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I want to determine whether or not $\cos x$ has an asymptotic expansion of the form $\sum_{k=0}^{\infty} \frac{a_k}{x^k}$ as $x \to \infty$ in $\mathbb{R^+}$.

This means we are taking the asymptotic series $\psi_k(x) = x^{-k}$, but I don't really know how to go about starting this.

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    $\begingroup$ $\cos\left(\frac{1}{x}\right)$ is not a holomorphic function in a neighbourhood of zero: it has an essential singularity there. $\endgroup$
    – Jack D'Aurizio
    Commented May 18, 2017 at 18:59
  • $\begingroup$ @JackD'Aurizio But my function is $\cos(x)$, why can we look at $\cos\left(\frac{1}{x}\right)$? Oh, is it that we can look at $\cos(1/x)$ as x tends to 0 as that's the same as $\cos(x)$ as x tends to infinity? $\endgroup$
    – user112495
    Commented May 18, 2017 at 19:01
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    $\begingroup$ @user112495 $\lim_{x\to \infty}\cos(x)$ does NOT exist. $\endgroup$
    – Mark Viola
    Commented May 18, 2017 at 19:01
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    $\begingroup$ Apply the change of variable $x\to\frac{1}{x}$. Assuming that $\cos(x)$ has such expansion, it follows that $\cos\frac{1}{x}$ has a Taylor series at the origin. $\endgroup$
    – Jack D'Aurizio
    Commented May 18, 2017 at 19:02
  • $\begingroup$ @JackD'Aurizio very instructive argument! $\endgroup$
    – tired
    Commented May 18, 2017 at 19:02

1 Answer 1

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$$\lim_{n\to+\infty}\cos (2n\pi)=1$$

$$\lim_{n\to+\infty}\cos (\frac {\pi}{2}+n\pi)=0$$

thus $\lim_{x\to+\infty}\cos (x)$ doesn't exist and so does $\lim_{x\to+\infty}x^n\cos (x) $.

therefore the asymptotic expansion doesn't exist.

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