The problem involves finding the Taylor series expansion for the function $f(x) = \sqrt{1+x}$ around $x_0 = 0$. Let's solve step by step:

Step 1: The Taylor series formula

The Taylor series for a function $f(x)$ around $x_0 = 0$ is given by: $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \cdots$

Step 2: Compute derivatives of $f(x)$

The function is $f(x) = \sqrt{1+x} = (1+x)^{1/2}$.

1. Zeroth derivative (function itself): $f(x) = (1+x)^{1/2}, \quad f(0) = \sqrt{1} = 1$

2. First derivative: $f'(x) = \frac{1}{2}(1+x)^{-1/2}, \quad f'(0) = \frac{1}{2}$

3. Second derivative: $f''(x) = \frac{-1}{4}(1+x)^{-3/2}, \quad f''(0) = \frac{-1}{4}$

4. Third derivative: $f^{(3)}(x) = \frac{3}{8}(1+x)^{-5/2}, \quad f^{(3)}(0) = \frac{3}{8}$

Step 3: Substitute into the Taylor series

Using the derivatives, the Taylor expansion becomes: $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \cdots$

Substitute the computed values: $f(x) = 1 + \frac{1}{2}x - \frac{1}{4}\frac{x^2}{2} + \frac{3}{8}\frac{x^3}{6}$

Simplify each term: $f(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3$

Final Answer:

The Taylor expansion is: $f(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3$

Let me know if you want more details about any step or further clarifications! Here are 5 related questions to expand your understanding:

1. What is the general formula for the $n$-th derivative of $(1+x)^n$?
2. How can we determine the radius of convergence for this Taylor series?
3. Why do we divide by $n!$ in the Taylor series formula?
4. How does the remainder term in the Taylor series affect the approximation?
5. Can you approximate $\sqrt{1.1}$ using this Taylor series?

Tip: For Taylor series, always compute a few derivatives carefully to ensure accuracy!