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Prove that the $n$th Fibonacci number $f_n$ is the integer that is closest to the number $$\frac{1}{\sqrt{5}}\left( \frac{1 + \sqrt{5}}{2} \right)^n.$$

Hi everyone, I don't really understand the problem. I have the following hint, but I don't know how to work it.

Hint: Show that the absolute value of $\frac{1}{\sqrt{5}}\left( \frac{1 - \sqrt{5}}{2} \right)^n$ is less than $1/2$.

Chris Culter
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okie
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    What do you know about the Fibonacci numbers? Do you know an explicit formula for the $n^{\mathrm{th}}$ Fibonacci number? – rogerl Oct 27 '14 at 02:57
  • The question is asking you to prove that if you take a positive integer $n$ and plug it into that formula, then you get something really close to the $n$th Fibonacci number. – NoName Oct 27 '14 at 02:58
  • https://www.youtube.com/watch?v=dTWKKvlZB08 Numberphile video touching on this topic. – JMoravitz Oct 27 '14 at 02:59
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    http://math.stackexchange.com/questions/65011/prove-this-formula-for-the-fibonacci-sequence – lab bhattacharjee Oct 27 '14 at 02:59
  • If you are familiar with difference equations that may help with this problem. – EhBabay Oct 27 '14 at 03:32

2 Answers2

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The question seems to assume that you know the exact formula for $f_n$, namely $$f_n = \frac{1}{\sqrt{5}}\left( \frac{1 + \sqrt{5}}{2} \right)^n - \frac{1}{\sqrt{5}}\left( \frac{1 - \sqrt{5}}{2} \right)^n.$$ From here the reason the hint is useful should be clear: the expression in the hint is, up to sign, the distance between $f_n$ and the expression it's supposed to be close to.

user187373
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We prove the hint by induction on $n$. For $n = 1$, $LHS < \dfrac{1}{\sqrt{5}} < \dfrac{1}{2}$, so the statement is true for $n = 1$. Assume it is true for $n = k$, that is: $|R_k|=\left|\dfrac{1}{\sqrt{5}}\cdot \left(\dfrac{1-\sqrt{5}}{2}\right)^k\right| < \dfrac{1}{2}$, we have:

$|R_{k+1}| = \dfrac{1}{\sqrt{5}}\cdot \left(\dfrac{\sqrt{5}-1}{2}\right)^{k+1} < |R_k| < \dfrac{1}{2}$, and the hint is proved. From this the answer follows.

DeepSea
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  • The base case should show that the absolute value $\left|\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{1}\right|<\frac{1}{2}$ – chharvey Oct 27 '14 at 03:25
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    You might not need to use induction for this one. You could just show that $\left|R_1\right| < \frac{1}{2}$ and that the sequence $R_n$ is geometric and convergent. That would be sufficient. (All converging geometric sequences are monotone and converge to $0$. Not counting the constant sequence $a, a, a, \dots$.) – chharvey Oct 27 '14 at 03:33
  • @chharvey Also $a,-a,a,\ldots$. – user236182 Jul 24 '17 at 18:30