Problem 5
Question
Zeigen Sie, daß für die reelle Potenzreihe $$ \sum_{n=O}^{\infty}(-1)^{n} x^{2 n} $$ der Konvergenzradius \(R\) gleich 1 ist. Zeigen Sie weiterhin, daß für \(x \in]-R, R[\) gilt $$ \frac{1}{1+x^{2}}=\sum_{n=O}^{\infty}(-1)^{n} x^{2 n} $$
Step-by-Step Solution
Verified Answer
The convergence radius is 1, and \(\frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}\) for \(-1 < x < 1\).
1Step 1: Identify the Formula for Convergence Radius
The convergence radius \(R\) of a power series \(\sum_{n=0}^{\infty} a_n x^n\) can be found using the formula \(R = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}}\). For the given series \(\sum_{n=0}^{\infty} (-1)^n x^{2n}\), the terms are given by \(a_n = (-1)^n\).
2Step 2: Calculate Radius of Convergence
Find the \(n\)-th root of the absolute value of the coefficients: \(\sqrt[n]{|a_n|} = \sqrt[n]{1} = 1\). The limit superior as \(n\) approaches infinity is \(1\). Thus, the convergence radius \(R = 1\).
3Step 3: Simplify the original power series
Notice that the power series is given by \(\sum_{n=0}^{\infty} (-1)^n x^{2n}\), which is a geometric series of the form \(\sum_{n=0}^{\infty} r^n\) where \(r = -x^2\).
4Step 4: Evaluate the Geometric Series
The sum of a geometric series \(\sum_{n=0}^{\infty} r^n\) converges to \(\frac{1}{1-r}\) for \(|r| < 1\). Here \(r = -x^2\), so the series sum is \(\frac{1}{1+x^2}\) for \(|x^2| < 1\) or \(-1 < x < 1\).
5Step 5: Verify the Function Equality
We need to show that the function \(\frac{1}{1+x^2}\) is equal to the series \(\sum_{n=0}^{\infty} (-1)^n x^{2n}\) within the interval \(x \in (-1, 1)\). Since the geometric series converges to \(\frac{1}{1+x^2}\), the equality holds for \(-1 < x < 1\).
Key Concepts
Power SeriesGeometric SeriesConvergence Criteria
Power Series
Power series are mathematical expressions in the form of \( \sum_{n=0}^{\infty} a_n x^n \). Each term in the series involves a coefficient \( a_n \) and a power \( x^n \). The main idea is to represent functions as infinite sums of power terms.
For example, consider the power series \( \sum_{n=0}^{\infty} (-1)^n x^{2n} \), where the coefficients \( a_n \) alternate in sign.
Understanding power series is fundamental in calculus as it allows approximating complex functions through polynomial expressions.
For example, consider the power series \( \sum_{n=0}^{\infty} (-1)^n x^{2n} \), where the coefficients \( a_n \) alternate in sign.
- A power series can converge or diverge depending on the value of \( x \).
- If the series converges for some values of \( x \), it means that as you add more terms, the sum approaches a specific value.
- The set of all \( x \) values that will make the series converge is determined by its convergence radius.
Understanding power series is fundamental in calculus as it allows approximating complex functions through polynomial expressions.
Geometric Series
A geometric series is a special type of series where each term is a constant multiple of the previous one. It is expressed in the form \( \sum_{n=0}^{\infty} r^n \), where \( r \) is the common ratio.
In our example, the series \( \sum_{n=0}^{\infty} (-1)^n x^{2n} \) behaves like a geometric series with common ratio \( r = -x^2 \).
This property is used to equate the given power series with the function \( \frac{1}{1+x^2} \) in the specified domain. Recognizing when a power series is also a geometric series can simplify complex expressions significantly.
In our example, the series \( \sum_{n=0}^{\infty} (-1)^n x^{2n} \) behaves like a geometric series with common ratio \( r = -x^2 \).
- A geometric series converges if the absolute value of the common ratio, \( |r| \), is less than 1.
- When it converges, the sum of a geometric series is given by \( \frac{1}{1-r} \).
This property is used to equate the given power series with the function \( \frac{1}{1+x^2} \) in the specified domain. Recognizing when a power series is also a geometric series can simplify complex expressions significantly.
Convergence Criteria
Convergence criteria are rules to determine if a series converges or not. For power series, the key criterion involves finding the radius of convergence.
The convergence radius \( R \) for a series \( \sum_{n=0}^{\infty} a_n x^n \) is calculated using the formula \( R = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}} \). This tells us the range of \( x \) values for which the series converges.
Verifying convergence and understanding these criteria is crucial in ensuring that your calculations lead to meaningful results. They help in analyzing if and where the function representations through series are valid.
The convergence radius \( R \) for a series \( \sum_{n=0}^{\infty} a_n x^n \) is calculated using the formula \( R = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}} \). This tells us the range of \( x \) values for which the series converges.
- For geometric series, the criterion \( |r| < 1 \) can be used to determine convergence.
- In our context, it means the series converges for \( -1 < x < 1 \) because \( r = -x^2 \).
Verifying convergence and understanding these criteria is crucial in ensuring that your calculations lead to meaningful results. They help in analyzing if and where the function representations through series are valid.
Other exercises in this chapter
Problem 3
Zeigen Sie, daß die reelle Potenzreihe \(\sum_{n=O}^{\infty} n^{n} x^{n}\) für keine reelle zahl x konvergiert, abgesehen von dem Fall \(x=O\).
View solution Problem 4
Uberprüfen Sie noch einmal mit der Formel von Hadamard, für welche \(x \in \mathbb{R}\) die geometrische Reihe $$ \begin{gathered} \sum_{n=O}^{\infty} x^{n} \\
View solution Problem 2
Zeigen Sie mit Hilfe des Quotientenkriteriums, da\beta die komplexe Potenzreihe $$ \sum_{n=O}^{\infty} \frac{z^{n}}{n !} $$ absolut konvergent ist für alle \(z
View solution