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Consider the sequence defined by $a_0=a_1=1$ and $a_{n}=2a_{n-1}-3a_{n-2}$.

Here, a proof is given that $|a_n|\to +\infty $. I would like to prove that $|a_n|>100$ when $n>10$. How can we do it ? (Also, is there an explicit lower bound for the sequence $|a_n|$ ?)

The first terms :

1, 1, -1, -5, -7, 1, 23, 43, 17, -95, -241, -197, 329, 1249, 1511, -725, -5983, -9791, -1633, 26107, 57113, 35905, -99529, -306773, -314959, 290401, 1525679, 2180155, -216727, -6973919, -13297657, -5673557, 28545857, 74112385, 62587199, -97162757, -382087111, -472685951, 200889431, 1819836715, 3037005137, 614500129, -7882015153, -17607530693, -11569015927, 29684560225, 94076168231, 99098655787, -84031193119, -465358353599, -678623127841, 38828805115, 2113526993753, 4110567572161, 1880554163063, -8570594390357, -22782851269903, -19853919368735, 28640715072239, 116843188250683, 147764231284649, -55001102182751, -553294898219449, -941586489890645, -223288285122943, 2378182899426049, 5426230654220927, 3717912610163707, -8842866742335367, -28839471315161855, -31150342403317609, 24217729138850347

Explicit formulas :

\begin{align} a_n& =\frac{(1+i\sqrt2)^n+(1-i\sqrt 2)^n}{2}\\ &=(\sqrt3)^n\cdot \cos(n\cdot \theta) \end{align} where $\theta = \tan^{-1}(\sqrt 2)$.

Paolo Leonetti
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Friedrich
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  • maybe help: ${ cos(n \cdot tan^{-1}(x)) = \frac{1}{(x^2+1)^{\frac{n}{2}}} \cdot \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} (-1)^k \binom{n}{2k} x^{2k}} $, so when $x=\sqrt2$, $${a_n=\cdot \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} (-1)^k \binom{n}{2k} 2^{k}}$$ – serg_1 Jun 26 '17 at 14:57
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    A not simple idea. We have $\cos(n\theta)=\pm \sin(n\theta+k\pi+\pi/2)$ ($k$ is in $\mathbb{Z}$). Let $k=k_n$ such that $|n\theta+k_n\pi+\pi/2|\leq \pi/2$. If $x\in [-\pi/2,\pi/2]$ we have $|\sin(x)|\geq \frac{2}{\pi}|x|$ so $|\cos(n\theta)|\geq \frac{2}{\pi}|n\theta+(2k_n+1)\frac{\pi}{2}|=\frac{2}{\pi}A$. Now use a determination of the logarithm, $A=|n\log (\exp(i\theta))+(2k_n+1)\log (i)|$. By Baker's Theorem (Google it...) there exists constants $c_1,c_2$ such that $|\cos(n\theta)|\geq c_1n^{-c_2}$, but there is a lot of work to finish, ie to find explicitely $c_1$ and $c_2$... – Kelenner Jun 26 '17 at 15:23

1 Answers1

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hint

$$T_n (\cos (\theta))=\cos (n\theta) $$ where $T_n $ is the Tchebychev polynomial.

hamam_Abdallah
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