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Knuth has a neat algorithm that uses matrix exponentiation to compute the $n$th Fibonacci number in $O(\log_2 n)$-time 1. However, there doesn't seem to be a lot of resources on generalizing his idea to other linear recurrences. Is it a generalizable idea, or is it just a special technique only applicable to the Fibonacci recurrence?

(I'm not asking for a closed-form solution, but an efficient method similar to Knuth's technique for Fibonacci numbers).

As a related question, can we speed up other recurrences by rewriting them as linear transformations, and how far can we go in applying this technique?

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Yes. This generalizes to any linear recurrence. Suppose we have the linear recurrence

$$x_{n+1} = a_0 x_n + a_1 x_{n-1} + \dots + a_k x_{n-k}.$$

Define the column vector $v_n = (x_n,x_{n-1},\dots,x_{n-k})^\top$. Then we have the equation

$$v_{n+1} = M v_n$$

where the matrix $M$ is given by

$$M = \begin{pmatrix} a_0 &a_1 &a_2 &\dots &a_k\\ 1 &0 &0 &\dots &0\\ 0 &1 &0 &\dots &0\\ &&&\vdots\\ \end{pmatrix}.$$

It follows that

$$v_n = M^n v_0.$$

Therefore you can compute $v_n$ using $O(\log n)$ matrix multiplications, using a fast exponentiation algorithm for computing $M^n$. This gives an efficient way to compute $x_n$: first compute $v_n$ using matrix exponentiation, then extract the first coefficient of $v_n$ to get $x_n$.

In fact, we can take this farther. Suppose $M$ diagonalizes, say $M=PDP^{-1}$ where $D$ is a diagonal matrix. Then

$$v_n = M^n v_0 = (PDP^{-1})^n v_0 = P D^n P^{-1} v_0,$$

so we don't need to perform matrix exponentiation; it suffices to exponentiate each of the diagonal entries of $D$. In fact, expanding terms, you can get an algebraic closed-form expression for $x_n$. If you apply this to the Fibonacci sequence, you get the closed-form expression in https://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression (in terms of powers of $(1\pm \sqrt{5})/2$, because those are the elements of $D$); for a general sequence, you get a closed-form expression of the form

$$x_n = c_1 \lambda_1^n + \dots + c_k \lambda_k^n,$$

where $\lambda_1,\dots,\lambda_k$ are the diagonal entries of $D$ (i.e., the eigenvalues of $M$). This also gives a potentially more efficient algorithm for computing $x_n$: instead of $O(\log n)$ multiplications of $k\times k$ matrices, we get $O(k \log n)$ operations on numbers.

If $M$ does not diagonalize, you may still be able to use the Jordan normal form of $M$.

See also https://en.wikipedia.org/wiki/Recurrence_relation#Solving_homogeneous_linear_recurrence_relations_with_constant_coefficients, which does something similar using the characteristic polynomial of the linear recurrence (which I think coincides with the characteristic polynomial of $M$, or something like that).

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  • $\begingroup$ Wow! This is so cool! Thank you! $\endgroup$ – BearAqua Mar 11 '19 at 18:32

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