### HOMEWORK 1 SOLUTIONS Problem 1 Let F(n) be

```HOMEWORK 1 SOLUTIONS
Problem 1
Let F (n) be ”all the polynomials p(x) with degree less than n can be written as
(x − b)q(x) + r where degree q(x) is one less than p(x)”
F (1): a degree zero polynomial is just an integer which has the desired form with
q(x) = 0
F (n) → F (n + 1): without loss of generality we can assume p(x) is a degree n
polynomial.
p(x) = an xn + an−1 xn−1 + ... + a0
= an xn − an bxn−1 + an bxn−1 + an−1 xn−1 + ... + a0
= an xn−1 (x − b) + (an bxn−1 + an−1 xn−1 + ... + a0 )
by using F (n) for the second term we get:
p(x) = an xn−1 (x − b) + (an bxn−1 + an−1 xn−1 + ... + a0 )
= an xn−1 (x − b) + (x − b)q0 (x) + r
= (x − b)(an xn−1 + q0 (x)) + r
for some q0 (x). So we have proved F (n + 1) with q(x) = an xn−1 + q0 (x).
1. Problem 2
Let Sk (n) be the sum of the first n k-th powers. S4 (n) is the size of the set
A := {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, k, l ≤ m, m ≤ n}
We introduce symmetry by defining sets
B := {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, k, m ≤ l, l ≤ n}
C := {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, l, m ≤ k, k ≤ n}
D := {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, k, l, m ≤ j, j ≤ n}
E := {(i, j, k, l, m) : i, j, k, l, m ∈ N, j, k, l, m ≤ i, i ≤ n}
By the principle of inclusion-exclusion,
|A ∪ B ∪ C ∪ D ∪ E|
5
5
5
5
5
=
|A| −
|A ∩ B| +
|A ∩ B ∩ C| −
|A ∩ B ∩ C ∩ D| +
|A ∩ B ∩ C ∩ D ∩ E|
1
2
3
4
5
Date: October 11, 2013.
1
2
HOMEWORK 1 SOLUTIONS
Note that
A ∪ B ∪ C ∪ D ∪ E = {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, k, l, m ≤ n}
⇒|A ∪ B ∪ C ∪ D ∪ E| = n5
A ∩ B = {(i, j, k, l, m) : i, j, k, l, m ∈ N, i, j, k ≤ l, l ≤ n, l = m}
⇒|A ∩ B| = S3 (n)
and similarly, |A ∩ B ∩ C| = S2 (n), |A ∩ B ∩ C ∩ D| = S1 (n), |A ∩ B ∩ C ∩ D ∩ E| =
S0 (n) = n. So
n5 = 5S4 (n) − 10S3 (n) + 10S2 (n) − 5S1 (n) + n
which gives us after rearranging, the desired formula
1 5
(n + 10S3 (n) − 10S2 (n) + 5S1 (n) − n)
5
1
1
1
1
1
= n5 + n2 (n + 1)2 − n(n + 1)(2n + 1) + n(n + 1) − n
5
2
3
2
5
1 5 1 4 1 3
1
= n + n + n − n
5
2
3
30
S4 (n) =
Let us now prove the above formula by induction. We define the polynomial
1
1
function p(n) = 15 n5 + 12 n4 + 13 n3 − 30
n. For n = 1, p(1) = 15 + 12 + 13 − 30
=1=
5
1 = S4 (1), so the formula holds for n = 1.
Now suppose that S4 (k) = p(k) for some k ≥ 1. Then
S4 (k + 1) = S4 (k) + (k + 1)4
1
1
1
1
= k 5 + k 4 + k 3 − k + (k + 1)4
5
2
3
30
1
1
1
1
= k 5 + k 4 + k 3 − k + k 4 + 4k 3 + 6k 2 + 4k + 1
5
2
3
30
1
3
13
119
= k 5 + k 4 + k 3 + 6k 2 +
k+1
5
2
3
30
and
1
1
1
1
(k + 1)5 + (k + 1)4 + (k + 1)3 − (k + 1)
5
2
3
30
1 5
1
4
3
2
= (k + 5k + 10k + 10k + 5k + 1) + (k 4 + 4k 3 + 6k 2 + 4k + 1)
5
2
1 3
1
2
+ (k + 3k + 3k + 1) − (k + 1)
3
30
1 5 3 4 13 3
119
2
= k + k + k + 6k +
k+1
5
2
3
30
p(k + 1) =
so indeed we have S4 (k + 1) = p(k + 1). By mathematical induction, the formula
S4 (n) =
holds for all n ∈ N.
1 5 1 4 1 3
1
n + n + n − n
5
2
3
30
HOMEWORK 1 SOLUTIONS
3
Problem 3
Let Sk (n) denote the sum of the First n-th powers as in the notes for lecture
1. Prove by induction that Sk (n) is a polynomial (whose coefficients are rational
numbers) of degree k + 1 in n. (Hint: You should prove this by induction on k. You
should use as your induction hypothesis that Sj (n) is a polynomial of degree j + 1
for all j smaller than k. [This is sometimes called strong induction.] The last page
of the notes for lecture 1 give you a good guess for what the leading term of Sk (n)
should be. Express the rest of it as a combination of Sj (n)’s for smaller j.)
Let’s prove the statement of the problem for all integers k ≥ 0 by induction in
k.
To check the base case for k = 0, we have to check, that S0 (n) is a polynomial
in n of degree 1. Indeed,
S0 (n) = 10 + 20 + · · · + n0 = 1 + 1 + · · · + 1 = n.
To do the induction step we start with the guess “the highest term in Sk (n)
is nk+1 /(k + 1)” and define Rk (n) to be Sk (n) − nk+1 /(k + 1). It is enough to
prove, that Rk (n) is a polynomial in n of degree not greater than k, because then
Sk (n) = nk+1 /(k + 1) + Rk (n) is a polynomial of degree k + 1. Note, that from the
definition of Sk (n) we have S(0) = 0 and for n ≥ 1 we have Sk (n) = Sk (n − 1) + nk .
Using this we compute
Rk (0) = Sk (0) − 0k+1 /(k + 1) = 0
and
Rk (n)−Rk (n−1) = Sk (n)−
(n − 1)k+1
(n − 1)k+1 − nk+1
nk+1
−Sk (n−1)+
= nk +
= pk (n).
k+1
k+1
k+1
Here pk (n) is a polynomial of degree less or equal than k + 1, so it can be written
as
pk (n) = nk +
(n − 1)k+1 − nk+1
= ak+1 nk+1 + ak nk + ak−1 nk−1 + · · · + a0 .
k+1
Note, that we can compute
(1)
ak+1 = 0 +
1−1
= 0,
k+1
ak = 1 +
−(k + 1) + 0
= 0.
k+1
Therefore pk (n) is a polynomial of degree less or equal than k − 1, so
pk (n) = ak−1 nk−1 + ak−2 nk−2 + · · · + a0 .
Thus Rk satisfies the following relations:
(2)
Rk (0) = 0,
Rk (n) = Rk (n − 1) + ak−1 nk−1 + ak−2 nk−2 + · · · + a0 .
We claim, that this means, that
(3)
Rk (n) = ak−1 Sk−1 (n) + ak−2 Sk−2 (n) + · · · + a0 S0 (n),
so Rk (n) is indeed a polynomial in n of degree less or equal than k, because by
induction hypothesis Sj (n) for j < k is a polynomial in n of degree j + 1. We can
check equation (3), for example, by induction in n (yes, this is an induction inside
4
HOMEWORK 1 SOLUTIONS
the induction step of another induction). We have Rk (0) = 0 = ak−1 0 + ak−2 0 +
· · · + a0 0 for the base case n = 0. For induction step we have
Rk (n) = Rk (n − 1) + ak−1 nk−1 + ak−2 nk−2 + · · · + a0 =
ak−1 Sk−1 (n−1)+ak−2 Sk−2 (n−1)+· · ·+a0 S0 (n−1)+ak−1 nk−1 +ak−2 nk−2 +· · ·+a0 =
ak−1 (Sk−1 (n − 1) + nk−1 ) + ak−2 (Sk−2 (n − 1) + nk−2 ) + · · · + a0 (S0 (n − 1) + 1) =
ak−1 Sk−1 (n) + ak−2 Sk−2 (n) + · · · + a0 S0 (n)
as desired.
Alternate solution to Homework 1 Problem 3.
Our goal is to show Sk (n) is a polynomial in n by induction on k. The base case
.
k = 1 is true since we know that S1 (n) = n(n+1)
2
Now we get to assume S1 (n), S2 (n), . . . , Sk−1 (n) are polynomials. We have a
k+1
guess that the leading term of Sk (n) is nk+1 . Thus we write a sum which equals
the guess by telescoping.
n
X j k+1
nk+1
(j − 1)k+1
=
−
k + 1 j=1 k + 1
k+1
n
k+1
k+1
X
k
k−1
l−1
2
l
=
j −
j
+ · · · + (−1)
j k−l+1 + . . .
k
+
1
k
+
1
j=1
k+1
k+1
2
Sk−1 (n) + · · · + (−1)l−1 l Sk−l+1 (n) + . . .
k+1
k+1
Now since every term in the equality between the first and fourth lines is known
to be a polynomial except for Sk (n), we determine that Sk (n) is a polynomial.
= Sk (n) −
Problem 4
Sx,y = {p − q | p < x, q > y, p ∈ Q, q ∈ Q}.
So, ∀r ∈ Sx,y , r = p − q with p < x, q > y, p ∈ Q, q ∈ Q.
=⇒ r = p − q < x − q < x − y.
So, x − y is an upper bound of Sx,y and therefore Sx,y is bounded above.
Take any real number > 0. ∃N ∈ N such that N > 2 . (why?)
=⇒ N 2 > 1.
=⇒ N (x − (x − 2 )) > 1 and N ((y + 2 ) − y) > 1.
=⇒ N x − N (x − 2 ) > 1 and N (y + 2 ) − N y > 1.
=⇒ ∃P, M ∈ N such that N x > M > N (x − 2 ) and N (y + 2 ) > P > N y. (why?)
P
=⇒ x > M
N > x − 2 and y + 2 > N > y.
M
P
M
P
P
=⇒ N − N ∈ Sx,y and N − N > (x − 2 ) − N
> (x − 2 ) − (y + 2 ) = x − y − .
Therefore, x − y is the least upper bound of Sx,y . (Make sure you understand
why this follows from the previous sentence)
[Something to think about: We have been using the preposition “the” before
“least upper bound”, but is that justified? Can a set of real numbers have more
than one least upper bound?]
HOMEWORK 1 SOLUTIONS
5
Problem 5
There are many ways to construct examples with the desired properties, and
here is one of them.
Take x1 = 0.111 · · · 1333 · · · 3 (n 1’s follwed by m−n 3’s) and y1 = 0.111 · · · 1666 · · · 6
(n 1’s follwed by m − n 6’s).
Now add an extra digit 5 at the end of the decimal expansion for x1 and y1 to
get x2 and y2 , respectively.
So x2 = 0.111 · · · 1333 · · · 35, y2 = 0.111 · · · 1666 · · · 65.
It is clear tm (x1 ) = x1 = tm (x2 ) and tm (y1 ) = y1 = tm (y2 )
On the other hand, we compute
x1 + y1 = 0.222 · · · 22999 · · · 9 (n 2’s follwed by m − n 9’s)
x2 + y2 = 0.222 · · · 23 (n − 1 2’s follwed by a 3).
Therefore we have tn (x1 +y1 ) = 0.222 · · · 22 (n 2’s) and tn (x2 +y2 ) = 0.222 · · · 23
(n − 1 2’s follwed by a 3).
Hence x1 , x2 , y1 and y2 have the desired properties.
```