I recognized the fact that the powers of this matrix gives the Fibonacci numbers and you can kinda just reference which Fibonacci number is on the diagonals. Using diagonalization you can get a closed form in terms of n and solve for n more explicitly.
Yes, I deliberately put that in as an alternative pathway to a solution, in case anyone didn't remember their linear algebra. (The diagonalization approach is nice in that it is substantially more general.)
You seem to have a couple of typos on the powers of the matrix h. The bottom right coefficient of h^2 should be 1 and the bottom right coefficient of h^4 should be 2.
I didn't read the article yet. I just wanted to inform you that I am able to see the post from the note you published.
“You can do it by brute force, but that will take you over a thousand years.”
Well that’s much better than the 10^296 years mentioned before 😂
It's from the same ~2000 year bound that assumes that we get unlimited exponential growth from Moore's Law.
i love this series youre doing! would absolutely love for you to do one on topology and graph theory at some point!
Thank you! I haven’t taught either one, so it would be a bit more of an undertaking. But I am very open to the idea.
I'm not sure how to do spoilers in Substack comments, but it was pretty easy.
Did you use diagonalization, or another method?
I recognized the fact that the powers of this matrix gives the Fibonacci numbers and you can kinda just reference which Fibonacci number is on the diagonals. Using diagonalization you can get a closed form in terms of n and solve for n more explicitly.
I used recognising the sequence of elements.
Yes, I deliberately put that in as an alternative pathway to a solution, in case anyone didn't remember their linear algebra. (The diagonalization approach is nice in that it is substantially more general.)
Hi, thank you for the article.
You seem to have a couple of typos on the powers of the matrix h. The bottom right coefficient of h^2 should be 1 and the bottom right coefficient of h^4 should be 2.