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 lim⁡n→∞cos⁡(1−x02x1x2x3⋯xn) \displaystyle \lim_{n\to\infty} \cos \left(\dfrac{\sqrt{1-x_0^2 }}{x_1 x_2 x_3\cdots x_n} \right) .

#Calculus

Note by Aaron Jerry Ninan
4 years, 10 months ago

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Comments

Let x0=cos⁡θ0x_0=\cos\theta_0 for some θ0∈R\theta_0\in\Bbb R and note that −1≤xn≤1  ⟹  0≤xn+1≤1 ∀ n≥0-1\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≤xn≤1 ∀ n≥10\leq x_n\leq 1~\forall~n\geq 1.

Then, substitute xn=cos⁡θn ∀ n≥1x_n=\cos\theta_n~\forall~n\geq 1 and using the recurrence relation, we get that xn+1=∣cos⁡(θn/2)∣ ∀ n≥0x_{n+1}=|\cos(\theta_n/2)|~\forall~n\geq 0 from which we conclude using the base case that xn=∣cos⁡(θ0/2n)∣ ∀ n≥1x_n=|\cos(\theta_0/2^n)|~\forall~n\geq 1. So, the limit in question becomes,

L=lim⁡n→∞cos⁡(∣sin⁡θ0∣∏k=1n∣cos⁡(θ0/2k)∣)=lim⁡n→∞cos⁡(∣sin⁡θ0∏k=1ncos⁡(θ0/2k)∣)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⁡(θ0/2n)\sin(\theta_0/2^n) and using the sine double-angle formula repeatedly, the limit reduces to,

L=lim⁡n→∞cos⁡(∣sin⁡θ0sin⁡θ02nsin⁡(θ0/2n)∣)=lim⁡n→∞(∣θ0⋅sin⁡(θ0/2n)θ0/2n∣)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⁡(∣lim⁡n→∞θ0⋅sin⁡(θ0/2n)θ0/2n∣)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=θ0/2nu=\theta_0/2^n, so since θ0\theta_0 is a finite real number, u→0u\to 0 as n→∞n\to\infty. Then, the limit becomes,

L=cos⁡(∣θ0lim⁡u→0sin⁡uu∣)=cos⁡(∣θ0∣)=cos⁡θ0=x0L=\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⁡x→0sin⁡xx=1\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→∞cos⁡(1−x02x1x2⋯xn)=x0\lim_{n\to\infty}\cos\left(\frac{\sqrt{1-x_0^2}}{x_1x_2\cdots x_n}\right)=x_0

Prasun Biswas - 4 years, 10 months ago

please help me to solve this question. thanx in advance

Aaron Jerry Ninan - 4 years, 10 months ago
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