已知 $\text{f}(x) = \text{e}^{\tan^{-1} x}$，其中 $\tan^{-1} x$ 表示主值。

1. Differentiate $\text{f}(x)$ using the chain rule and the standard derivative of the inverse tangent function.

$\text{f}(x) = \text{e}^{\tan^{-1} x}$

Using the chain rule:
$\text{f}'(x) = \text{e}^{\tan^{-1} x} \cdot \frac{\text{d}}{\text{d}x}(\tan^{-1} x)$

$\text{f}'(x) = \text{e}^{\tan^{-1} x} \cdot \frac{1}{1 + x^2}$

2. Substitute $\text{f}(x)$ back into the derivative and rearrange to show the required result.

Since $\text{f}(x) = \text{e}^{\tan^{-1} x}$, substitute this into the equation for $\text{f}'(x)$:
$\text{f}'(x) = \text{f}(x) \cdot \frac{1}{1 + x^2}$

Multiply both sides by $(1 + x^2)$:
$(1 + x^2)\text{f}'(x) = \text{f}(x)$

1. Differentiate the given identity $(1 + x^2)\text{f}'(x) = \text{f}(x)$ with respect to $x$ to find an expression involving $\text{f}''(x)$.

$\frac{\text{d}}{\text{d}x} [(1 + x^2)\text{f}'(x)] = \frac{\text{d}}{\text{d}x} [\text{f}(x)]$
$(1 + x^2)\text{f}''(x) + 2x\text{f}'(x) = \text{f}'(x)$

2. Differentiate the equation again with respect to $x$ to obtain an expression involving $\text{f}'''(x)$.

Applying the product rule again to both terms on the LHS:
$[(1 + x^2)\text{f}'''(x) + 2x\text{f}''(x)] + [2x\text{f}''(x) + 2\text{f}'(x)] = \text{f}''(x)$
$(1 + x^2)\text{f}'''(x) + 4x\text{f}''(x) + 2\text{f}'(x) = \text{f}''(x)$

3. Evaluate the function and its derivatives at $x = 0$.

When $x = 0$:
$\text{f}(0) = \text{e}^{\tan^{-1}(0)} = \text{e}^0 = 1$
Substitute $x = 0$ into $(1 + x^2)\text{f}'(x) = \text{f}(x)$:
$(1 + 0)\text{f}'(0) = 1 \implies \text{f}'(0) = 1$
Substitute $x = 0$ into $(1 + x^2)\text{f}''(x) + 2x\text{f}'(x) = \text{f}'(x)$:
$(1 + 0)\text{f}''(0) + 0 = 1 \implies \text{f}''(0) = 1$
Substitute $x = 0$ into $(1 + x^2)\text{f}'''(x) + 4x\text{f}''(x) + 2\text{f}'(x) = \text{f}''(x)$:
$(1 + 0)\text{f}'''(0) + 0 + 2(1) = 1 \implies \text{f}'''(0) + 2 = 1 \implies \text{f}'''(0) = -1$

4. Substitute the derivative values into the standard Maclaurin series expansion formula.

$\text{f}(x) = \text{f}(0) + x\text{f}'(0) + \frac{x^2}{2!}\text{f}''(0) + \frac{x^3}{3!}\text{f}'''(0) + \ldots$
$\text{f}(x) = 1 + x(1) + \frac{x^2}{2}(1) + \frac{x^3}{6}(-1) + \ldots$
$\text{f}(x) = 1 + x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \ldots$

1. Find the Maclaurin series for $\text{f}(x)$ up to the $x^3$ term (from part ii).

From part (i), $(1+x^2)\text{f}'(x) = \text{f}(x)$. Evaluating at $x=0$:
$\text{f}(0) = \text{e}^0 = 1 \implies \text{f}'(0) = 1$

Differentiating implicitly:
$(1+x^2)\text{f}''(x) + 2x\text{f}'(x) = \text{f}'(x) \implies (1+x^2)\text{f}''(x) + (2x-1)\text{f}'(x) = 0$
At $x=0$: $1\cdot\text{f}''(0) + (-1)(1) = 0 \implies \text{f}''(0) = 1$

Differentiating again:
$(1+x^2)\text{f}'''(x) + 2x\text{f}''(x) + (2x-1)\text{f}''(x) + 2\text{f}'(x) = 0$
$(1+x^2)\text{f}'''(x) + (4x-1)\text{f}''(x) + 2\text{f}'(x) = 0$
At $x=0$: $1\cdot\text{f}'''(0) - 1(1) + 2(1) = 0 \implies \text{f}'''(0) = -1$

$\text{f}(x) \approx \text{f}(0) + x\text{f}'(0) + \frac{x^2}{2!}\text{f}''(0) + \frac{x^3}{3!}\text{f}'''(0) = 1 + x + \frac{1}{2}x^2 - \frac{1}{6}x^3$

2. Integrate the polynomial approximation over the interval $[0, 0.5]$.

$\int_0^{0.5} \text{f}(x) \text{ d}x \approx \int_0^{0.5} \left(1 + x + \frac{x^2}{2} - \frac{x^3}{6}\right) \text{ d}x$

$= \left[ x + \frac{x^2}{2} + \frac{x^3}{6} - \frac{x^4}{24} \right]_0^{0.5}$

3. Evaluate the definite integral and round to 4 significant figures.

$= 0.5 + \frac{(0.5)^2}{2} + \frac{(0.5)^3}{6} - \frac{(0.5)^4}{24}$

$= \frac{1}{2} + \frac{1}{8} + \frac{1}{48} - \frac{1}{384}$

$= \frac{247}{384} \approx 0.643229...$

To 4 significant figures, the value is $0.6432$.

1. State the reason for unsuitability based on the nature of Maclaurin series approximations.

A truncated Maclaurin series gives a polynomial approximation that is most accurate near $x = 0$.
Since $x = 1$ is not small (or not close to $0$), substituting $x = 1$ into the series up to $x^3$ will not give an accurate estimate for $\text{e}^{\frac{\pi}{4}}$.
Therefore, the substitution is not suitable.