 xn − Rx n^2.
Thus the Newton’s method is simplified as xn+ 1 = xn( 2 − Rxn). Now suppose R is a positive number.
- Notice that if we choose x 0 = 0, then next iterate will be 0, and this keeps going. So we cannot choose x 0 = 0.
- If we choose an initial iterate that is a negative number, the expression in parenthesis will be positive, and as a result the next iterate will be negative, and by the same token, we will create a sequence of negative numbers. So negative initial iterate is not an option.
- If we choose a positive number x 0 that makes Rx 0 > 2, then the term in parenthesis will be negative, which would in turn result in a negative number for the next iterate. Thus we may not choose x 0 that makes Rx 0 > 2.
- Our only option then is to choose x 0 > 0 and 2 − Rx 0 > 0. This would result in 0 < x 0 < 2 /R. Note: the inequality is strict, i.e., x 0 6 = 0 and x 0 6 = 2 /R.
Problem 3.
If Newton’s method is used with f (x) = x^2 − 1 and x 0 = 1010 , how many steps are required to obtain the root with accuracy 10−^8? Answer: Newton’s method applied to this f reads
xn+ 1 = xn − x
(^2) n − 1 2 xn^ =^
x^2 n + 1 2 xn^.
We know that there are r = −1 and r = 1 are zeros of f (x) that are simple, so we may expect that if the iterates are close enough to the neighborhood of one of the zero, then Newton method will exhibit a quadratic convergence, namely |en+ 1 | ≤ C|en|^2 ,
where C = 12 f^
′′(r) f ′(r). Suppose now that x 0 = 1010. Since this is a positive number, we may expect that Newton’s method will converge to r = 1. However, initially it will not exhibit a quadratic convergence because x 0 is clearly much far away from r = 1. In particular, writing en = xn − 1, we have
en+ 1 = x
(^2) n + 1 2 xn^ −^1 =^
x^2 n + 1 − 2 xn 2 xn^ =
(xn − 1 )^2 2 xn^ =^
e^2 n 2 xn^.
At the initial stage when the iterates are still very large, we know that en ≈ xn and thus,
|en+ 1 | = |en|
2 2 |xn| ≤
|en| 2 ,^ (0.1)
i.e., we only have a first order convergence. The question is how close xn should be when it will gain quadratic convergence. We think that xn is close enough already if |en| < 1. Thus before |en| < 1, we should expect the method to have first order convergence. By induction argument we see that (0.1) yields
|en| < |e 20 n|.
Setting this inequality to be less than 2, yields
|e 0 | 2 n^ <^1 ,
which gives 1010 < 2 n.
Problem 4.
If we attempt to find a fixed point of F(x) by using Newton’s method, what iteration formula would you get? Answer: A number x is a fixed point of F if x = F(x). Now choose G(x) = x − F(x). Then finding a fixed point of F is equivalent to finding a zero of G. Notice that G′(x) = 1 − F′(x). Application of Newton’s method gives xn+ 1 = xn − x 1 n −^ − F^ F′((xxnn)) = xn^ −^ xnF
′(xn) − xn + F(xn) 1 − F′(xn) = F(xn)^ −^ xnF
′(xn) 1 − F′(xn)
