Question

While studying Unit 2 of this course it occurred to me there must be a relationship between linear approximation and Newton's method, but I haven't seen any discussion of this. Is anyone aware of commentary on this relationship? I'll post an example below.

Answer 1

Prof. Jerison shows how Newton's method can be used to calculate the square root of 5 by working toward the root of the equation 0 = x^2 - 5, with an initial guess of 2. The first iteration produces a result of 9/4, and the second one produces a result of 161/72, which is pretty close.

Suppose we start instead with the equation f(x) = sqrt(x) and find the linear approximation near 4 with a difference of 1. Here's one way to do that, substituting 4 + h for x so the linear approximation near 0 can be used:\[(4+h)^\frac{ 1 }{ 2 }=2(1+\frac{ h }{ 4 })^\frac{ 1 }{ 2 }\approx 2(1+\frac{ h }{ 8 })\]In this case, h = 1, so we get the 9/4, the same result as in the first iteration of Newton's method.

Now we can do a second linear approximation, this time at a base value of 81/16 (which is (9/4) squared) and with h = -1/16 (because 81/16 is 1/16 more than 5). If you "do the math" you get 161/72, which is exactly the same as the result produced in the second application of Newton's method. It would appear that Newton's method is less cumbersome than this approach, but at least for this type of formula both approaches are equally powerful.