This problem explores how to use the floating-point representation to obtain a good starting point for Newton’s method. To illustrate this idea, the equation to solve is 2 −  = 0, where  > 0, which is...

This problem explores how to use the floating-point representation to obtain a good starting point for Newton’s method. To illustrate this idea, the equation to solve is

²
−
 = 0, where
 > 0, which is used to evaluate

(a) Show that the formula for Newton’s method is

(b) The floating-point approximation of
 has the form


_f

=
 × 2
^E
, where

Also, recall that for 0 ≤

Assuming
 is even, and

₁
or

₂
are nonzero, use the above information to show that

In what follows it is assumed that this formula is used to determine

₀. Note that this expression only involves additions and multiplications (and not a square root).

(c) What does the approximation in part (b) reduce to for
 = 28 = (1 + 1/2+1/2²) × 2⁴? How close does

₀
come to

(d) Modify the derivation in part (b) to find a starting value in the case of when
 = 50 = ( 1+1/2+1/2⁴)  × 2⁵. Make sure to compare

₀
with the exact value. Also, you can assume the value of
 is known.

(e) Write a MATLAB program that uses this idea to calculate
 for any given
 > 1. The stopping condition for Newton’s method should be
 With this, compute
 and