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.