Problem 7
Question
Find a power series for the function, centered at \(c,\) and determine the interval of convergence. $$ g(x)=\frac{1}{2 x-5}, \quad c=-3 $$
Step-by-Step Solution
Verified Answer
The power series representation of the function, centered at \(c = -3\), is: \( g(x) = \sum_{n=0}^{\infty} ((2/5)^n * (x+3)^n) \). The interval of convergence is \( -13/2 < x < -3/2 \).
1Step 1: Represent the function in the form of geometric series
Firstly, rewrite the function in a form that fits the geometric series format, \(1/(1-r)\). The function can be rewritten as \( g(x) = 1/((2/5)(x+3)-1) \). This can be represented as a geometric series with \(r=(2/5)(x+3)\) which results in \( g(x) = \sum_{n=0}^{\infty} ((2/5)^n * (x+3)^n) \).
2Step 2: Determine the radius of convergence
The radius of convergence, R, for a geometric series is determined by \(|r| < 1\). So, \((2/5)|x+3| < 1\). Solving this inequality gives the radius of convergence, \(R = 5/2\). This means the power series converges for all x within this radius.
3Step 3: Determine the interval of convergence
The interval of convergence is determined by \( |x+3| < R \). This leads to the inequality \( -5/2 < x+3 < 5/2 \). Solving this yields the interval of convergence, \( -13/2 < x < -3/2 \).
Key Concepts
Geometric SeriesRadius of ConvergenceInterval of Convergence
Geometric Series
A geometric series is a sum of the form \( \sum_{n=0}^{\infty} ar^n \) where \( a \) is the first term, and \( r \) is the common ratio. The series converges if the absolute value of \( r \) is less than 1, i.e., \( |r| < 1 \). In other words, for a geometric series to sum to a finite value, the ratio between consecutive terms must shrink in absolute value.
In the context of the problem provided, the function \( g(x)=\frac{1}{2x-5} \) is transformed into a geometric series by rewriting it in the form \( 1/(1-r) \) with \( r \) being \( (2/5)(x+3) \). This restructuring allows us to use the properties of geometric series to find a power series representation of \( g(x) \).
In the context of the problem provided, the function \( g(x)=\frac{1}{2x-5} \) is transformed into a geometric series by rewriting it in the form \( 1/(1-r) \) with \( r \) being \( (2/5)(x+3) \). This restructuring allows us to use the properties of geometric series to find a power series representation of \( g(x) \).
Radius of Convergence
In power series, the radius of convergence is a measure of the interval around the center point \( c \) where the series converges absolutely. It is denoted by \( R \). For a geometric series with ratio \( r \) based on the variable \( x \) around a center \( c \) the radius of convergence is the distance from \( c \) to the nearest \( x \) where \( |r(x)| \geq 1 \).
The radius of convergence can be calculated by solving the inequality \( |r|<1 \). In our example, we find that \( R = 5/2 \) by setting the condition \( (2/5)|x+3| < 1 \) and solving for \( x \). The radius of convergence tells us that our series for \( g(x) \) will reliably converge when \( x \) is within \( 5/2 \) units of the center \( -3 \) on the number line.
The radius of convergence can be calculated by solving the inequality \( |r|<1 \). In our example, we find that \( R = 5/2 \) by setting the condition \( (2/5)|x+3| < 1 \) and solving for \( x \). The radius of convergence tells us that our series for \( g(x) \) will reliably converge when \( x \) is within \( 5/2 \) units of the center \( -3 \) on the number line.
Interval of Convergence
The interval of convergence describes the exact range of \( x \) values for which a power series converges. It can be an interval like \( (a, b) \) or it could extend to infinity. To find the interval of convergence, one must test the endpoints of the interval determined by the radius of convergence to see if the series converges at these points as well.
For our example, the radius of convergence gives us a preliminary interval \( (-13/2, -3/2) \). However, it's crucial to plug these endpoints back into the original series to confirm whether the series converges at \( -13/2 \) and \( -3/2 \). Doing this provides us with the definitive interval of convergence, ensuring that whenever we plug a number within this interval into our power series, it will converge to a finite sum, which is the value of our original function \( g(x) \).
For our example, the radius of convergence gives us a preliminary interval \( (-13/2, -3/2) \). However, it's crucial to plug these endpoints back into the original series to confirm whether the series converges at \( -13/2 \) and \( -3/2 \). Doing this provides us with the definitive interval of convergence, ensuring that whenever we plug a number within this interval into our power series, it will converge to a finite sum, which is the value of our original function \( g(x) \).
Other exercises in this chapter
Problem 7
Use the definition to find the Taylor series (centered at \(c\) ) for the function. $$ f(x)=\sin 2 x, \quad c=0 $$
View solution Problem 7
Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n+1)}{\ln (n+1)} $$
View solution Problem 8
Verify that the infinite series diverges. $$ \sum_{n=0}^{\infty}\left(\frac{4}{3}\right)^{n} $$
View solution Problem 8
Write the first five terms of the sequence. \(a_{n}=10+\frac{2}{n}+\frac{6}{n^{2}}\)
View solution