To approximate the function $e^{x^2}$ at the point $x = 0$ using a computer, we can rely on its Taylor series expansion around $x = 0$. The Taylor series of $e^{x^2}$ can be written as:

$e^{x^2} = 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6} + \cdots$

For a numerical approximation at $x = 0$ using terms up to order 2 and 3, we truncate the series.

Order 2 Approximation:

Here, we keep terms up to $x^2$:

$e^{x^2} \approx 1 + x^2$

Order 3 Approximation:

Since there is no term involving $x^3$ directly in the expansion (it is a function of $x^2$), the order 3 approximation also looks like:

$e^{x^2} \approx 1 + x^2$

This means that both the 2nd and 3rd-order approximations are effectively the same: $1 + x^2$, since the next higher term involves $x^4$, which isn't included in either approximation.

Approximation Error (for $0 < x < 1$):

The error in the approximation can be determined by comparing the exact function $e^{x^2}$ to the truncated Taylor series.

Using the remainder term from the Taylor series, the error can be approximated by the next term not included in the expansion. In this case, the next term is $\frac{x^4}{2}$. So the error $E(x)$ for $0 < x < 1$ can be bounded as:

$E(x) = e^{x^2} - (1 + x^2) \approx \frac{x^4}{2}$

For $0 < x < 1$, this error is small since $x^4$ is quite small for values of $x$ in this range. Specifically, $E(x) \leq \frac{1}{2}$ when $x = 1$, and decreases as $x$ gets closer to 0.

Conclusion:

• The 2nd and 3rd-order approximations for $e^{x^2}$ at $x = 0$ yield the same result: $1 + x^2$.
• The error in this approximation, for $0 < x < 1$, is bounded by $\frac{x^4}{2}$, which is relatively small.

Would you like more details on this or have any further questions?

Here are some related questions to expand on this topic:

1. What are higher-order approximations of $e^{x^2}$ using more Taylor series terms?
2. How does the approximation error change for larger values of $x$?
3. How can you calculate the remainder term rigorously in the Taylor series?
4. What is the role of machine precision in numerical approximations for small values of $x$?
5. Can we use other methods like Padé approximants to improve the approximation of $e^{x^2}$?

Tip: For better numerical accuracy, consider using higher-order terms in the Taylor expansion, especially when $x$ is not close to zero.