Problem 2 - Single Variable Calculus

Question

Use Equation (1) to find the Taylor series of \(f\) at the given value of \(c .\) Then find the radius of convergence of the series. \(f(x)=e^{-3 x}, \quad c=0\)

Step-by-Step Solution

Answer

The Taylor series of \(f(x) = e^{-3x}\) at \(c = 0\) is \(f(x) = \sum_{n=0}^{\infty} \frac{(-3)^n}{n!}x^n\), and the radius of convergence of the series is \(\infty\).

Step 1: Find the first few derivatives of \(f(x)\)

Start by finding the first few derivatives of \(f(x)\): \(f'(x) = -3e^{-3x}, f''(x) = 9e^{-3x}, f'''(x) = -27e^{-3x}, \cdots\) Notice the pattern; the \(n\)-th derivative will be: \(f^{(n)}(x) = (-3)^n e^{-3x}\)

Step 2: Evaluate the derivatives at c = 0

Now evaluate the derivatives at \(c = 0\): \(f(0) = e^0 = 1\) \(f'(0) = -3e^0 = -3\) \(f''(0) = 9e^0 = 9\) \(f'''(0) = -27e^0 = -27\) And in general: \(f^{(n)}(0) = (-3)^n e^0 = (-3)^n\)

Step 3: Substitute the derivatives into the Taylor series formula

Use the Taylor series formula to write the Taylor series for \(f(x)\) at \(c = 0\): \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n = \sum_{n=0}^{\infty} \frac{(-3)^n}{n!}x^n\) This is the Taylor series of \(f(x) = e^{-3x}\) at \(c = 0\).

Step 4: Determine the radius of convergence

To find the radius of convergence of the series, we will use the Ratio Test: \(\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{\frac{(-3)^{n+1}}{(n+1)!}x^{n+1}}{\frac{(-3)^n}{n!}x^n}\) Simplify the expression: \(\lim_{n\to\infty} \frac{(-3)^{n+1}(n!)}{(-3)^n(n+1)!}x = \lim_{n\to\infty} \frac{-3}{n+1}x\) The limit is 0 when \(n\) goes to infinity for any value of \(x\). This means the radius of convergence is infinite: Radius of convergence = \(\infty\) #Conclusion#: So, the Taylor series of \(f(x) = e^{-3x}\) at \(c = 0\) is: \(f(x) = \sum_{n=0}^{\infty} \frac{(-3)^n}{n!}x^n\) The radius of convergence of the series is \(\infty\).

Key Concepts

Understanding the Radius of Convergence

When you encounter a power series, it's essential to know where the series converges, or in simple terms, where it 'works'. The span or extent in which the series converges is known as the "radius of convergence". If the radius of convergence is \(R\), then the series converges absolutely for all \(x\) such that \[ |x-c| < R \]. For some series, this radius can be finite, meaning the series converges only within a specified interval around \(c\). However, some series can have an infinite radius of convergence, meaning the series converges for every real number \(x\). In the case of the series \( e^{-3x} \), calculated using the Taylor series at \( c=0 \), the radius of convergence is infinite. This could be confirmed through the Ratio Test, which shows that the necessary condition for convergence holds for all \(x\). Hence, the function behaves well and converges everywhere on the real line.

The Role of Derivatives in Taylor Series

Derivatives play a crucial role in constructing a Taylor series. The Taylor series is essentially built using the derivatives of a function at a particular point. For the function \(f(x) = e^{-3x}\), you start by computing the first few derivatives:

• The first derivative, \(f'(x) = -3e^{-3x}\), tells us how the function is changing at any given point.
• Finding the second derivative, \(f''(x) = 9e^{-3x}\), then tells us how this change rate itself is changing, and so forth.

Subsequently, by evaluating these derivatives at the point \(c=0\), we acquire terms used in forming the Taylor series. Each derivative evaluated at \(c\) contributes to the series' terms in a specific way, as they are divided by factorials. This series offers a powerful way to approximate or represent functions, especially for calculations requiring estimations near the point \(c\).

Applying the Ratio Test to Determine Convergence

The Ratio Test is a handy tool to find the radius of convergence of a Taylor series. This test works by examining the limit of the ratio of successive terms in a series. For a series with general term \(a_n\), the Ratio Test considers:\[ \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| \]If this limit is less than 1, the series converges absolutely. If it is greater than 1, the series diverges.In the problem, for \(f(x) = e^{-3x}\), the series \(\sum_{n=0}^{\infty} \frac{(-3)^n}{n!}x^n\) gives trigonometric ratios that simplify through algebraic manipulation to reveal a limit of 0. This finding informs us that the radius of convergence is \(\infty\), indicating convergence for all real numbers \(x\). The thorough simplification underlines the convergence behavior, making the Ratio Test an excellent method for assessing where a Taylor series remains valid.