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Material Type: Notes; Class: SI Calculus I; Subject: Math; University: Weber State University; Term: Unknown 2008;
Typology: Study notes
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Newton’s method does not always work. The following result gives sufficient conditions for when it does.
Theorem 1 (Newton). Let f : [a, b] → R be continuous. Suppose that f (c) = 0 for some c in (a, b). Suppose that f ′′^ exists and is bounded on (a, b); that is, there exists M > 0 such that
f ′′(x) ≤ M
for every x ∈ (a, b). Suppose that f ′^ is bounded away from zero; that is, there exists ≤ > 0 such that
f ′(x) ≥ ≤
for every x ∈ (a, b). Then there is a closed interval I := [d, e] ⊂ (a, b) containing c such that the sequence given recursively by
xn = xn− 1 − f (xn− 1 ) f ′(xn− 1 )
converges to c. Moreover, xn is in I for each n.
For this to make sense it helps if we have a notion of convergence of sequences. We will do this in second semester calculus. For now, we will just use the intuitive statement “as n gets large, xn gets arbitrarily close to c” and write lim n→∞ xn = c. Let us now analyze the various conditions in the statement. Bounding f ′^ away from zero should seem reasonable from the definition of xn but it is not necessary. The following exercise show that Newton’s method may work even if f ′^ is not bounded away from zero.
Exercise 2. Let f (x) = x^2. Adopt the notation of Newton’s theorem. Show that if xn− 1 6 = 0 then xn = xn 2 − 1. Suppose that x 0 = 1. What is x 1? How about x 2? How about xn? Argue that xn converges to zero as n goes to ∞ when x 0 = 1.
In the above example, any x 0 except zero will do. The next exercise illustrates what can go wrong if we do not bound f ′^ away from zero.
Exercise 3. Let f (x) = x^4 − 4 x − 11. Then f (0) < 0 < f (3) so by the intermediate value theorem there exists a root between x = 0 and x = 3. What happens if we apply Newton’s method when we start at x 0 = 1? Sketch the graph of f on the interval [0, 3].
We also require a bound on the second derivative. The following exercise shows what can go wrong.
Exercise 4. Let f (x) = x (^13)
. Observe that x = 0 is a root of f. Compute f ′′(x) and explain why f ′′^ is unbounded near x = 0. Attempt to use Newton’s method with x 0 = 1. What happens? Illustrate your conclusion with a diagram. 1
2 ADDITIONAL NOTES AND EXERCISES FOR 6.
Newton’s method, when it works, is extremely fast. The number of decimal places of accuracy approximately doubles with each successive iteration. The statement of Newton’s theorem does not tell us how to how to find the closed interval I. The proof gives some indication but is beyond the scope of the course. For our purposes, we can use a graphing calculator to give us an idea of where to look for the roots of a given function and then apply Newton’s method to get a good decimal approximation.
Exercise 5. Use Newton’s method to find the coordinates of the inflection point of y = x
3 12 + sin^ x^ correct to six decimal places.
Exercise 6. Let a be a non-zero number. Suppose we want to compute a decimal expansion of (^) a^1. This could be difficult if a is big or has a large number of decimal places. Newton’s method gives us a way of computing a without division. Let f (x) = (^) x^1 − a. Notice that f has a root at x = (^1) a. Using the notation of Newton’s theorem, show that
(1) xn = 2xn− 1 − ax^2 n− 1.