Use the binomial series to expand $\sqrt{\dfrac{1 + x}{1 - x}}$ as a series of ascending powers of $x$ up to and including the terms in $x^2$, where $|x| < 1$. By putting $x = \tfrac{1}{10}$ in your result, show that $\sqrt{11} \approx \dfrac{663}{200}$.

1. Rewrite the given expression as a product of two binomial terms with rational powers.

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

2. Expand $(1 + x)^{\frac{1}{2}}$ up to the term in $x^2$.

$(1 + x)^{\frac{1}{2}} = 1 + \frac{1}{2}x + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}x^2 + \dots = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \dots$

3. Expand $(1 - x)^{-\frac{1}{2}}$ up to the term in $x^2$.

$(1 - x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-x) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}(-x)^2 + \dots = 1 + \frac{1}{2}x + \frac{3}{8}x^2 + \dots$

4. Multiply the two expansions, discarding terms with powers higher than $x^2$.

$\begin{aligned} (1 + x)^{\frac{1}{2}} (1 - x)^{-\frac{1}{2}} &\approx \left(1 + \frac{1}{2}x - \frac{1}{8}x^2\right)\left(1 + \frac{1}{2}x + \frac{3}{8}x^2\right) \\ &= 1\left(1 + \frac{1}{2}x + \frac{3}{8}x^2\right) + \frac{1}{2}x\left(1 + \frac{1}{2}x\right) - \frac{1}{8}x^2(1) \\ &= 1 + \frac{1}{2}x + \frac{3}{8}x^2 + \frac{1}{2}x + \frac{1}{4}x^2 - \frac{1}{8}x^2 \\ &= 1 + x + \frac{1}{2}x^2 \end{aligned}$

5. Substitute $x = \frac{1}{10}$ into both the original expression and the polynomial approximation.

$\text{When } x = \frac{1}{10},$
$\sqrt{\frac{1 + \frac{1}{10}}{1 - \frac{1}{10}}} \approx 1 + \left(\frac{1}{10}\right) + \frac{1}{2}\left(\frac{1}{10}\right)^2$
$\sqrt{\frac{\frac{11}{10}}{\frac{9}{10}}} \approx 1 + \frac{1}{10} + \frac{1}{200}$

6. Simplify the resulting numerical equation to prove the target approximation for $\sqrt{11}$.

$\sqrt{\frac{11}{9}} \approx \frac{200}{200} + \frac{20}{200} + \frac{1}{200}$
$\frac{\sqrt{11}}{3} \approx \frac{221}{200}$
$\sqrt{11} \approx \frac{663}{200} \quad \text{(shown)}$