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Chapter 3: Problem 3

Establish that the function \(\sum_{n=1}^{\infty} \frac{1}{e^{n} n}\) is an analytic function of \(z\) for all \(z\) that is, it is an entire function.

Short Answer

Expert verified
The function \(\sum_{n=1}^{\infty} \frac{1}{e^n n}\) is entire because its power series representation converges for all z.

Step by step solution

01

Understand the Given Series

The series given is \(\sum_{n=1}^{\infty} \frac{1}{e^{n} n}\). Set this series as a function of z: \(f(z) = \sum_{n=1}^{\infty} \frac{z^n}{e^{n} n}\).
02

Analyze Convergence

Determine whether the series converges for all complex numbers z. This can be achieved using the ratio test: \(\left|\frac{a_{n+1}}{a_n}\right| = \left| \frac{\frac{z^{n+1}}{e^{n+1} (n+1)}}{\frac{z^n}{e^n n}} \right| = \left| \frac{z}{e (n+1)} \right|\).
03

Applying the Ratio Test

Since \(\left| \frac{z}{e (n+1)} \right| \to 0\) as \ n \to \infty, the ratio test shows that the series converges absolutely for all z. Hence, the series \( \sum_{n=1}^{\infty} \frac{z^n}{e^n n} \) converges for all z.
04

Conclusion about Analyticity

A power series that converges absolutely everywhere (for all z) defines an entire function. Therefore, the function \(f(z) = \sum_{n=1}^{\infty} \frac{z^n}{e^n n}\) is an entire function because its series representation converges for all complex numbers z.

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Key Concepts

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

entire function
An entire function is a type of complex function that is analytic at every point in the complex plane \( \mathbb{C} \). To say a function \( f(z) \) is analytic means it is both differentiable and smooth over its entire domain without any singularities.
Such functions are special because they have series expansions that converge everywhere on the complex plane.
Because of this, entire functions are also referred to as holomorphic functions across \( \mathbb{C} \).
Examples of entire functions include polynomials, the exponential function \( e^z \), and sine and cosine functions.
power series
A power series is an infinite series of the form \( \sum_{n=0}^{\infty} a_n z^n \) where \( a_n \) are coefficients and \( z \) is a variable. This kind of series is significant in complex analysis because it can represent many types of functions. When you look at a power series, you're essentially expressing a function as an infinite sum of its variables raised to increasing powers, each multiplied by a coefficient.
Power series are powerful because they can often represent complex functions in a simpler form. They are also at the heart of many convergence tests, like the ratio test, which can tell us whether the series converges (produces a finite value) or diverges (does not produce a finite value).
ratio test
The ratio test helps us determine if a series converges or diverges. It works by examining the limit of the ratio of successive terms in a series as the number of terms goes to infinity, i.e., \[ \lim_\{n \to \infty\} \left|\frac{a_\{n+1\}}{a_n}\right| \].
If this limit is less than 1, the series converges absolutely. If it's greater than 1, the series diverges. If it equals 1, the test is inconclusive. This test is especially useful for power series because of their specific structure of terms. In the context of our given series \( \sum_{n=1}^{\infty} \frac{z^n}{e^n n}\), applying the ratio test shows that the series converges for all \( z \), hence suggesting the series is absolutely convergent and defines an entire function.
absolute convergence
A series \( \sum a_n \) is said to converge absolutely if the series of absolute values \( \sum \left| a_n\right| \) converges. Absolute convergence is stronger than ordinary convergence because if a series converges absolutely, it converges in any order, preventing issues with reordering terms that might alter the sum.
For our given series \( \sum_{n=1}^{\infty} \frac{z^n}{e^n n}\), we use both the ratio test and the property of absolute convergence to conclude that the series converges for all complex numbers \( z \). This convergence across the entire complex plane implies that the function defined by the series is an entire function.

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Most popular questions from this chapter

Show that about any point \(z=x_{0}\) $$ e^{z}=e^{x_{0}} \sum_{n=0}^{\infty} \frac{\left(z-x_{0}\right)^{n}}{n !} $$

(a) Let \(f(z)\) have simple poles at \(z=z_{n}, n=1,2,3, \ldots, N\), with strengths \(a_{n}\), and be analytic everywhere else. Show by contour integration that (the reader may wish to consult Theorem 4.1.1) $$ \frac{1}{2 \pi i} \oint_{C_{N}} \frac{f\left(z^{\prime}\right)}{z^{\prime}-z} d z^{\prime}=f(z)+\sum_{n=1}^{N} \frac{a_{n}}{z_{n}-z} $$ where \(C_{N}\) is a large circle of radius \(R_{N}\) enclosing all the poles. Evaluate (1) at \(z=0\) to obtain $$ \frac{1}{2 \pi i} \oint_{C_{N}} \frac{f\left(z^{\prime}\right)}{z^{\prime}} d z^{\prime}=f(0)+\sum_{n=1}^{N} \frac{a_{n}}{z_{n}} $$ (b) Subtract equation (2) from equation (1) of part (a) to obtain $$ \frac{1}{2 \pi i} \oint_{C_{N}} \frac{z f\left(z^{\prime}\right)}{z^{\prime}\left(z^{\prime}-z\right)} d z^{\prime}=f(z)-f(0)+\sum_{n=1}^{N} a_{n}\left(\frac{1}{z_{n}-z}-\frac{1}{z_{n}}\right) $$ (c) Assume that \(f(z)\) is bounded for large \(z\) to establish that the left- hand side of Equation (3) vanishes as \(R_{N} \rightarrow \infty\). Conclude that if the sum on the right-hand side of Equation (3) converges as \(N \rightarrow \infty\), then $$ f(z)=f(0)+\sum_{n=1}^{\infty} a_{n}\left(\frac{1}{z-z_{n}}+\frac{1}{z_{n}}\right) $$ (This is a special case of the Mittag-Leffler Theorems 3.6.2-3.6.3) (d) Let \(f(z)=\pi \cot \pi z-1 / z\), and show that $$ \pi \cot \pi z-\frac{1}{z}=\sum_{n=-\infty}^{\infty}\left(\frac{1}{z-n}+\frac{1}{n}\right) $$ where the prime denotes the fact that the \(n=0\) term is omitted. (Equation (5) is another derivation of the result in this section.) We see that an infinite series of poles can represent the function \(\cot \pi z\). Section \(3.6\) establishes that we have other series besides Taylor series and Laurent series that can be used for representations of functions.

Let the Euler numbers \(E_{n}\) be defined by the power series $$ \frac{1}{\cosh z}=\sum_{n=0}^{\infty} \frac{E_{n}}{n !} z^{n} $$ (a) Find the radius of convergence of this series. (b) Determine the first six Euler numbers.

Suppose we are given the equation \(d^{2} w / d z^{2}=2 w^{3}\). (a) Let us look for a solution of the form $$ w=\sum_{n=0}^{\infty} a_{n}\left(z-z_{0}\right)^{n-r}=a_{0}\left(z-z_{0}\right)^{-r}+a_{1}\left(z-z_{0}\right)^{1-r}+\cdots $$ for \(z\) near \(z_{0}\). Substitute this into the equation to determine that \(r=1\) and \(a_{0}=\pm 1\) (b) "Linearize" about the basic solution by letting \(w=\pm 1 /\left(z-z_{0}\right)+v\) and dropping quadratic terms in \(v\) to find \(d^{2} v / d z^{2}=6 v /\left(z-z_{0}\right)^{2}\). Solve this equation (Cauchy-Euler type) to find $$ v=A\left(z-z_{0}\right)^{-2}+B\left(z-z_{0}\right)^{3} $$ (c) Explain why this indicates that all coefficients of subsequent powers in the following expansion (save possibly \(a_{4}\) ) $$ w=\frac{\pm 1}{\left(z-z_{0}\right)}+a_{1}+a_{2}\left(z-z_{0}\right)+a_{3}\left(z-z_{0}\right)^{2}+a_{4}\left(z-z_{0}\right)^{3}+\cdots $$ can be solved uniquely. Substitute the expansion into the equation for \(w\), and find \(a_{1}, a_{2}\), and \(a_{3}\), and establish the fact that \(a_{4}\) is arbitrary. We obtain two arbitrary constants in this expansion: \(z_{0}\) and \(a_{4}\). The solution to \(w^{\prime \prime}=2 w^{3}\) can be expressed in terms of elliptic functions; its general solution is known to have only simple poles as its movable singular points. (d) Show that a similar expansion works when we consider the equation $$ \frac{d^{2} w}{d z^{2}}=z w^{3}+2 w $$ (this is the second Painlev茅 equation (Ince, 1956)), and hence that the formal analysis indicates that the only movable algebraic singular points are poles. (Painlev茅 proved that there are no other singular points for this equation.) (e) Show that this expansion fails when we consider $$ \frac{d^{2} w}{d z^{2}}=2 w^{3}+z^{2} w $$ because \(a_{4}\) cannot be found. This indicates that a more general expansion is required. (In fact, another term of the form \(b_{4}\left(z-z_{0}\right)^{3} \log \left(z-z_{0}\right)\) must be added at this order, and further logarithmic terms must be added at all subsequent orders in order to obtain a consistent formal expansion.)

Given the equation $$ \frac{d w}{d z}=p(z) w^{2}+q(z) w+r(z) $$ where \(p(z), q(z), r(z)\) are (for convenience) entire functions of \(z\) (a) Letting \(w=\alpha(z) \phi^{\prime}(z) / \phi(z)\), show that taking \(\alpha(z)=-1 / p(z)\) eliminates the term \(\left(\phi^{\prime} / \phi\right)^{2}\), and find that \(\phi(z)\) satisfies $$ \phi^{\prime \prime}-\left(q(z)+\frac{p^{\prime}(z)}{p(z)}\right) \phi^{\prime}+p(z) r(z) \phi=0 $$ (b) Explain why the function \(w(z)\) has, as its only movable singular points, poles. Where are they located? Can there be any fixed singular points? Explain.
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