Problem 4 - Advanced Engineering Mathematics

Question

Find the radius of convergence and interval of convergence for the given power series. $$ \sum_{k=0}^{\infty} k !(x-1)^{k} $$

Step-by-Step Solution

Answer

The radius of convergence is 0, and the interval of convergence is \( x = 1 \).

Step 1: Understanding the Problem

We need to find the radius of convergence and interval of convergence for the power series \( \sum_{k=0}^{\infty} k !(x-1)^{k} \). This involves determining the values of \( x \) for which this series converges.

Step 2: Using the Ratio Test for Convergence

To find the radius of convergence, we can apply the Ratio Test to the series \( a_k = k! (x-1)^k \). The Ratio Test requires that we compute the limit \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \).

Step 3: Compute the Ratio

Calculate \( \frac{a_{k+1}}{a_k} = \frac{(k+1)! (x-1)^{k+1}}{k! (x-1)^k} = (k+1)|x-1| \).

Step 4: Finding the Limit for the Ratio Test

The limit therefore becomes \( \lim_{k \to \infty} (k+1)|x-1| \). As \( k \to \infty \), this limit tends to infinity for any non-zero \( |x-1| \).

Step 5: Conclusion of the Ratio Test

Since the limit tends to infinity whenever \( x eq 1 \), the series converges only for \( x = 1 \). The radius of convergence \( R \) is therefore 0.

Step 6: Interval of Convergence

Given that the series converges only at the point \( x = 1 \), it follows that the interval of convergence is \( x = 1 \).

Key Concepts

Understanding Power Series

A power series is incredibly useful in mathematics and engineering. It's a series of the form:\[\sum_{k=0}^{\infty} c_k (x-a)^{k}\]where \(c_k\) are coefficients and \(a\) is the center of the series. It represents a function similar to a polynomial but with infinitely many terms. The power series is centered at \(a\), meaning all terms are in the form \((x-a)^k\). This centering allows us to model functions as if they are polynomials near that specific point \(a\).

• Coefficients \(c_k\): Each term of the power series has a coefficient which influences the shape and convergence of the series.
• Center \(a\): This is the point around which the series is expanded. For our exercise's series, the center is \(x=1\).

Understanding the structure of a power series helps simplify complex functions into manageable algebraic expressions.

Discovering the Interval of Convergence

The interval of convergence is the set of values for which a power series converges. For our purposes, finding this interval involves determining the values of \(x\) such that the series converges to a finite sum. For series centered at \(x=1\), we identify points where changing \(x\) leads to the series converging or diverging.

To check convergence at different \(x\) values, we calculate the radius of convergence \(R\):

• Inside the interval where \(|x-a| < R\), the series converges.
• Outside, where \(|x-a| > R\), it diverges.

In the given series from the exercise:

• The series only converges at \(x=1\), meaning the radius of convergence is \(R=0\).
• This results in a convergent interval of just the single point \(x=1\).

Understanding the interval helps determine how effectively we can use series to approximate functions in certain regions.

Utilizing the Ratio Test for Convergence

The Ratio Test is a potent method for assessing the convergence of series, especially power series. To employ the test, we evaluate the limit:\[L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right|\]Here \(a_k\) is the general term of the power series. This test helps understand how quickly the terms of the series are approaching zero.

• Convergence: If \(L < 1\), the series converges.
• Divergence: If \(L > 1\), the series diverges.
• If \(L = 1\), the test is inconclusive; other methods might be necessary.

For the exercise, the limit \( L = \lim_{k \to \infty} (k+1)|x-1|\) was obtained. As \(k\) approaches infinity, this limit tends to infinity unless \(x = 1\). Therefore, the radius of convergence \(R\) is \(0\), leading to a convergence at \(x=1\) only. This practical approach simplifies solving convergence problems.