Mathematics Asked by jeongbyulji on January 18, 2021
I’m studying dynamical systems and I’ve found myself trying to solve this limit:
$$ lim_{k to +infty} frac{log_{2}(frac{1}{sqrt{5}}(frac{1+sqrt{5}}{2})^k – frac{1}{sqrt{5}}(frac{1-sqrt{5}}{2})^k)}{k}$$
Till now, I changed the base of the logarithm and converted it to a natural logarithm:
$$ frac{1}{ln(2)} lim_{k to +infty} frac{ln(frac{1}{sqrt{5}}(frac{1+sqrt{5}}{2})^k – frac{1}{sqrt{5}}(frac{1-sqrt{5}}{2})^k)}{k}$$
But I’m not sure how should I continue. I know that the result is $$frac{ln(frac{1+sqrt{5}}{2})}{ln(2)}$$
Thanks!
begin{align*} &frac{ lnleft( frac{1}{sqrt{5}} left( frac{1+sqrt{5}}{2} right)^k - frac{1}{sqrt{5}} left( frac{1-sqrt{5}}{2} right)^k right) }{k} \ &qquad = frac{ lnleft( frac{1}{sqrt{5}} left( frac{1+sqrt{5}}{2} right)^k left( 1 - left( frac{1-sqrt{5}}{1+sqrt{5}} right)^k right) right) }{k} \ &qquad = frac{ lnleft( frac{1}{sqrt{5}} left( frac{1+sqrt{5}}{2} right)^k right) + ln left( 1 - left( frac{1-sqrt{5}}{1+sqrt{5}} right)^k right)}{k} \ &qquad = frac{ lnleft( frac{1}{sqrt{5}} right) + k ln left( frac{1+sqrt{5}}{2} right) + ln left( 1 - left( frac{1-sqrt{5}}{1+sqrt{5}} right)^k right)}{k} \ &qquad = ln left( frac{1+sqrt{5}}{2} right) + frac{ lnleft( frac{1}{sqrt{5}} right) + ln left( 1 - left( frac{1-sqrt{5}}{1+sqrt{5}} right)^k right)}{k} text{.} end{align*} The first term is a constant, so survives the limit. In the numerator, the first term is a constant, and the second term goes to $ln 1 = 0$ because $-1 < frac{1 - sqrt{5}}{1 + sqrt{5}} < 1$, so its $k^text{th}$ power goes to $0$. That is, the wide fraction on the right vanishes in the limit.
Correct answer by Eric Towers on January 18, 2021
A variation on @Eric Towers solution, for the last term, you can use Maclaurin series for the logarithm function expansion around $0$ after noticing that the term $a^k to 0$: $$ frac{log (1-a^k)}{k} sim -frac{a^k}{k} to0< s_k < frac{log (1+a^k)}{k} sim frac{a^k}{k} to0 $$ where $a_k = frac{1-sqrt{5}}{1+sqrt{5}}$
Answered by Alex on January 18, 2021
Remember that Binet's formula for the Fibonacci numbers is $$F_n=frac{phi^n-{tilde{phi}}^n}{sqrt{5}}$$ Here $phi=(1=sqrt{5})/2$ and $tilde{phi}=(1-sqrt{5})/2$
So, $$L=frac{1}{ln(2)}lim_{ktoinfty}frac{ln(F_k)}{k}$$ But we know that $F_kasymp phi^k$, $$L=frac{1}{ln(2)}lim_{ktoinfty}frac{ln(phi^k)}{k}=frac{1}{ln(2)}lim_{ktoinfty}frac{kln(phi)}{k}=frac{ln(phi)}{ln(2)}.$$
Answered by K.defaoite on January 18, 2021
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