I am trying to find a two term expansion for the root of $$1+\sqrt{(x^2+\epsilon)}=e^x$$. Since $$\epsilon << 1$$, I can tell that this equation behaves like $$1+x=e^x$$ which has a root close to zero. That gave me the idea that the root might look like $$x \approx x_0 + \epsilon^{\alpha} x_1 + \dots $$. I used Taylor expansion for both functions $\sqrt{(x^2+\epsilon)}$ and $e^x$ and then plugged my guess into the Taylor expanded equation and balance of O(1) terms gave me $$x_0=0$$ as expected. My problem begins now. For the higher orders of $\epsilon$. Long story short this expansion ends up with some inconsistencies. So I thought maybe $x \approx x_0+\mu(\epsilon)$. But then again I get nowhere. I am getting frustrated and I would really appreciate some help.

Thank you in advance.

  • $\begingroup$ $$ 1+x\sqrt{1+\frac{\epsilon}{x^2}}\approx 1+ x\left(1 + \frac{1}{2x^2}\epsilon - \frac{1}{8x^4}\epsilon^2\right) =1 + x +\frac{1}{2x}\epsilon - \frac{1}{8x^3}\epsilon^2 $$ as we can see that we have $1/x$ terms that do not vanish (maybe with very specific conditions on $\epsilon$ with respect to $x$ and the number of terms in the expansion. $\endgroup$ – Chinny84 Sep 10 '17 at 2:21
  • $\begingroup$ $x \geq 0$, as $\sqrt{x^2+\epsilon} \geq 0$ by definition. Then by the fact $f(x)=x^2$ is injective on $[0,\infty)$, the following equation is made equivalent to your's: $\epsilon=(e^x-1)^2-x^2$. All that remains is to use inversion to find the series for the inverse of $(e^x-1)^2-x^2$. $\endgroup$ – Ahmed S. Attaalla Sep 10 '17 at 3:46

Here's a hint from my instructor. Rearranging the given equation as follows \begin{align*} x^2 + \varepsilon & = (e^x - 1)^2 \\ \varepsilon & = (e^x - 1)^2 - x^2 \\ & = (e^x - 1 - x)(e^x - 1 + x) \end{align*} You can then Taylor expand $e^x$ around 0, substitute the given asymptotic expansion and try to balance term. Note that for $\varepsilon=0$, $x=0$ is the solution to the equation $1 + \sqrt{x^2} = e^x$, so you may choose $x_0 = 0$.


enter image description here

The first thing is to solve it numerically to vizualize how $x_\varepsilon$ varies in function of $\varepsilon$.

Above is the graph of :$\displaystyle\quad\frac{\ln(\varepsilon)}{\ln(x_\epsilon)}$ when $\varepsilon=10^{-a}\quad a\in[1,10]$

So we can consider that $\varepsilon\ll 1\implies \varepsilon=O(x^3)$ or similarly $x=O(\sqrt[3]\varepsilon)$

Now we can compare the development of $\displaystyle \sqrt{1+\frac{\varepsilon}{x^2}}$ and the one of $e^x$ since $\displaystyle \frac{\varepsilon}{x^2}=O(x)$.

  • $\displaystyle e^x=1+x+\frac 12 x^2+\frac 16 x^3+O(x^4)$

  • $\displaystyle 1+\sqrt{x^2+\varepsilon}=1+x\left(1+\frac 12 \frac{\varepsilon}{x^2} -\frac 18 \frac{\varepsilon^2}{x^4}+O(x^3)\right)=1+x+\underbrace{\frac 12 \frac{\varepsilon}{x}}_{O(x^2)} -\underbrace{\frac 18 \frac{\varepsilon^2}{x^3}}_{O(x^3)}+O(x^4)$

If we consider the development up to $O(x^2)$ we get :

$\displaystyle \frac 12 x^2\sim\frac 12\frac{\varepsilon}x\iff \bbox[5px,border:2px solid]{x\sim\sqrt[3]{\varepsilon}}\quad$ this is more precise than just $x=O(\sqrt[3]{\varepsilon})$.

If we consider now the development up to $O(x^3) $ we get :

$\displaystyle \frac 12 x^2+\frac 16x^3=\frac 12\frac{\varepsilon}x-\frac 18\frac{\varepsilon^2}{x^3}+o(\varepsilon)$

Let's have $x^3=\varepsilon+u\quad$ with $u=o(\varepsilon)\quad$

(we could have instead $x=\sqrt[3]{\varepsilon}+u$, but it complicates the calculation)

$\begin{array}{l}\require{cancel} 12x^5+4x^6=12x^2\varepsilon-3\varepsilon^2+o(\varepsilon^2)\\ 12x^2(x^3-\varepsilon)+4(u+\varepsilon)^2+3\varepsilon^2=o(\varepsilon^2)\\ 12x^2u+\cancel{4u^2}+\cancel{8u\varepsilon}+7\varepsilon^2=o(\varepsilon^2) & u^2,u\varepsilon=o(\varepsilon^2)\text{ so we cancel them } \end{array}$

In the same way $12x^2u\sim 12\varepsilon^\frac 23u$ other terms are negligible.

So we get $\quad\displaystyle 12\varepsilon^\frac 23u\sim-7\varepsilon^2\iff u\sim -\frac 7{12}\varepsilon^\frac 43$

Now reporting in $x^3=\varepsilon+u\quad$ we get $\quad \bbox[5px,border:2px solid]{x=\varepsilon^\frac 13-\frac 7{36}\varepsilon^\frac 23+o(\varepsilon^\frac 23)}$

We could continue the development, but it becomes delicate to know which terms we can keep and which ones we can dismiss.


Starting from Chee Han's answer, we have $$\epsilon=(e^x - 1 - x)(e^x - 1 + x)$$ Now, using Taylor expansion around $x=0$ $$(e^x - 1 - x)=\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+O\left(x^6\right)$$ $$(e^x - 1 + x)=2 x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+O\left(x^6\right)$$ making $$\epsilon=x^3+\frac{7 x^4}{12}+\frac{x^5}{4}+\frac{31 x^6}{360}+O\left(x^7\right)$$ Now, use series reversion to get $$x= \epsilon ^{1/3}-\frac{7}{36} \epsilon ^{2/3}+\frac{13 }{432}\epsilon-\frac{733 }{699840}\epsilon ^{4/3}+O\left(\epsilon ^{5/3}\right)$$ Let us try with $\epsilon=\frac 1 {27}$ (which is quite large); the above expansion will give $x=\frac{17733407}{56687040}\approx 0.312830$ while the "exact" solution would be $\approx 0.312826$

  • $\begingroup$ I wasn't aware of this series reversion process, seems powerful. Glad it agrees with my result though :-) $\endgroup$ – zwim Sep 10 '17 at 6:38
  • 1
    $\begingroup$ @zwim. It is a very important thing to know about and extremely useful when trying to make approximations. Moreover, as you probably noticed, it is extremely simple. Cheers. $\endgroup$ – Claude Leibovici Sep 10 '17 at 6:52
  • $\begingroup$ If you don't mind may I ask you how you perform series reversion here when $a_1=0$, every time I try something like In the link provided I end up with some nonsense like $0=1$. Furthermore the exponents of $\epsilon$ are not integers, how did you get the series? $\endgroup$ – Ahmed S. Attaalla Sep 11 '17 at 21:04
  • $\begingroup$ @AhmedS.Attaalla. I do not see what you mean. Could you be more explicit ? For the problem of the powers, let $\epsilon=\delta^3$ from the start. $\endgroup$ – Claude Leibovici Sep 12 '17 at 4:22
  • 1
    $\begingroup$ @AhmedS.Attaalla. Consider $\delta^3=x^3+\frac{7 x^4}{12}$. Solve the quartic for $x$ and use Taylor expansion around $\epsilon=0$. You will see which powers are coming. $\endgroup$ – Claude Leibovici Sep 12 '17 at 4:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.