Tessellate S.T.E.M.S - Mathematics - College - Set 2 - Subjective Problem

Let \(\displaystyle \{ a_n \}_{n=0}^\infty\) be any real valued sequence such that \(a_{n+1} = \cos (a_n)\) for \(n \geq 1\).

Does the sequence {an}\{ a_n \} always converge? Justify.

This problem is a part of Tessellate S.T.E.M.S.

#Calculus

Note by Aditya Raut
3 years, 5 months ago

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Comments

Yes. Let α\alpha be the unique solution to cos⁡(x)=x.\cos(x) = x. We'll show the sequence converges to α.\alpha.

First, note that a2∈[−1,1],a_2 \in [-1,1], so a3∈[cos⁡(1),1],a_3 \in [\cos(1),1], and so a4∈[cos⁡(1),cos⁡(cos⁡(1))].a_4 \in [\cos(1),\cos(\cos(1))]. (This is because cos⁡(x)\cos(x) is decreasing on [cos⁡(1),1].[\cos(1),1].)

So a4a_4 lies in the interval [0.5403,0.8576],[0.5403, 0.8576], and α≈0.7391.\alpha \approx 0.7391. So ∣a4−α∣<0.2.|a_4-\alpha| < 0.2.

Now one way to proceed is via the mean value theorem: for some cnc_n between ana_n and α,\alpha, cos⁡(an)−cos⁡(α)an−α=−sin⁡(cn)an+1−αan−α=−sin⁡(cn). \begin{aligned} \frac{\cos(a_n)-\cos(\alpha)}{a_n-\alpha} &= -\sin(c_n) \\ \frac{a_{n+1}-\alpha}{a_n-\alpha} &= -\sin(c_n). \end{aligned} This shows that ∣an+1−α∣≤∣an−α∣.|a_{n+1}-\alpha| \le |a_n-\alpha|. But we can strengthen this: since ∣a4−α∣<0.2,|a_4-\alpha| < 0.2, ∣an−α∣<0.2|a_n-\alpha| < 0.2 for all n≥4n \ge 4 by induction. In particular, an∈[α−0.2,α+0.2],a_n \in [\alpha-0.2,\alpha+0.2], and so cn∈[α−0.2,α+0.2]c_n \in [\alpha-0.2,\alpha+0.2] as well. Now let MM be the maximum value of ∣sin⁡(x)∣|\sin(x)| on the interval [α−0.2,α+0.2].[\alpha-0.2,\alpha+0.2]. We can use the above MVT computation to get ∣an+1−α∣≤M∣an−α∣≤M2∣an−1−α∣≤⋯≤Mn−3∣a4−α∣<0.2Mn−3. |a_{n+1}-\alpha| \le M |a_n - \alpha| \le M^2 |a_{n-1}-\alpha| \le \cdots \le M^{n-3} |a_4 - \alpha| < 0.2M^{n-3}. Now 0≤M<1,0 \le M < 1, so the distance between an+1a_{n+1} and α\alpha gets arbitrarily close to 00 for nn sufficiently large. This shows that an→α.a_n \to \alpha.

Patrick Corn - 3 years, 5 months ago

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