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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
5 months, 2 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,


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


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,


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,

\[\lim_{n\to\infty}\cos\left(\frac{\sqrt{1-x_0^2}}{x_1x_2\cdots x_n}\right)=x_0\] Prasun Biswas · 5 months, 1 week ago

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please help me to solve this question. thanx in advance Aaron Jerry Ninan · 5 months, 2 weeks ago

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