1

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.

Lee David Chung Lin
  • 7,043
user112495
  • 384
  • 4
  • 13
  • 2
    $\cos\left(\frac{1}{x}\right)$ is not a holomorphic function in a neighbourhood of zero: it has an essential singularity there. – Jack D'Aurizio May 18 '17 at 18:59
  • @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? – user112495 May 18 '17 at 19:01
  • 2
    @user112495 $\lim_{x\to \infty}\cos(x)$ does NOT exist. – Mark Viola May 18 '17 at 19:01
  • 2
    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. – Jack D'Aurizio May 18 '17 at 19:02
  • @JackD'Aurizio very instructive argument! – tired May 18 '17 at 19:02
  • @JackD'Aurizio If we were considering $\sqrt{x}$ instead, then would this not have an expansion in terms of the series $\psi_k(x) = x^{-k}$ ($x \rightarrow \infty$) or $\phi_k(x) = x^k$ ($x \rightarrow 0$) as it isn't differentiable at 0 (or similar for $1/\sqrt{x}$)? – user112495 May 18 '17 at 19:08

1 Answers1

1

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

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

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.

hamam_Abdallah
  • 62,951