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I'm trying to solve an exercise whose text reads thus:

Show that for $|x|\ll a$, to within $(\frac{x}{a})^2$, we have the approximate equality $$e^{\frac{x}{a}}\simeq \sqrt{\frac{a+x}{a-x}}$$

What I think the text means is that I should prove that $e^{\frac{x}{a}}- \sqrt{\frac{a+x}{a-x}}<(\frac{x}{a})^2 $ for sufficiently small values of $|x|$. Is it right? I don't really undestand what "to within" is supposed to mean (not a native english-speaker). In case my interpreation is correct, is my solution fine as well?

By using Taylor expansion, we derive the following equalities $$e^{\frac{x}{a}}=1+\frac{x}{a}+\frac{g''(\xi_1)}{2}x^2$$ $$ f(x)=\sqrt{\frac{a+x}{a-x}}=1+\frac{x}{a}+\frac{f''(\xi_2)}{2}x^2 $$ with $|\xi_{1,2}|<|x|$ Subtracting the two quantities we get $$ e^{\frac{x}{a}}- \sqrt{\frac{a+x}{a-x}}=\frac{g''(\xi_1)-f''(\xi_2)}{2}x^2 $$. Thus we have to prov that $\frac{g''(\xi_1)-f''(\xi_2)}{2}<\frac{1}{a^2} $. Considering that $$g''(\xi_2)=\frac{a(2\xi_2 +a)}{(a+\xi_2)^{\frac{3}{2}}(a-\xi_2)^{\frac{5}{2}}}>0$$, since $|\xi_2| \ll a$, it is sufficient to prove that $$\frac{g''(\xi_1)}{2}= \frac{e^{\frac{\xi_1}{a}}}{2a^2}\leq \frac{1}{a^2} $$, which holds for $\xi_1<a\log{2}$, hence the thesis.

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  • $\begingroup$ $$e^{\frac{x}{a}}- \sqrt{\frac{a+x}{a-x}} = -\frac{x^3}{3a^3} + o(x^3)$$ As a result $e^{\frac{x}{a}}- \sqrt{\frac{a+x}{a-x}}<0$ for sufficiently small $x$. $\endgroup$
    – Gabriel Romon
    Sep 30, 2016 at 12:19
  • $\begingroup$ Does this entail in some way that $e^{\frac{x}{a}}\simeq \sqrt{\frac{a+x}{a-x}}$? $\endgroup$
    – Nicol
    Sep 30, 2016 at 12:31
  • $\begingroup$ yes it does. It says exactly $e^{\frac{x}{a}}- \sqrt{\frac{a+x}{a-x}} \sim -\frac{x^3}{3a^3}$ in the sense of asymptotic equivalence. Since $-\frac{x^3}{3a^3}$ goes to $0$, that implies $e^{\frac{x}{a}}\sim\sqrt{\frac{a+x}{a-x}}$ $\endgroup$
    – Gabriel Romon
    Sep 30, 2016 at 12:37
  • $\begingroup$ @LeGrandDODOM how did you derive that equality though? $\endgroup$
    – Nicol
    Sep 30, 2016 at 14:41
  • $\begingroup$ It's pure asymptotic calculus. If you have never done this, practice a lot with simpler functions $\endgroup$
    – Gabriel Romon
    Sep 30, 2016 at 19:38

2 Answers 2

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$$\log\left(\frac{1+y}{1-y}\right)=\int_0^y\frac{2\,dt}{1-t^2}=\int_0^y\left(2+\frac{2t^2}{1-t^2}\right)\,dt=\int_0^y(2+O(t^2))\,dt=2y+O(y^3)$$

Let $y=x/a\implies$

$$ \log\frac{1+x/a}{1-x/a}=\frac{2x}{a}+O((x/a)^3)\implies\frac{1}{2}\log\frac{a+x}{a-x}=\frac{x}{a}+O((x/a)^3)$$

Exponentiate both sides to see the desired result.

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It is enough to prove that in a neighbourhood of the origin we have $e^{2x}\approx\frac{1+x}{1-x}$, or $$ x\approx \frac{1}{2}\log\left(\frac{1+x}{1-x}\right) = \text{arctanh}(x) \tag{1}$$ that simply follows from $$ \text{arctanh}(x)=\sum_{n\geq 0}\frac{x^{2n+1}}{2n+1}=x+O(x^3).\tag{2} $$

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