om1429888's blog

By om1429888, history, 6 weeks ago, In English

Hi guys. We all know that matrix exponentiation can be used to find Nth term of a recurrent relation in logN time, this is generally used to find Nth fibonacci in logN time where N can be as large as 1e18(obviously it is modulo something), now fibonacci is represented as f(n-1)+f(n-2),but what is relation isnt linear, it can be for example, f(n-1)+f(n-2)+f(n-1)*f(fn-2). i tried and could'nt find a solution, so any suggestions>

 
 
 
 
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6 weeks ago, # |
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Newbies
Me in a round
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    6 weeks ago, # ^ |
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    lmao, thanks for the advice.

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    6 weeks ago, # ^ |
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    Me solving those problems
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6 weeks ago, # |
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Mathematical savvy should help you. For example Solving Non Linear Recurrence Relation. In our case we can rewrite the recurrence as

$$$f[n] + 1 = f[n-1] * f[n-2] + f[n-1] + f[n-2] + 1 = (f[n-1] + 1) * (f[n-2] + 1)$$$

Or if we denote

$$$g_n = f[n] + 1$$$

we have

$$$ g_n = g_{n-1} g_{n-2}$$$

If our initial conditions are

$$$ f[0] = b, f[1] = a$$$

(and nonzero) then

$$$f[n] = (a + 1)^{F_n} (b + 1)^{F_{n-1}} - 1$$$

where $F_n$ is $$$n$$$-th Fibonacci number ($$$F_2 = F_1 = 1$$$). The case with zero $$$a$$$ or $$$b$$$ is simple.

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6 weeks ago, # |
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maybe this video can give u an intution