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) + 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]