Solution 1 (c).

Answer.   No.  

Solution.   No.  Since    is analytic at  ,  there exists a neighborhood     in which     is analytic,

and we can choose an integer      so that      and   .  

Apply Corollary 7.10 for the sequences  ,  respectively.

For the sequence   ,   and the function      we have

                    ,   

and Corollary 7.10 implies that   .  

For the sequence   ,   and the function      we have

                    ,   

and Corollary 7.10 implies that   .  

Since   ,   this is impossible.

Therefore, there does not exist a function    that is analytic at    such that  .  

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell