Write out the first five terms of the Taylor series for √x centered at x = 1.

To find the Taylor series for \( f(x) = \sqrt{x} \) centered at \( x = 1 \), we use the formula for a Taylor series expansion:

\[
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (x - 1)^n
\]

We need to calculate the first few derivatives of \( f(x) = \sqrt{x} \) at \( x = 1 \) and evaluate them at \( x = 1 \).

1. **First, the function itself:**
   \[
   f(x) = \sqrt{x} = x^{1/2}
   \]
   \[
   f(1) = \sqrt{1} = 1
   \]

2. **First derivative:**
   \[
   f'(x) = \frac{1}{2} x^{-1/2}
   \]
   \[
   f'(1) = \frac{1}{2} \cdot 1^{-1/2} = \frac{1}{2}
   \]

3. **Second derivative:**
   \[
   f''(x) = \frac{-1}{4} x^{-3/2}
   \]
   \[
   f''(1) = \frac{-1}{4} \cdot 1^{-3/2} = \frac{-1}{4}
   \]

4. **Third derivative:**
   \[
   f^{(3)}(x) = \frac{3}{8} x^{-5/2}
   \]
   \[
   f^{(3)}(1) = \frac{3}{8} \cdot 1^{-5/2} = \frac{3}{8}
   \]

5. **Fourth derivative:**
   \[
   f^{(4)}(x) = \frac{-15}{16} x^{-7/2}
   \]
   \[
   f^{(4)}(1) = \frac{-15}{16} \cdot 1^{-7/2} = \frac{-15}{16}
   \]

Now, we can plug these values into the Taylor series formula:

\[
f(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 + \frac{f^{(3)}(1)}{3!}(x - 1)^3 + \frac{f^{(4)}(1)}{4!}(x - 1)^4 + \cdots
\]

Substituting the calculated values:

\[
\sqrt{x} = 1 + \frac{1}{2}(x - 1) + \frac{-1}{4} \frac{(x - 1)^2}{2!} + \frac{3}{8} \frac{(x - 1)^3}{3!} + \frac{-15}{16} \frac{(x - 1)^4}{4!} + \cdots
\]

Simplifying:

\[
\sqrt{x} = 1 + \frac{1}{2}(x - 1) - \frac{1}{8}(x - 1)^2 + \frac{1}{16}(x - 1)^3 - \frac{5}{128}(x - 1)^4 + \cdots
\]

Thus, the first five terms of the Taylor series for \( \sqrt{x} \) centered at \( x = 1 \) are:

\[
1 + \frac{1}{2}(x - 1) - \frac{1}{8}(x - 1)^2 + \frac{1}{16}(x - 1)^3 - \frac{5}{128}(x - 1)^4
\]

Do you want more details or have any questions?

Here are some related questions to expand your understanding:
1. How do you find higher-order terms in a Taylor series?
2. What conditions ensure that a Taylor series converges to the original function?
3. How is a Taylor series related to a Maclaurin series?
4. Can you compute the Taylor series for \( \sqrt{x} \) around a different center, say \( x = 4 \)?
5. How can you use the Taylor series to approximate \( \sqrt{1.1} \)?

**Tip:** A Taylor series approximates a function around a specific point. The more terms you use, the more accurate the approximation near that point.

Explore the first five terms of the Taylor series for the square root function √x centered at x = 1. This solution walks through the step-by-step process of deriving the Taylor series, using derivatives and key mathematical concepts like Taylor expansion. Suitable for undergraduate calculus students.

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