Calculus II
Quiz #11 
May 15, 2002
Name____________________
R.  Hammack 
Score ______

(1) Find the interval of convergence of the power series [Graphics:Images/quiz12sol_gr_1.gif]
Notice that this series will be alternating if x happens to be negative. Therefore, we need to test for absolute convergence. Using the ratio test for absolute convergence, we get

ρ [Graphics:Images/quiz12sol_gr_2.gif][Graphics:Images/quiz12sol_gr_3.gif]=[Graphics:Images/quiz12sol_gr_4.gif]

Therefore  ρ = [Graphics:Images/quiz12sol_gr_5.gif]and the series will converge if [Graphics:Images/quiz12sol_gr_6.gif], or if   |x|< 4,   or rather if  -4 < x < 4.

What about the endpoints? If we plug x=4 into [Graphics:Images/quiz12sol_gr_7.gif], we get the convergent p-series[Graphics:Images/quiz12sol_gr_8.gif]
If we plug x=-4 into [Graphics:Images/quiz12sol_gr_9.gif], we get the convergent alternating series[Graphics:Images/quiz12sol_gr_10.gif].
  
  Therefore, the interval of convergence is [-4, 4].



(2)   Find a power series representation of the function  [Graphics:Images/quiz12sol_gr_11.gif]

We know:     cos(u) = [Graphics:Images/quiz12sol_gr_12.gif]
Plugging in u=[Graphics:Images/quiz12sol_gr_13.gif] to this, we get:   [Graphics:Images/quiz12sol_gr_14.gif] =[Graphics:Images/quiz12sol_gr_15.gif]



(3) Use your answer from part (2) to express [Graphics:Images/quiz12sol_gr_16.gif]as an infinite series.

[Graphics:Images/quiz12sol_gr_17.gif][Graphics:Images/quiz12sol_gr_18.gif]dx =[Graphics:Images/quiz12sol_gr_19.gif]
  
  [Graphics:Images/quiz12sol_gr_20.gif]
  
[Graphics:Images/quiz12sol_gr_21.gif] = [Graphics:Images/quiz12sol_gr_22.gif]

Therefore [Graphics:Images/quiz12sol_gr_23.gif][Graphics:Images/quiz12sol_gr_24.gif]dx  =  [Graphics:Images/quiz12sol_gr_25.gif]

 

(4) Consider the function f(x) = [Graphics:Images/quiz10sol_gr_1.gif] . Find the formula for the [Graphics:Images/quiz10sol_gr_2.gif]Maclaurin polynomial [Graphics:Images/quiz10sol_gr_3.gif].


We know  f(x) = [Graphics:Images/quiz10sol_gr_7.gif], f '(x) =[Graphics:Images/quiz10sol_gr_8.gif],  f ''(x) =[Graphics:Images/quiz10sol_gr_9.gif], f '''(x) =[Graphics:Images/quiz10sol_gr_10.gif], ... ,[Graphics:Images/quiz10sol_gr_11.gif](x) =[Graphics:Images/quiz10sol_gr_12.gif], ...
and thus   f(0) = 1,      f '(0) =[Graphics:Images/quiz10sol_gr_13.gif],     f ''(0) =1    , f '''(0) =[Graphics:Images/quiz10sol_gr_14.gif],   ...   ,[Graphics:Images/quiz10sol_gr_15.gif](0) =[Graphics:Images/quiz10sol_gr_16.gif], ...

Consequently,  [Graphics:Images/quiz10sol_gr_17.gif]= [Graphics:Images/quiz10sol_gr_18.gif]