Riemann Zeta Function


Today we provided an overview to the Riemann Zeta Function. We started by discussing the man of the hour — Bernhard Riemann. We then defined the Riemann Zeta function and performed a calculation of $\zeta(2)$, which is known as the Basel problem. We then proved the Euler Product Formula and described its importance in relative primality of integers. We also provided a few equations which relate the Zeta Function to other functions we discussed throughout the class. Finally, we concluded with a very important hypothesis: The Riemann Hypothesis.


Bernhard Riemann was born in 1826 and died in 1866 — he was 39 years old. His theories contributed to Riemannian geometry, algebraic geometry, complex manifolds, and mathematical physics. He is best known for his work in analysis, for defining the Riemann integral using Riemann sums. In the field of number theory, Riemann only wrote one paper, establishing the importance of the Riemann Zeta function and its relation to prime numbers. Interesting enough, upon hearing of his death in 1866, Riemann's housekeeper started throwing out papers in his study, possibly destroying a proof he was working on of the Riemann Hypothesis.

We then defined Zeta, Urban Dictionary style:

"a concept, heretics refer to it as a "function", in mathematics. the meaning of life and the universe and all. zeta will show the limits of the human mind. god. "1

The Riemann Zeta Function

The Riemann Zeta Function can be expressed in many ways. Without any fancy complex analysis, the Zeta Function is defined as:

\begin{align} \zeta(s) = \sum_{n=1} ^\infty \frac{1}{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + ... \end{align}

such that $s \in \mathbb{C}$ and $\Re(s) >1$. It is important the real component of $s$ is greater than one, because if it is not, then the infinite sum diverges because of the p-test in calculus. Since the Zeta Function is defined in the complex plane, we briefly reviewed complex numbers. We noted that $s= a + i b$ such that $a,b \in \mathbb{R}$ and that $i$ is imaginary. On the complex plane, it looks something like this:



The Basel Problem

To show just how awesome the Zeta Function is, we calculated $\zeta(2)$. This seems like an easy task, but in actuality, several famous mathematicians tried calculating the sum… but failed. The Basel problem was first proposed in 1644, and then solved by Euler in 1735. He was 27! We know the sum converges because of the p-test mentioned before. All we have to do is calculate the sum, as in, we must calculate:

\begin{align} \zeta(2) = \sum_{n=1} ^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + ... \end{align}


First, we made the note this is proof is not entirely rigorous since we use the assumption below; however, it outlines the concept behind the proof nicely. You can find several proofs of this problem online — they typically use Fourier series.

Lemma 1
Consider the finite polynomial $p(x)$ with degree $n$ and non-zero roots $a_1, a_2, ... , a_n$. Finally, assume that $p(0)=1$. We can write:

\begin{align} p(x) = \left(1-\frac{x}{a_1}\right)\left(1-\frac{x}{a_2}\right)\times...\times\left(1-\frac{x}{a_n}\right) \end{align}

Although we did not supply a proof this is true. Here is an example you can work out:

\begin{align} p(x) = x^3 - \frac{7x^2}{6}-\frac{35x}{6} + 1 \end{align}

Then, $p(0)=1$ and the roots of $p(x)$ are $-2, 3$ and $\frac{1}{6}$. The Lemma states that we can write $p(x)$ as:

\begin{align} \left(1-\frac{x}{3}\right)\times\left(1+\frac{x}{2}\right)\times\left(1-\frac{x}{\frac{1}{6}}\right) \end{align}

Euler's Assumption
Euler's bold assumption was that this property which applies to finite polynomials also applies to infinite polynomials.

Euler was very tricky in formulating this proof. He assigned:

\begin{align} x \cdot p(x) = sin(x) =\sum_{n=0} ^\infty \frac{{(-1)^n}}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!}+ \frac{x^5}{5!}-... \end{align}
\begin{align} \Rightarrow p(x) = \frac{sin(x)}{x} = 1 - \frac{x^2}{3!}+ \frac{x^4}{5!}-... \end{align}

Now, we know that $p(x)$ has roots at every $x=\pm k\pi$ where $k \in \mathbb{Z}$. Also, note that $p(0) = 1$. Using Euler's Assumption, we write:

\begin{align} p(x) = \left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\times\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\times... \end{align}

Now we need another lemma…

Lemma 2

Again, consider $p(z)$ with degree $n$ and non-zero roots $r_1, r_2, ... , r_n$. We can write (because of Lemma 1):

\begin{align} p(z) = \left(1-\frac{z}{r_1}\right)\left(1-\frac{z}{r_2}\right)\times...\times\left(1-\frac{z}{r_n}\right) \end{align}
\begin{align} \Rightarrow (9) = \left(1 - \left(\frac{1}{r_1}+\frac{1}{r_2}\right)z + \frac{z^2}{r_1 r_2}\right) \left(1 - \frac{z}{r_3}\right) \cdots \left( 1- \frac{z}{r_n}\right) \end{align}
\begin{align} Now \hspace{.05in} (10) = 1 - \frac{z}{r_3} - \left(\frac{1}{r_1}+\frac{1}{r_2}\right)z + \left(\frac{1}{r_1}+\frac{1}{r_2}\right)\frac{z^2}{r_3}+ \frac{z^2}{r_1 r_2} - \frac{z^3}{r_1 r_2 r_3} \cdots \end{align}
\begin{align} (11) \Rightarrow p(z) = 1 - \left(\frac{1}{r_1}+\frac{1}{r_2}+ \frac{1}{r_3} \cdots \right)z +\left( \frac{1}{r_1 r_3}-\frac{1}{r_2 r_3}+ \frac{1}{r_1 r_2} \cdots\right) z^2 + \cdots \end{align}

Basically, given the assumptions above, we can write the coefficient of the linear term as the sum of the reciprocal of the roots of $p(z)$ We apply this Lemma to our original problem such that $z=x^2$.

Proof … continued

We rewrite (8) by multiplying out the like terms (each of the terms separated by $\times$). This gives us:

\begin{align} p(x) = \left[ 1-\frac{x^2}{\pi^2}\right]\left[ 1-\frac{x^2}{4\pi^2}\right]\left[ 1-\frac{x^2}{9\pi^2}\right] \cdots \end{align}
\begin{align} \Rightarrow p(x) = 1 - \frac{x^2}{3!}+ \frac{x^4}{5!}-\frac{x^6}{7!} \cdots = 1-x^2\left( \frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}+... \right) \cdots = 1-x^2\left(\frac{1}{\pi^2}\sum_{n=1} ^\infty \frac{{1}}{n^2}\right) \cdots \end{align}

However, we only care about the $x^2$ terms, so we equate coefficients to get:

\begin{align} -\frac{1}{3!}= - \frac{1}{\pi^2}\sum_{n=1} ^\infty \frac{{1}}{n^2} \end{align}
\begin{align} \Rightarrow \sum_{n=1} ^\infty \frac{{1}}{n^2} = \frac{\pi^2}{6} \hspace{.25in} \Box \end{align}

Yeah! However, to date, there do not exist precise sums for odd integers. We just know that the infinite sum leads to an irrational number.

Euler Product Formula

Next, we considered the Euler Product Formula. This is the first connection of the Zeta Function to prime numbers. We showed:

\begin{align} \zeta(s) = \sum_{n=1} ^\infty \frac{1}{n^s} = \prod_{p} \frac{1}{1 - \frac{1}{p^s}} = \frac{1}{2^s}}\cdot \frac{1}{3^s}}\cdots \frac{1}{p_n^s}} \end{align}

such that p are primes and that $\Re(s)>1$.

We briefly reviewed geometric series since it was also gone over on Friday. Note that $\sum_{n=1} ^\infty a r^k = \frac{a}{1-r}$. If $a=1$, then $\sum_{n=1} ^\infty r^k = \frac{1}{1-r}$. In our proof, let $r = \frac{1}{p^s}$

We then have:

\begin{align} \prod_{p} \frac{1}{1 - \frac{1}{p^s}} = \prod_{p}\left( \sum_{k=0} ^\infty {\left(\frac{1}{p_i^s}\right)}^k \right)= \sum_{k=0} ^\infty {\left(\frac{1}{p_1^s}\right)}^k \times \sum_{k=0} ^\infty {\left(\frac{1}{p_2^s}\right)}^k \times ... \times \sum_{k=0} ^\infty {\left(\frac{1}{p_n^s}\right)}^k \times ... \end{align}

since we can expand the infinite product of the geometric series. Now, let's expand out a few of the individual geometric series…

\begin{align} \prod_{p} \frac{1}{1 - \frac{1}{p^s}} = \left(1 + \frac{1}{p_1^s} + \frac{1}{p_1^{2s}} + \frac{1}{p_1^{3s}}+...\right) \left(1 + \frac{1}{p_2^s} + \frac{1}{p_2^{2s}} + \frac{1}{p_2^{3s}}+...\right)... \end{align}

Let's plug in some values:

\begin{align} \prod_{p} \frac{1}{1 - \frac{1}{p^s}} = \left(1 + \frac{1}{2^s} + \frac{1}{2^{2s}} + \frac{1}{2^{3s}}+...\right) \left(1 + \frac{1}{3^s} + \frac{1}{3^{2s}} + \frac{1}{3^{3s}}+...\right)... \end{align}

But for example $2^{2s}$ is really just $4^s$, etc. We can be smart in choosing how to multiply these series to get

\begin{align} \prod_{p} \frac{1}{1 - \frac{1}{p^s}} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + ... = \sum_{n=1} ^\infty \frac{1}{n^s} = \zeta(s) \hspace{.25in} \Box \end{align}

The Euler Product Formula is pretty awesome. One application using it is the probability of randomly selecting $s$ integers and calculating the probability that the $s$ many integers are relatively prime. It terms out that this probability is exactly:

\begin{align} {\left(\prod_{p} \frac{1}{1 - \frac{1}{p^s}}\right)}^{-1} = \frac{1}{\zeta(s)} \end{align}

For example, if we selected two integers randomly, there is a $\frac{1}{\zeta(2)}$ chance that the 2 integers are relatively prime. Well, we know $\zeta(2) = \pi^2 / 6$. That means that the chances that the two integers are relatively prime is $6 / \pi^2 = 0.607927102$. Gnarly!

Other formulations using the Riemann Zeta Function.

Although we did not explore these functions fully, we briefly mentioned how the Riemann Zeta Function relates to other functions we have seen in class. Here they are!

Thm: Let $f$ be a multiplicative function, then we have:

\begin{align} \sum_{n=1} ^\infty \frac{f(n)}{n^s} = \prod_p \left( 1 + \frac{f(p)}{p^2}+\frac{f(p^2)}{p^{2s}}+\frac{f(p^3)}{p^{3s}}+ \cdots\right) \end{align}

Use this for your homework!!!

Also, for $\Re(s)>1$:

\begin{align} \frac{1}{\zeta(s)} = \sum_{n=1} ^\infty \frac{\mu(n)}{n^s} \end{align}


\begin{align} \frac{\zeta(2s)}{\zeta(s)} = \sum_{n=1} ^\infty \frac{\lambda(n)}{n^s} \end{align}

where $\lambda$ is the Liouville function that is discussed in the forum.

Riemann Hypothesis

We ended class with a brief discussion of the Riemann Hypothesis — one of the most important unsolved problems in pure mathematics today. In fact, according to the Millennium Prize Problems website, this is the most importance unsolved problem in pure mathematics. The Millennium Prize Problems are a collection of now six awesome math problems. If you solve the Riemann Hypothesis, you get $1,000,000! Start proving!

The Riemann Hypothesis states: All the non-trivial zeros to $\zeta(s)$ occur on the $\Re(s)= 1/2$ line.

Although this hypothesis is truly awesome, without a deep understanding of complex analysis, it is hard to grasp. First, we have to extend the domain of the Zeta Function to include the entire complex plane. This can be done. First, the Zeta Function is an analytic function. This means that it behaves smoothly in analysis. It is characterized by a convergent power series. Luckily, this is exactly what how the Zeta Function is defined. Since the Zeta Function is an analytic function, we can extend the domain via an analytic continuation. We drew on the board circles where the diameter is the $\Re(s) = 1$ line. The analytic continuation works like that. We can define the analytic continuation of $\zeta$ as:

\begin{align} \zeta(s) = \frac{1}{1-2^{1-s}} \sum_{n=1} ^\infty \frac{(-1)^{n-1}}{n^s} \end{align}

This gives us the functional equation for $\zeta$:

\begin{align} \zeta(s) = 2^s \pi^{s-1} sin(\frac{\pi s}{2})\Gamma(1-s) \zeta(1-s) \end{align}

Yikes! However, we need this formulation. A few things to note: $\Gamma$ (Gamma) is the function you have seen if you have taken STAT 400. What is important is that it is also an analytic function and is defined for all $s \in \mathbb{C}$. In fact, the functional equation is defined for all $s \in \mathbb{C}$, except when $s=1$.

We then discussed zeros of the Zeta Function. Using the functional equation, we see that negative even integers are trivial zeros. This is true because then the sine function is zero. However, positive even integers are not zeros because then $\zeta(1-s)$ is not well-defined. The Riemann Hypothesis is concerned with non-trivial zeros. We know that these zeros are found in the critical strip, that is, for $0 < \Re(s) < 1$. We also know that the zeros are symmetric around the $\Re(s) = 1/2$ line. The Hypothesis is saying that all the zeros are in fact on the line.

But that is not the only possibility. We mentioned the Siegel Zero. These are potential zeros near the $\Re(s) = 1$ line. If proved, this would imply that there exist infinitely many twin primes! With that said, the Riemann Hypothesis seems to have a lot of support. In fact, the first 250 billion non-trivial zeros are all located on the $\Re(s) = 1/2$ line!

What are some non-trivial zeros? Here are the first few:




In the math world, this hypothesis is über important. Recall that $\pi(x)$ is the Prime Counting Function, it counts the number of prime numbers less than $x$. At the age of 15, Gauss proposed that $\pi(x) \approx x / ln(x)$. In fact, he is right, since this later became the Prime Number Theorem, defined as:

\begin{align} \lim_{x \rightarrow \infty} \frac{\pi(x)\cdot ln(x)}{x} = 1 \end{align}

Note that the PNT does not state anything about the difference between the two, just that the ratio converges to 1. The Riemann Hypothesis concerns this distance. But first, we defined a better approximation for $\pi(x)$:

\begin{align} \pi(x) \approx Li(x) = \int_2^x \frac{dt}{ln(t)} \end{align}

This is actually the best estimate available. Another version of the PNT states:

\begin{align} \lim_{x \rightarrow \infty} \frac{\pi(x)}{\int_2^x \frac{dt}{ln(t)}} = 1 \end{align}

While $\frac{x}{ln(x)}$ is within 10% of $\pi(x)$, $Li(x)$ is very accurate. For example, if $x=3\times 10^6$, $Li(x)$ is only off by 155.

Here is another example: Consider $x=10^{23}$. In this case, $\pi(x) = 1,925,320,391,606,803,968,923$, while $\pi(x) - x / ln(x)= 37,083,513,766,578,631,309$ and $\pi(x) - Li(x) = 7,250,186,216$.4

Riemann is proposing that:

\begin{align} | \pi(x) - Li(x) | \leq \sqrt(x) \cdot ln(x) \end{align}

This is then the upper-bound on the difference between the estimate and the actual number of primes less than a given number. But notice the difference between the Riemann Hypothesis and the PNT. While, the PNT proves that the limit of the ratio is equal to one, the Hypothesis is suggesting that the absolute value of the difference grows without bound. This is saying that primes act in somewhat of a random fashion; while we have testing to determine if a number of prime, the difference between prime numbers is somewhat stochastic.

The square root is the key part of the hypothesis. As an example, recall a possibly familiar rule of thumb from statistics. If you flip a coin 100 times, the square root law says that you should get within $\sqrt(n)$. Since $\sqrt(100)= 10$, this is suggesting that you should be within plus or minus 5… as in, 45-55 is a good guess. Well, using the binomial distribution, the standard deviation of this experiment is given exactly by $\sqrt{(n \cdot p \cdot (1-p))}$, where $p$ is just .5 since that is the probability of getting heads. That means the standard deviation is 5, so there is a 68% of being within 45-55. In this case, the rule of thumb is exact. Go Riemann!

Wonder where the square root comes from? Well, $\sqrt{x} = x^{1/2}$. That 1/2 is precisely the $\Re(s) = 1/2$ line. Woah!


What if the Riemann Hypothesis is wrong? While we doubt the world will end, many things will need to change. Mathematical literature is full of conditional proofs that rest upon the Riemann Hypothesis being true. If proved false, a lot of established mathematics will be wrong… dun dun dun.

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