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When I learned sums and sequences in algebra II with trig I learned about recursive rules and explicit rules.

A recursive rule written with the formula of: $$a_n = r * a_{n-1}$$ Or as: $$a_n = a_{n-1} + d$$ And an explicit rule written with the formula of: $$a_n = a_1 + (n – 1)d$$ Or as: $$a_n = a_1 * r^{n-1}$$ My math teacher told me that every recursive rule can be written as an explicit rule too and I found that to hold true through all of the math problems I did for homework. However, when I thought of the Fibonacci sequence, $$a_n = a_{n-1} + a_{n-2};\ \ \ a_0 = 0,\ a_1 = 1$$ I couldn't figure out how to turn it into an explicit formula. Is there any way to do this with just the mathematical skills of a pre-calculus student?

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    $\begingroup$ Yes: the one usually presented is called Binet’s formula. An even nicer one, in my opinion, is $$F_n=\left\lfloor\frac{\varphi^n}{\sqrt5}+\frac12\right\rfloor\;,$$ where $\varphi=\frac12(1+\sqrt5)$ is the so-called golden ratio. You can these in the Wikipedia article on the Fibonacci numbers. $\endgroup$ – Brian M. Scott Aug 30 '15 at 21:10
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    $\begingroup$ This pdf has a derivation of it, and is written in a very simple easy-to-understand way: faculty.mansfield.edu/hiseri/Old%20Courses/SP2012/MA1115/… $\endgroup$ – Brenton Aug 30 '15 at 21:13
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    $\begingroup$ Your teacher is quite optimistic: there are many recurrence relations for which no closed form is known. $\endgroup$ – Rob Arthan Aug 30 '15 at 21:16
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    $\begingroup$ @RobArthan Perhaps she meant linear recursive rules. $\endgroup$ – Akiva Weinberger Aug 30 '15 at 21:59
  • $\begingroup$ @columbus8myhw: maybe, but guessing how to fix the teacher's alleged statement is a waste of time (and, unlike you, I wouldn't dare to make any assumptions about the teacher's gender). $\endgroup$ – Rob Arthan Aug 30 '15 at 22:08
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You could use Binet's formula: $$F_n=\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n\sqrt{5}}$$

A good derivation is given here, and it should be easily accessible to a pre-calculus student.

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