Ho do you approximate $4 \epsilon x + x^2 -1=0 $ using $x=x_0+x_1 \epsilon + x_2 \epsilon^2 $?

Question

Consider the quadratic $4 \epsilon x + x^2 -1=0 $ and suppose we want to approximate the solution to the equation using

$$x=x_0+x_1 \epsilon + x_2 \epsilon^2 $$

Why does substituting the line directly above into the quadratic yield $x_0^2 -1=0$?

I know that after substituting into the quadratic the term independent of epsilon is $x_0^2 -1$ but why must it be zero? thanks

Answer 1

The substitution gives

$$4\epsilon(x_0+x_1 \epsilon + x_2 \epsilon^2)+(x_0+x_1 \epsilon + x_2 \epsilon^2)^2-1=0$$

or, by increasing powers of $\epsilon$

$$x_0^2-1+(4x_0+2x_0x_1)\epsilon+(4x_1+x_1^2+2x_0x_2)\epsilon^2+(4x_2+2x_1x_2)\epsilon^3+x_2^2\epsilon^4=0.$$

To obtain a good approximation, you will cancel the terms of the lowest degree, as they give the largest contribution to the sum. As there are three parameters, you can cancel three coefficients.

$$\begin{cases}x_0^2-1=0,\\4x_0+2x_0x_1=0,\\4x_1+x_1^2+2x_0x_2=0.\end{cases}$$

Answer 2

NOTE: answer valid before the editing by the OP

By binomial expansion for $\epsilon<<1$

$$4 \epsilon x + x^2 -1=0\implies x=\frac{-4\epsilon\pm\sqrt{16\epsilon^2+4}}{2}=-2\epsilon\pm\sqrt{1+4\epsilon^2}\approx-2\epsilon\pm(1+2\epsilon^2)$$

then

$$x\approx\pm1-2\epsilon\pm2\epsilon^2$$

Answer 3

(Variables renamed because lazy.)

$x=u+vc+ wc^2 $ in $ 4cx + x^2 -1=0 $ gives, assuming that $c$ is small,

$\begin{array}\\ 0 &=4cx + x^2 -1\\ &=4c(u+vc+ wc^2) + (u+vc+ wc^2)^2 -1\\ &=4cu+4vc^2+ 4wc^3 + u^2+2uvc+v^2c^2+2uwc^2+O(c^3) -1\\ &=u^2-1+c(4u+2uv)+c^2(4v+v^2+2uw)+O(c^3)\\ \end{array} $

For this to hold, $u^2 = 1$, $4u+2uv = 0$, and $4v+v^2+2uw = 0$.

From the first, $u = \pm 1$.

From the second, $0 = 4u+2uv =2u(2+v) $. Since $u \ne 0$, $v = -2$.

From the third, $0 = -8+4\pm 2w $ so $w =\mp 2 $.

Therefore $x =\pm 1 -2c\mp 2c^2 $.