The exercise examines various choices that can be made for the approximate solution using the bisection method. Assume that the subinterval ( i , i ) has just been calculated, and the goal is to now...

The exercise examines various choices that can be made for the approximate solution using the bisection method. Assume that the subinterval (

_i
,


_i
) has just been calculated, and the goal is to now determine what point


_i

should be selected from this subinterval as the approximation for the solution
.

(a) Whatever choice is made, the error is


_i

−
. Sketch this as a function of
, for


_i

≤
 ≤


_i
. Explain why the minimum error occurs when


_i

−


_i

=


_i

−


_i
, and from this conclude that


_i

= (

_i

+


_i
)/2. In other words, one should select the midpoint of the subinterval.

(b) Suppose one instead uses the relative error (

_i

−
)/. This requires a nonzero solution, so it is assumed here that 0


<sub>i



i</sub>
. By sketching (
=
−
)/, for


_i

≤
 ≤


_i
, explain why the minimum error occurs when


_i

= 2

<sub>i



i</sub>
/(

_i

+


_i
).

(c) Does it make much difference which choice is made for


_i
? In answering this, assume that the stopping condition for the loop in Table 2.1 is the same irrespective of the choice for


_i
.