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Here

What is the result of this equation?

I mentioned the approximation $$\arccos(e^{-x})\approx \sqrt{2x}\ for\ \ x\approx 0$$

but I could not give a reason, why this is the case , with calculation by hand.

How can I justify the given approximation by hand ?

I tried the taylor-series but the function $\arccos(e^{-x})$ is not differentiable at $x=0$.

Do I need a substitution, or the expansion of $\arccos^2(e^{-x})$ ? Or can I expand $\arccos(e^{-x})$ directly ?

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Peter
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5 Answers5

4

The reason why you cannot use a Taylor expansion around $0$ is the fact that $\arccos \Bbb e ^{-x}$ is not defined to the left of $0$ (where $\Bbb e ^{-x} > 1$, and $\arccos$ is defined only on $[-1,1]$).

One way is to show that

$$\lim \limits _{x \to 0, x > 0} \frac {\arccos \Bbb e ^{-x}} {\sqrt {2x}} = 1 .$$

Since this is a case of $\frac 0 0$, the usual procedure goes through l'Hospital's theorem:

$$\lim \limits _{x \to 0, x > 0} \frac {\arccos \Bbb e ^{-x}} {\sqrt {2x}} = \lim \limits _{x \to 0, x > 0} \frac {\sqrt {2x}} {-\sqrt {1 - \Bbb e ^{-2x}}} (- \Bbb e ^{-x}) = \lim \limits _{x \to 0, x > 0} \frac {\sqrt {2x}} {\sqrt {\Bbb e ^{2x} -1}} = \\ \sqrt {\lim \limits _{x \to 0, x > 0} \frac {2x} {\Bbb e ^{2x} -1}} = \dots$$

and applying l'Hospital's theorem again under the square root gives

$$\dots = \sqrt {{\lim \limits _{x \to 0, x > 0} \frac 2 {2 \Bbb e ^{2x}}}} = 1 .$$

Alex M.
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If $y=\arccos e^{-x}$ then $e^{-x}=\cos(y)=1-2\sin^2\frac y2$ and from that $$ y=2\arcsin\sqrt{\frac{1-e^{-x}}2} $$ Now employ Taylor approximations $2\arcsin(u)=2u+\frac13u^3+O(u^5)$ and $$ \sqrt{\frac{1-e^{-x}}2}=\sqrt{\frac x2}\sqrt{1-\frac12x+O(x^2)} =\sqrt{\frac x2}(1-\frac14 x+O(x^2)) $$ to get approximatively \begin{align} y&=\sqrt{2x}(1-\frac14 x+O(x^2))+\frac13\frac{x\sqrt{2x}}4+O(x^{5/2})\\ \arccos e^{-x}&=\sqrt{2x}(1-\frac16 x+O(x^2)) \end{align}

$$\scriptsize \arccos e^{-x}=\sqrt{2x}·\left(1 - \frac16 x + \frac1{120}x^2 + \frac1{336}x^3 - \frac1{5760}x^4 - \frac{19}{126720}x^5 + O(x^6)\right) $$

Lutz Lehmann
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \arccos\pars{\expo{-x}} & = y \implies \expo{-x} = \cos\pars{y} \end{align} Since $\ds{\pars{~x \to 0 \implies \cos\pars{y} \to 1~}}$: \begin{align} &\pars{~1 - x \sim 1 - {1 \over 2}\,y^{2}\quad\mbox{as}\quad x \to 0~} \implies \pars{~y \sim \root{2x}\quad\mbox{as}\quad x \to 0~} \\[5mm] &\implies \bbx{\ds{\arccos\pars{\expo{-x}} \sim \root{2x}\quad\mbox{as}\quad x \to 0}} \end{align}

Felix Marin
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You can compare the series $\enspace\displaystyle \cos\sqrt{2x}=\sum\limits_{k=0}^\infty (-1)^k\frac{(2x)^k}{(2k)!}\enspace$ and $\enspace \displaystyle e^{-x}=\sum\limits_{k=0}^\infty (-1)^k\frac{x^k}{k!}$ .

They have the linear part $\enspace 1-x\enspace$ common. The difference begins with the quadratic terms.

user90369
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  • Can I estimate the difference of $\arccos(e^{-x})$ and $\sqrt{2x}$ based on these expansions ? – Peter Nov 05 '16 at 19:35
  • @Peter Sure, for small $x$ it is dominated by the difference of the quadratic terms. – Ian Nov 05 '16 at 19:39
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    @Ian: That gives you the difference between $e^{-x}$ and $\cos\sqrt{2x}$, not the difference between $\arccos(e^{-x})$ and $\sqrt{2x}$. – TonyK Nov 05 '16 at 19:41
  • @TonyK Fair point, but you could run the same thing with $\arccos$ instead. You can't directly Taylor expand, but you can do the analogous thing (keeping in mind that $\arccos$ is not differentiable at $1$). – Ian Nov 05 '16 at 20:33
  • @Ian: It gets complicated, doesn't it? Just saying "the analogous thing" is dodging the issue. – TonyK Nov 05 '16 at 21:30
  • @TonyK It doesn't really get complicated. The issue is just that you have to expand in non-integer powers because you don't have differentiability of $\arccos$ at $1$. But since we know $\cos''(0)$ is nonzero, it suffices to expand in half-integer powers, so that the problem reduces to finding the second coefficient in the power expansion of $\arccos$ at $1$. – Ian Nov 05 '16 at 21:34
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PRIMERS:

In THIS ANSWER, I showed using only elementary geometry that the arccosine function satisfies the inequalities

$$\sqrt{1-x^2}\le \arccos(x)\le \frac{1-x^2}{x} \tag 1$$

for $0<x\le 1$.

And in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$1+x\le e^x\le \frac{1}{1-x} \tag 2$$

for $x<1$.

Putting $(1)$ and $(2)$ together, we have for $x<1$

$$\frac{1}{\sqrt{1+2x}} \le \frac{\arccos(e^{-x})}{\sqrt{2x}}\le \frac{1}{1-x}$$

whereupon applying the squeeze theorem we obtain

$$\lim_{x\to 0}\frac{\arccos(e^{-x})}{\sqrt{2x}}=1$$

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Mark Viola
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