F13 MA 122 Study Guide/Practice Exam 1
This study guide/practice exam is longer and harder than the actual exam.
You should, on average, be able to answer each question in 5 minutes.
Problem 1: Know the general formula for the nth Taylor approximation
to a differentiable function f based at x = a.
n
P
f (k) (x)
Solution: Pn (x) =
(x − a)k
k!
k=0
Problem 2: Know the formulae for the nth MacLaurin polynomials for
1
.
the functions ln(1 + x), ex , 1−x
Solution:
P
ln(1 + x) : Pn (x) = Pnk=1 (−1)k (1/k)xk
k
ex : Pn (x) = P∞
k=0 x /k!
∞
1
: Pn (x) = k=0 xk
1−x
Problem 3: Find the formula for the 2n + 1st MacLaurin polynomial for
sin(x)
Solution: Let f (x) = sin(x). The first few derivatives of f are
f (x)
f 0 (x)
f 00 (x)
f 000 (x)
f iv (x)
= sin(x)
= cos(x)
= − sin(x)
= − cos(x)
= sin(x)
and the other derivatives repeat in a similar fashion. Thus, f (n) (0) = 0
whenever n is even. Consequently
P2n+1 (x) = x −
x3 x5
(−1)n x2n+1
+
− ... +
.
3!
5!
(2n + 1)!
In summation notation this is:
P2n+1 (x) =
n
X
(−1)i x2i+1
i=0
Problem 4:
cos(x).
(2i + 1)!
.
Find the formula for the 2nth MacLaurin polynomial for
Solution: The work is similar to the previous problem. The ultimate answer
is:
1
2
P2n (x) =
n
X
(−1)k
k=0
x2k
(2k)!
Problem 5: What is the nth Taylor polynomial Pn (x) for the function
f (x) = x12 based at a = 1?
Solution: Here are the first few derivatives of f (x):
f (x) = 1/x2
f 0 (x) = −2/x3
f 00 (x) = 6/x4
In general,
f (n) (x) = (−1)n−1 (n + 1)!/xn+2
So at x = 1, we have:
f (1)
f 0 (1)
f 00 (1)
f (n) (1)
=
1
=
−2
=
6
= (−1)n (n + 1)!
Thus,
Pn (x) = 1 − 2(x − 1) + 3(x − 1)2 − 4(x − 1)3 + . . . + (−1)n (n + 1)(x − 1)n .
Problem 6: What is the 3rd MacLaurin polynomial P3 (x) for the function
f (x) = sin(x2 )?
Solution: You can do this by plugging in x2 for x in the MacLaurin polynomials for sin(x) or you can do it brute-force. Here’s the brute-force method.
Here are the first 3 derivatives:
f (x) =
sin(x2 )
0
f (x) =
2x cos(x2 )
f 00 (x) =
−4x2 sin(x2 ) + 2 cos(x2 )
000
3
f (x) = −8x cos(x2 ) − 8x sin(x2 ) − 4x sin(x2 )
Thus, at x = 0:
f (0)
f 0 (0)
f 00 (0)
f 000 (0)
=
=
=
=
0
0
2
0
Thus,
P3 (x) = x2 .
3
Problem 7: What is a general formula for all the MacLaurin polynomials
of f (x) = sin(x2 )?
Solution: We need only plug x2 into the MacLaurin polynomials for sin(x)
to get:
∞
X
(−1)k 4k+2
2
P2n1 (x ) =
x
(2k + 1)!
k=0
Problem
8: Find the MacLaurin polynomials for the function g(x) =
Rx
2
sin(t ) dt.
0
Solution: Plug in t2 for x in the MacLaurin polynomials for sin(x) and then
integrate from 0 to x to obtain
n
X
k=0
(−1)k
x4k+3
(4k + 3)(2k + 1)!
Problem 9: Suppose that f (x) is a function such that |f 00 (x)| ≤ 3 for all x
values. Find an upper bound on the error of the tangent line approximation
to f (x) at 0. (Your answer will be in terms of x.)
Solution: Taylor’s theorem tells us that
|f (x) − P1 (x)| ≤
3|x|2
.
2
Problem 10: Let f (x) = x3 + x + 1.
(1) Find the tangent line approximation L(x) to f based at x = 1.
(2) Use L(x) to estimate f (1/2).
(3) Use Taylor’s theorem to give an upperbound on the error of the approximation at x = 1/2.
Problem 11: Carefully state the definition of the limit of a sequence (an ).
Solution: The sequence (an ) converges to a number L if, for each > 0,
there exists a number N such that whenever n ≥ N , then
|an − L| < .
Problem 12: Use the definition of limit, to prove that
1
lim √ = 0.
n→∞
n
4
Solution: Let > 0 be given. The following inequalities are all equivalent:
| √1n − 0|
√1
n
1
1
2
< < √
<
n
<n
√
Thus, whenever n > 1/2 , the terms of the sequence (1/ n) are within of 0. This means that the sequence is converging to 0.
r
q
p
√
Problem 13: Let x = 1 + 1 + 1 + · · ·.
(1) Write x as a sequence.
Solution:
√
r
q
q
√
√
1, 1 + 1, 1 + 1 + 1, . . .
. We can write this recursively as a1 =
√
1 and an =
√
1 + an−1 .
(2) Show (in detail) that the sequence is both monotonic and bounded.
Solution: We begin by showing that the sequence is bounded.
The inequality
an ≤ 2
is equivalent to
p
1 + an−1 ≤ 2.
Algebra shows this inequality is equivalent to
an−1 ≤ 3
Repeating these steps shows that is equivalent to
an−2 ≤ 8
We keep going until we see that all these inequalities are equivalent
to the inequality
a1 ≤ M
√
for some large number M (which is at least 2). Since a1 = 1 = 1,
this last inequality is obviously true and so an ≤ 2. Hence the
sequence (an ) is bounded.
5
Now we show that the sequence is montonic by showing that it is
increasing. The inequality an ≥ an−1 is equivalent to the inequalities:
√
√
1 + an−1
≥
1 + an−2
1 + an−1
≥
1 + an−2
an−1
≥
an−2
..
..
.
.
a
≥
a1
p
√2
√
1 + √1 ≥ 1
2
≥
1
Since the last inequality is clearly true, the sequence is increasing.
(3) Explain why the sequence must converge (you may appeal to a theorem).
Solution: The sequence converges by Dedekind’s Theorem since it
is increasing and bounded.
(4) Determine what number the sequence converges to.
√
Solution: Note that x = 1 + x. Algebra gives us the equation
x2 − x − 1. Solving this using the quadratic equation and discarding
the negative result we get
√
1+ 5
x=
2
(This is the golden ratio)
Problem 14: Use Riemann sums and an integral to prove that the series
P
∞ 1
k=1 k diverges. (You may not just appeal to the integral test.)
Solution: Refer to your homework (or the solutions for WHW1), or come
see me. You really need a picture to explain this well.
Problem
15: Suppose that for all n, 0 ≤ an ≤ bn and that the series
P∞
n=0 an diverges.
P By examining the sequences of partial sums, explain
why the series ∞
n=0 must also diverge. (You are being asked to prove part
of the comparison test. You may not appeal to that test in your answer.)
Solution: We will prove that if 0 ≤ ak ≤
P
Pb∞k for all k and if the series
∞
b
converges,
then
so
does
the
series
k=0 k
k=0 ak .
Suppose that 0 ≤ ak ≤ bk for all k. Since each ak is positive, the sequence
of partial sums is an increasing sequence. Each of those partial sums is less
6
Pn
P∞
than the corresponding
partial sum
P
∞ k=0 bk . Since the series k=0 bk conn
verges, the sequence
is bounded. The upper bound for that
k=0 bk
n=0
sequence is also an upper
Pbound for
∞ the sequence of partial sums for the
n
ak . Thus, the sequence
is a monotonic, bounded sequence.
k=0 ak
n=0
Therefore, it converges P
by Dedekind’s Theorem. Since that sequence converges, we say the sum ∞
k=0 ak converges.
P
k
Problem 16: Show that the sequence ∞
k=0 r (for a fixed r) converges
if |r| < 1 and diverges if |r| > 1. (You are being asked to prove part of
the geometric series test. You may not appeal to that test in your answer.
You also may not appeal to the ratio test as that depends logically on the
geometric series test.)
Solution: Let Sn = 1 + x + x2 + . . . + xn . As we found in class, we have:
(1 − x)Sn = 1 − xn+1 .
Thus if x 6= 1:
1
xn+1
Sn =
−
.
1−x 1−x
If |x| < 1, then the sequence (xn+1 ) converges to 0 and so if |x| < 1 the
series
1
1 + x + x2 + . . . =
.
1−x
P
k
If |x| ≥ 1, the terms of the series ∞
k=0 x do not coverge to zero and so the
series diverges.
P
k 2k+1
Problem 17: What does the series ∞
k=0 (−1) 5k converge to?
P
−2 k
Solution: We can rewrite it as 2 ∞
k=0 ( 5 ) . By the geometric series test it
converges to
2
= 10/7
1 + 2/5
Problem 18: For each of the following series, determine if they converge
or diverge. Be sure to provide a thorough explanation for each.
P
−5/4
(1) ∞
i=1 i
(2)
Solution: Converges by the integral test.
P∞ 3
(3)
Solution: Converges by the geometric series test
P∞ 3i
i=0 5i
i=0 5i
7
(4)
Solution: It is a geometric series and it converges.
P∞ 3i
(5)
Solution: Use a comparison test with the previous series.
P∞
3
i=1 5i (i!)
n=1 n2 −5
Solution: Use the integral test to see that
∞
X
n=3
(6)
3
n2 − 5
converges. Or use a comparison test using the fact that n21−5 ≤
1
for n ≥ 3. The comparison test method is really better.
(n2 )/3
P∞
2
k=1 3k +4k
Solution: Use the comparison test and compare to a geometric series.
P
nn
(7) ∞
n=1 n! .
n
Solution: This diverges since ( nn! ) does not converge to 0.
Problem 19: Consider the series:
∞
X
(−1)k xk
k=1
k2
.
Determine all values of x for which the series converges.
Solution: Use the ratio test:
|x|k+1 /(k + 1)2
k2
=
lim
|x|
= |x|.
k→∞
|x|k /k 2
(k + 1)2
lim
The series converges when −1 < x < 1 and diverges for x < −1 and
x > 1. At x = −1 it converges by the integral test and at x = 1 it converges
by the alternating series test.
Problem 21: Let Pn (x) be the MacLaurin polynomials of f (x) = ex .
Show that for every real number x, the sequence (Pn (x)) converges.
Solution: The sequence (Pn (x)) converges if and only if the series P (x) =
P∞ xk
k=0 k! converges, since the MacLaurin polynomials are simply the partial
sums of the series P (x).
The series P (x) converges by the ratio test, as we now show.
8
|x|k+1 /(k + 1)!
|x|
= lim
=0
k
k→∞
k→∞ (k + 1)
|x| k!
Since this is less than one no matter what x is, the ratio test says the series
converges no matter what x is.
lim