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Chapter 3: Problem 4

Show that the following functions are analytic functions of \(z\) for all \(z\); that is, they are entire: (a) \(\sum_{n=1}^{\infty} \frac{z^{n}}{(n !)^{2}}\) (b) \(\sum_{n=1}^{\infty} \frac{\cosh n z}{n !}\) (c) \(\sum_{n=1}^{\infty} \frac{z^{2 n+1}}{[(2 n+1) !]^{1 / 2}}\)

Short Answer

Expert verified
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Step by step solution

01

- Determining Radius of Convergence for Series (a)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radius of Convergence
When dealing with power series, understanding their **radius of convergence** is crucial. The radius of convergence is the distance from the center of the series within which the series converges.
To determine the radius of convergence, we often use the Root Test or the Ratio Test.

For a power series \(\sum_{n=0}^{\infty}a_n z^n\) centered at 0, the radius of convergence \(R\) is given by:
\[ R = \frac{1}{\limsup_{n \rightarrow \infty}\sqrt[n]{|a_n|}} \]

The series converges if \( |z| < R \) and diverges if \( |z| > R \). When \( |z| = R \), additional tests are needed to determine convergence.

For series (a): \ \sum_{n=1}^{\infty} \frac{z^n}{(n!)^2} \
The ratio test gives:
\ \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \rightarrow \infty} \frac{|z|^{n+1}}{(n+1)!^2} \cdot \frac{n!^2}{|z|^n} = \lim_{n \rightarrow \infty} \frac{|z|}{(n+1)^2} = 0 \ for all \ |z| < \infty, meaning \( R = \infty \).
Thus, the radius of convergence is infinite, indicating the series converges everywhere in the complex plane.
Power Series
A **power series** is an infinite series of the form: \ \sum_{n=0}^{\infty} a_n (z - c)^n \ where \( a_n \) are the coefficients and \( c \) is the center of the series.
Power series can represent various types of functions within their radius of convergence. Analyzing power series is essential in complex analysis.

Convergence of a power series depends on:
  • Radius of Convergence \( R \)
  • Behavior when \( |z - c| < R \) and \( |z - c| = R \)
In our examples:
For series (b): \ \sum_{n=1}^{\infty} \frac{\cosh(nz)}{n!} \
The ratio test for the series:
\ \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \rightarrow \infty} \frac{\cosh((n+1)z)}{(n+1)!} \cdot \frac{n!}{\cosh(nz)} = \lim_{n \rightarrow \infty} \frac{\cosh((n+1)z)}{(n+1)\cosh(nz)} = 0 \ for all \ |z| < \infty.
This means the radius of convergence is again infinite.
Entire Functions
An **entire function** is a complex function that is analytic everywhere in the complex plane. This means it can be represented by a power series that converges everywhere.

To show that a function is entire, demonstrate that its power series has an infinite radius of convergence.

For series (c): \ \sum_{n=1}^{\infty} \frac{z^{2n+1}}{[(2n+1)!]^{1/2}} \
Using the ratio test:
\ \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \rightarrow \infty} \frac{|z|^{2(n+1) + 1}}{[(2(n+1) + 1)!]^{1/2}} \cdot \frac{[(2n+1)!]^{1/2}}{|z|^{2n+1}} = \lim_{n \rightarrow \infty} \frac{|z|^2}{[(2(n+1) + 1)/(2n+1)]^{1/2}} \ which also tends to 0.
Consequently, the radius of convergence is infinite and the series represents an entire function.

By confirming each series converges for all \( z \) in the complex plane, we have shown that each function in the exercise is indeed entire.

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Most popular questions from this chapter

Suppose we know a function \(f(z)\) is analytic in the finite \(z\) plane apart from singularities at \(z=i\) and \(z=-i\). Moreover, let \(f(z)\) be given by the Taylor series $$ f(z)=\sum_{j=0}^{\infty} a_{j} z^{j} $$ where \(a_{j}\) is known. Suppose we calculate \(f(z)\) and its derivatives at \(z=\) \(3 / 4\) and compute a Taylor series in the form $$ f(z)=\sum_{j=0}^{\infty} b_{j}\left(z-\frac{3}{4}\right)^{j} $$ Where would this series converge? How could we use this to compute \(f(z)\) ? Suppose we wish to compute \(f(2.5)\); how could we do this by series methods?

Expand the function \(f(z)=1 /\left(1+z^{2}\right)\) in (a) a Taylor series for \(|z|<1\) (b) a Laurent series for \(|z|>1\)

Consider the equation $$ \frac{d y}{d z}+2 z y=1, \quad y(1)=1 $$ (a) Solve this equation using the series method. Evaluate the solution as we traverse the unit circle. Show that the solution is single valued. (b) Evaluate an approximation to \(y(-1)\) from the series. (c) Show that an exact representation of the solution in terms of integrals is $$ y(z)=\int_{1}^{z} e^{t^{2}-z^{2}} d t+e^{1-z^{2}} $$ and verify that, by evaluating \(y(z)\) by a Taylor series (i.e., use \(e^{r^{2}}=\) \(\left.1+t^{2}+t^{4} / 2 !+t^{6} / 3 !+t^{8} / 4 !+\cdots\right)\), the answer obtained from this series is a good approximation to that obtained in part (b).

Discuss whether the following series converge uniformly in the given domains: (a) \(\sum_{j=1}^{n} z^{j}, \quad|z|<1\) (b) \(\sum_{j=0}^{n} e^{-j z}, \quad \frac{1}{2}<|z|<1\) (c) \(\sum_{j=1}^{n} j ! z^{2 j}, \quad|z|0\)

Define \(F(z)=\int_{-\infty}^{\infty} f(t) e^{i z t} d t\), where \(|f(t)| \leq e^{-a|t|}, a>0\). Find where \(F(z)\) is analytic. Explain.
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