# A nested $\sin(x)$ sequence

This question probably sounds quite strange as I'm inexperienced in mathematics, and am not familiar with things like this.

I'm trying to prove that a sequence converges to a particular value that I call "$y$".

The sequence is defined as follows: $$x_0 = 1,\\ x_{n+1} = \frac{\sin(x_n)}{x_n}$$

For example, $$x_1 = \frac{\sin(1)}{1}, \text{or just} \sin(1) \\ x_2 = \frac{\sin(\sin(1))}{\sin(1)} \\$$

And so on. I understand this question may be quite confusing and non-sensical, but any help pointing me in the correct direction would be greatly appreciated.

Even better, if somebody could provide a closed/explicit definition of the nth term, I could try and work out if it converges or not by myself.

Regardless, any help would be very usefull.

• The sequence is doing fixed point iteration to solve the equation $x=\frac{\sin(x)}{x}$, starting from $x_0=1$. From the way the graph of $y=x$ and $y=\frac{\sin(x)}{x}$ look like, it looks like the iteration will have the even-numbered terms decreasing and the odd-numbered terms increasing. You can prove this by induction. Then show that the difference between consecutive terms tends to zero. For this last part use the mean value theorem and that the derivative of $\sin(x)/x$ is $<1$ in absolute value in the interval $[x_1,x_0]$. – user561777 May 18 '18 at 21:03
• Unfortunately, I am mainly self-taught and have yet to learn about the mean value theorem or study proof by induction, however thanks for your comment – Olly Britton May 18 '18 at 21:05
• Then you can put aside this exercise and study induction. The mean value theorem can also wait. [A funny thing is that since the mean value theorem is equivalent to continuous induction, essentially you are lacking training in induction. Both on the natural numbers and on the reals. Don't worry about this last comment. I am just thinking aloud. But do study induction.]. – user561777 May 18 '18 at 21:17
• You mean like these? bit.ly/thisisajokebytheway – Olly Britton May 18 '18 at 21:18
• Like these – user561777 May 18 '18 at 21:28