3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ It is monotonic since $a_{n+1}\ge a_n$. $\endgroup$ – xpaul May 12 '18 at 12:11
  • 1
    $\begingroup$ @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
0
$\begingroup$

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 $a_0 < b_0$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.