To find the Taylor series for f(x) = e^x² centered at x = 0, we substitute u = x² into the Maclaurin series for e^u. This gives us the series in terms of x.

ind the Taylor series expansion of f(x) = e^(x²) centered at $x = 0$ and determine its interval of convergence.

Answer: The Taylor series expansion of $f(x) = e^{x^2}$ centered at $x = 0$ is $\sum_{n=0}^{\infty} \frac{x^{2n}}{n!} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots$ . The interval of convergence is $(-\infty, \infty)$ .

Step-by-step solution

Step 1: Recall the Maclaurin series for $e^u$

The Maclaurin series is a special case of the Taylor series where the expansion is centered at $x = 0$ . We know the Maclaurin series for $e^u$ is given by the sum from $n=0$ to infinity of $u^n$ divided by $n!$ . This series converges for all real values of $u$ .

Step 2: Substitute $u = x^2$ into the series

To find the Taylor series for $f(x) = e^{x^2}$ centered at $x = 0$ , we substitute $u = x^2$ into the Maclaurin series for $e^u$ . This gives us the series in terms of $x$ .

Step 3: Write out the first few terms of the series

Expanding the series, we can write out the first few terms to better visualize the pattern. This gives us $1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots$ .

Step 4: Determine the interval of convergence

We use the Ratio Test to find the interval of convergence. Let $a_n = \frac{x^{2n}}{n!}$ . We calculate the limit of the absolute value of the ratio of consecutive terms, $\frac{a_{n+1}}{a_n}$ . Since the limit is $0$ , which is less than $1$ , the series converges for all real values of $x$ .

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ind the Taylor series expansion of f(x) = e^(x²) centered at x=0x = 0x=0 and determine its interval of convergence.