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# Infinite cosine!

Let $$x_0 , x_1, x_2,\ldots$$ be a sequence of numbers satisfying the recursion $$x_{n+1} =\sqrt{\dfrac{1+x_n}2}$$ for $$n=0,1,2,\ldots$$ and $$-1 \leq x_0 \leq 1$$.

Find the value of $$\displaystyle \lim_{n\to\infty} \cos \left(\dfrac{\sqrt{1-x_0^2 }}{x_1 x_2 x_3\cdots x_n} \right)$$.

Note by Aaron Jerry Ninan
7 months, 3 weeks ago

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Let $$x_0=\cos\theta_0$$ for some $$\theta_0\in\Bbb R$$ and note that $$-1\leq x_n\leq 1\implies 0\leq x_{n+1}\leq 1~\forall~n\geq 0$$ using the recurrence relation which shows that $$0\leq x_n\leq 1~\forall~n\geq 1$$.

Then, substitute $$x_n=\cos\theta_n~\forall~n\geq 1$$ and using the recurrence relation, we get that $$x_{n+1}=|\cos(\theta_n/2)|~\forall~n\geq 0$$ from which we conclude using the base case that $$x_n=|\cos(\theta_0/2^n)|~\forall~n\geq 1$$. So, the limit in question becomes,

$L=\lim_{n\to\infty}\cos\left(\frac{|\sin\theta_0|}{\prod\limits_{k=1}^n|\cos(\theta_0/2^k)|}\right)=\lim_{n\to\infty}\cos\left(\left|\frac{\sin\theta_0}{\prod\limits_{k=1}^n\cos(\theta_0/2^k)}\right|\right)$

Multiplying both numerator and denominator by $$\sin(\theta_0/2^n)$$ and using the sine double-angle formula repeatedly, the limit reduces to,

$L=\lim_{n\to\infty}\cos\left(\left|\frac{\sin\theta_0}{\dfrac{\sin\theta_0}{2^n\sin(\theta_0/2^n)}}\right|\right)=\lim_{n\to\infty}\left(\left|\theta_0\cdot\frac{\sin(\theta_0/2^n)}{\theta_0/2^n}\right|\right)$

Because of the continuity of the cosine and the modulus function, we can shift the limit operator inside the function body, hence, the limit becomes,

$L=\cos\left(\left|\lim_{n\to\infty}\theta_0\cdot\frac{\sin(\theta_0/2^n)}{\theta_0/2^n}\right|\right)$

Now, substitute $$u=\theta_0/2^n$$, so since $$\theta_0$$ is a finite real number, $$u\to 0$$ as $$n\to\infty$$. Then, the limit becomes,

$L=\cos\left(\left|\theta_0\lim_{u\to 0}\frac{\sin u}{u}\right|\right)=\cos\left(|\theta_0|\right)=\cos\theta_0=x_0$

In the second step in the above line, we used the well-known result that $$\lim\limits_{x\to 0}\frac{\sin x}{x}=1$$ which can be easily proved using the squeeze theorem.

So, we conclude that,

$\lim_{n\to\infty}\cos\left(\frac{\sqrt{1-x_0^2}}{x_1x_2\cdots x_n}\right)=x_0$ · 7 months, 1 week ago