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Math Help - fibonnaci - inequality

  1. #1
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    fibonnaci - inequality

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

    Solve for n

    F(n-1)< googol <F(n)
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  2. #2
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    Quote Originally Posted by austinvishal View Post
    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
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  3. #3
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    fibonnaci inequality

    Last edited by austinvishal; April 1st 2009 at 06:53 AM. Reason: repeated two times
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  4. #4
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    fibonnaci inequality

    but googol being 10^100 the n might be generalised solution
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by austinvishal View Post
    but googol being 10^100 the n might be generalised solution
    What does that mean?

    CB
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    answer

    is this solution right
    480.19<n<481.19
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  7. #7
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    solution

    can you explain me the steps? i made a calculation error i suppose
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  8. #8
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    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.
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  9. #9
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    Quote Originally Posted by Opalg View Post
    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
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