Prove that for every $a_0\in(0,2\pi)$ the following sequence $$ a_{n+1} = \int_0^{a_n}(1+\frac{1}{4}\cos^{2n+1}t) \,\mathrm{d}t $$ converges and find a limit of this sequence.

It is evident that this sequence is bounded. But it isn't monotonic. Also I have tried to show that this sequence is fundamental, but my attempts failed.

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    $\begingroup$ It is monotonic since $a_{n+1}\ge a_n$. $\endgroup$ – xpaul May 12 '18 at 12:11
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    $\begingroup$ @xpaul consider $a_0 = 1.5 \pi$. $$a_1 = \int_0^{a_n}(1+\frac{1}{4}\rm{cos}t)dt = a_0 + \int_0^{1.5\pi}(\frac{1}{4}\rm{cos}t)dt = 1.5\pi - \frac14 < a_0$$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 12 '18 at 12:27
  • $\begingroup$ Perhaps it helps to consider the indefinite integral $C_m = \int \cos^m(t)dt$. We have $m C_m = \sin(x) \cos^{m-1}(x) + (m-1) C_{m-2}$. Starting with $C_1 = \sin(x)$ we can recursively determine all $C_{2n+1}$. In particular it is easy to see that $C_{2n+1}(0) = 0$. $\endgroup$ – Paul Frost May 12 '18 at 15:37
  • $\begingroup$ After experimenting, I can propose that the answer will be $\pi$. And for $a_0<\pi$ the sequence will increase, for $a_0=\pi$ the sequence will be constant. And for $a_0>\pi$ the sequence will decrease. $\endgroup$ – Mikhail Goltvanitsa May 12 '18 at 16:27
  • $\begingroup$ All $C_{2n+1}(\pi) = 0$. Therefore if $a_0 = \pi$, then all $a_n = \pi$. I conjecture that $(a_n)$ converges to $\pi$ for any $a_0$. $\endgroup$ – Paul Frost May 12 '18 at 16:34

This is only a partial answer - it shows that the sequence converges, but does not give the limit.

Define $$I_n(x) = \int_0^x \cos^{2n+1}(t)dt .$$ We have $I_n(\pi) = 0$ because $\cos^{2n+1}(\pi/2 + t) = -\cos^{2n+1}(\pi/2 - t)$. For $a \in (0,2\pi)$ define $$f(n,a) = \int_0^a (1 + 1/4\cos^{2n+1}(t))dt = a + 1/4 I_n(a) .$$ Then $a_{n+1} = f(n,a_n)$. For $a_0 = \pi$ we get $a_n = \pi$ for all $n$; this sequence trivially converges to $\pi$. We claim that $$a < f(n,a) <\pi \text{ for } 0 < a < \pi .$$ This implies that $(a_n)$ is bounded and strictly increasing, i.e. is convergent. The case $\pi < a < 2\pi$ can be treated similarly (we get $\pi < f(n,a) < a$ so that $(a_n)$ is bounded and strictly decreasing).

Let us prove the above claim. For $0 < a \le \pi/2$ we have $0 < I_n(a) < a \le \pi/2$ which holds because $0 \le \cos^{2n+1}(t) \le 1$ for $0 \le t \le \pi/2$. For $\pi/2 < a < \pi$ we have $I_n(a) = I_n(\pi) - \int_a^\pi \cos^{2n+1}(t)dt = - \int_a^\pi \cos^{2n+1}(t)dt = \int_a^\pi \lvert \cos^{2n+1}(t) \rvert dt \in (0, \pi - a)$.

Added: For $a < b$ we have $f(n,a) < f(n,b)$ because $f(n,b) - f(n,a) = b - a + 1/4\int_a^b \cos^{2n+1}(t)dt \ge b -a -1/4\int_a^b \lvert \cos^{2n+1}(t) \rvert dt >$ $ b -a - 1/4(b-a) > 0$.

Letting denote $\overline{a}_0$ the limit of the sequence $(a_n)$ starting with $a_0$ we see that $\overline{a}_0 \le \overline{b}_0$ when $a_0 < b_0$.

  • $\begingroup$ As I think we must investigate a function $f(n,a) = \int_0^{a_{n-1}} (1 + 1/4\cos^{2n+1}(t))dt = a_{n-1} + 1/4 I_n(a_{n-1}) .$. Don't we? $\endgroup$ – Mikhail Goltvanitsa May 13 '18 at 18:21
  • $\begingroup$ @MikhailGoltvanitsa Yes, that seems reasonable. It is easy to see that these functions are strictly increasing (I added this to my above answer), but I don't see how we can get any information about the limit. $\endgroup$ – Paul Frost May 13 '18 at 22:40
  • $\begingroup$ @MikhailGoltvanitsa Doing some calculations, one comes to the conjecture that the factor $1/4$ in the recursive definiton of $(a_n)$ can be replaced by any constant $c \in (0,1]$ - we always get $a_n \to \pi$. It is clear, however, that if $a_n(c) \to \pi$, then $a_n(c') \to \pi$ for any $c'$ between $c$ and $1$. By the way, in which context does this sequence appear? $\endgroup$ – Paul Frost May 15 '18 at 16:01

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