Prove that $\lim _{n \rightarrow \infty} \sqrt{n} \cdot\sin^{\circ n}(\frac{1}{\sqrt{n}})=\frac{\sqrt{3}}{2}$ [duplicate]

Question

It's known that $\lim _{n \rightarrow \infty} \sqrt{n} \cdot \sin^{\circ n}(x)=\sqrt{3}$ for any $x>0$.

And I found a new conclusion $\lim _{n \rightarrow \infty} \sqrt{n} \cdot \sin^{\circ n}(\frac{1}{\sqrt{n}})=\frac{\sqrt{3}}{2}$ by Python. But I can't prove it because I can't use the same method to solve.

Can anyone give me some methods?

Answer 1

Define $f_N(x)$ and $\varphi(x)$ by

$$ f_N(x) = \sqrt{N} \sin\left(\frac{x}{\sqrt{N}}\right) \qquad\text{and}\qquad \varphi(x) = \sqrt{\frac{3}{x}}. $$

In this answer, we establish the following result.

Theorem. Let $x, y > 0$. Then $$\lim_{N\to\infty} f_N^{\circ\lfloor yN \rfloor}(\varphi(x)) = \varphi(x+y). $$

Here, $f^{\circ k}$ denotes the $k$-fold composition of $f$. Below are some remarks:

• OP's question corresponds to the case $x = 3$ and $y = 1$.

• User @Anne Bauval pointed out that I (!) actually proved an almost identical result 8 years ago, the existence of which I completely forgot until Anne conjured this up in the comment. At least my answer is not a duplicate, since the main idea of my former proof is quite different from what we have here.

• Although not necessary, let me explain the intuition behind the proof below. In a numerical experiment, I noticed that the graph of $y = f_N^{\circ \lfloor xN \rfloor}(a)$, as a function of $x$, approach some deterministic curve, say $y = g(x)$, as $N\to\infty$. To determine the formula for $g(x)$, I assumed $f_N^{\circ k}(a) \approx g(k/N)$ and then noticed that \begin{align*} g'(\tfrac{k}{N}) &\approx N[g(\tfrac{k+1}{N}) - g(\tfrac{k}{N})] \approx N[f_N^{\circ (k+1)}(a) - f_N^{\circ (k)}(a)] \\ &\approx -\frac{1}{6}f_N^{\circ (k)}(a)^3 \approx -\frac{1}{6}g(\tfrac{k}{N})^3. \end{align*} This suggests that $g$ has to solve the differential equation $g' = -\frac{1}{6}g^3$. Solving this equation then gives $g(x) = \sqrt{\frac{3}{x + c}}$ for some constant $c$. This naturally leads us to consider the function $\varphi(x)$.

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Proof of Theorem. To begin with, we note that the following inequalities hold for any $N \geq 1$ and $a, t > 0$.

$$ \left| f_N(a) - \left(a - \frac{a^3}{6N}\right) \right| \leq \frac{a^5}{120N^2}. \tag{1} $$

$$ \left| \varphi(t+N^{-1}) - \left( \varphi(t) - \frac{\varphi(t)^3}{6N}\right) \right| \leq \frac{\varphi(t)^5}{24N^2}. \tag{2} $$

Indeed, both inequalities are easy consequences of Taylor's theorem.

Now we fix $x > 0$ and $N \geq \varphi(x)^2$, and define the sequences $(a_k)$, $(b_k)$, and $(\varepsilon_k)$ by

$$ a_k = f^{\circ k}(\varphi(x)), \qquad b_k = \varphi\left(x + \frac{k}{N}\right), \qquad \varepsilon_k = a_k - b_k. $$

Since $0 \leq f_N(t) \leq t$ whenever $t \in [0, \sqrt{N}]$, it follows that $(a_k)$ is positive and monotone decreasing. In particular, we have $a_k, b_k \in [0, \varphi(x)]$. Now, define $R_k$ by

$$ R_k = \biggl[ a_{k+1} - \biggl( a_k - \frac{a_k^3}{6N} \biggr) \biggr] - \biggl[ b_{k+1} - \biggl( b_k - \frac{b_k^3}{6N} \biggr) \biggr]. $$

Then by $(1)$ and $(2)$, we have $ |R_k| \leq \frac{\varphi(x)^5}{20N^2} $ for all $k \geq 0$. Then

\begin{align*} \left| \varepsilon_{k+1} \right| &= \left| \varepsilon_{k} - \frac{a_k^3 - b_k^3}{6N} + R_k \right| \\ &= \left| \varepsilon_{k}\left( 1 - \frac{a_k^2 + a_k b_k + b_k^2}{6N}\right) + R_k \right| \leq |\varepsilon_k| + |R_k|, \end{align*}

hence it follows that

$$ \left| \varepsilon_{k} \right| \leq \frac{\varphi(x)^5}{20N^2}k. $$

So, plugging $k = \lfloor yN\rfloor$ for $y > 0$ gives

$$ \left| \varepsilon_{\lfloor yN\rfloor} \right| \leq \frac{\varphi(x)^5 y}{20N} \xrightarrow{N\to\infty} 0. $$

Combining this with the fact $b_{\lfloor yN\rfloor} \to \varphi(x + y)$, the proof is complete. $\square$