One of these is known as Newton's method or the Newton-Raphson method. It
has been developed further since its introduced in the 17th Century. We take a ...
1
The Newton-Raphson Method
It is frequently important to know if and where a given function, f : R → R takes a specified value, b. Defining F by F (x) := f (x) − b, we see that this is equivalent to the problem Find all solutions x ∈ R of the equation F (x) = 0. This is one of the central difficulties in applying mathematical theory and theorems to concrete problems. For example, we have seen that if a function is differentiable, then its extreme values only where its derivative is 0. Optimisation problems which can be expressed in terms of finding extreme values of a differentiable function reduce to the above problem. There is no single formula, no general method for solving the above problem. Nevertheless techniques are available for some cases which have broad enough application to be extremely useful, especially when seeking numerical solutions. One of these is known as Newton’s method or the Newton-Raphson method . It has been developed further since its introduced in the 17th Century. We take a brief look at it, because it illustrates applications of the theory we have developed, as well as some of the limitations, and fits our strategy of approximating when absolute precision is not readily available. The basic idea is to geometrically approximate a zero, a, of the function, f , using a succession of points beginning with a first guess, x0 . If the nth approximation is xn , then the next approximation, xn+1 is obtained by taking the tangent to the graph of f at (xn , f (xn )) and finding its intersection, (xn+1 , 0), with the x-axis. (xn , f (xn ))
a
xn+1
xn
The tangent to the graph of f at the point xn , f (xn ) has equation y − f (xn ) = f 0 (xn )(x − xn ) and meets the x-axis at (xn+1 , 0), so that f 0 (xn )(xn+1 − xn ) + f (xn ) = 0
(1)
Observation 1. If f 0 (xn ) = 0, then either f (xn ) = 0, whence xn is a zero of f , or f (xn ) 6= 0, in which case Equation (1) has no solution xn+1 . If f 0 (xn ) 6= 0, then we have the more common representation xn+1 = xn −
f (xn ) . f 0 (xn )
(N-R)
The Newton-Raphson method reduces finding the sought zero of the function, f , to the problem of finding the limit of the sequence f (xn ) xn − 0 f (xn ) n∈N 1
Observation 2. This sequence need not converge, or it may converge to the “wrong” zero of f , as the next examples illustrate. Example 3. Take f : R → R, x 7→ x2 − x + 1 and x0 = 1. As f 0 (x) = 2x − 1, x1 = 1 −
f (1) 1 =1− =0 0 f (1) 1
and 1 f (0) =0− = 1 = x0 0 f (0) −1 ( 1 if n is even, It follows that xn = 0 if n is odd. Thus (xn )n∈N does not converge. x2 = 0 −
Example 4. Suppose we are trying to find the zero a of the function f whose graph is below, and we apply the Newton-Raphson method commencing with x0 as indicated. Then, a the graph shows, x1 lies near another zero of f . As the graph suggests, (xn )n∈N converges in this case, but not to a. (x0 , f (x0 ))
x1
a
x0
Having indicated some of the difficulties with the Newton-Raphson method, we show when the method can be applied. Theorem 5. Let f : R → R be a continuous function which has precisely one zero, a, in the open interval ]u, v[. If f 00 , the second derivative of f , is continuous on ]u, v[ and f 0 (a) 6= 0, then there is a δ > 0 such that if a − δ < x0 < a + δ, then (xn )n∈N , the sequence of successive approximations generated by the Newton-Raphson method, converges to a. Proof. Since f has a second derivative on ]u, v[, its first derivative, f 0 , is continuous on ]u, v[. Since u < a < v and f 0 (a) 6= 0, there is a µ > 0 with u < a − µ < a + µ < v, such that f 0 (x) 6= 0 whenever a − µ < x < a + µ. The function g : ]a − µ, a + µ[ −→ R,
x 7−→ x −
f (x) . f 0 (x)
is continuously differentiable because f has a continuous second derivative and f 0 (x) 6= 0 on ]a − µ, a + µ[. 2
Moreover, g(a) = a, and g 0 (x) = 1 −
(f 0 (x))2 − f 00 (x)f (x) , (f 0 (x))2
whence g 0 (a) = 0. Since g 0 is continuous, there is a δ with 0 < δ ≤ µ and |g 0 (x)| < 21 , whenever a − δ < x < a+δ Take such an x. Then |g(x) − a| = |g(x) − g(a)|
as g(a) = a
0
= |g (c)| |x − a| 1 ≤ δ 2
