# fibonnaci - inequality

• March 31st 2009, 11:38 PM
austinvishal
fibonnaci - inequality
If F(0) = 0 , F(1)= 1....

Solve for n

F(n-1)< googol <F(n)
• April 1st 2009, 06:43 AM
CaptainBlack
Quote:

Originally Posted by austinvishal
If F(0) = 0 , F(1)= 1....

Solve for n

F(n-1)< googol <F(n)

Use Binet's formula for the $n$-th Fibonacci number:

$F(n)=\frac{\varphi^n-(-1/\varphi)^n}{\sqrt{5}}$

where $\varphi=(1+\sqrt{5})/2 \approx 1.618..$ is the Golden number

As $n$ will be very large $-(-1/\varphi)^n$ will be negligible so you need to find $n$ such that:

$\frac{\varphi^{n-1}}{\sqrt{5}}<\text{googol }<\frac{\varphi^{n}}{\sqrt{5}}$

CB
• April 1st 2009, 06:49 AM
austinvishal
fibonnaci inequality
(Happy)
• April 1st 2009, 06:50 AM
austinvishal
fibonnaci inequality
but googol being 10^100 the n might be generalised solution
• April 1st 2009, 08:20 AM
CaptainBlack
Quote:

Originally Posted by austinvishal
but googol being 10^100 the n might be generalised solution

What does that mean?

CB
• April 1st 2009, 08:23 AM
austinvishal
is this solution right
480.19<n<481.19
• April 1st 2009, 08:30 AM
austinvishal
solution
can you explain me the steps? i made a calculation error i suppose
• April 1st 2009, 12:12 PM
Opalg
I agree with austinvishal on this one. Taking logs to base 10 (as we old-timers used to do all the time before they came up with these newfangled electronic thingies), we want to find $\frac{\log(\sqrt5\times10^{100})}{\log\varphi} \approx \frac{100.349485}{0.2089876}\approx480.17$. So we should take n = 481.
• April 1st 2009, 12:41 PM
CaptainBlack
Quote:

Originally Posted by Opalg
I agree with austinvishal on this one. Taking logs to base 10 (as we old-timers used to do all the time before they came up with these newfangled electronic thingies), we want to find $\frac{\log(\sqrt5\times10^{100})}{\log\varphi} \approx \frac{100.349485}{0.2089876}\approx480.17$. So we should take n = 481.

Yes, arithmetic error.

CB