Ex03 - Cs.ioc.ee

Concrete Mathematics, Lesson 3: Exercises
Exercise 2.3
Silvio Capobianco
Revised: October 3, 2012
Demonstrate your understanding of σ-notation by writing out the sums
ak and
ak2 .
0≤k≤5
0≤k 2 ≤5
in full. (Watch out—the second sum is a bit tricky.)
Solution. Recall that
P (k) af(k) is the sum of precisely those values aj such
that j = f(k) and the property P is satisfied by the index k. Then the first
sum is clearly a0 + a1 + a2 + a3 + a4 + a5.
And what is the second sum? It is the sum of ak2 for k satisfying 0 ≤
k2 ≤ 5. But k is an integer value, and there are five integers whose square
is between 0 and 5: that is, 0, 1, 2, −1, and −2. For f(x) = x2 the second
sum is thus a (−2) 2 + a (−1) 2 + a02 + a12 + a22 = a0 + 2a1 + 2a4.
Exercise 2.13 (as we did it)
Find a closed form for n
k=0 (−1)k · k 2 .
Solution. The solution can be found by considering the cases n even and n
odd separately. If n = 2m + 1 is odd, then
n
(−1) k · k 2 =
k=0
=
m
((2j) 2 − (2j + 1) 2 )
j=0
m
(−4j − 1) :
j=0
1

this is the opposite of the sum of the first m + 1 terms of an arithmetic
sequence with initial value a = 1 and step b = 4, therefore
n
(−1) k · k 2
m(m + 1)
= − (m + 1) · 1 + · 4
2
k=0
If n is even, then n + 1 is odd and
n
(−1) k · k 2 =
k=0
= −(m + 1) · (2m + 1)
(2m + 1)(2m + 2)
= −
2
n(n + 1)
= − .
2
n+1
(−1) k · k 2 + (n + 1) 2
k=0
(n + 1)(n + 2)
= − + (n + 1)
2
2
−n − 2 + 2n + 2
= (n + 1) ·
2
= (n + 1) · n
2 .
In conclusion, n
k=0 (−1)k · k 2 = (−1) n n(n + 1)/2 = (−1) n n
k=0 k.
Exercise 2.13 (as it was supposed to be done)
Use the repertoire method to find a closed form for n
k=0 (−1)k k 2 .
Solution. The function g(n) = n
k=0 (−1)k k 2 is a special solution of the
recurrence equation
R0 = α ,
Rn = Rn−1 + (−1) n (β + γn + δn 2 ) for n ≥ 1
for the special values α = β = γ = 0, δ = 1: the recurrence equation has the
form Rn = Φ(Rn−1) + Ψ(n; β, γ, δ) with Φ(g) = g linear and Ψ(n; β, γ, δ) =
(−1) n (β + γn + δn 2 ) linear in β, γ and δ. As we know that we can express
Rn = A(n)α + B(n)β + C(n)γ + D(n)δ for special functions A(n), B(n),
2

C(n) and D(n), if we manage to find D(n) in closed form, then that will be
the closed form of g(n).
Let us use the repertoire method. First of all, for α = 1, β = γ = δ = 0
we find A(n) = 1 for every n ≥ 0. The next step should not be to put
Rn = 1 for every n ≥ 0, as we already know that this is associate to the
special values α = 1, β = γ = δ = 0. Instead, we put Rn = (−1) n , which
corresponds to α = 1, β = 2, γ = δ = 0 and yields A(n) + 2B(n) = (−1) n :
as we know that A(n) = 1 for every n ≥ 0, this means 2B(n) = (−1) n − 1
and thus B(n) = ((−1) n − 1)/2 = −[n is odd].
The third step will be to put Rn = (−1) n · n. This corresponds to α = 0
and the recurrence equation
(−1) n n = (−1) n−1 (n − 1) + (−1) n β + (−1) n γn
= (−1) n−1 n − (−1) n−1 + (−1) n β + (−1) n γn + (−1) n δn 2 ,
which is satisfied for every n ≥ 1 if and only if δ = 0, β = −1, and γ = 2.
We thus get the equation −B(n) + 2C(n) = (−1) n n.
The fourth step will be to put Rn = (−1) n n 2 . This corresponds to α = 0
and the recurrence equation
(−1) n n 2 = (−1) n−1 (n − 1) 2 + (−1) n (β + γn + δn 2 )
= (−1) n−1 (n 2 − 2n + 1) + (−1) n (β + γn + δn 2 )
= ((−1) n−1 + (−1) n β)
+((−1) n−1 · (−2) + (−1) n γ)n
+((−1) n−1 + (−1) n δ)n 2 ,
which is satisfied for every n ≥ 1 if and only if β = 1, γ = −2, and δ = 2.
We thus get B(n) − 2C(n) + D(n) = (−1) n n 2 .
At this point, we have a full system of equations:
A(n) = 1
A(n) +2B(n) = (−1) n
−B(n) +2C(n) = (−1) n n
B(n) −2C(n) +2D(n) = (−1) n n 2
from which we want to find D(n). But by adding together the third and
fourth equation we immediately find 2D(n) = (−1) n · (n + n 2 ). Then g(n) =
D(n) = (−1) n (n 2 + n)/2 = (−1) n Sn.
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