To find $T_2(x)$, which represents the second-degree Taylor polynomial for $y = e^x$ around $x = 0$, we can use the Taylor series formula:

$T_2(x) = f(0) + f'(0) \cdot x + \frac{f''(0)}{2} \cdot x^2$

For $y = e^x$:

1. $f(x) = e^x$
2. $f'(x) = e^x$
3. $f''(x) = e^x$

Since all derivatives of $e^x$ are $e^x$, we have: $f(0) = e^0 = 1, \quad f'(0) = e^0 = 1, \quad f''(0) = e^0 = 1$

Substitute into the Taylor polynomial formula:

$T_2(x) = 1 + 1 \cdot x + \frac{1}{2} \cdot x^2 = 1 + x + \frac{x^2}{2}$

Step 1: Compute $T_2(0.5)$

Substitute $x = 0.5$:

$T_2(0.5) = 1 + 0.5 + \frac{(0.5)^2}{2} = 1 + 0.5 + \frac{0.25}{2} = 1 + 0.5 + 0.125 = 1.625$

Step 2: Calculate the error $|e^{1.5} - T_2(1.5)|$

Now, we need to evaluate $T_2(1.5)$ and find the error compared to the actual value of $e^{1.5}$.

1. Compute $T_2(1.5)$:

$T_2(1.5) = 1 + 1.5 + \frac{(1.5)^2}{2} = 1 + 1.5 + \frac{2.25}{2} = 1 + 1.5 + 1.125 = 3.625$

2. Use a calculator to find $e^{1.5}$:

$e^{1.5} \approx 4.4817$

3. Calculate the error:

$|e^{1.5} - T_2(1.5)| = |4.4817 - 3.625| \approx 0.8567$

Summary

• $T_2(0.5) = 1.625$
• The error $|e^{1.5} - T_2(1.5)| \approx 0.8567$

Would you like further details or have any questions?

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Here are some related questions you might explore:

1. How does the accuracy of the Taylor polynomial improve with higher-degree terms?
2. What would the third-degree Taylor polynomial $T_3(x)$ for $e^x$ look like?
3. Can the error for a Taylor polynomial be bounded or estimated using known formulas?
4. How does the Taylor polynomial approximation compare at different points, such as $x = 2$ or $x = -0.5$?
5. How does the error change if we expand the Taylor series around a different center, like $x = 1$ instead of $x = 0$?

Tip: Taylor polynomials provide good approximations near the expansion point $a$. The further $x$ is from $a$, the larger the approximation error typically becomes.