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) + [tex]\frac{h^{2} }{2!}[/tex]f''(a) = 0
[tex]h = \frac{-f'(a) }{f''(a)}[/tex] ± [tex]\frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}[/tex]
So, we get the succeeding equation for Newton's method as
[tex]x_{i+1} = x_{i} + \frac{1}{f''x_{i}} [-f'(x_{i})[/tex] ± [tex]\sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ][/tex]
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 [tex]x_{i+1}[/tex] is close to [tex]x_{i}[/tex], the choice that shouldbe made instead of ± in part A is:
f'(x) = [tex]\sqrt{f'(x)^{2} - 2f(x)f''(x)}[/tex] ⇔ [tex]x_{i+1} = x_{i}[/tex]
Part D.
As given [tex]x_{i+1}[/tex] = [tex]x_{i}[/tex] = h
or h = [tex]x_{i+1}[/tex] - [tex]x_{i}[/tex]
We get,
f(a) + hf'(a) +(h²/2)f''(a) = 0
or h² = -hf(a)/f'(a)
Also, ([tex]x_{i+1}[/tex]-[tex]x_{i}[/tex])² = -([tex]x_{i+1}[/tex]-[tex]x_{i}[/tex])(f([tex]x_{i}[/tex])/f'([tex]x_{i}[/tex]))
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, [tex]x_{i+1}[/tex] = [tex]x_{i}[/tex] -f([tex]x_{i}[/tex])/f'([tex]x_{i}[/tex]) + [([tex]x_{i+1}[/tex] - [tex]x_{i}[/tex])f''([tex]x_{i}[/tex])f([tex]x_{i}[/tex])]/[2(f'([tex]x_{i}[/tex]))²]
