Problem 6
Question
Find a power series representation for the function and determine the interval of convergence. \( f(x) = \frac {4}{2x + 3} \)
Step-by-Step Solution
Verified Answer
The power series is \( \sum_{n=0}^{\infty} \frac{4}{3} \left(-\frac{2}{3}\right)^n x^n \) with interval of convergence \( -\frac{3}{2} < x < \frac{3}{2} \).
1Step 1: Rewrite the Function
First, we rewrite the function in a form that will allow us to eventually express it as a geometric series. Start by factoring out the constant from the denominator:\[ f(x) = \frac{4}{3} \cdot \frac{1}{1 - (-\frac{2x}{3})} \]This allows us to put the function in a form that can be expressed as a geometric series.
2Step 2: Identify the Geometric Series
We recall that the geometric series formula is:\[ \sum_{n=0}^{\infty} r^n = \frac{1}{1-r} \]Here, we identify that our function already resembles this form with \( r = -\frac{2x}{3} \). Thus:\[ \frac{1}{1 - (-\frac{2x}{3})} = \sum_{n=0}^{\infty} \left(-\frac{2x}{3}\right)^n \]
3Step 3: Construct the Power Series
Now that we have identified the geometric series, we multiply it by the factor we factored out previously:\[ f(x) = \frac{4}{3} \cdot \sum_{n=0}^{\infty} \left(-\frac{2x}{3}\right)^n = \sum_{n=0}^{\infty} \frac{4}{3} \cdot \left(-\frac{2}{3}\right)^n x^n \]Thus, the power series representation for \( f(x) \) is:\[ \sum_{n=0}^{\infty} \frac{4}{3} \left(-\frac{2}{3}\right)^n x^n \]
4Step 4: Determine the Interval of Convergence
The interval of convergence can be determined by finding where the series converges. For a geometric series \( \sum_{n=0}^{\infty} r^n \) to converge, the condition is \( |r| < 1 \). Here, \( r = -\frac{2x}{3} \), so:\[ \left| -\frac{2x}{3} \right| < 1 \]Simplifying this inequality gives:\[ \left| \frac{2x}{3} \right| < 1 \]\[ \left| x \right| < \frac{3}{2} \]Thus, the interval of convergence is \( -\frac{3}{2} < x < \frac{3}{2} \).
Key Concepts
Understanding Geometric SeriesExploring the Interval of ConvergenceWhat Makes a Series Convergent?
Understanding Geometric Series
A geometric series is a type of series where each term is derived by multiplying the previous term by a fixed, non-zero number called the ratio. This fixed number is key in identifying and expressing a series as geometric.
The standard form of a geometric series is \( \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio.
For example, the series \( 1 + 2 + 4 + 8 + ... \) is a geometric series where \( a = 1 \) and \( r = 2 \).
The standard form of a geometric series is \( \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio.
For example, the series \( 1 + 2 + 4 + 8 + ... \) is a geometric series where \( a = 1 \) and \( r = 2 \).
- If \( |r| < 1 \), the series converges, meaning it adds up to a finite value.
- If \( |r| \ge 1 \), the series diverges, continuing indefinitely without reaching a sum.
Exploring the Interval of Convergence
The interval of convergence of a power series is the range of values for which the series converges, or in other words, sums to a finite value.
This interval is critical in understanding the behavior of functions represented by power series.
To determine an interval of convergence for a geometric series, we focus on the ratio \( r \). The series converges if \( |r| < 1 \).
For our example, we have \( r = -\frac{2x}{3} \), so:
\( \left| \frac{2x}{3} \right| < 1 \) simplifies to \( |x| < \frac{3}{2} \), giving us the interval \( -\frac{3}{2} < x < \frac{3}{2} \). Understanding this can help solve various problems in analysis.
This interval is critical in understanding the behavior of functions represented by power series.
To determine an interval of convergence for a geometric series, we focus on the ratio \( r \). The series converges if \( |r| < 1 \).
For our example, we have \( r = -\frac{2x}{3} \), so:
\( \left| \frac{2x}{3} \right| < 1 \) simplifies to \( |x| < \frac{3}{2} \), giving us the interval \( -\frac{3}{2} < x < \frac{3}{2} \). Understanding this can help solve various problems in analysis.
- An interval of convergence can be finite as in the example, or infinite, extending from negative to positive infinity.
- The endpoints of the interval need to be checked individually for convergence.
What Makes a Series Convergent?
A convergent series is one that converges to a specific value. This means that as you add more terms, the series approaches a finite sum.
This concept is central in mathematics, especially in calculus and analysis.
To check if a series converges, we often use tests. The geometric series test is a straightforward method for series with constant ratios.
This concept is central in mathematics, especially in calculus and analysis.
To check if a series converges, we often use tests. The geometric series test is a straightforward method for series with constant ratios.
- If the ratio of each term \( r \) is less than 1 in absolute value, the series converges.
- Otherwise, it diverges, meaning it continues indefinitely without settling on a finite value.
Other exercises in this chapter
Problem 6
Find the Taylor polynomials \( T_3(x) \) for the function \( f \) centered at the number \( a \) Graph \( f \) and \( T_3 \) on the same screen. \( f(x) = e^{-x
View solution Problem 6
Use the definition of a Taylor series to find the first four nonzero terms of the series for \( f(x) \) centered at the given value of \( a. \) \( f(x) = \frac
View solution Problem 6
Find the radius of convergence and interval of convergence of the series. \( \sum_{n = 1}^{\infty} \frac {( - 1)^n x^n}{n^2} \)
View solution Problem 6
Test the series for convergence or divergence. \( \displaystyle \sum_{n = 1}^{\infty} \frac {n^{2n}}{(1 + n)^{3n}} \)
View solution