▪Μυστηριώδης ακολουθία

Μια ακολουθία θετικών πραγματικών αριθμών ορίζεται ως εξής:
▪ $a_0 = 1$,
▪ $a_{n+2} = 2a_n − a_{n+1}$, για $n = 0, 1, 2, ...$.
Να βρεθεί ο όρος $a_{2005}$.

ΑΠΑΝΤΗΣΗ:

(του μαθηματικού Νick Hobson)

It may at first seem that the sequence is not uniquely defined!  However, the constraint that the sequence consists of positive numbers allows us to deduce the value of a₁.  We will show that, if a₁  1, the sequence will eventually contain a negative number.

Letting a₁ = x, we find

a0 = 1 + 0x,	a1 = 0 + x    x > 0,
a2 = 2 − x    x < 2,	a3 = −2 + 3x    x > 2/3, (1)
a4 = 6 − 5x    x < 6/5,	a5 = −10 + 11x    x > 10/11, ...

It seems clear that, as we calculate more and more terms, x will be "squeezed" between two fractions, both of which are part of a sequence which tends to 1 as n tends to infinity.  (It would follow that x = 1.)  We verify this intuition below.

Setting x = 0, to isolate the constant terms in (1), we obtain the sequence {b_n}: 1, 0, 2, −2, 6, −10, ...
We conjecture that, from b₂ = 2, b₃ = −2 onwards, the sequence alternates in sign, with |b_n+2| > |b_n|.
We prove this conjecture by mathematical induction.

Consider b_2n = r, b_2n+1 = −s, where n, r, s are positive integers.
If n = 1, r = s = 2, which alternates in sign, as per the inductive hypothesis.
If n = k, b_2k+2 = b_2(k+1) = 2r + s > r > 0, and b_2k+3 = b_2(k+1)+1 = −(2r + 3s) < 0, so that |−(2r + 3s)| > s.
That is, b_2k+2 > b_2k > 0, and b_2k+3 < b_2k+1 < 0.
The result follows by induction; sequence {b_n} alternates in sign, with |b_n+2| > |b_n|.

Setting x = 1, we know from the recurrence relation that a_n = 1, for all n  0.
Therefore, the absolute value of the coefficient of x in sequence (1) must always differ by 1 from the absolute value of the constant term.

More specifically, we have
a_2n = b_2n − (b_2n − 1)x    x < b_2n/(b_2n − 1), and
a_2n+1 = b_2n+1 − (b_2n+1 − 1)x    x > b_2n+1/(b_2n+1 − 1).  (Note: b_2n+1 < 0.)

Since |b_2n| and |b_2n+1| are strictly increasing with n, the limit as n tends to infinity of both {b_2n/(b_2n − 1)} and {b_2n+1/(b_2n+1 − 1)} is 1.
Hence x = 1.
(For any x  1, there exists n such that x > b_2n/(b_2n − 1) or x < b_2n+1/(b_2n+1 − 1), and hence a_2n < 0 or a_2n+1 < 0.)

Therefore a_n = 1, for all n  0.  Specifically, a₂₀₀₅ = 1

The above proof is from first principles.  We can also use the theory of recurrence relations.

This recurrence relation may be written as a_n+2 + a_n+1 − 2a_n = 0.  It is linear and homogenous, and has characteristic equation
m² + m − 2 = (m − 1)(m + 2) = 0, with roots m = 1, m = −2.
Hence the general solution is a_n = A·1ⁿ + B·(−2)ⁿ, where A, B are constants, to be determined.

Then a_n > 0, for all n  0    B = 0.
And a₀ = 1    A = 1.

Therefore the solution to the recurrence relation is a_n = 1, for all n  0.