Last Update: 22 November 2010
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Index to all lecture notes
…Previous lecture
Let a_{0} = C_{0} and a_{1} = C_{1} be
the initial conditions for a recurrence relation;
i.e., the first 2
terms of a sequence.
Let c_{1}, c_{2} ∈ R be such that a_{n} = c_{1}a_{n–1} + c_{2}a_{n–2} is the recurrence relation.
Let r_{1} ≠ r_{2} be 2 distinct roots of the "characteristic equation"
of the recurrence relation.
Then:
–b – √(b² – 4ac)
r_{2} = ______________
2a
where ar² + br + c = 0
i.e.)
∴
–c_{1} –
√(c_{1}²
– 4c_{2})
r_{2} = ______________
2
∴ a_{1} = (a_{0} –
α_{2})r_{1} + α_{2}r_{2}
(from 3(a)(ii))
= a_{0}r_{1} – α_{2}r_{1} +
α_{2}r_{2}
= a_{0}r_{1} + α_{2}(r_{2} –
r_{1})
∴ α_{2}(r_{2} – r_{1}) = a_{1} – a_{0}r_{1}
∴ α_{2} = (a_{1} – a_{0}r_{1}) / (r_{2} – r_{1})
∴ α_{1} = a_{0} – α_{2}
= a_{0} – (a_{1} –
a_{0}r_{1}) / (r_{2} –
r_{1})
= (a_{0}r_{2} – a_{0}r_{1} +
a_{0}r_{1} – a_{1}) / (r_{2}
–
r_{1})
= (a_{0}r_{2} –
a_{1}) / (r_{2} – r_{1})
f_{n} = f_{n–1} + f_{n–2}
0,1,1,2,3,5,…
—allegedly the most aesthetically pleasing rectangle:
∴ r² = r + 1
∴ r² – r – 1 = 0 [does that look
familiar?]
= α_{1}((1+√5)/2) + α_{2}((1–√5)/2)
= α_{2}( (–1–√5)/2 + (1–√5)/2 )
= α_{2}( (–1–√5+1–√5)/2 )
= α_{2}( (–2√5)/2 )
= –α_{2}√5
And the answer is: 1/3 of the way from the edge!
The puzzle is: How does a sequence of foldings-in-half yield the number 1/3?
a_{n} = (a_{n–1} + a_{n–2})/2
= ½a_{n–1} + ½a_{n–2}
Where did the "3" come from?
∴ α_{2} = –α_{1}
∴ ½ = α_{1} + α_{1}/2 = 3α_{1}/2
∴ 1 = 3α_{1}
∴ α_{1} = 1/3 [there's the 1/3!!!]
∴ α_{2} = –1/3
Consider this sequence:
a_{n} = 3a_{n–1} – 2a_{n–2}, ∀ n ≥ 2
Try it! Use the algorithm to show that a_{n} = 2^{n} – 1, as suggested in the table in the previous lecture.
(answer will be given in the next lecture)