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10.1.12 (Newton's method) Suppose you want to find the roots of an equation g(x) = 0. Then Newton's method says you should consider the map x_{n+1} = f(x_n), where f(x_n) = x_n - frac{g(x_n)}{g'(x_n)}. a) To calibrate the method, write down the "Newton map" x_{n+1} = f(x_n) for the equation g(x) = x^2 - 4 = 0. b) Show that the Newton map has fixed points at x* = ± 2. c) Show that these fixed points are superstable. d) Iterate the map numerically, starting from x_0 = 1. Notice the extremely rapid convergence to the right answer!

          10.1.12 (Newton's method) Suppose you want to find the roots of an equation g(x) = 0. Then Newton's method says you should consider the map x_{n+1} = f(x_n), where f(x_n) = x_n - frac{g(x_n)}{g'(x_n)}. a) To calibrate the method, write down the "Newton map" x_{n+1} = f(x_n) for the equation g(x) = x^2 - 4 = 0. b) Show that the Newton map has fixed points at x* = ± 2. c) Show that these fixed points are superstable. d) Iterate the map numerically, starting from x_0 = 1. Notice the extremely rapid convergence to the right answer!

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$10.1.12 (Newton's method) Suppose you want to find the roots of an equation g(x) = 0. Then Newton's method says you should consider the map xn+1 = f(xn), where f(xn) = xn - fracg(xn)g'(xn). a) To calibrate the method, write down the "Newton map" xn+1 = f(xn) for the equation g(x) = x^2 - 4 = 0. b) Show that the Newton map has fixed points at x* = ± 2. c) Show that these fixed points are superstable. d) Iterate the map numerically, starting from x0 = 1. Notice the extremely rapid convergence to the right answer!$

Added by Raquel H.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert William Fairburn

03/09/2024

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### a) Writing down the "Newton map" for the given equation ** Show more…

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10.1.12 (Newton's method) Suppose you want to find the roots of an equation g(x) = 0. Then Newton's method says you should consider the map xn+1 = f(xn), where f(xn) = xn - g(xn)/g'(xn). a) To calibrate the method, write down the "Newton map" xn+1 = f(xn) for the equation g(x) = x^2 - 4 = 0. b) Show that the Newton map has fixed points at x* = ±2. c) Show that these fixed points are superstable. d) Iterate the map numerically, starting from x0 = 1. Notice the extremely rapid convergence to the right answer!

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00:01 Okay, i've written down the function that describes the newton's method.

00:09 So the so -called map that we want to write down, i think they just mean put in the specific function that we're talking about.

00:19 So the specific function we're trying to find the roots for is g of x is equal to x squared minus 4.

00:31 And of course we're trying to find the roots is where this is equal to zero.

00:38 So the map in this case, and i've never really heard it referred to as a map, i guess it is, is x sub n minus g of x, so g of x is x squared minus 4 and g prime of x is 2x.

01:03 So that's our map and we want to show that this map has fixed points at plus or minus 2.

01:15 So the idea here is this map, when we have found a root then the map would stop the, so so, i mean, this f of xn is the next, it's x of n plus 1, it's the next proposed root.

01:37 So when f of xn is equal to xn, then you've found a root.

01:46 So what they're saying is set x equals x minus x squared minus 4 over 2x.

01:54 This is where the process stops.

01:58 And so subtracting x from both sides we have 0 equals minus x squared minus 4 over 2x.

02:12 And this of course will be 0 where x squared minus 4 is 0.

02:22 0, and so that's where x is equal to plus or minus 2, which we knew to begin with.

02:33 We could have just answered the question, but that's not the point.

02:36 Okay, show that these fixed points are super stable.

02:44 I believe that means show that this f prime at plus and minus 2 is equal to 0.

02:56 So f prime is equal to 1 minus x squared minus 4 over 2x...

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