Proof $\lim_{n\to\infty} \cos^{2n}\left(\frac{x}{\sqrt{n}}\right) = e^{-x^2}$

Question

I have it in the notes of an old project that $\cos^{2n}\left(\frac{x}{\sqrt{n}}\right) \to e^{-x^2}$ pointwise as $n\to \infty$. The proof I gave at the time sounds bogus, but it's quite close empirically for $n\ge 10$.

The bogus argument is basically that

$$\cos^{2}\left(\frac{x}{\sqrt{n}}\right)\approx \left(1-\frac{x^2}{n}\right)$$ and $$ \lim_{n\to\infty} \left(1-\frac{x^2}{n}\right)^n = \exp(-x^2) $$ by the definition of $\exp$. But unless I'm forgetting something nice about limits, I see no reason why the quality of the approximation improves if I take both sides to a power.

A promising argument that got nasty is to show that $\lim_{n\to\infty} \cos^{2n}\left(\frac{x}{\sqrt{n}}\right)$ has the same Taylor series expansion as \begin{align} e^{-x^2}&=\sum_{k=0}^\infty \frac{(-x^2)^k}{k!}\\ &=\sum_{\text{$k$ even}}\frac{(-1)^{k/2}k!}{(k/2)!}\frac{x^k}{k!}.\end{align}

Well, consider

$$ \left. \frac{d^k}{dx^k} \cos^{2n}\left(\frac{x}{\sqrt{n}}\right)\right|_{x=0}.$$

Each differentiation operation on a product of $2n$ sines and cosines yields a sum of $2n$ such products with one of the factors differentiated. So we have a sum of $(2n)^k$ terms of the form

$$\frac{\pm 1}{\sqrt{n}^k} \cos^{2n-m}\left(\frac{x}{\sqrt{n}}\right)\sin^m\left(\frac{x}{\sqrt{n}}\right)$$

where for each term, $m\le k$. At $x=0$, nonvanishing terms have $m=0$ (which is only possible for $k$ even) and take the value $\pm1/\sqrt{n}^k$. For $n\gg k$, the overwhelmingly typical term will have each cosine differentiated either twice or never. So for $k$ even,

$$\frac{d^k}{dx^k} \lim_{n\to\infty} \left.\cos^{2n}\left(\frac{x}{\sqrt{n}}\right)\right|_{x=0} = \lim_{n\to\infty} \frac{(-1)^{k/2}}{n^{k/2}} {2n \choose k/2} \alpha,$$

where $\alpha$ is the number of words of length $k$ drawn from an alphabet of $k/2$ letters in which each letter appears exactly twice.

I confess to running out of steam at this point. I actually have the sneaking suspicion I'm embarrassingly close with both approaches. I just don't want to get caught in a sunk cost fallacy just because I feel close. I don't need this, empirical confirmation is fine for my application, it's just vexing.

Does the $k$th derivative somehow somehow converge to $(-1)^{k/2}\frac{k!}{(k/2)!}$? Or was the argument from the definition of $\exp$ just in need of the right application of something silly like l'Hospital's rule that I can't quite see?

Answer 1

Another approach, and a standard trick for dealing with limits of the indeterminate form "$1^{\infty}$", is to take a log to convert it to "$\infty \cdot 0$", then use another trick (series or L'Hopitals' rule) to evaluate that.

Write $L_n(x) = \ln\left(\cos^{2n} \left(\frac{x}{\sqrt n}\right)\right)$. Then \begin{align*} \lim_{n\to \infty} L_n(x) & = \lim_{n\to\infty} 2n \ln \cos\left(\frac{x}{\sqrt n}\right) \\&= \lim_{n\to\infty} \frac{2\ln \cos\left(\frac{x}{\sqrt n}\right)}{n^{-1}} \\&= \lim_{n\to\infty} \frac{2\cdot \frac{-\sin(x/\sqrt n)}{\cos(x/\sqrt{n})} \cdot (-1/2)n^{-3/2}x}{(-1)n^{-2}} \\&= \lim_{n\to\infty} -\tan(x/\sqrt{n}) \cdot n^{1/2}x \\&= \lim_{t \to 0^+} \frac{-\tan(tx)}{t}\cdot x \\&= \lim_{t \to 0^+} \frac{-\sec^2(tx) \cdot x}{1}\cdot x \\&= -x^2. \end{align*}

Above, I'm using the substitution $t=1/\sqrt{n}$, along with L'Hopital's rule a couple of times.

Now,

$$ \lim_{n\to\infty} \cos^{2n} \left(\frac{x}{\sqrt n}\right) = \lim_{n\to\infty} e^{L_n(x)} = e^{\lim_{n\to\infty} L_n(x)} = e^{-x^2}. $$

Answer 2

Your first idea is basically correct. Fix $x$. We know $\cos(x/\sqrt{n})=1-\frac{x^2}{2n}+\mathcal{O}(1/n^2)$, that $\log(1+y)=y+\mathcal{O}(y^2)$ as $y\to0$, and therefore that $\log\cos(x/\sqrt{n})=-\frac{x^2}{2n}+\mathcal{O}(1/n^2)$ as $n\to\infty$ (we can chain it because the error $\cos(x/\sqrt{n})-1=:y\to0$ as $n\to\infty$).

Now it follows $\log\cos^{2n}(x/\sqrt{n})=-x^2+\mathcal{O}(1/n)$ tends to $-x^2$ as $n\to\infty$, which is what you wanted.

The Big-$\mathcal{O}$ notation is rigorous. Replace all instances of it with explicit bounds if you so wish.

Answer 3

Let $L= \lim_{n\to\infty} \cos^{2n}\left(\frac{x}{\sqrt{n}}\right)$

$logL=\lim_{n\to\infty} log(\cos^{2n}\left(\frac{x}{\sqrt{n}}\right))$

Using the Taylor expansion of $cosx$, we get $cos^{2n}(\frac{x}{\sqrt{n}})=(1-\frac{x^2}{2!n}+\frac{x^4}{4!n^2}+...)^{2n}$.

Thus we get $logL=\lim_{n\to\infty} (2n)$log$(1-\frac{x^2}{2!n}+\frac{x^4}{4!n^2}+...)$. Since $n\to\infty$, we can neglect the terms starting from $\frac{x^4}{4!n^2}$ to get the following $logL=\lim_{n\to\infty} (2n)$log$(1-\frac{x^2}{2!n})$.

Using series expansion of log($1-x$),

$logL=\lim_{n\to\infty} (2n)(-\frac{x^2}{2!n}-\frac{x^4}{2!^2n^2}...)= -x^2$ after neglecting the other terms.

$logL=-x^2$ so $L=e^{-x^2}$ which is the required result.