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