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Recurrence Relations
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| import math | |
| """ | |
| Fibonacci recurrence relation of the form | |
| fsubn - fsub(n-1) - fsub(n-2) = 0 | |
| could be solved by figuring out a value | |
| q^n - q^(n-1) - q^(n-2) = 0 | |
| Then q^(n-2) * (q^2 - q - 1) = 0 | |
| We thus obtain roots | |
| q1=(1+sqrt(5))/2 | |
| q2=(1-sqrt(5))/2 | |
| """ | |
| def fn(n): | |
| return ((1 + math.sqrt(5))/2) ** n + ((1 - math.sqrt(5))/2) ** n | |
| def fn_with_constants(n, c1, c2): | |
| return c1 * (((1 + math.sqrt(5))/2) ** n) + c2 * (((1 - math.sqrt(5))/2) ** n) | |
| """ | |
| fn(100) == fn(99) + fn(98) | |
| fn_with_constants(50, 3, 4) == fn(49, 3, 4) + fn(48, 3, 4) | |
| """ |
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