Taylor Expansion for very small and very big arguments

[deuteron]

Homework Statement

Expand the given functions for very small and very big arguments $x$ giving $3$ terms

Relevant Equations

$T_nf[x;a]= \displaystyle\sum_k \frac 1 {k!}\ f^{(k)}|_a\ (x-a)^k$

The function is
$$ f(x)=\sqrt{1-x}$$

and we are expected to expand it using Taylor expansion for very small $(x<<1)$ and very big $(x>>1)$ arguments

My thought process was the following:

$$T_2f[x;x_0]=\sqrt{1-x_0} -\frac 12 \frac 1{\sqrt{1-x_0}}(x-x_0) -\frac 14 \frac 1 {\sqrt{1-x_0}^3}(x-x_0)^2$$

and then assuming small arguments, namely assuming $x_0<<1$, I did the following approximation:

$$ T_2f[x;x_0]\approx \sqrt{1} -\frac 12 \frac 1 {\sqrt{1}} (x) -\frac 1 4 \frac 1 {\sqrt{1}^3} (x)^2 = 1-\frac 12 x -\frac 14 x^2$$

For big arguments, I assumed $x_0>>1$ and made the following approximation:

$$T_2f[x;x_0]\approx i\sqrt{a} +\frac 12 \frac 1 {i\sqrt{a}} (a) +\frac 14 \frac 1 {i\sqrt{a}^3} a^2$$

However, the real answers are, for small arguments:

$$f(x<<1)= 1-\frac 12 x - \frac 18 x^2$$

and for big arguments:

$$f(x>>1)= i\sqrt{a} \sqrt{1-\frac 1x} = i\sqrt{x} \ (1-\frac 1 {2x} -\frac 1{8x^2})$$

I see that my mistake is to assume the expansion point to be very big instead of assuming big $x$, but I don't know how to do that, does anyone understand the in between steps for the solution?

 

[pasmith]

Use the binomial expansion: $$(1 + x)^{\alpha} = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!}x^2 + \dots + \frac{\alpha(\alpha - 1) \cdots (\alpha - n + 1)}{n!}x^n + \dots,\qquad |x| < 1.$$ For small $|x|$, $$f(x) = (1 - x)^{1/2}$$ can be expanded as is. A Taylor series of $x^{1/2}$ about $1$ should also yield the same result, bearing in mind that we are looking at $(1 - x)^{1/2}$ not $(1 + x)^{1/2}$. For large negative $x$, $$(1 - x)^{1/2} = (1 + |x|)^{1/2} = |x|^{1/2}(1 + |x|^{-1})^{1/2}$$ can be expanded. For large positive $x$, $$f(x) = ix^{1/2}(1 - x^{-1})^{1/2}$$ can be expanded.