# Explain the determine red and the eqautions are here Transcribed Image Text: This last inequality is just the condition of equation (7

Explain the determine red and the eqautions are here Transcribed Image Text: This last inequality is just the condition of equation (7.131) given by the

geometric analysis.

Finally, we discuss, for the computation of reciprocals, the choice of the

initial value yo. Our selection process must be such that the condition of

equation (7.131) is always satisfied. Let the number x be written in binary

format; i.e.,

Cx = 2″ x1,

(7.140)

where m is an integer and

< x1 < 1.
(7.141)
Then select yo to have the value
Yo = 2-m
(7.142)
For this choice of yo, we have
(2" x1)(2-m) = X1,
(7.143)
XYO =
and
< xyo < 1.
(7.144)
Therefore, the iteration scheme will always converge. In fact, using equation
(7.144), we have from equation (7.137) the result
2k
Yk < 2y0
(7.145)
which has extremely fast convergence.
Next, we will calculate approximations to the square root of a positive
number x. Putting m = 2 into equation (7.129) gives
() ►
Yk +
Yk
(7.146)
Yk+1 =
Again, the geometric methods of Section 2.7 may be used to analyze the
behavior of the iteration scheme of equation (7.146).
Examination of the graph in Figure 7.3 leads to the following conclusions:
(i) Yk+1 has the same sign as yk.
(ii) The two fixed points are located at (Va, V) and (-Va, - V®).
(iii) Both fixed points are stable.
(iv) For yo > 0, the iterates yk converge to +Vx; i.e.,

lim yk =

+Vx, yo > 0.

(7.147)

For yo < 0, the iterates yk converge to – Vx; i.e.,
lim yk =
-Væ, yo < 0.
(7.148) Transcribed Image Text: (-)
т — 1
Ук+1
Yk +
(7.129)
m(yk)m
m-1
m
1
Yk =
() u
🙂 [1- ayo)2".
Zk
(7.137)
2
0 < yo <
(7.131)
||
||

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