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by andrewp123
761 days ago
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A powerful method in math is to assume the thing you're looking for is continuous, in which case you can write it F(n) = a0 + a1 n + a2 n^2 + ... . You can solve Fibonacci by writing it as a0 + a1 n + a2 n^2 + ... = a0 + a1 (n-1) + a2 (n-1)^2 + ... + a0 + a1 (n-2) + a2 (n-2)^2 + ... Expand terms, then match the n, n^2, n^3 etc terms and solve for a0, a1, a2, etc. This works in solving differential equations too - it's called "Method of Frobenius" (easier in diffeq because you don't have to do annoying binomial coefficient math). |
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