Exercise 16.  Show that the function  [Graphics:Images/MapElementaryFunModHome_gr_935.gif]  in Equation (10-22) is analytic on the ray  [Graphics:Images/MapElementaryFunModHome_gr_936.gif].  

Solution 16.

        In Section 10.3.1 we looked at the double-valued function  [Graphics:../Images/MapElementaryFunModHome_gr_937.gif],   

and considered expressing  [Graphics:../Images/MapElementaryFunModHome_gr_938.gif]  as the function  [Graphics:../Images/MapElementaryFunModHome_gr_939.gif]  which is defined by the equation   

(10-22)             [Graphics:../Images/MapElementaryFunModHome_gr_940.gif],  

where the principal branch of the square root function is used in both factors.  

Furthermore, it was shown that [Graphics:../Images/MapElementaryFunModHome_gr_941.gif] is continuous on the ray  [Graphics:../Images/MapElementaryFunModHome_gr_942.gif].

        Now use implicit differentiation to determine the derivative of  [Graphics:../Images/MapElementaryFunModHome_gr_943.gif].   Square both sides of (10-22) and obtain

                    [Graphics:../Images/MapElementaryFunModHome_gr_944.gif].

Take the derivatives of both sides

                    [Graphics:../Images/MapElementaryFunModHome_gr_945.gif],   

then obtain

                    [Graphics:../Images/MapElementaryFunModHome_gr_946.gif].

For the point  [Graphics:../Images/MapElementaryFunModHome_gr_947.gif]  on the ray  [Graphics:../Images/MapElementaryFunModHome_gr_948.gif]  we showed that we have

                    [Graphics:../Images/MapElementaryFunModHome_gr_949.gif],   

and the derivative is defined to be

                    [Graphics:../Images/MapElementaryFunModHome_gr_950.gif].

Hence,  [Graphics:../Images/MapElementaryFunModHome_gr_951.gif]  is also continuous on the ray  [Graphics:../Images/MapElementaryFunModHome_gr_952.gif].

Therefore,  [Graphics:../Images/MapElementaryFunModHome_gr_953.gif]  is analytic on the ray  [Graphics:../Images/MapElementaryFunModHome_gr_954.gif].  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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