From ‘s definition:
we have
and,
To obtain other values of , we may simply solve
for where
For example (see Fig. 1), solving for gives It is in agreement with the fact that
Fig. 1
In Fig. 2, we compute from repeatedly solving for where
Fig. 2
Since is a periodic function, has infinitely many solutions. It is possible that the solution obtained by Newton’s method lies outside of , the range of by its definition. Such solution cannot be considered the value of
Fig. 3
Exercise-1 Compute by solving for
Exercise-2 Explain Fig. 3.