To solve this, let's focus on approximating $\sqrt{28}$ using two Taylor series expansions: one centered at 25 and the other at 36. We will compare their accuracies based on how well each approximation aligns with the actual value of $\sqrt{28}$.

1. Taylor Series Expansion for $\sqrt{x}$ at $x = 25$

The general form of the Taylor series for $f(x) = \sqrt{x}$ centered at $x_0$ is:

$f(x) = \sqrt{x_0} + \frac{1}{2\sqrt{x_0}}(x - x_0) - \frac{1}{4(x_0^{3/2})}(x - x_0)^2 + \frac{3}{8(x_0^{5/2})}(x - x_0)^3 + \dots$

For $x_0 = 25$:

$f(x) = 5 + \frac{1}{10}(x - 25) - \frac{1}{200}(x - 25)^2 + \frac{3}{4000}(x - 25)^3$

Substitute $x = 28$ into this series:

$f(28) \approx 5 + \frac{1}{10}(28 - 25) - \frac{1}{200}(28 - 25)^2 + \frac{3}{4000}(28 - 25)^3$

Simplifying:

$f(28) \approx 5 + \frac{1}{10}(3) - \frac{1}{200}(9) + \frac{3}{4000}(27)$

$f(28) \approx 5 + 0.3 - 0.045 + 0.02025$

$f(28) \approx 5.27525$

2. Taylor Series Expansion for $\sqrt{x}$ at $x = 36$

Similarly, for $x_0 = 36$:

$f(x) = 6 + \frac{1}{12}(x - 36) - \frac{1}{432}(x - 36)^2 + \frac{1}{15552}(x - 36)^3$

Substitute $x = 28$ into this series:

$f(28) \approx 6 + \frac{1}{12}(28 - 36) - \frac{1}{432}(28 - 36)^2 + \frac{1}{15552}(28 - 36)^3$

Simplifying:

$f(28) \approx 6 + \frac{1}{12}(-8) - \frac{1}{432}(64) + \frac{1}{15552}(-512)$

$f(28) \approx 6 - 0.66667 - 0.14815 - 0.03292$

$f(28) \approx 5.15226$

3. Compare Accuracy

The actual value of $\sqrt{28}$ is approximately:

$\sqrt{28} \approx 5.291502622$

• Using the Taylor series centered at 25: $f(28) \approx 5.27525$ (error $\approx 0.01625$)
• Using the Taylor series centered at 36: $f(28) \approx 5.15226$ (error $\approx 0.13924$)

Conclusion:

The Taylor series centered at $x = 25$ gives a more accurate approximation of $\sqrt{28}$ than the one centered at $x = 36$. Thus, it is better to use the approximation centered at 25.

Final Answers:

• The accuracy of the approximation centered at 25 is about $5.27525$.
• The accuracy of the approximation centered at 36 is about $5.15226$.
• It is better to use the approximation centered at 25.

Would you like any more details, or do you have further questions?

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Here are 5 related questions to expand your understanding:

1. How do higher-order terms in Taylor series affect the accuracy of approximations?
2. How can we determine the optimal center point for a Taylor series approximation?
3. Why does the error increase as we move further from the center of the Taylor series?
4. What are the limitations of using Taylor series for approximating square roots?
5. How does the radius of convergence affect the accuracy of Taylor series approximations?

Tip: Taylor series approximations are most accurate near the center point $x_0$, so choose a center point close to the value you're approximating.