To solve these problems, we'll follow these steps:

(a) Find the Taylor polynomial of $f(x) = \sqrt{1 + \frac{x}{2}}$ of order 2 about the origin.

First, we find the derivatives needed:

1. $f(x) = (1 + \frac{x}{2})^{1/2}$

   The Taylor series expansion about 0 is given by:

   $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots$

2. Evaluate $f(0)$:

   $f(0) = \sqrt{1 + \frac{0}{2}} = 1$

3. First derivative $f'(x)$:

   Using the chain rule,

   $f'(x) = \frac{1}{2}(1 + \frac{x}{2})^{-1/2} \cdot \frac{1}{2} = \frac{1}{2 \sqrt{1 + \frac{x}{2}}} \cdot \frac{1}{2} = \frac{1}{4} (1 + \frac{x}{2})^{-1/2}$

4. Evaluate $f'(0)$:

   $f'(0) = \frac{1}{4}$

5. Second derivative $f''(x)$:

   Differentiating $f'(x)$,

   $f''(x) = -\frac{1}{8} (1 + \frac{x}{2})^{-3/2} \cdot \frac{1}{2} = -\frac{1}{16} (1 + \frac{x}{2})^{-3/2}$

6. Evaluate $f''(0)$:

   $f''(0) = -\frac{1}{16}$

7. Taylor polynomial of order 2:

   $T_2(x) = 1 + \frac{1}{4}x - \frac{1}{32}x^2$

(b) Use the result of (a) to find an approximation for $\sqrt{103}$. Estimate the error of your result by computing the remainder and finding an upper bound for it.

Express $\sqrt{103}$ in terms of $f(x)$:

• Let $x = 204$, then $\sqrt{1 + \frac{204}{2}} = \sqrt{103}$.

Using the polynomial:

$f(x) \approx 1 + \frac{1}{4}(204) - \frac{1}{32}(204)^2$

Calculate the approximation:

$f(x) \approx 1 + \frac{204}{4} - \frac{204^2}{32}$

Estimate the error using the remainder term in the Taylor series (third derivative):

The third derivative and calculating the error isn't explicitly done here, but you can compute it using

$R_2(x) = \frac{f'''(c)}{3!} x^3$

where $c$ is between 0 and $x$. You would need $f'''(x)$ for a precise bound.

Once you have the calculation, you'll get an approximation for $\sqrt{103}$ and can estimate the error.