No calculators. No partial credit.
1. The series
infinity
----- n
\ (-1)
) -----
/ 1/4
----- n
n = 1
is
a) harmonic b) divergent c) absolutely convergent d) geometric e) conditionally convergent.
SOLUTION: e) The terms 1/n^(1/4) monotonically converge to zero; so
this series converges by the alternating series test. This series does not
converge absolutely because the exponent 1/4 is less than or equal to one
(the integral test implies that the series involving 1/n^p converges for p>1).
2. The curve parametrized by x= e^t y=e^{-t} for 0 infinity).
Since arctan(infinity) = Pi/2 and arctan(1) = Pi/4, the answer is
Pi/2 - Pi/4 = Pi/4.
6. Find the radius of convergence of the series
> Sum((x-4)^n/n^n,n=1..infinity);
infinity
----- n
\ (x - 4)
) --------
/ n
----- n
n = 1
a) 0 b) 1/2 c) 1 d) 2 e) infinity
SOLUTION: e) Use the root test - - the nth root of the nth term
is |x-4|/n which converges to zero for any value of x. Therefore the
root test implies that this series converges for any value of x. In other
words, the radius of convergence is infinity.
7. Find the coefficient of (x-2)^2 in the Taylor series expansion of
g(x) = 1/x about x=2.
a) 1/2 b) 1/4 c) 1/8 d) 1/16 e) 1/32
SOLUTION: c) The coefficient is g''(2)/2; since g''(x) = 2/x^3 inserting
x = 2 yields g''(2) = 2/8 = 1/4. Therefore g''(2)/2 = 1/8.
8. The line tangent to the curve x = t^3-t y = t^2 at the point (6,4) has slope
a) 8/107 b) 4/11 c) 2/3 d) 3/2 e) 47
SOLUTION: b) The point (6,4) corresponds to t = 2. The slope of the tangent
at (6,4) is therefore equal to y'(2)/x'(2); since x' = 2t and y' = 3t^2 - 1,
the slope is 4/11.
9. Find the coefficient of x^3 in the Maclaurin series expansion
(Taylor series about x=0) of
> g(x) := (1-x^2/2)*exp(x);
2
g(x) := (1 - 1/2 x ) exp(x)
a) -1/3 b) 0 c) 1/2 d) 1 e) 3/2
SOLUTION: a) Expanding e^x, we get
g(x) = (1 - x^2/2) (1 + x + x^2/2! + x^3/3! + ...)
The x^3 term of this product equals x^3/3! - x^3/2 = -x^3/3.
Therefore the coefficient of x^3 is -1/3.
10. On which of the following intervals does the series
infinity
-----
\ n
) (x - 4)
/
-----
n = 1
converge?
a) (-1,1) b) (0,8) c) (3,5) d) (-1/2, 1/2) e) (-infinity, infinity)
SOLUTION: c) This is a geometric series, which converges provided
|x-4| < 1 or another words, 3 < x < 5.
11. Suppose that
infinity
-----
\
) a_n
/
-----
n = 1
infinity
-----
\
) b_n
/
-----
n = 1
are series with positive terms such that a_n/b_n --> 0 ans n - -> infinity.
Which statement is true?
a) If
infinity
-----
\
a) If ) b_n converges then
/
-----
n = 1
infinity
-----
\
) a_n converges
/
-----
n = 1
infinity
-----
\
b) If ) b_n diverges, then
/
-----
n = 1
infinity
-----
\
) a_n diverges
/
-----
n = 1
infinity
-----
\
c) If ) a_n converges, then
/
-----
n = 1
infinity
-----
\
) b_n diverges
/
-----
n = 1
infinity
-----
\
d) If ) b_n converges, then
/
-----
n = 1
infinity
-----
\
) a_n diverges.
/
-----
n = 1
infinity
-----
e) If \
) a_n converges, then
/
-----
n = 1
infinity
-----
\
) b_n converges.
/
-----
n = 1
SOLUTION: a) If a_n/b_n --> 0, then eventually, the a_n
are all smaller than the b_n. This means that if the series involving
b_n converges, then the series involving a_n must also converge.
Calculators are permitted. Partial credit may be given.
12. Find the Maclaurin series for 1/(1+x^5). Use this series
to approximate
.1
/
| 66
| ------ dx
| 5
/ 1 + x
0
SOLUTION: The Maclaurin series is
infinity
-----
\ n (5 n)
) (-1) x
/
-----
n = 0
using the formula for a geometric series.
To approximate the integral, the idea is to break up this series into
a sum of the first N terms and a remainder which is a sum from N+1 to infinity.
Then we integrate the first N terms (which is a polynomial). The trick is to find
a convenient value of N so that the integral of the remainder is less than 10^(-14).
Now the remainder is a geometric series whose value is
(5 N + 5)
x
66 ----------
5
1 + x
which is less than 66 x^(5N+5). Integrating this over 0 to 0.1, gives
66 10^(-5N-6)/(5N+6). By trial and error, we see that N=2 gives
(66/16)(10)^(-16) which is less than 10^(-14). Therefore, the approximate
value for the integral is obtained by computing
1/10
/
| 5 10
| 66 - 66 x + 66 x dx
|
/
0
which is 6.59998900006.
13. Find the radius of convergence of the series
infinity
----- (2 n)
\ x
) ------
/ n
----- 2
n = 1
SOLUTION: Write the series as
infinity
-----
\ 2 n
) (1/2 x )
/
-----
n = 1
This is a geometric series which converges when x^2/2 < 1,
or x^2 < 2 which is equivalent to -sqrt(2) < x < sqrt(2).
So the radius of convergence is sqrt(2).
14. Determine whether the series
infinity
-----
\ 1
) ----------
/ 2 n
----- n + (-1)
n = 2
converges. Justify your answer.
SOLUTION: This series converges by comparison with the series
infinity
-----
\ 1
) ----
/ 2
----- n
n = 2
which converges by the integral test.
15. What is the smallest value of n for which we can be sure that
the nth partial sum
n
-----
\ / 5 \
) |----|
/ | 6 |
----- \ k /
k = 1
approximates
infinity
-----
\ / 5 \
) |----|
/ | 6 |
----- \ k /
k = 1
to within 10^(-10)?
HINT: Use the integral test to bound the remainder.
SOLUTION: By the integral test, the remainder
infinity
-----
\ / 5 \
) |----|
/ | 6 |
----- \ k /
k = n + 1
is less than
infinity
/
| 5
| ---- dx
| 6
/ x
n
which is
1
----
5
n
This remainder is within 10^(-10) provided n is at least 10^2 = 100.
16. The point A=(a,0) is moving on the positive x - axis
and the point B = (0,b) is moving on the positive y - axis so
that the distance between A and B is always 5. Let C be the
midpoint of the line segment from A to B.. Find parametric equations
for the curve traced out by C, using b as the parameter.
SOLUTION: The midpoint from A to B is x = a/2 , y = b/2.
We must describe these equations in terms of the parameter b.
From the given information, a^2+b^2 = 25 (Pythagorean Theorem). So
2 1/2
b = (25 - a )
and so the point is
2 1/2
x = 1/2 (25 - a )
y = 1/2 b
17. What is the length of the curve parametrized by
x = t^2/2 - t y = (4/3) t^(3/2)
between t = 0 and t = 2?
SOLUTION: The arclength is
2
/
| //d \2 /d \2\1/2
| ||-- x(t)| + |-- y(t)| | dt
| \\dt / \dt / /
/
0
Inserting x = t^2/2 - t y = (4/3) t^(3/2), the integrand becomes
sqrt( (t-1)^2 +4t) = sqrt(t^2+2t+1) = t+1. Therefore the length is
2
/
|
| 1 + t dt
|
/
0
which is 4.