在以 $A$、$B$ 和 $C$ 为顶点的三角形中，角 $BAC$ 是直角，且角 $ABC = \frac{\pi}{3} - x$。 证明 $\frac{AB}{AC} = \frac{1+\sqrt{3}\tan x}{\sqrt{3}-\tan x}.$ 由此证明，当 $x$ 足够小，可以忽略 $x^2$ 及更高次项时，$\frac{AB}{AC} \approx a + b x,$ 其中 $a$ 和 $b$ 是待确定的精确常数。 一条曲线由方程 $(1+x^2)\frac{dy}{dx}+xy=\sqrt{1+x^2}$ 定义，且 $(0,1)$ 是该曲线上的一个点。 求 $y$ 的麦克劳林展开式，展开到包含 $x^2$ 的项。 由此求 $e^y$ 的级数展开式，展开到包含 $x^2$ 的项。

1. Relate the ratio of the sides to the given angle using the tangent function.

In the right-angled triangle $ABC$ with the right angle at $A$, the side opposite to angle $ABC$ is $AC$, and the adjacent side is $AB$.

$\tan(\angle ABC) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AC}{AB}$

Given $\angle ABC = \frac{\pi}{3} - x$, we have:
$\tan\left(\frac{\pi}{3} - x\right) = \frac{AC}{AB}$

Therefore, by taking the reciprocal:
$\frac{AB}{AC} = \frac{1}{\tan\left(\frac{\pi}{3} - x\right)}$

2. Apply the tangent compound angle formula and substitute standard exact values.

Using the compound angle identity $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:

$\tan\left(\frac{\pi}{3} - x\right) = \frac{\tan\frac{\pi}{3} - \tan x}{1 + \tan\frac{\pi}{3} \tan x}$

Since $\tan\frac{\pi}{3} = \sqrt{3}$:
$\tan\left(\frac{\pi}{3} - x\right) = \frac{\sqrt{3} - \tan x}{1 + \sqrt{3}\tan x}$

Substituting this back into the expression for $\frac{AB}{AC}$:
$\frac{AB}{AC} = \frac{1}{\left( \frac{\sqrt{3} - \tan x}{1 + \sqrt{3}\tan x} \right)} = \frac{1+\sqrt{3}\tan x}{\sqrt{3}-\tan x} \quad \text{(shown)}$

1. Substitute the small angle approximation for $\tan x$.

For small values of $x$ in radians, $\tan x \approx x$.
Substituting this into the result from (a)(i):
$\frac{AB}{AC} \approx \frac{1+\sqrt{3}x}{\sqrt{3}-x}$

2. Rewrite the expression as a product and expand the denominator using the binomial series.

We can rewrite the fraction as a product with a negative index:
$\frac{1+\sqrt{3}x}{\sqrt{3}-x} = (1+\sqrt{3}x)(\sqrt{3}-x)^{-1}$
Factor out $\sqrt{3}$ from the second bracket to use the standard binomial expansion $(1+u)^n$:
$= (1+\sqrt{3}x) \left[ \sqrt{3}\left(1-\frac{x}{\sqrt{3}}\right) \right]^{-1}$
$= \frac{1}{\sqrt{3}}(1+\sqrt{3}x)\left(1-\frac{x}{\sqrt{3}}\right)^{-1}$
Expanding $\left(1-\frac{x}{\sqrt{3}}\right)^{-1}$ up to the term in $x$:
$\approx \frac{1}{\sqrt{3}}(1+\sqrt{3}x)\left( 1 + (-1)\left(-\frac{x}{\sqrt{3}}\right) \right)$
$= \frac{1}{\sqrt{3}}(1+\sqrt{3}x)\left( 1 + \frac{x}{\sqrt{3}} \right)$

3. Multiply out the brackets, neglect $x^2$, and deduce $a$ and $b$.

$= \frac{1}{\sqrt{3}} \left( 1 + \frac{x}{\sqrt{3}} + \sqrt{3}x + x^2 \right)$
Since $x^2$ and higher powers can be neglected:
$\approx \frac{1}{\sqrt{3}} \left( 1 + \left(\frac{1}{\sqrt{3}} + \sqrt{3}\right)x \right)$
$= \frac{1}{\sqrt{3}} \left( 1 + \frac{4}{\sqrt{3}}x \right)$
$= \frac{1}{\sqrt{3}} + \frac{4}{3}x$
$= \frac{\sqrt{3}}{3} + \frac{4}{3}x$
Comparing this to $a + bx$, the exact constants are:
$a = \frac{\sqrt{3}}{3}$ and $b = \frac{4}{3}$

1. Find the value of $\frac{dy}{dx}$ at $x=0$.

Substitute $x=0$ and $y=1$ into the given differential equation $(1+x^2)\frac{dy}{dx} + xy = \sqrt{1+x^2}$:
$(1+0^2)\frac{dy}{dx} + (0)(1) = \sqrt{1+0^2}$
$\frac{dy}{dx} = 1$

2. Differentiate the given equation implicitly with respect to $x$.

Differentiating both sides of $(1+x^2)\frac{dy}{dx} + xy = (1+x^2)^{\frac{1}{2}}$ with respect to $x$ using the product and chain rules:
$\left[ (1+x^2)\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} \right] + \left[ x\frac{dy}{dx} + y \right] = \frac{1}{2}(1+x^2)^{-\frac{1}{2}}(2x)$
$(1+x^2)\frac{d^2y}{dx^2} + 3x\frac{dy}{dx} + y = \frac{x}{\sqrt{1+x^2}}$

3. Evaluate the second derivative at $x=0$.

Substitute $x=0$, $y=1$, and $\frac{dy}{dx}=1$ into the differentiated equation:
$(1+0^2)\frac{d^2y}{dx^2} + 3(0)(1) + 1 = \frac{0}{\sqrt{1+0^2}}$
$\frac{d^2y}{dx^2} + 1 = 0$
$\frac{d^2y}{dx^2} = -1$

4. Formulate the Maclaurin series expansion up to the term in $x^2$.

The Maclaurin series is given by $y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \dots$
$y = 1 + (1)x + \frac{-1}{2}x^2$
$y = 1 + x - \frac{1}{2}x^2$

1. Calculate the Maclaurin series for $y$ up to the term in $x^2$, which is the prerequisite result from part (b)(i).

Given $(1+x^2)\frac{dy}{dx}+xy=\sqrt{1+x^2}$ and $y=1$ when $x=0$.
At $x=0$:
$(1)y'(0) + 0 = 1 \implies y'(0) = 1$.

Differentiating implicitly with respect to $x$:
$2x\frac{dy}{dx} + (1+x^2)\frac{d^2y}{dx^2} + y + x\frac{dy}{dx} = \frac{1}{2}(1+x^2)^{-\frac{1}{2}}(2x)$
$(1+x^2)\frac{d^2y}{dx^2} + 3x\frac{dy}{dx} + y = \frac{x}{\sqrt{1+x^2}}$

At $x=0$:
$(1)y''(0) + 0 + 1 = 0 \implies y''(0) = -1$.

Thus, $y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \dots \approx 1 + x - \frac{1}{2}x^2$.

2. Substitute the expansion of $y$ into $e^y$ and factorise out the constant factor $e$.

From part (b)(i), we have $y \approx 1 + x - \frac{1}{2}x^2$.
Substitute this into $e^y$:
$e^y \approx e^{1 + x - \frac{1}{2}x^2}$
$e^y \approx e^1 \cdot e^{x - \frac{1}{2}x^2}$

3. Use the standard Maclaurin series for $e^X$ to expand and simplify up to the $x^2$ term.

Using the standard expansion $e^X \approx 1 + X + \frac{X^2}{2!}$ where $X = x - \frac{1}{2}x^2$:
$e^{x - \frac{1}{2}x^2} \approx 1 + \left(x - \frac{1}{2}x^2\right) + \frac{1}{2}\left(x - \frac{1}{2}x^2\right)^2$

Expanding and ignoring terms with powers of $x$ higher than 2:
$e^{x - \frac{1}{2}x^2} \approx 1 + x - \frac{1}{2}x^2 + \frac{1}{2}(x^2)$
$e^{x - \frac{1}{2}x^2} \approx 1 + x$

Therefore,
$e^y \approx e(1 + x) = e + ex$