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I want to find a two term expansion of the form $x\sim x_0 + \epsilon^\alpha x_1 + \epsilon^\beta x_2 + \ldots$, with $\alpha < \beta < \ldots$, for small $\epsilon$, of each solution $x$ of the following equation: $$ x^2 - 2x + (1 - \epsilon^2)^{25} = 0 $$ I've substituted $x\sim x_0 + \epsilon^\alpha x_1 + \epsilon^\beta x_2 + \ldots$ into the equation to arrive at: $$ (x_0 + \epsilon^\alpha x_1 + \epsilon^\beta x_2 + \ldots)^2 -2(x_0 + \epsilon^\alpha x_1 + \epsilon^\beta x_2 + \ldots) + (1 - \epsilon^2)^{25} = 0 $$ Since I want to find a two term expansion for the equation I want to find $x_0, x_1$ and $\alpha$. I will try to do so by inspecting the terms in the above equation.

$\mathcal{O}(1)$: If I look at the $\mathcal{O}(1)$ terms I find that $x_0$ needs to be chosen such that it satisfies $$ x_0^2 - 2x_0 + 1 = 0 $$ The $1$ in this equation comes from the fact that $(1 - \epsilon^2)^{25} = 1+ \text{terms involving $\epsilon$}$. Solving this equation gives that $x_0 = 1$. Also since the $(1 - \epsilon^2)^{25}$ involves a term that is of order $\epsilon^2$, I know that I need $\alpha = 2$ for balance.

$\mathcal{O}(\epsilon^2)$: If I look at the $\mathcal{O}(\epsilon^2)$ terms I find that I need to choose $x_1$ such that it satisfies $$ 2x_0x_1 - 2x_1 - 1 = 0 $$ Where the $-1$ comes from the fact that $(1 - \epsilon^2)^{25}$ expansion contains $-\epsilon^2$. Since $x_0 = 1$ this means that I need to find $x_1$ such that $$ 2x_1 - 2x_1 -1 = 0 $$ But this is not possible..

Question: How do I find a two term expansion for the given equation? What am I doing wrong in my approach?

Edit: As Maxim pointed out, I did not consider the possibility that $\alpha = 1$. If $\alpha = 1$ I get $2x_0x_1 - 2x_1 = 0$ which is fine since $x_0= 1$. If $\alpha = 1$ then $\beta = 2$ because I need a term to balance the $-\epsilon^2$ in $(1 - \epsilon^2)^{25}$. With $\alpha = 1$ and $\beta = 2$ I get the following equation: $$ 2x_0x_2 -2x_2 + x_1 -1 = 0\Leftrightarrow x_1 = 1 $$ Since I now know $x_0$ and $x_1$ I think I have a two term asymptotic expansion that looks like this: $$ x = 1 + \epsilon + \mathcal{O}(\epsilon^3) $$ I'm not sure that this expansion is the result that I want though since I know that the equation $$ x^2 - 2x + (1 - \epsilon^2)^{25} = 0 $$ has two solutions. As $\epsilon$ goes to zero, the first solution approaches one from below while the second solution approaches one from above. It seems that the expansion that I found is only an approximation of the second solution. Is this correct?

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  • $\begingroup$ You haven't considered the possibility $\alpha = 1$. Then the $\epsilon^1$ terms in $x^2$ and in $2 x$ cancel each other out. $\endgroup$ – Maxim Mar 24 at 22:29
  • $\begingroup$ I'm not sure how you're equating the terms. What you should get is $$x = 1 + x_1 \epsilon + x_2 \epsilon^2 + O(\epsilon^3), \\ x^2 - 2 x + (1 - \epsilon^2)^{25} = (x_1^2 - 25) \epsilon^2 + O(\epsilon^3).$$ $\endgroup$ – Maxim Apr 1 at 13:46
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The solutions to this equation are given by the quadratic formula as $$x_{1,2}=\frac{2\pm\sqrt{(-2)^2-4(1)(1-\epsilon^2)^{25}}}{2}=1\pm\sqrt{1-(1-\epsilon^2)^{25}}$$ Now $$(1-\epsilon^2)^{25}=\sum_{k=0}^{25}\binom{25}{k}(-\epsilon^2)^k=1-25\epsilon^2+300\epsilon^4-...-\epsilon^{50}$$ So the roots are given by $$x_{1,2}=1\pm\sqrt{1-(1-25\epsilon^2+300\epsilon^4-...-\epsilon^{50})}$$ $$=1\pm\sqrt{25\epsilon^2-300\epsilon^4+...+\epsilon^{50}}$$ $$=1\pm\epsilon\sqrt{25-300\epsilon^2+...+\epsilon^{48}}$$ Now for small $\epsilon$ we can ignore terms of $\epsilon^4$ and greater giving the following approximation $$x_{1,2}\approx1\pm\epsilon\sqrt{25-300\epsilon^2}$$ $$=1\pm5\epsilon\sqrt{1-12\epsilon^2}$$ The binomial expansion of the latter term is given by $$\sqrt{1-12\epsilon^2}=1+\Big(\frac12\Big)(-12\epsilon^2)+...=1-6\epsilon^2+...$$ $$\therefore x_{1,2}\approx1\pm5\epsilon(1-6\epsilon^2)=1\pm5\epsilon\mp30\epsilon^3$$

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