Analytic Continuation Of Zeta Function Vs 1/(1-x)^2 At X=1

Introduction

The fascinating world of analytic continuation allows us to extend the domain of a function beyond its initial definition. In complex analysis, this technique is particularly powerful when dealing with functions defined by infinite series or integrals that converge only within a limited region. A classic example is the Riemann zeta function, denoted by $\zeta(s)$, which is initially defined for complex numbers $s$ with a real part greater than 1 by the infinite series:

$$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$

This series converges nicely for $\Re(s) > 1$, but the magic happens when we use analytic continuation to extend $\zeta(s)$ to the entire complex plane, except for a simple pole at $s = 1$. This extended function reveals deep connections to number theory, particularly the distribution of prime numbers. Many popular videos and resources delve into the analytic continuation of the Riemann zeta function, often focusing on intriguing values like $\zeta(-1) = -\frac{1}{12}$, which arises from the analytic continuation despite the divergent nature of the original series at $s = -1$.

In this article, we will explore a thought-provoking question: How does the analytic continuation of $\zeta(-x)$ at $x = 1$ compare to that of another function, namely $\frac{1}{(1-x)^2}$, which also possesses a series expansion? This comparison highlights the nuances of analytic continuation and provides insights into how different functions behave when extended beyond their initial domains of convergence. We'll delve into the series expansion of $\frac{1}{(1-x)^2}$, its region of convergence, and how its analytic continuation differs from that of $\zeta(-x)$. This exploration will not only enhance our understanding of analytic continuation but also shed light on the unique properties of the Riemann zeta function and its profound significance in mathematics.

Series Expansion and Analytic Continuation of $\frac{1}{(1-x)^2}$

To understand the behavior of $\frac{1}{(1-x)^2}$, let's first examine its series expansion. We can derive this expansion using the geometric series formula and differentiation. Recall that the geometric series is given by:

$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$

This series converges for $|x| < 1$. Now, let's differentiate both sides of the equation with respect to $x$:

$$\frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{d}{dx} \left( \sum_{n=0}^{\infty} x^n \right)$$

Applying the differentiation, we get:

$$\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} nx^{n-1} = \sum_{n=0}^{\infty} (n+1)x^n$$

Thus, the series expansion of $\frac{1}{(1-x)^2}$ is $\sum_{n=0}^{\infty} (n+1)x^n$. This series, like the geometric series, converges for $|x| < 1$. The radius of convergence can be easily determined using the ratio test. Let $a_n = (n+1)x^n$. Then:

$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(n+2)x^{n+1}}{(n+1)x^n} \right| = \lim_{n \to \infty} \left| \frac{n+2}{n+1} x \right| = |x|$$

For convergence, we require $|x| < 1$, which confirms that the radius of convergence is 1. The interval of convergence is therefore $-1 < x < 1$.

The function $\frac{1}{(1-x)^2}$ is already defined for all complex numbers except $x = 1$. Therefore, the analytic continuation is simply the function itself. There is no need to extend the domain beyond its natural domain of definition. The function has a pole of order 2 at $x = 1$, which is evident from its form. This behavior contrasts with the Riemann zeta function, where the analytic continuation is crucial to understanding its behavior beyond $\Re(s) > 1$.

In summary, the series expansion of $\frac{1}{(1-x)^2}$ is $\sum_{n=0}^{\infty} (n+1)x^n$, which converges for $|x| < 1$. The analytic continuation of this function is the function itself, and it has a pole of order 2 at $x = 1$. This straightforward analytic continuation provides a valuable point of comparison when we consider the more intricate analytic continuation of the Riemann zeta function.

Analytic Continuation of the Riemann Zeta Function $\zeta(-x)$

The Riemann zeta function, initially defined by the series $\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$ for complex numbers $s$ with $\Re(s) > 1$, requires a more sophisticated approach for analytic continuation. Unlike $\frac{1}{(1-x)^2}$, the series definition of $\zeta(s)$ diverges for $\Re(s) \leq 1$, necessitating the use of other methods to extend its domain. One common technique involves the Dirichlet eta function, also known as the alternating zeta function, defined as:

$$\eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s} = 1 - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \cdots$$

This series converges for $\Re(s) > 0$, a larger domain than the original series for $\zeta(s)$. The connection between the Dirichlet eta function and the Riemann zeta function is given by:

$$\eta(s) = (1 - 2^{1-s}) \zeta(s)$$

This relationship allows us to express $\zeta(s)$ in terms of $\eta(s)$:

$$\zeta(s) = \frac{\eta(s)}{1 - 2^{1-s}}$$

While this expression extends the definition of $\zeta(s)$ to $\Re(s) > 0$ (excluding the points where $1 - 2^{1-s} = 0$), it still doesn't provide the full analytic continuation. Further techniques, such as integral representations or functional equations, are needed to extend $\zeta(s)$ to the entire complex plane, except for a simple pole at $s = 1$.

One crucial tool for understanding the analytic continuation of $\zeta(s)$ is the functional equation, which relates $\zeta(s)$ to $\zeta(1-s)$:

$$\zeta(s) = 2^s \pi^{s-1} \sin\left( \frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s)$$

where $\Gamma(s)$ is the gamma function. This equation is vital because it connects the values of $\zeta(s)$ in the region $\Re(s) > 1$ to its values in the region $\Re(s) < 0$. The gamma function itself has an analytic continuation to the entire complex plane, except for poles at non-positive integers.

Now, let's consider $\zeta(-x)$. We are interested in the behavior of this function at $x = 1$, which corresponds to evaluating $\zeta(-1)$. Using the functional equation, we can find $\zeta(-1)$:

Setting $s = -1$ in the functional equation, we get:

$$\zeta(-1) = 2^{-1} \pi^{-2} \sin\left( -\frac{\pi}{2} \right) \Gamma(2) \zeta(2)$$

We know that $\sin(-\frac{\pi}{2}) = -1$, $\Gamma(2) = 1! = 1$, and $\zeta(2) = \frac{\pi^2}{6}$. Substituting these values, we obtain:

$$\zeta(-1) = \frac{1}{2} \cdot \frac{1}{\pi^2} \cdot (-1) \cdot 1 \cdot \frac{\pi^2}{6} = -\frac{1}{12}$$

Thus, $\zeta(-1) = -\frac{1}{12}$. This result is a classic example of the surprising values that arise from analytic continuation. It's important to note that this value is obtained through the analytic continuation and not from the original series definition, which diverges at $s = -1$.

In summary, the analytic continuation of the Riemann zeta function is a complex process that involves techniques like the Dirichlet eta function and the functional equation. Evaluating $\zeta(-x)$ at $x = 1$ yields $\zeta(-1) = -\frac{1}{12}$, a value obtained through the functional equation and not the divergent series definition. This contrasts sharply with the behavior of $\frac{1}{(1-x)^2}$, where the analytic continuation is simply the function itself.

Comparison at $x = 1$

Now, let's directly compare the behavior of $\zeta(-x)$ and $\frac{1}{(1-x)^2}$ at $x = 1$. We have already established that the analytic continuation of $\zeta(-x)$ at $x = 1$ yields $\zeta(-1) = -\frac{1}{12}$. This value is finite and well-defined, arising from the functional equation and the analytic continuation process.

On the other hand, the function $\frac{1}{(1-x)^2}$ has a pole of order 2 at $x = 1$. This means that as $x$ approaches 1, the function tends to infinity. We can express this as:

$$\lim_{x \to 1} \frac{1}{(1-x)^2} = \infty$$

This divergence at $x = 1$ is a crucial difference between the two functions. While $\zeta(-x)$ has a finite value at $x = 1$ due to its analytic continuation, $\frac{1}{(1-x)^2}$ has a pole, indicating a singularity.

The series expansion perspective also provides valuable insights. The series expansion of $\frac{1}{(1-x)^2}$ is $\sum_{n=0}^{\infty} (n+1)x^n$, which converges only for $|x| < 1$. At $x = 1$, the series becomes $\sum_{n=0}^{\infty} (n+1)$, which clearly diverges to infinity. This divergence aligns with the pole at $x = 1$ for the function $\frac{1}{(1-x)^2}$.

In contrast, the series definition of the Riemann zeta function, $\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$, diverges for $s = -1$. However, the analytic continuation provides a finite value. This highlights the power of analytic continuation in assigning values to functions beyond their initial domains of convergence.

The comparison at $x = 1$ underscores the distinct behaviors of these two functions. The Riemann zeta function, through its intricate analytic continuation, manages to have a finite value at $x = 1$ (when considering $\zeta(-x)$), while $\frac{1}{(1-x)^2}$ exhibits a singularity, tending to infinity. This difference stems from the unique properties of the Riemann zeta function and the methods used to extend its domain.

Conclusion

In this article, we have explored the analytic continuation of two functions: the Riemann zeta function $\zeta(-x)$ and $\frac{1}{(1-x)^2}$, with a particular focus on their behavior at $x = 1$. We've seen that while both functions have series representations, their analytic continuations and behaviors at $x = 1$ differ significantly.

The function $\frac{1}{(1-x)^2}$ has a straightforward analytic continuation, as the function itself is defined for all complex numbers except $x = 1$, where it has a pole of order 2. Its series expansion, $\sum_{n=0}^{\infty} (n+1)x^n$, converges for $|x| < 1$, and the function tends to infinity as $x$ approaches 1.

The Riemann zeta function, on the other hand, requires a more complex approach for analytic continuation. Techniques like the Dirichlet eta function and the functional equation are crucial in extending its domain beyond $\Re(s) > 1$. Evaluating $\zeta(-x)$ at $x = 1$ yields $\zeta(-1) = -\frac{1}{12}$, a finite value obtained through the functional equation. This result showcases the power of analytic continuation in assigning values to functions beyond their initial domains of convergence.

The comparison at $x = 1$ highlights the distinct behaviors of these functions. $\zeta(-1)$ is finite, while $\frac{1}{(1-x)^2}$ has a pole, tending to infinity. This difference underscores the unique properties of the Riemann zeta function and the importance of analytic continuation in understanding its behavior.

This exploration into the analytic continuation of $\zeta(-x)$ and $\frac{1}{(1-x)^2}$ provides a valuable insight into the fascinating world of complex analysis. The Riemann zeta function, with its deep connections to number theory, continues to be a subject of intense study, and understanding its analytic continuation is paramount to unlocking its secrets.