Fibonacci sequence: Difference between revisions

Revision as of 13:00, 24 July 2025

Revision as of 22:46, 28 July 2025

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To see the relation between the sequence and these constants,{{Sfn | Ball | 2003 | pp = 155–156}} note that <math>\varphi</math> and <math>\psi</math> are both solutions of the equation <math display=inline>x^2 = x + 1</math> and thus <math>x^n = x^{n-1} + x^{n-2},</math> so the powers of <math>\varphi</math> and <math>\psi</math> satisfy the Fibonacci recursion. In other words,

<math display=block>\begin{align}

\varphi^n &= \varphi^{n-1} + \varphi^{n-2}, \\[3mu]

\psi^n &= \psi^{n-1} + \psi^{n-2}.

\end{align}</math>

It follows that for any values {{mvar|a}} and {{mvar|b}}, the sequence defined by

<math display=block>U_n=a \varphi^n + b \psi^n</math>

satisfies the same recurrence,

<math display=block>\begin{align}

Line 167 ⟶ 158:

<math display=block>

a = \frac{1}{\varphi-\psi} = \frac{1}{\sqrt 5},\quad b = -a,

</math>

producing the required formula.

Taking the starting values {{math|''U''₀}} and {{math|''U''₁}} to be arbitrary constants, a more general solution is:

<math display=block> U_n = a\varphi^n + b\psi^n </math>

where

<math display=block>\begin{align}

a&=\frac{U_1-U_0\psi}{\sqrt 5}, \\[3mu]

b&=\frac{U_0\varphi-U_1}{\sqrt 5}.

\end{align}</math>

=== Computation by rounding ===