c. Determine the interval of convergence of the series.
f(x) = e²ˣ, a = 0
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Identify the given function and the center of the Taylor series expansion. Here, the function is \(f(x) = e^{2x}\) and the center is \(a = 0\).
Recall the Taylor series expansion for \(e^x\) centered at 0, which is \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\). For \(e^{2x}\), replace \(x\) by \$2x\( to get \)\sum_{n=0}^{\infty} \frac{(2x)^n}{n!}$.
Write the general term of the series as \(a_n = \frac{(2x)^n}{n!}\). To find the interval of convergence, apply the Ratio Test, which involves computing the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
Calculate the ratio inside the limit: \(\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(2x)^{n+1} / (n+1)!}{(2x)^n / n!} \right| = \left| \frac{2x}{n+1} \right|\). Then, find the limit as \(n\) approaches infinity.
Since \(\lim_{n \to \infty} \left| \frac{2x}{n+1} \right| = 0\) for all real \(x\), the Ratio Test tells us the series converges for all real numbers. Therefore, the interval of convergence is \((-\infty, \infty)\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Taylor Series Expansion
A Taylor series represents a function as an infinite sum of terms calculated from the derivatives of the function at a single point, called the center (a). For f(x) = e^(2x) at a = 0, the series is formed by evaluating derivatives at 0 and expressing f(x) as a power series in x.
The radius of convergence defines the distance from the center within which the Taylor series converges to the function. The interval of convergence is the set of x-values for which the series converges, determined by tests like the Ratio Test, and may include or exclude endpoints.
The Ratio Test is used to find the radius of convergence by examining the limit of the absolute value of the ratio of consecutive terms. If this limit is less than one, the series converges; if greater, it diverges. This test helps identify the interval where the Taylor series for e^(2x) converges.