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

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) + f''(a) = 0

±

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

±

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   is close to , the choice that shouldbe made instead of ± in part A is:

f'(x) =  ⇔

Part D.

As given = = h

or                 h = -

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             (-)² = -(-)(f()/f'())

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,             = -f()/f'() + [( - )f''()f()]/[2(f'())²]