MTH 312 Week 11

Alec Bonnell Week 11

1 Exercise 1

th Let SHn denote the n Schroder-Hipparchus√ number. SHn+1 Prove that limn→∞ = 3 + 8. SHn

1.1 Proof √ The number 3 + 8 seems to be the solution to a√ quadratic equation. We can ﬁnd this equation by setting the value of t = 3 + 8. √ √ t = 3 + 8 → t − 3 = 8 → (t − 3)2 = 8 → t2 − 6t + 1 = 0

1 We can then see that t = 6 − t .

The recurrence relation to ﬁnd Schroder-Hipparchus numbers is the following with SH1 = 1 and SH2 = 1: 1 SH = ((6n − 9)SH − (n − 3)SH ) n n n−1 n−2

SHn+1 Using the ratio Rn = , we can ﬁnd that this ratio is Sn

6(n + 1) − 9 (n − 2) SHn−1 Rn = − n + 1 n + 1 SHn 6n − 3 n − 2 1 = − n + 1 n + 1 Rn−1

1 This looks to be tending towards Rn = 6 − as n → ∞. The equation Rn−1 1 looks very similar to t = 6 − t . We can subtract t from Rn to ﬁnd if these are equivalent. 6n − 3 n − 2 1 1 Rn − t = − − (6 − ) n + 1 n + 1 Rn−1 t 6n − 3 n − 2 1 1 = − 6 − + n + 1 n + 1 Rn−1 t

1 There are two main terms we should look at in this equation. When we take the limit, we can see for the ﬁrst term 6n − 3 lim − 6 = 0 n→∞ n + 1 For the second term n − 2 1 1 1 1 lim − + = lim − n→∞ n + 1 Rn−1 t n→∞ t Rn−1

1 Thus, if Rn → R, then R = 6 − R = t.

2 Exercise 2

th Let Mn denote the n Motzkin number, satisfying the recurrence: M1 = 1, M2 = 2 2n + 1 3n − 3 M = M + M n n + 2 n−1 n + 2 n−2

Mn+1 Prove that limn→∞ = 3. Mn

2.1 Proof

Mn+1 The ratio will be denoted as: = Rn. Mn

2(n + 1) + 1 3(n + 1) − 3 Mn−1 Rn = + (n + 1) + 2 (n + 1) + 2 Mn 2n + 2 3n − 2 1 = + n + 3 n + 3 Rn−1

3 3 It seems as though when n → ∞, then Rn = 2+ . We can set t = 3 = 2+ . Rn−1 t 2n + 2 3n − 2 1 3 Rn − t = + − (2 + ) n + 3 n + 3 Rn−1 t 2n + 2 3n − 2 1 3 = − 2 + − n + 3 n + 3 Rn−1 t Similar to example 1, 2n + 2 lim − 2 = 0 n→∞ n + 3 and 3n − 2 1 3 3 3 lim − = lim − n→∞ n + 3 Rn−1 t n→ Rn−1 t 3 Thus, if Rn → R, then R = 2 + R = t.

2 3 Exercise 3

The sequence a(n) is deﬁned with a(1) = 1, a(2) = 1, and

a(n) = a(a(n − 1)) + a(n − a(n − 1))

a(n) 1 The sequence has the property limn→∞ n = 2

3.1 Graph

a(n) First, we can see what the graph of n looks like:

As we can see, the graph looks to be hopping on on the value of 1/2. One way to see if the limit approaches 1/2 is to take the average of the values.

3.2 Proof

This is the proof that if a sequence xn converges to a limit x, then the sequence of meansx ¯n deﬁned by x + x + ... + x x¯ = 1 2 n n n also converges to x. Another way of saying this is

lim xn = lim x¯n = x n→∞ n→∞

To prove this, we can ﬁrst assume that xn is a convergent sequence and converges to x. This means for all > 0, there exists a number N such that |xn − x| < if n ≥ N.

3 We need to show that |x¯n − x| < .

x1 + x2 + ... + xn |x¯n − x| = − x n x x x = 1 + 2 + ... + n − x n n n x x x x x x = 1 + 2 + ... + n − + + ... + n n n n n n

x1 − x x2 − x xn − x = + + ... + n n n |x − x| |x − x| |x − x| ≤ 1 + 2 + ... + n n n n <

3.3 Converse

The converse of the statement is: if a sequence of meansx ¯n converges to a limit x, then the sequence xn also converges to x.

n A counter-example to this statement is the sequence xn = (−1) .

The function jumps from -1 to 1 and back forever. This oscillating sequence does not converge to a value. However, its sequence of means does converge to 0. Therefore, the statement must be false due to this counter-example.

3.4 Means of a(n) We now know that if a(n) converges to 1/2, thena ¯(n) will also converge to 1/2. The given deﬁnition states that a(n) does converge to 1/2, so let’s see what the graph ofa ¯(n) looks like

4 This graph converges much more quickly to 1/2 than a(n). This means that if it is diﬃcult to tell where a sequence is converging to, it may be useful to look at the graph of its means.