Taylor series and interval of convergence

c. Determine the interval of convergence of the series.

f(x)=3ˣ, a=0

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Recall that the Taylor series for the function \(f(x) = 3^x\) centered at \(a=0\) (Maclaurin series) can be expressed using the general formula for the Taylor series: \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\]

Find the \(n\)th derivative of \(f(x) = 3^x\). Since \(3^x = e^{x \ln 3}\), the \(n\)th derivative is \[f^{(n)}(x) = (\ln 3)^n 3^x\] Therefore, at \(x=0\), \[f^{(n)}(0) = (\ln 3)^n 3^0 = (\ln 3)^n\]

Substitute the derivatives into the Taylor series formula to get \[f(x) = \sum_{n=0}^{\infty} \frac{(\ln 3)^n}{n!} x^n\]

To determine the interval of convergence, apply the Ratio Test to the series terms \[a_n = \frac{(\ln 3)^n}{n!} x^n\] Calculate \[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(\ln 3)^{n+1}}{(n+1)!} x^{n+1} \cdot \frac{n!}{(\ln 3)^n x^n} \right| = \lim_{n \to \infty} \left| \frac{(\ln 3) x}{n+1} \right|\]

Since \[\lim_{n \to \infty} \frac{|(\ln 3) x|}{n+1} = 0\] 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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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) = 3^x at a = 0, the series expresses 3^x as a power series in terms of (x - 0), allowing approximation of the function near zero.

Radius and Interval of Convergence

The interval of convergence is the set of x-values for which the Taylor series converges to the function. It is determined by finding the radius of convergence, often using the Ratio or Root Test, and then checking the endpoints to see if the series converges there.

Ratio Test for Convergence

The Ratio Test is a method to determine the convergence of an infinite series by examining the limit of the absolute value of the ratio of consecutive terms. If this limit is less than one, the series converges absolutely; if greater than one, it diverges; if equal to one, the test is inconclusive.

Taylor series and interval of convergencec. Determine the interval of convergence of the series.f(x)=3ˣ, a=0