If F(0) = 0 , F(1)= 1....

Solve for n

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

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- Mar 31st 2009, 11:38 PMaustinvishalfibonnaci - inequality
If F(0) = 0 , F(1)= 1....

Solve for n

F(n-1)< googol <F(n) - Apr 1st 2009, 06:43 AMCaptainBlack
Use Binet's formula for the $\displaystyle n $-th Fibonacci number:

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

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

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

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

CB - Apr 1st 2009, 06:49 AMaustinvishalfibonnaci inequality
(Happy)

- Apr 1st 2009, 06:50 AMaustinvishalfibonnaci inequality
but googol being 10^100 the n might be generalised solution

- Apr 1st 2009, 08:20 AMCaptainBlack
- Apr 1st 2009, 08:23 AMaustinvishalanswer
is this solution right

480.19<n<481.19 - Apr 1st 2009, 08:30 AMaustinvishalsolution
can you explain me the steps? i made a calculation error i suppose

- Apr 1st 2009, 12:12 PMOpalg
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 $\displaystyle \frac{\log(\sqrt5\times10^{100})}{\log\varphi} \approx \frac{100.349485}{0.2089876}\approx480.17$. So we should take n = 481.

- Apr 1st 2009, 12:41 PMCaptainBlack