第一部分：纯粹数学 1 (a) 证明 $\dfrac{1}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}=\dfrac{1}{2x^{2}}\big(\sqrt{1+x^{2}}+\sqrt{1-x^{2}}\big).$

1. Multiply the numerator and the denominator by the conjugate of the denominator.

$\begin{aligned} \text{LHS} &= \frac{1}{\sqrt{1+x^2}-\sqrt{1-x^2}} \\ &= \frac{1}{\sqrt{1+x^2}-\sqrt{1-x^2}} \times \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}} \end{aligned}$

2. Simplify the denominator using the difference of two squares identity to complete the proof.

$\begin{aligned} &= \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{(\sqrt{1+x^2})^2-(\sqrt{1-x^2})^2} \\ &= \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{(1+x^2)-(1-x^2)} \\ &= \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{2x^2} \\ &= \frac{1}{2x^2}\big(\sqrt{1+x^2}+\sqrt{1-x^2}\big) \\ &= \text{RHS} \end{aligned}$

1. Use the result from part (a) to rewrite the given expression in a form suitable for series expansion.

From part (a), we know:
$\dfrac{1}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} = \dfrac{1}{2x^{2}}\big(\sqrt{1+x^{2}}+\sqrt{1-x^{2}}\big)$
Multiply both sides by $x^2$:
$\dfrac{x^{2}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} = \dfrac{1}{2}\big((1+x^{2})^{\frac{1}{2}}+(1-x^{2})^{\frac{1}{2}}\big)$

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

Using the expansion $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \dots$ from MF26:

For $(1+x^2)^{\frac{1}{2}}$ (with $y=x^2, n=\frac{1}{2}$):
$(1+x^2)^{\frac{1}{2}} = 1 + \frac{1}{2}(x^2) + \frac{(\frac{1}{2})(-\frac{1}{2})}{2}(x^2)^2 + \dots$
$= 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \dots$

For $(1-x^2)^{\frac{1}{2}}$ (with $y=-x^2, n=\frac{1}{2}$):
$(1-x^2)^{\frac{1}{2}} = 1 + \frac{1}{2}(-x^2) + \frac{(\frac{1}{2})(-\frac{1}{2})}{2}(-x^2)^2 + \dots$
$= 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 + \dots$

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

Substituting the expansions into the rewritten expression:
$\dfrac{1}{2}\big((1+x^{2})^{\frac{1}{2}}+(1-x^{2})^{\frac{1}{2}}\big)$
$= \dfrac{1}{2} \left[ \left(1 + \frac{1}{2}x^2 - \frac{1}{8}x^4\right) + \left(1 - \frac{1}{2}x^2 - \frac{1}{8}x^4\right) + \dots \right]$
$= \dfrac{1}{2} \left( 2 - \frac{1}{4}x^4 + \dots \right)$
$= 1 - \frac{1}{8}x^4 + \dots$

The first two non-zero terms are $1$ and $-\frac{1}{8}x^4$.

1. Determine the condition for the binomial expansions to be valid.

The series expansions for $(1+x^2)^{\frac{1}{2}}$ and $(1-x^2)^{\frac{1}{2}}$ require the terms being added to $1$ to have a magnitude strictly less than $1$:
$|x^2| < 1 \quad \text{and} \quad |-x^2| < 1$
$x^2 < 1 \implies -1 < x < 1$

2. Combine with the given restriction to state the set of values.

The question specifies $x \neq 0$ as the original expression is undefined at $x=0$. Therefore, the combined valid set of values is:
$\{x \in \mathbb{R} : -1 < x < 1, x \neq 0\}$

1. Determine the first two non-zero terms from part (b) to use in the comparison.

From part (a), we have $\frac{x^{2}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} = \frac{1}{2}\left(\sqrt{1+x^{2}}+\sqrt{1-x^{2}}\right)$.

Expanding both terms using the standard binomial expansion $(1+u)^n \approx 1 + nu + \frac{n(n-1)}{2}u^2$:
$(1+x^2)^{\frac{1}{2}} \approx 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4$
$(1-x^2)^{\frac{1}{2}} \approx 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4$

Adding these together and multiplying by $\frac{1}{2}$:
$\frac{1}{2}\left(2 - \frac{1}{4}x^4\right) = 1 - \frac{1}{8}x^4$

2. Find the first two non-zero terms of the series expansion of $\cos(ax^b)$.

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

3. Equate the terms and solve for the constants $a$ and $b$.

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

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

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