Problem 50 - Calculus

Question

Give an example of: A Taylor series that is convergent at \(x=-1\).

Step-by-Step Solution

Answer

The Taylor series for \( e^x \) converges at \( x = -1 \).

Step 1: Define a known function with a Taylor series

A common choice for a Taylor series is the exponential function, \( e^x \), which has a Taylor series centered at \( x = 0 \):\[e^x = \sum_{n=0}^{fty} \frac{x^n}{n!}.\]

Step 2: Evaluate the convergence at x = -1

To determine if the series converges at \( x = -1 \), substitute \(-1\) into the Taylor series:\[e^{-1} = \sum_{n=0}^{fty} \frac{(-1)^n}{n!}.\]This series is known to be convergent for all \( x \) because the exponential function \( e^x \) converges everywhere.

Step 3: Verify using the Ratio Test

To confirm the convergence at \( x = -1 \), apply the Ratio Test:- Consider \( a_n = \frac{(-1)^n}{n!} \).- Compute the ratio:\[\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(-1)^{n+1}}{(n+1)!} \cdot \frac{n!}{(-1)^n}\right| = \frac{1}{n+1}.\]- The limit as \( n \to \infty \) of this ratio is 0, which is less than 1, indicating convergence by the Ratio Test.

Key Concepts

Exponential Function

The exponential function, denoted as \( e^x \), is one of the most important functions in mathematics. It is often chosen for Taylor series examples due to its nice and well-behaved properties.
The Taylor series for the exponential function centered at \( x = 0 \) can be expressed as:

• \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \)

This means we can represent \( e^x \) as an infinite sum of terms, where each term is \( \frac{x^n}{n!} \). The world of exponential functions becomes even more fascinating because their Taylor series converges for all real numbers \( x \).
This convergence property makes the exponential function a handy tool in math, especially when needing a function that behaves nicely under integration and differentiation.

Convergence

Convergence is a fundamental concept when dealing with series like Taylor series. In simple terms, it refers to whether the sum of the series approaches a fixed number as more terms are added. If it does, we say the series is convergent. If not, it is divergent.
For the Taylor series of \( e^x \), convergence at a specific point like \( x = -1 \) can be checked by substituting \( -1 \) into the series expression, resulting in:

• \( e^{-1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \)

The exponential function has the remarkable property of converging for all real values of \( x \). This means that the infinite series will sum to a finite number no matter what real number you substitute for \( x \), such as \( -1 \) in this exercise. This property is beneficial in proving that a Taylor series converges at a specific point.

Ratio Test

The Ratio Test is a method used to determine the convergence of an infinite series. It's a simple yet powerful tool in calculus and analysis.
To use the Ratio Test on a series with terms \( a_n \), compute the limit of the absolute value of the ratio of successive terms as \( n \) approaches infinity:

• \( \left| \frac{a_{n+1}}{a_n} \right| \)

If this limit is less than 1, the series converges.
For the series \( e^{-1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \), applying the Ratio Test involves calculating:

• \( \left| \frac{(-1)^{n+1}}{(n+1)!} \cdot \frac{n!}{(-1)^n} \right| = \frac{1}{n+1} \)
• As \( n \rightarrow \infty \), this becomes 0, which is less than 1

Therefore, by the Ratio Test, the series converges at \( x = -1 \). The Ratio Test is invaluable for quickly assessing the behavior of a series without summing an infinite number of terms.