Download Undergraduate Level

Undergraduate Level
Calculus I
Suppose we have a function f which is both continuous and differentiable. We also know that
f (6) = −2 and f 0 (x) ≤ 10. What is the largest possible value for f (15)?
Solution. Since f is continuous and differentiable, by the Mean Value Theorem there exists c ∈
(6, 15) such that
f (15) − f (6) = f 0 (c)(15 − 6)
Substituting what we know and rearranging we arrive at
f (15) = 9f 0 (c) − 2
Since f 0 (x) ≤ 10 we know that f 0 (c) ≤ 10. To maximize the equation above we need to choose f 0 (c)
as large as possible, i.e., we will set f 0 (c) = 10 to obtain
max f (15) = 9(10) − 2
⇒
max f (15) = 88.
Calculus
Evaluate
4 · lim
n→∞
n
X
j=1
q
¡ ¢2
1 − nj
n
Solution. Since this sum has an upper limit that tends to infinity we suspect that this problem can
be phrased as an integral. If we choose a uniform partition of length 1/n, then
s
µ ¶2 µ ¶ Z 1 p
n
X
j
1
lim
1−
·
=
1 − x2 dx
n→∞
n
n
0
j=1
The integral is simply the area of the unit circle restricted to the first quadrant, which is π/4. So
four times the sum will be π.
Linear Algebra
Prove that every monic polynomial over the reals is the characteristic polynomial of some matrix
over the reals.
Solution. Consider any monic polynomial
p(x) = xn + an−1 xn−1 + · · · + a1 x + a0
We can form the following n × n matrix

0
 1

 0


A :=  0



 0
0
0
0
1
0
..
.
0
0
0
1
···
···
···
···
0 0
0 0
0 0
0 0
..
.
0
0
0
0
···
···
1
0
0
1
−a0
−a1
−a2
−a3
..
.
−an−2
−an−1











You can quickly verify that p(x) = det(xI − A).
Algebra
(a) Let G be a finite group. Prove that if the set of elements of g ∈ G with the property that g 2 = 1
form a subgroup, then that subgroup is a normal Abelian subgroup of G.
Solution. Let H = {g ∈ G : g 2 = 1}. First notice that every element in H is its own inverse. Now
consider any two elements h1 , h2 ∈ H. Since H is a subgroup it follows that the product h1 h2 ∈ H.
By definition of H we get
1 = (h1 h2 )2 = h1 h2 h1 h2
−1
Multiplying first by h−1
2 then h1 on the right yields
−1
h−1
2 h1 = h1 h2
−1
Using the fact that h−1
1 = h1 and h2 = h2 we arrive at
h2 h1 = h1 h2
Thus H is an Abelian subgroup of G. To show that H is normal in G consider the product ghg −1
where g ∈ G and h ∈ H. Observe that
(ghg −1 )2 = (ghg −1 ghg −1 ) = gh(g −1 g)hg −1 = ghhg −1 = gg −1 = 1 ∈ H
(b) Let R be a ring with 1 ∈ R. Suppose that r ∈ R has the property that rn = 0 for some
positive integer n. Prove that u = 1 + r + r2 + · · · + r2004 is invertible in R.
Solution. First suppose that n − 1 ≤ 2004. Then
u = 1 + r + r2 + · · · + r2004 = 1 + r + r2 + · · · + rn−1
Multiplying u by 1 − r we find that
(1 + r + r2 + · · · + rn−1 )(1 − r) = 1 + r + r2 + · · · + rn−1
− r − r2 − · · · − rn−1 − rn = 1 − rn = 1.
Thus 1 − r is the multiplicative inverse of u. Now suppose that n > 2005. Set d = blog2
n
) and define
is, d is the largest integer ≤ log2 2005
v = (1 − r)(1 + r2005 )(1 + r4010 ) · · · (1 + r2
d
·2005
n
2005 c
(that
)
Then
uv = (1 + r + r2 + · · · + r2004 )(1 − r)(1 + r2005 )(1 + r4010 ) · · · (1 + r2
= (1 − r2005 )(1 + r2005 )(1 + r4010 ) · · · (1 + r2
= (1 − r
4010
= (1 − r
2d ·2005
= 1 − r2
But
)(1 + r
(d+1)
4010
)(1 + r
) · · · (1 + r
2d ·2005
2d ·2005
d
·2005
d
·2005
)
)
)
)
·2005
³ n ´
2005
Therefore uv = 1, and so u is invertible.
2 · 2d ≥
⇒
2(d+1) · 2005 ≥ n.
Analysis
Prove that supn∈Z+ cos(n) = 1 using the fact that for any irrational number α and any ε > 0 there
exist nonzero integers a and b such that | aα − b | < ε.
Solution. We know that cos(x) = 1 if and only if x = 2π · k for some k ∈ Z. It follows from the
irrationality of 2π that cos(n) 6= 1 for any n ∈ Z+ .
However, since 2π is irrational we can always find integers a and b such that | a · 2π − b | < ε for
any given ε > 0. Using the fact that cos(−x) = cos(x) and the difference formula for cosine we find
that
cos(|2aπ − b|) = cos(2aπ − b) = cos(2aπ) cos(−b) − sin(2aπ) sin(−b) = cos(−b)
= cos(b).
Since cosine is increasing on (− π2 , 0) and decreasing on (0, π2 ), it follows that
cos(b) = cos(|2aπ − b|) ≥ cos(ε)
for 0 < ε < π/2. The result follows by letting ε tend to zero.
Probability
(a) Let m and n be positive integers. Show that
(m + n)!
m! n!
< m n
m+n
(m + n)
m n
Solution. Let X, Y be independent with
X ∼ Poisson(m) and
Then
P (X = m) =
mm e−m
m!
Y ∼ Poisson(n)
and P (Y = n) =
nn e−n
n!
But X + Y ∼Poisson with mean (m + n). So
P (X + Y = m + n) =
(m + n)m+n e−(m+n)
(m + n) !
We know that P (X + Y = m + n) ≥ P (X = m, Y = n) and that
P (X = m, Y = n) ≥
Thus
mm e−m nn e−n
·
m!
n!
(m + n)m+n e−(m+n)
mm e−m nn e−n
≥
·
(m + n) !
m!
n!
The result follows by routine simplification.
(b) “Let’s Make a Deal” TV show problem with Monty Hall.
There are three doors. Behind one of them is prize of a car. Behind the other two there is
nothing. A contestant chooses a door. Monty always picks one of the other doors, opens it, and
always there is nothing behind the door Monty opened (since Monty knows where the car is). The
contestant is given an opportunity to change his/her choice to the other remaining door. Should the
contestant switch or not? What is the probability of the contestant winning the car, if the contestant
picks the optimal strategy?
Solution. The contestant has a probability of 1/3 of choosing the door with the car initially. Since
Monty can always open an empty door, the probability of the contestant being correct after the door
is opened is still 1/3. Thus the probability that the car is behind the last door is 1 − 1/3 = 2/3.
Hence the contestant should switch doors since the car is behind the third door with probability 2/3
(which is twice the probability that it is behind the door the contestant chose originally). If you
don’t believe it, make up a simulation experiment and test it out!
Graduate Level
(a) What is the largest number which divides every number of the form p4 − 1 for primes p > 5?
Solution. Since p is odd and
(p4 − 1) = (p2 − 1)(p2 + 1) = (p − 1)(p + 1)(p2 + 1)
it follows that 2 divides each of (p − 1), (p + 1) and (p2 + 1). However, 4 must divide one of (p − 1)
and (p+1) since they are consecutive even numbers. Together these conditions imply that 16 divides
p4 − 1.
Because p is a prime number greater than 3, it follows that either p − 1 or p + 1 is divisible by 3.
Using the factorization of p4 − 1 above, it follows that 3 divides p4 − 1. By Fermat’s Little Theorem,
5 divides p4 − 1.
We claim that 16 · 3 · 5 = 240 is the largest number that divides all primes of the given form.
Observe that (74 − 1)/240 = 10 and (114 − 1)/240 = 61. Our claim follows from the fact that the
first two number of the form p4 − 1 where p > 5 share no common prime factors aside from those
appearing in the prime factorization of 240.
(b) Show that
n−1
Y
k=0
is divisible by n! for each positive integer n.
(2n − 2k )
Solution. Let F = GF (2), i.e., the Galois field of 2 elements, and define the vector space V = Fn
as the canonical n-dimensional vector space over F. Thus a typical element of V is an n-tuple
v = (a1 , a2 , . . . , an )
Qn
where each ai ∈ {0, 1}. Then, k=0 (2n − 2k ) is the number of ordered bases of V . If we let M stand
for the number of unordered bases of V we obtain the relationship
n=1
Y
(2n − 2k ) = M · n!
k=0
which proves our result.
(c) Determine
Ã
lim
n→∞
n
Y
4k + 4n − 3
4n
! 4n+4k−3
2
4n
k=1
Solution. Taking the natural logarithm of the product we obtain
¶
4k + 4n − 3
1
·
4n
4n
n
k=1
¶ µ
¶
n µ
X
4k − 3
1
4k − 3
=
ln 1 +
· .
1+
4n
4n
n
Sn =
¶
n µ
X
4k + 4n − 3
µ
ln
k=1
Observe that
k−1
4k − 3
k
<
<
for all k = 1, 2, . . . , n.
n
4n
n
Consequently, Sn is a Riemann sum for f (x) = (1 + x) ln(1 + x) over the interval [0, 1] using the
partition (0, n1 , . . . , n−1
n , 1). Since f (x) is continuous over the region of integration it follows that
Z
1
lim Sn =
n→∞
0
3
(1 + x) ln(1 + x) dx = 2 ln 2 − .
4
Accounting for the natural log we find that,
Ã
lim
n→∞
n
Y
4k + 4n − 3
4n
k=1
! 4n+4k−3
2
4n
3
3
= e2 ln 2− 4 = 4e− 4 .