Question

Suppose that Newton’s method is applied to find the solution p
= 0 of the equation

e^x −1−x− (1/2)x^2 = 0. It is known that, starting with any p0
> 0, the sequence {pn} produced by the Newton’s method is
monotonically decreasing (i.e., p0 >p1 >p2 >···)and
converges to 0.

Prove that {pn} converges to 0 linearly with rate 2/3. (hint:
You need to have the patience to use L’Hospital rule repeatedly. )
Please do the proof.

Answer #1

Suppose that r is a double zero of the C2 function f, i.e., f(r)
= f′(r) = 0 but f′′(r) is not 0. Show that Newton’s method applied
to f converges linearly with the asymptotic constant 1/2, i.e.,
show that
lim n->infinity | x(n+1)−r | / | x(n)−r | = 1/2.

Suppose we modify the bisection method into the following
variation: for each step, with bracketing interval [a,b],
approximations are chosen at the location (2a + b)/3, but the
interval is cut into two at the diﬀerent location (a + 3b)/4.
(a) Calculate the ﬁrst 2 approximations p1,p2 for this variation
when f(x) = cosx−x with starting interval [0,π/2].
(b) Bound the absolute errors of the approximations pn for a
starting interval of length L.

Newton's method: For a function ?(?)=ln?+?2−3f(x)=lnx+x2−3
a. Find the root of function ?(?)f(x) starting with
?0=1.0x0=1.0.
b. Compute the ratio |??−?|/|??−1−?|2|xn−r|/|xn−1−r|2, for
iterations 2, 3, 4 given ?=1.592142937058094r=1.592142937058094.
Show that this ratio's value approaches
|?″(?)/2?′(?)||f″(x)/2f′(x)| (i.e., the iteration converges
quadratically). In error computation, keep as many digits as you
can.

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