Question

In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order Taylor approximation centered at xi. This problem investigates what happens when you try to use a second-order Taylor approximation

Answer 1

Answer:

Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

By second order Taylor approximation, we get

f(a) + hf'±(a) + $\frac{h^{2} }{2!}$f''(a) = 0

$h = \frac{-f'(a) }{f''(a)}$ ± $\frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}$

So, we get the succeeding equation for Newton's method as

$x_{i+1} = x_{i} + \frac{1}{f''x_{i}} [-f'(x_{i})$ ± $\sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]$

Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)    

Part C.

In case  $x_{i+1}$ is close to $x_{i}$, the choice that shouldbe made instead of ± in part A is:

f'(x) = $\sqrt{f'(x)^{2} - 2f(x)f''(x)}$  ⇔ $x_{i+1} = x_{i}$

Part D.

As given $x_{i+1}$ = $x_{i}$ = h

or                 h = $x_{i+1}$ - $x_{i}$

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             ($x_{i+1}$-$x_{i}$)² = -($x_{i+1}$-$x_{i}$)(f($x_{i}$)/f'($x_{i}$))

So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

Also,             $x_{i+1}$ = $x_{i}$ -f($x_{i}$)/f'($x_{i}$) + [($x_{i+1}$ - $x_{i}$)f''($x_{i}$)f($x_{i}$)]/[2(f'($x_{i}$))²]