已知 $y = \sqrt{1+\ln(1+x)}$，求使 $y$ 有定义时 $x$ 取值范围的精确值。 证明 $2y\,\dfrac{dy}{dx} = e^{-y^2}$。 由此求出 $\sqrt{1+\ln(1+x)}$ 的麦克劳林级数，直到包含 $x^2$ 的项。 验证使用公式表（MF15）中给出的标准级数展开式可以得到相同的结果。

1. Set up the conditions for the natural logarithm and square root to be valid

For $y$ to be well defined, the argument of the logarithm must be strictly positive and the expression inside the square root must be non-negative:
$1+x > 0 \implies x > -1$
$1+\ln(1+x) \ge 0$

2. Solve the inequality to find the exact range of $x$

$\ln(1+x) \ge -1$
$1+x \ge e^{-1}$
$x \ge e^{-1} - 1$
Since $e^{-1} - 1 \approx -0.632 > -1$, this condition automatically satisfies $x > -1$.

1. Find y and dy/dx at x = 0 using the result of part (ii)

When $x=0$, $y=\sqrt{1+\ln 1}=1$.

From part (ii), $2y\dfrac{dy}{dx}=e^{1-y^2}$. Substituting $x=0,\;y=1$:
$2(1)\dfrac{dy}{dx}=e^{1-1}=1 \;\Rightarrow\; \left.\dfrac{dy}{dx}\right|_{x=0}=\tfrac{1}{2}.$

2. Differentiate the relation again to find d^2y/dx^2 at x = 0

Differentiate $2y\dfrac{dy}{dx}=e^{1-y^2}$ with respect to $x$:
$2\left(\dfrac{dy}{dx}\right)^2+2y\dfrac{d^2y}{dx^2}=e^{1-y^2}\left(-2y\dfrac{dy}{dx}\right).$
Since $e^{1-y^2}=2y\dfrac{dy}{dx}$, the right side $=-4y^2\left(\dfrac{dy}{dx}\right)^2$.

At $x=0$ ($y=1,\;\dfrac{dy}{dx}=\tfrac12$):
$2\left(\tfrac12\right)^2+2\dfrac{d^2y}{dx^2}=-4(1)\left(\tfrac12\right)^2$
$\tfrac12+2\dfrac{d^2y}{dx^2}=-1 \;\Rightarrow\; \left.\dfrac{d^2y}{dx^2}\right|_{x=0}=-\tfrac34.$

3. Assemble the Maclaurin series up to x^2

$y=y(0)+y'(0)\,x+\dfrac{y''(0)}{2!}x^2+\cdots$
$=1+\tfrac12 x+\dfrac{-\tfrac34}{2}x^2+\cdots=1+\tfrac12 x-\tfrac38 x^2+\cdots$

1. Write the standard series for $\ln(1+x)$

From MF15, $\ln(1+x) = x - \tfrac{1}{2}x^2 + \cdots$ so $1+\ln(1+x) = 1 + \left(x - \tfrac{1}{2}x^2\right) + \cdots$

2. Apply the binomial expansion $(1+u)^{1/2}$

Let $u = x - \tfrac{1}{2}x^2$. Then $\sqrt{1+\ln(1+x)} = (1+u)^{1/2} = 1 + \tfrac{1}{2}u - \tfrac{1}{8}u^2 + \cdots$ Since $u^2 = x^2 + \cdots$ (terms up to $x^2$), $= 1 + \tfrac{1}{2}\left(x - \tfrac{1}{2}x^2\right) - \tfrac{1}{8}x^2 + \cdots$

3. Collect terms and confirm agreement with (iii)

$= 1 + \tfrac{1}{2}x - \tfrac{1}{4}x^2 - \tfrac{1}{8}x^2 + \cdots = 1 + \tfrac{1}{2}x - \tfrac{3}{8}x^2 + \cdots$ This is identical to the series obtained in part (iii), as required.