第一部分：纯粹数学 1 (a) 证明 $\displaystyle \frac{1}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}=\frac{1}{2x^{2}}\left(\sqrt{1+x^{2}}+\sqrt{1-x^{2}}\right)$。 (b) 由此，利用公式表 (MF26) 中适当的展开式，求出 $\displaystyle \frac{x^{2}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}},\; x\ne 0$ 按 $x$ 的升幂排列的级数展开式中的前两个非零项。 (c) 写出使该级数展开式成立的 $x$ 的取值集合。 (d) 已知在 (b) 部分中求得的两项等于 $\cos\left(a x^{b}\right)$ 的级数展开式中的前两项。求常数 $a$ 和 $b$ 的所有可能值。

1. Multiply the numerator and denominator by the conjugate of the denominator to rationalise it.

$\frac{1}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \times \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}$

2. Simplify the denominator using the difference of two squares identity to arrive at the required result.

$= \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{(1+x^{2}) - (1-x^{2})}$
$= \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{2x^{2}}$
$= \frac{1}{2x^{2}}\left(\sqrt{1+x^{2}}+\sqrt{1-x^{2}}\right) \quad \text{(shown)}$

1. Use the result from part (a) to simplify the given expression.

$\frac{x^{2}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} = x^2 \left[ \frac{1}{2x^2}\left(\sqrt{1+x^{2}}+\sqrt{1-x^{2}}\right) \right]$
$= \frac{1}{2} \left( (1+x^2)^{\frac{1}{2}} + (1-x^2)^{\frac{1}{2}} \right)$

2. Expand $(1+x^2)^{\frac{1}{2}}$ and $(1-x^2)^{\frac{1}{2}}$ using the standard binomial series expansion up to the $x^4$ term.

Using $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \dots$
$(1+x^2)^{\frac{1}{2}} = 1 + \frac{1}{2}(x^2) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(x^2)^2 + \dots = 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \dots$
$(1-x^2)^{\frac{1}{2}} = 1 + \frac{1}{2}(-x^2) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(-x^2)^2 + \dots = 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 + \dots$

3. Substitute the expansions back into the simplified expression and collect terms to find the first two non-zero terms.

$\frac{1}{2} \left( (1+x^2)^{\frac{1}{2}} + (1-x^2)^{\frac{1}{2}} \right)$
$= \frac{1}{2} \left[ \left( 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4 \right) + \left( 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 \right) \right]$
$= \frac{1}{2} \left( 2 - \frac{1}{4}x^4 \right)$
$= 1 - \frac{1}{8}x^4$

1. State the validity condition for the binomial expansions and combine it with the given domain restriction.

The series expansions used in part (b) are for $(1+x^2)^{\frac{1}{2}}$ and $(1-x^2)^{\frac{1}{2}}$.
By the standard binomial expansion, these are valid when $|x^2| < 1$ and $|-x^2| < 1$.
$x^2 < 1 \implies -1 < x < 1$
Given the restriction $x \ne 0$ for the original expression in part (b), we must exclude $0$.
Therefore, the set of values is $-1 < x < 1, \; x \ne 0$.

1. Determine the result from part (b) (as it is required to solve this part).

From part (a), the expression simplifies to $\frac{1}{2}\left(\sqrt{1+x^2}+\sqrt{1-x^2}\right)$.
Using the standard binomial expansion:
$(1+x^2)^{1/2} = 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \dots$
$(1-x^2)^{1/2} = 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 + \dots$
Adding them and multiplying by $\frac{1}{2}$, the first two non-zero terms from part (b) are:
$\frac{1}{2}\left(2 - \frac{1}{4}x^4\right) = 1 - \frac{1}{8}x^4$

2. Expand $\cos(ax^b)$ using the standard series.

Using the standard series expansion $\cos u \approx 1 - \frac{u^2}{2!}$, substitute $u = ax^b$:
$\cos(ax^b) \approx 1 - \frac{(ax^b)^2}{2} = 1 - \frac{a^2 x^{2b}}{2}$

3. Equate the expansions to find the possible values of $a$ and $b$.

Equating the first two terms of both expansions:
$1 - \frac{a^2 x^{2b}}{2} \equiv 1 - \frac{1}{8}x^4$

Comparing the powers of $x$:
$2b = 4 \implies b = 2$

Comparing the coefficients of the second term:
$-\frac{a^2}{2} = -\frac{1}{8}$
$a^2 = \frac{1}{4} \implies a = \pm \frac{1}{2}$