# Find a two term expansion of the following equation

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?

• You haven't considered the possibility $\alpha = 1$. Then the $\epsilon^1$ terms in $x^2$ and in $2 x$ cancel each other out. – Maxim Mar 24 at 22:29
• I'm not sure how you're equating the terms. What you should get is $$x = 1 + x_1 \epsilon + x_2 \epsilon^2 + O(\epsilon^3), \\ x^2 - 2 x + (1 - \epsilon^2)^{25} = (x_1^2 - 25) \epsilon^2 + O(\epsilon^3).$$ – Maxim Apr 1 at 13:46

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 $$\sqrt{1-12\epsilon^2}=1+\Big(\frac12\Big)(-12\epsilon^2)+...=1-6\epsilon^2+...$$ $$\therefore x_{1,2}\approx1\pm5\epsilon(1-6\epsilon^2)=1\pm5\epsilon\mp30\epsilon^3$$