Download MATH 2202 Exam 4 Solutions: Ratio Test, Power Series, Taylor Series, and Integration - Pro and more Exams Calculus in PDF only on Docsity!
MATH 2202 - Exam 4 (Version 1) Solutions November 23, 2009 S. F. Ellermeyer Name
Instructions. Show all of your work! You will not get full credit if you donít include correct notation. In particular, you must write ì=îwhere needed. Also, when discussing sequences and series, take care not to write the word ìitî. Special Thanksgiving Bonus!: There are six questions on this exam but I will only use your best Öve to compute your grade.
- Try to use the Ratio Test to determine whether or not the series X^1
n=
p n 1 + n^2
converges or diverges. Does the Ratio Test succeed of fail? If the test succeeds, then does it tell you that the series converges or diverges (and why)? If the test fails, then why does it fail? Solution: Since an =
p n 1 + n^2
then an+1 =
p n + 1 1 + (n + 1)^2 and an+ an
p n + 1 1 + (n + 1)^2
1 + n^2 p n
=
r n + 1 n
n^2 + 1 n^2 + 2n + 2
Since
n^ lim!
an+ an = lim n!
r n + 1 n
n^2 + 1 n^2 + 2n + 2
p 1 1 = 1,
the Ratio Test fails to give any information about the convergence or divergence of the series. (The series can be shown to converge by the Limit Comparison Test.)
- Find the radius of convergence of the power series X^1
n=
n^2 xn 2 4 6 (2n)
Solution: We have bn = n^2 xn 2 4 6 (2n)
and bn+1 = (n + 1)^2 xn+ 2 4 6 (2n) (2n + 2)
Thus
bn+ bn
(n + 1)^2 xn+ 2 4 6 (2n) (2n + 2)
2 4 6 (2n) n^2 xn
=
n + 1 n
2 n + 2
jxj
and we see that
n^ lim!
bn+ bn = lim n!
n + 1 n
2 n + 2
jxj = (1)^2 (0) jxj = 0.
By the Ratio Test, we conclude that the given series converges no matter what x is. In other words, the radius of convergence is 1.
- Begin with the known power series representation
1 1 x = 1 + x + x^2 + x^3 + x^4 + x^5 +
which holds for all x 2 ( 1 ; 1) and use substitutions and algebra to obtain the power series representation of x^2 1 + 9x^2
What is the interval of convergence of the power series that you obtain? (Be detailed in your solution.) Solution: Note that x^2 1 + 9x^2 = x^2
1 ( 9 x^2 )
Since 1 1 x
X^1
n=
xn^ for all x such that jxj < 1 ,
then 1 1 ( 9 x^2 )
X^1
n=
9 x^2
n for all x such that 9 x^2 < 1.
The inequality j 9 x^2 j < 1 is equivalent to jxj^2 < 1 = 9 and this is equivalent to jxj < 1 = 3. The radius of convergence of the latter series is thus 1 = 3. Finally
x^2 1 + 9x^2 = x^2 1 +
X^1
n=
9 x^2
n
= x^2 +
X^1
n=
9 x^2
n x^2
and we obtain Z (^2)
1
ex x dx
ln (x) + x +
x^2 +
x^3 +
x^4
x=
x=
ln (2) + 2 +
(2)^2 +
(2)^3 +
(2)^4
ln (1) + 1 +
- Since
lim x! 0 cos (x) 1 x is a limit problem of the indeterminate form 0 = 0 , we can use LíHospitalís Rule to Önd that
x^ lim! 0
cos (x) 1 x = lim x! 0 sin (x) 1
sin (0) 1
Another way to show that
x^ lim! 0
cos (x) 1 x
is to use the MacClaurin series of cos (x). Show how this is done. Solution: Since
cos (x) = 1
x^2 +
x^4
x^6 + for all x,
then cos (x) 1 =
x^2 +
x^4
x^6 + for all x, and cos (x) 1 x
x +
x^3
x^5 + for all x 6 = 0. Since power series are continuous functions, we obtain
x^ lim! 0
cos (x) 1 x = (^) xlim! 0
x +
x^3
x^5 +
