Problem 8 - Essential Calculus

Question

Find a power series for the function, centered at \(c,\) and determine the interval of convergence. $$ h(x)=\frac{1}{2 x-5}, \quad c=0 $$

Step-by-Step Solution

Answer

The power series for the given function \( h(x)=\frac{1}{2x-5} \) centered at \( c=0 \) is \( h(x)=\frac{1}{5}\sum_{n=0}^\infty \left(\frac{-x}{5}\right)^n \) and the interval of convergence is (-5,5).

Step 1: Rewrite the function

First, rewrite the function \(h(x)=\frac{1}{2x-5}\) as \(h(x)=\frac{1}{5}(1+\frac{x}{5})^{-1}\), by dividing each term in the denominator by 5.

Step 2: Conversion into power series

Next, use the standard geometric series formula \( \frac{1}{1-r} = \sum_{n=0}^\infty r^n \), which converges for \( |r|<1 \), and where \( r=\frac{2x}{5} \). Thus, the power series is \( h(x)=\frac{1}{5}\sum_{n=0}^\infty \left(\frac{-x}{5}\right)^n \).

Step 3: Find Interval of Convergence

For convergence, the absolute value of \( r=\frac{x}{5} \) must be less than 1, i.e, \( |-x/5|<1 \). Solving this gives the interval of convergence (-5,5).

Key Concepts

Interval of Convergence

When dealing with power series, one of the fundamental concepts to understand is the interval of convergence. Essentially, this interval represents the set of all real numbers for which the series converges, or in simpler terms, adds up to a finite value. To determine this interval, we look at the ratio within the series, generally denoted by r. For the power series to converge, the absolute value of this ratio must be less than one.

In the given exercise, after applying the ratio of convergence condition, \( |-x/5|<1 \) yields the interval \( (-5,5) \). It is within this range that we can confidently declare the power series as convergent. It's essential for students to not only memorize this rule but to understand it, as this comprehension will assist them in solving convergence problems for any power series they encounter.

Geometric Series

Understanding the geometric series is crucial when transforming functions into power series. A series is geometric if each term after the first is obtained by multiplying the previous one by a constant ratio. The series can be summed up using the formula \( \frac{1}{1-r} = \sum_{n=0}^\infty r^n \), where \( r \) is the common ratio between consecutive terms.

However, this elegant expression only holds true when \( |r| < 1 \), which is why it's known as the condition for convergence. For instance, in the provided example, \( h(x) \) is expressed through such a series, with the terms changing by a factor of \( \frac{-x}{5} \) each time. Notably, this power series representation hinges on the principle of the geometric series, exemplifying how well-established mathematical concepts lay the foundation for broader applications.

Power Series Representation

The power series representation of a function is akin to expanding the function into an infinite polynomial. It is a potent tool that lies at the heart of many areas of calculus, analysis, and applied mathematics. This representation is useful for simplifying complicated functions and for performing computations that would otherwise be daunting.

In the step-by-step solution, the function \( h(x) \) is converted into a power series by identifying a comparable geometric series. After recognizing the appropriate format, we rewrite \( h(x) \) in terms of a series with an exponentiated term, which in our case is \( \left(\frac{-x}{5}\right)^n \). Grasping this component of power series can truly enhance a student's mathematical toolkit and allow them to tackle a vast range of problems with newfound ease and flexibility.