Taylor series $\sqrt{1-2x+x^2+o(x^3)}$

Question

I need to solve this asymptotic expansion for $x\to0$

$$\sqrt{1-2x+x^2+o(x^3)}$$

This is the expansion to use:

$$\sqrt{1+t} = 1+ \frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+o(x^3)$$

let: $$t = -2x+x^2+o(x^3)$$

My question is, should I stop to the first order or keep going? How can I decide when to stop?

I have tried first and second order and I get:

$$\sqrt{1+t}= 1+ \frac{-2x+x^2+o(x^3)}{2}+o(-2x+x^2+o(x^3) = 1-x+\frac{1}{2}x^3+o(x^3)$$

The second grade:

$$\sqrt{1+t}= 1+ \frac{-2x+x^2+o(x^3)}{2} - \frac{\big(-2x+x^2+o(x^3)\big)^2}{8} +o\big(-2x+x^2+o(x^3)\big)^2 = 1-x+4x^2+\frac{1}{2}x^3-4x^3-\frac{1}{8}x^4+x^4+o(x^4)$$

Answer 1

Since we have an $o(x^3)$ term we can use the following

$$\sqrt{1+t} = 1+ \frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+o(t^3)$$

that is

$$\sqrt{1-2x+x^2+o(x^3) }=1+ \frac{-2x+x^2+o(x^3)}{2}-\frac{(-2x+x^2+o(x^3))^2}{8}+\frac{(-2x+x^2+o(x^3))^3}{16}+o((-2x+x^2+o(x^3))^3)$$

keeping only the term with order less than $o(x^3)$, that is

$$1-x+\frac12x^2 -\frac12x^2+\frac12x^3 -\frac12x^3+o(x^3) =1-x+o(x^3)$$

Answer 2

You are being told to work in the ring $\Bbb Q[[x]]/(x^3)$, and to find there a square root of $1-2x+x^2$. I think that the answer is obvious.

Answer 3

Since $1-2x+x^2 = (1-x)^2$, it is more practical to check which coefficient $K$ ensures $$\left[1-x+K x^2+o(x^3)\right]^2 = 1-2x+x^2+o(x^3) $$ but the coefficient of $x^2$ in the LHS is $2K+1$, hence $K$ has to be $0$.
Just a different way for putting Lubin's argument.