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add addendum to a favourite proof of mine

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Rachel Lambda Samuelsson 2023-04-19 15:31:28 +02:00
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@ -129,4 +129,10 @@ $$ f_n = \frac{1}{\sqrt 5} \left(\left(\frac{1 + \sqrt 5}{2}\right)^n - \left(\f
Whenever I first happened upon the closed formula for the $n$-th Fibonacci number it seemed so shockingly random, a formula with bunch of square roots always giving me a recursively specified integer. After I learned this proof it doesn't feel as random anymore, instead, I feel it would be more surprising if we carried out the whole diagonalization process and ended up with no roots. Perhaps more importantly, it opened my eyes to the usage of linear algebra as a powerful mathematical tool, and not just something to be applied in geometry, flow balancing or computer graphics. Whenever I first happened upon the closed formula for the $n$-th Fibonacci number it seemed so shockingly random, a formula with bunch of square roots always giving me a recursively specified integer. After I learned this proof it doesn't feel as random anymore, instead, I feel it would be more surprising if we carried out the whole diagonalization process and ended up with no roots. Perhaps more importantly, it opened my eyes to the usage of linear algebra as a powerful mathematical tool, and not just something to be applied in geometry, flow balancing or computer graphics.
# Addendum
It was pointed out to me [on mastodon](https://mastodon.vierkantor.com/@Vierkantor/109978590835441958) that this technique is of interest even if it is not possible to diagonalize the stepping matrix. This is because using fast binary exponentiation one can perform matrix exponentiation in logarithmic time. Thus any linearly recursive algorithm can be calculated in logarithmic time!
Fast binary exponentiation uses the identity $A^{2k} = (A \cdot A)^k$, thus we can split the exponent in 2 when even, rather than performing $2k$ multiplications. Recursively doing this each time the exponent is even yields logarithmic time exponentiation.
{% endkatexmm %} {% endkatexmm %}