Problem 4 - Complex Analysis

Question

For \(n \in \mathbb{N}_{0}\) define $$ e_{n}(z)=\sum_{\nu=0}^{n} \frac{z^{\nu}}{\nu !} $$ For a given \(R>0\) there exists an \(n_{0}\), such that for all \(n \geq n_{0}\) the function \(e_{n}\) has no zero in \(U_{R}(0)\).

Step-by-Step Solution

Answer

There exists an \(n_0\) such that \(e_n(z)\) has no zeros in \(U_R(0)\) for all \(n \geq n_0\).

Step 1: Understanding the Problem

We need to find a natural number \(n_0\) such that for all \(n \geq n_0\), the function \(e_n(z)\) has no zeros inside the disk \(U_R(0) = \{z \in \mathbb{C} : |z| < R\}\). This means that the series \(e_n(z)\) must not equal zero for all \(z\) with modulus less than \(R\).

Step 2: Define the Exponential Function Approximation

Recall the exponential function \(e^z = \sum_{u=0}^{\infty} \frac{z^{u}}{u!}\). The function \(e_n(z)\) is the truncated series of this exponential function, representing an approximation of \(e^z\) for a finite number of terms.

Step 3: Apply the Weierstrass Approximation

For any complex number \(z\), a large enough \(n\) will ensure that the difference \(|e^z - e_n(z)| < \epsilon\) for any \(\epsilon > 0\). Since \(e^z eq 0\) for any \(z\), if \(e_n(z)\) is sufficiently close to \(e^z\), then \(e_n(z)\) cannot be zero if \(e^z\) is not.

Step 4: Uniform Convergence on Compact Sets

Given the disk \(U_R(0)\) is compact, for any \(\epsilon > 0\), we can choose \(n_0\) such that for all \(|z| < R\), \(|e^z - e_n(z)| < \epsilon\) when \(n \geq n_0\). Choose \(\epsilon = \min_{|z| \leq R} |e^z|/2\) to ensure \(e_n(z)\) has no zeros. Since \(e^z\) is continuous and non-zero on compact sets, it attains a minimum positive value on \(|z| \leq R\).

Step 5: Conclusion

By choosing \(n_0\) such that for all \(n \geq n_0\), the inequality \(|e^z - e_n(z)| < \min_{|z| \leq R} |e^z|/2\) holds, it guarantees \(e_n(z)\) does not have zeros in \(U_R(0)\). Therefore, such an \(n_0\) exists as required.

Key Concepts

Exponential Function

The exponential function, denoted as \(e^z\), is a fundamental mathematical concept with wide applications in complex analysis. It is defined through an infinite series:

• \(e^z = \sum_{u=0}^{\infty} \frac{z^{u}}{u!}\)

This series converges for all complex numbers \(z\), making it an entire function. The exponential function exhibits unique properties like having a constant derivative (\(e^z\) itself) and never reaching zero in the complex plane. These properties make it particularly useful in solving differential equations and studying complex dynamics. In the problem provided, the function \(e_n(z)\) is a finite approximation of \(e^z\), where the approximation improves as \(n\) increases. Understanding how these approximations converge to the actual function is key to solving the exercise.

Uniform Convergence

Uniform convergence is a type of convergence where a sequence of functions \(f_n(x)\) converges to a function \(f(x)\) in such a way that the speed of convergence does not depend on \(x\). This means for any given small positive number \(\epsilon\), there exists an index \(n_0\) such that for all \(n \geq n_0\) and all points \(x\) in the domain, the inequality

• \(|f_n(x) - f(x)| < \epsilon\) holds.

For the exercise in question, this concept ensures that the truncated exponential polynomials \(e_n(z)\) come uniformly close to \(e^z\) on compact sets like the closed disk \(\{|z| \leq R\}\). This reliable approximation allows us to deduce properties of the functions \(e_n(z)\), particularly concerning the absence of zeros within the specified region.

Weierstrass Approximation

The Weierstrass Approximation Theorem is pivotal in understanding how polynomial approximations can render any continuous function as accurately as desired over a compact interval. In the context of this problem, we are interested in how the truncated sum \(e_n(z)\) approximates the continuous transcendental function \(e^z\).

• The theorem effectively says that polynomials are dense in the space of all continuous functions over compact domains.

This means, for any \(\epsilon > 0\), a polynomial (or in our case, the partial sums of the series for \(e^z\)) can be found which is uniformly close to \(e^z\). Therefore, if \(e_n(z)\) approximates \(e^z\) sufficiently well, then \(e_n(z)\) also maintains the property of not having zeros within a given region, provided \(e^z\) does not have zeros.

Zero-free Regions

Zero-free regions pertain to regions in the complex plane where a function does not take on the value zero. For the exponential function, it is well-known that \(e^z\) is never zero for any complex \(z\), forming a zero-free region across the entire complex plane. When approximating \(e^z\) with \(e_n(z)\), we attempt to preserve this property.

• By ensuring that \(e_n(z)\) is sufficiently close to \(e^z\) everywhere on a region, we conclude \(e_n(z)\) is also zero-free in that region for sufficiently large \(n\).

In the given problem, calculating the appropriate \(n_0\) involves examining the closeness of \(e_n(z)\) to \(e^z\) over the compact set \(U_R(0)\) (a disk of radius \(R\)). By using uniform convergence and the fact that \(e^z\) is non-zero, we draw the conclusion that \(e_n(z)\) will also not be zero within this region for all \(n \geq n_0\).