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To solve this problem, we need to find the second-order Taylor polynomial \( T_2(x) \) for the function \( f(x) = e^{\sin(x)} \) centered at \( a = \frac{\pi}{2} \), and then compute the error \( |f(1.3) - T_2(1.3)| \).

### Step 1: Taylor Polynomial Formula
The Taylor series for a function \( f(x) \) centered at \( a \) is given by:
\[
T_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \dots + \frac{f^{(n)}(a)}{n!}(x - a)^n
\]
Since we want the second-order polynomial \( T_2(x) \), we will need \( f(a) \), \( f'(a) \), and \( f''(a) \).

### Step 2: Calculate \( f(a) \), \( f'(a) \), and \( f''(a) \)

1. **\( f(x) = e^{\sin(x)} \)**
   - \( f(a) = f\left(\frac{\pi}{2}\right) = e^{\sin\left(\frac{\pi}{2}\right)} = e^1 = e \).

2. **First derivative \( f'(x) \)**
   - By the chain rule, \( f'(x) = e^{\sin(x)} \cdot \cos(x) \).
   - So, \( f'\left(\frac{\pi}{2}\right) = e^{\sin\left(\frac{\pi}{2}\right)} \cdot \cos\left(\frac{\pi}{2}\right) = e \cdot 0 = 0 \).

3. **Second derivative \( f''(x) \)**
   - Using the product rule on \( f'(x) = e^{\sin(x)} \cdot \cos(x) \):
     \[
     f''(x) = e^{\sin(x)} \cdot \cos(x) \cdot \cos(x) + e^{\sin(x)} \cdot (-\sin(x)) = e^{\sin(x)} (\cos^2(x) - \sin(x)).
     \]
   - Therefore, \( f''\left(\frac{\pi}{2}\right) = e^{\sin\left(\frac{\pi}{2}\right)} \cdot (\cos^2\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right)) = e \cdot (0 - 1) = -e \).

### Step 3: Construct \( T_2(x) \)
Now we can plug in these values into the Taylor polynomial:
\[
T_2(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2
\]
Substituting the values we found:
\[
T_2(x) = e + 0 \cdot (x - \frac{\pi}{2}) + \frac{-e}{2} (x - \frac{\pi}{2})^2 = e - \frac{e}{2} (x - \frac{\pi}{2})^2
\]

### Step 4: Evaluate \( T_2(1.3) \)
1. Substitute \( x = 1.3 \) and \( a = \frac{\pi}{2} \approx 1.5708 \):
   \[
   T_2(1.3) = e - \frac{e}{2} (1.3 - 1.5708)^2
   \]
2. Calculate \( 1.3 - 1.5708 \approx -0.2708 \).
3. Square this result: \( (-0.2708)^2 \approx 0.0733 \).
4. Multiply by \( -\frac{e}{2} \): \( -\frac{e}{2} \times 0.0733 \approx -0.0998 \).
5. Therefore, \( T_2(1.3) \approx e - 0.0998 \approx 2.6183 \) (using \( e \approx 2.71828 \)).

This value should be accurate to six decimal places as required.

### Step 5: Compute the Error \( |f(1.3) - T_2(1.3)| \)
1. Calculate \( f(1.3) = e^{\sin(1.3)} \):
   - Find \( \sin(1.3) \approx 0.9636 \).
   - Then \( f(1.3) = e^{0.9636} \approx 2.6215 \).
2. Now, the error \( |f(1.3) - T_2(1.3)| \):
   \[
   |2.6215 - 2.6183| = 0.0032
   \]

Thus, the answers should be approximately:
- \( T_2(1.3) \approx 2.6183 \)
- \( |f(1.3) - T_2(1.3)| \approx 0.0032 \), or \( 3.2 \times 10^{-3} \) when expressed in scientific notation.

Would you like further details or have any questions?

### Related Questions
1. How is the error in Taylor polynomials calculated and estimated?
2. What happens to the accuracy of Taylor polynomials as the order increases?
3. How is the Taylor polynomial different from the Maclaurin polynomial?
4. Why does the Taylor polynomial use derivatives at a point \( a \) as the basis of approximation?
5. How can one calculate higher-order derivatives for more complex functions in Taylor series?

### Tip:
When approximating a function near a given point, using the Taylor polynomial centered at that point usually gives a more accurate approximation than using a polynomial centered elsewhere.

Find the Taylor polynomial T2(x) and compute the error |f(x) - T2(x)| for the given values of a and x.

f(x) = e^(sin(x)), a = π/2, x = 1.3

(Round your answer to six decimal places.)

Learn how to calculate the second-order Taylor polynomial T2(x) and the approximation error for the function f(x) = e^(sin(x)) centered at a = π/2, x = 1.3. This detailed solution involves Taylor series expansion and error analysis. Suitable for college calculus students.

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