Recurring Styles #6 - Trigonometric Recurrence!

Calculus Level 5

x n + 1 = sin ⁡ ( x n ) \Large{ x_{n+1} = \sin(x_n)}

Let x 0 ∈ ( 0 , π ) x_0 \in (0, \pi) and define the sequence ( x n ) n = 0 ∞ (x_n)_{n=0}^\infty by the recurrence relation given above. Find the value of the following upto 3 places of decimals:

lim ⁡ n → ∞ x n n = ? \large{\lim_{n \to \infty} x_n \sqrt{n} = \ ? }


The answer is 1.732.

Satyajit Mohanty by Satyajit Mohanty , 24, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

We have that x 1 ∈ ( 0 , 1 ) x_1\in(0,1) and x n + 1 = sin ⁡ ( x n ) < x n < 1 x_{n+1}=\sin\left(x_n\right)<x_n<1 for all n > 1 n>1 .

Hence x n x_n is strictly decreasing and tends to 0 + 0^+ .

Moreover, x n + 1 = sin ⁡ ( x n ) = x n − x n 3 6 + x n 5 120 + O ( x n 7 ) x_{n+1}=\sin\left(x_n\right)=x_n-\dfrac{x_n^3}{6}+\dfrac{x_n^5}{120}+O\left(x_n^7\right) .

This implies that 1 x n + 1 2 = 1 x n 2 + 1 3 + x n 2 15 + O ( x n 4 ) \dfrac{1}{x_{n+1}^2}=\dfrac{1}{x_n^2}+\dfrac{1}{3}+\dfrac{x_n^2}{15}+O\left(x_n^4\right) .

Therefore, by Stoltz-Cesaro, lim ⁡ n → ∞ 1 x n 2 n = lim ⁡ n → ∞ ( 1 x n + 1 2 − 1 x n 2 ) = 1 3 \displaystyle \lim_{n\to\infty}\dfrac{\dfrac{1}{x_n^2}}{n}=\lim_{n\to\infty}\left(\dfrac{1}{x_{n+1}^2}-\dfrac{1}{x_n^2}\right)=\dfrac{1}{3} .

So, lim ⁡ n → ∞ ( x n n ) = 3 ≈ 1.732 \displaystyle \lim_{n\to\infty}\left(x_n\sqrt{n}\right)=\sqrt3\approx\boxed{1.732} .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...