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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.

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  • $\begingroup$ $$ 1+x\sqrt{1+\frac{\epsilon}{x^2}}\approx 1+ x\left(1 + \frac{1}{2x^2}\epsilon - \frac{1}{8x^4}\epsilon^2\right) =1 + x +\frac{1}{2x}\epsilon - \frac{1}{8x^3}\epsilon^2 $$ 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
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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 $x_0 = 0$.

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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.

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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 $\approx 0.312826$

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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