Solution 7 (a).

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/TaylorSeriesModHome_gr_527.gif].  

Solution.   Given the power series   [Graphics:../Images/TaylorSeriesModHome_gr_528.gif].  

By Taylor's Theorem,  [Graphics:../Images/TaylorSeriesModHome_gr_529.gif].  

Therefore,   [Graphics:../Images/TaylorSeriesModHome_gr_530.gif],   so that   

                    [Graphics:../Images/TaylorSeriesModHome_gr_531.gif].  

We are done.   

We can expand the series in its even and odd terms

                    [Graphics:../Images/TaylorSeriesModHome_gr_532.gif]

We are really done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/TaylorSeriesModHome_gr_533.gif]

[Graphics:../Images/TaylorSeriesModHome_gr_534.gif]

[Graphics:../Images/TaylorSeriesModHome_gr_535.gif]


[Graphics:../Images/TaylorSeriesModHome_gr_536.gif]

[Graphics:../Images/TaylorSeriesModHome_gr_537.gif]


[Graphics:../Images/TaylorSeriesModHome_gr_538.gif]

[Graphics:../Images/TaylorSeriesModHome_gr_539.gif]


[Graphics:../Images/TaylorSeriesModHome_gr_540.gif]

[Graphics:../Images/TaylorSeriesModHome_gr_541.gif]


[Graphics:../Images/TaylorSeriesModHome_gr_542.gif]

[Graphics:../Images/TaylorSeriesModHome_gr_543.gif]


[Graphics:../Images/TaylorSeriesModHome_gr_544.gif]

[Graphics:../Images/TaylorSeriesModHome_gr_545.gif]


[Graphics:../Images/TaylorSeriesModHome_gr_546.gif]

[Graphics:../Images/TaylorSeriesModHome_gr_547.gif]


[Graphics:../Images/TaylorSeriesModHome_gr_548.gif]

[Graphics:../Images/TaylorSeriesModHome_gr_549.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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