1. ## fibonnaci - inequality

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

Solve for n

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

2. 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

4. ## fibonnaci inequality

but googol being 10^100 the n might be generalised solution

5. Originally Posted by austinvishal
but googol being 10^100 the n might be generalised solution
What does that mean?

CB

is this solution right
480.19<n<481.19

7. ## solution

can you explain me the steps? i made a calculation error i suppose

8. 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.

9. 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