Problem 1
Question
$$\frac{z}{1+z}=z\left[1-z+z^{2}-z^{3}+\cdots\right]=z-z^{2}+z^{3}-z^{4}+\cdots=\sum_{k=1}^{\infty}(-1)^{k+1} z^{k} ; \quad R=1$$
Step-by-Step Solution
Verified Answer
The series sums to \(\frac{z}{1+z}\) for \(|z| < 1\).
1Step 1: Understand the Given Series
We are given a series: \(\sum_{k=1}^{\infty}(-1)^{k+1}z^{k}\). This is an alternating series where the powers of \(z\) increase starting from \(z^1\).
2Step 2: Identify the Series Type
Recognize the series as a geometric series. The series is of the form \(z - z^2 + z^3 - z^4 + \cdots\). This is a geometric series with the first term \(a = z\) and common ratio \(r = -z\).
3Step 3: Find the Formula for the Sum of a Geometric Series
The sum \(S\) of an infinite geometric series with the first term \(a\) and common ratio \(r\) is given by \(S = \frac{a}{1-r}\) when \(|r| < 1\).
4Step 4: Apply the Formula
For this series, we have \(a = z\) and \(r = -z\). Applying the formula, the sum is \(S = \frac{z}{1 - (-z)} = \frac{z}{1 + z}\).
5Step 5: Validate the Radius of Convergence
The given radius of convergence \(R = 1\) is consistent with the condition \(|r| = |z| < 1\) for convergence, as \(r = -z\). This confirms that the series converges for \(|z| < 1\).
6Step 6: Conclusion
The expression \(\frac{z}{1+z}\) is indeed equal to the sum of the given series \(\sum_{k=1}^{\infty}(-1)^{k+1} z^{k}\) for \(|z| < 1\).
Key Concepts
Alternating SeriesRadius of ConvergenceInfinite Series
Alternating Series
An alternating series is a series where the terms alternate in sign. In mathematical notation, it is expressed through series like
The series begins with the first positive term and each subsequent term's sign alternates. This is why you see the pattern of
- \(a_1 - a_2 + a_3 - a_4 + \cdots\)
- or \((-1)^{k+1} a_k\) for each term \(a_k\)
The series begins with the first positive term and each subsequent term's sign alternates. This is why you see the pattern of
- \(z - z^2 + z^3 - z^4 + \cdots\)
Radius of Convergence
The radius of convergence is a crucial concept when analyzing power series similar to the one given in the exercise. It defines the distance from the center of convergence within which the series will converge. Simply put, it is the value \(R\) that dictates where the values of our variable \(z\) can lie:
Understanding the radius of convergence is essential for determining the values for which an infinite series representation is valid, ensuring that manipulation and use of the series are mathematically sound.
- Inside the radius (\(|z| < R\)), the series converges.
- On the boundary (\(|z| = R\)), convergence needs to be checked separately.
- Outside the radius (\(|z| > R\)), the series diverges.
Understanding the radius of convergence is essential for determining the values for which an infinite series representation is valid, ensuring that manipulation and use of the series are mathematically sound.
Infinite Series
An infinite series is simply a sequence of terms added together that continues indefinitely. These can represent functions, calculations, or even certain constants under conditions of convergence. The series given in the exercise, \(\sum_{k=1}^{\infty}(-1)^{k+1}z^{k}\), illustrates an infinite series.
One common type of infinite series is the geometric series, recognized by a constant ratio between consecutive terms. Our series is indeed geometric with a common ratio of \(-z\).
Infinite series find applications in approximations and deeper mathematical applications such as Fourier series, Taylor series, and other functions. Understanding their behavior enables predicting long-term results in many practical settings.
One common type of infinite series is the geometric series, recognized by a constant ratio between consecutive terms. Our series is indeed geometric with a common ratio of \(-z\).
- In the case of geometric series, it can sum to a finite value only if the absolute value of the common ratio, \(|r|\), is less than 1.
Infinite series find applications in approximations and deeper mathematical applications such as Fourier series, Taylor series, and other functions. Understanding their behavior enables predicting long-term results in many practical settings.
Other exercises in this chapter
Problem 1
$$\int_{0}^{2 \pi} \frac{d \theta}{1+\frac{1}{2} \sin \theta}=\oint_{C} \frac{4}{z^{2}+4 i z-1} d z=(4) 2 \pi i \operatorname{Res}(f(z),(\sqrt{3}-2) i)=\frac{4
View solution Problem 1
$$\text { Using } e^{2 z}=\sum_{k=0}^{\infty} \frac{2^{k} z^{k}}{k !} \text { we obtain }$$ \(\frac{e^{2 z}-1}{z}=\frac{\left(1+\frac{2}{1 !} z+\frac{2^{2}}{2 !
View solution Problem 1
$$f(z)=\frac{1}{z}\left(1-\frac{z^{2}}{2 !}+\frac{z^{4}}{4 !}-\frac{z^{6}}{6 !}+\cdots\right)=\frac{1}{z}-\frac{z}{2 !}+\frac{z^{3}}{4 !}-\frac{z^{5}}{6 !}+\cdo
View solution Problem 4
Write \(f(z)=z^{4}-16=\left(z^{2}-4\right)\left(z^{2}+4\right)=(z-2)(z+2)(z-2 i)(z+2 i)\) to see that \(2,-2,2 i,\) and \(-2 i\) are zeros of \(f .\) Now \(f^{\
View solution