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Unformatted text preview: METU
Department of Mathematics
Group Code
Acad. Year
Semester
Instructor : Math 120
: 20042005
: Spring
: Date
Time
Duration Calculus II
MidTerm 1
Last Name :
:
Name
Department :
:
Signature : April 2nd,2005
: 09:30
: 120 minutes 1 2 3 4 List No. Student No. :
:
Section 5 QUESTIONS ON 6 PAGES
TOTAL 50 POINTS 5 SOLUTION KEY
Question 1 (10 points) Find the radius of convergence and interval of convergence
∞ (−3)n xn
√
.
n+1
n=0 of the series Solution:
√
 an+1 
3n+1  x n+1
n+1
√
= lim
· n
= 3  x  lim
n→∞  an 
n→∞
n→∞
3  x n
n+2
lim n+1
=3x.
n+2 So if 3  x < 1, by ratio test the given power series is convergent. This gives  x < 1/3, so R = 1/3. For the interval of convergence, we must check the end points. If x = 1/3:
∞ ∞ (−1)n 3n
(−1)n
√
√
=
.
n + 1 3n
n+1
n=0
n=0
an = (−1)n
√
n+1 is alternating, • √1
n+1 > 0, • limn→∞ an = limn→∞
• an+1 
an  = √
√n+1
n+2 = (−1)n
√
n+1 n+1
n+2 = 0, < 1, so  an  is decreasing. So by AST, the series is convergent. If x = −1/3:
∞ ∞ √
n=0
∞ n=1 1
n1/2 1
(−3)n
√
=
.
n
n + 1 (−3)
n+1
n=0 is divergent by p−test, so the above series is divergent by limit comparison. So the interval of convergence is I = (−1/3, 1/3]. Question 2 (10 points)
1
; k ≥ 1 an integer.
1 + xk
a)Find the power series expansion of f (x) in powers of x.
Consider f (x) = Solution: Since ∞ 1
=
(−1)n xn
1 + x n=0
we have; for all  x < 1, ∞ 1
=
(−1)n xnk .
1 + xk
n=0 b)Use the series in part (a), to expand g(x) = x arctan(x) in powers of x. Solution:
x g(x) = x arctan x = x
0 1
du = x
1 + u2 c)Find the expansion of h(x) =
of convergence. √ ∞ x ∞ (−1)n u2n du =
0 (−1)n
n=0 n=0 x2n+2
2n + 1 for all  x < 1. √
x + 1 arctan( x + 1) around a = −1. Indicate the interval Solution:
h(x) = √ √
x + 1 arctan x + 1 = ∞ (−1)n
n=0 (x + 1)1/2
2n + 1 2n+2 ∞ = (−1)n (x + 1)n+1
2n + 1
n=0 for all  x + 1 < 1. Question 3 (9 points) Determine whether the followings are convergent or divergent.
Show tour work.
∞
(n!)2
a)
(2n)!
n=0 Solution: We apply ratio test, with an = (n!)2
:
(2n)! n+1
(n!)2 (n + 1)2
(2n)!
1
an+1
= lim
= lim
·
= < 1.
n→∞ 2(2n + 1)
n→∞ (2n)!(2n + 1)(2n + 2) (n!)2
n→∞ an
4
lim So by ratio test, the series converges.
∞ b) 1
2 + 3n
n=0 Solution: We have;
1
1
1
≤ n =
n
2+3
3
3 0≤ ∞
n=0 1 n
3 n . is a geometric series with r = 1/3, so it is convergent.
∞
1
n=0 2+3n Therefore, by comparison test, c)Approximate is convergent. 1
1
with an error less than
.
e
119 Solution: We have; ∞ −x e
Then; = (−1)n xn
.
n!
n=0 ∞ 1
(−1)n
1
1
1
1
= e−1 =
= 1 − 1 + − + − + ···.
e
n!
2! 3! 4! 5!
n=0
Since the series is alternating;
En (1) = 1
1
<
,
(n + 1)!
119 which gives n + 1 = 5. Therefore;
1
1
1
1
3
≈1−1+ − +
= .
e
2! 3! 4!
8 Question 4 (9 points)
∞ a)Prove that for an ≥ 0, if ∞ a2 is also convergent.
n an is convergent then
n=0 n=0 Solution:
∞ an is convergent, we have lim an = 0. This implies that there exists N ∈ I such
N Since n→∞ n=0 that for n ≥ N , we have 0 ≤ an ≤ 1. So for all n ≥ N , 0 ≤ a2 ≤ an .
n
∞ Since ∞ a2 is also convergent by comparison test.
n an is convergent,
n=0 n=0 b)The term of a sequence are deﬁned recursively by the equations
4n + 1
)an . Is (an )∞ convergent? (explain, and if so what is the limit?)
a1 = 2, an+1 = (
n=1
5n + 2 Solution:
∞ an . We have an ≥ 0. Apply ratio test to this series; Consider
n=1 lim n→∞ 4n + 1
4
an+1
= lim
= < 1.
n→∞ 5n + 2
an
5 So by ratio test, the series converges. This implies limn→∞ an = 0. c)Give an example of a power series which is convergent in (1, 5), and divergent on
(−∞, 1) ∪ (5, ∞). (No condition at x = 1 and x = 5) Solution:
∞ an (x − 3)n ; a power series around 3. We want the radius of convergence to be Consider,
n=0 2. So we want lim n→∞ Let an =  an+1 
1
= .
 an 
2 1
, then we get the desired limit and interval of convergence. So
2n
∞ (x − 3)n
2n
n=0
is a power series satisfying the conditions we want. Question 5 ( 12 points)
→ → → → → → → → a) Find all vectors v that satisfy the equation (− i +2 j +3 k )× v = i +5 j −3 k Solution:
→ Let v = (a, b, c). Then we should have; (−1, 2, 3) × (a, b, c) = i
−1
a j k
2 3
b c = (2c − 3b, 3a + c, −2a − b) = (1, 5, −3). This happens if and only if −3b + 2c = 1, 3a + c = 5, and −2a − b = −3 which gives
→ → → → v = t i +(3 − 2t) j +(5 − 3t) k for all t ∈ I
R. b)Show that the four points A(1, 2, −1), B(0, 1, 5), C(−1, 2, 1) and D(2, 1, 3) are coplanar
(i.e. are on the same plane). Solution:
→ →
→
− −
−
→ →
−→
−
Let the three vectors u , v and w represent the three segments AB, AC and AD, respectively.
→
→
→
Then u = (−1, −1, 6), v = (−2, 0, 2) and w= (1, −1, 4). The four points are coplanar if the volume of the parallelpiped spanned by the three vectors
is 0;
−1 −1
−2 0
1 −1 6
2
4 =− 0
1 2
4 + −2 2
1 −1 +6 −2 0
1 −1 = −2 − 10 + 12 = 0. c)Find the distance between the lines
L1 : intersection of the planes (x + 2y = 3 and y + 2z = 3) and
L2 : intersection of the planes (x + y + z = 6 and x − 2z = −5) Solution: L1 contains the points (1, 1, 1), (3, 0, 3/2), so it is parallel to the vector (3, 0, 3/2)−(1, 1, 1) =
→
(2, −1, 1/2), so it is parallel to the vector u 1 = (4, −2, 1). L2 contains the points (−5, 11, 0), (−1, 5, 2), so it is parallel to the vector (−1, 5, 2) −
→
(−5, 11, 0) = (4, −6, 2), so it is parallel to the vector u 2 = (2, −3, 1).
→ → → → Using the vectors r 1 = (1, 1, 1), r 2 = (−1, 5, 2), u 1 = (4, −2, 1) and u 2 = (2, −3, 1) we ﬁnd
the distance: → d= → → →
u1 →
u2 →  ( r 2 − r 1) · ( u 1 × u 2) 
 ×  =  (2, −4, −1) · (1, −2, −8) 
18
=√ .
 (1, −2, −8) 
69 ...
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