Lecture 29: Analytic Number Theory and the Riemann Zeta Function


Today we took a sneak peak at the kind of number theory which comes after a typical "elementary" course like ours. Much of our time was devoted to studying properties of the famous Riemann Zeta function. Along the way we saw the so-called Euler Product formula for $\zeta(s)$, and we used this to give a new proof of the infinitude of primes. We then discussed the kinds of properties of $\zeta(s)$ which mathematicians today are currently interested in.

Introducing the Riemann Zeta Function

So far in this class we've talked mainly about "elementary" number theory, meaning the kinds of number theoretic statements one can make and prove that are concerned only with basic definitions of divisibility. There is, however, an entire other branch of number theory which uses ideas from analysis to prove properties about numbers. Much of the work in this area started with the investigation of the function

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

Notice that good ol' fashioned p-series tests will show that $\zeta(s)$ is defined when $s>1$, and that $\zeta(1)$ is undefined (since it corresponds to the harmonic series).

Today this function is called the Riemann Zeta function, but it has been an object of interest for a long time. One of the first people who made significant progress in understanding this function was our good friend Euler. He was able to use this function to prove many classical results in number theory. For instance, the following theorem is a kind of analogue of the Fundamental Theorem of Arithmetic. We'll wind up using it to give a new proof of the infinitude of primes.

The Riemann Zeta function can be expressed as the following infinite product:

$\displaystyle\zeta(s) = \prod_{p \mbox{\tiny{ prime}}}\left(1-p^{-s}\right)^{-1}.$

Proof: We define the function $P(n)$ to return the largest prime factor of the input number n. Using the series identity

\begin{align} \frac{1}{1-x} = 1+x+x^2+x^3+x^4+\cdots \end{align}

then gives

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

Now notice that any number n which has $P(n)\leq N$ can be expressed as

\begin{align} n = p_1^{e_1}\cdots p_k^{e_k} \end{align}

where each prime $p_i\leq N$. Hence we have

\begin{align} n^s = p_1^{se_1}\cdots p_k^{se_k}, \end{align}

and so multiplying out the infinite sums above shows us

\begin{align} \prod_{p<N}\left(1-p^{-s}\right)^{-1} = \prod_{p<N} (1+p^{-s}+p^{-2s}+p^{-3s}+\cdots ) = \sum_{P(n)\leq N}\frac{1}{n^s}. \end{align}

Now the Riemann Zeta function can be split into two pieces:

\begin{align} \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \underbrace{\sum_{P(n)\leq N}\frac{1}{n^s}}_{T_1}+ \underbrace{\sum_{P(n)>N}\frac{1}{n^s}}_{T_2}. \end{align}

We've already seen that the $T_1$ term is given by a certain finite Euler product, so we consider the $T_2$. Note that this term is bounded from above:

\begin{align} T_2 = \sum_{P(n)>N}\frac{1}{n^s} \leq \sum_{n>N}\frac{1}{n^s}, \end{align}

where this upper bound is the tail of the (convergent) series $\zeta(s)$. Since the series is convergent, we know this tail must approach 0, and hence $T_2 \rightarrow 0$ as $N\rightarrow 0$. Recalling Equation (6) we get

\begin{split} \zeta(s) &= \lim_{N \rightarrow \infty} \left(\sum_{P(n)\leq N} \frac{1}{n^s} + \sum_{P(n)>N} \frac{1}{n^s}\right) \\ &=\lim_{N \rightarrow \infty}\left(\sum_{P(n)\leq N} \frac{1}{n^s}\right) \\ &= \lim_{N \rightarrow \infty} \left(\prod_{p \leq N} \left(1-p^{-s}\right)^{-1}\right) \\ &= \prod_{p} \left(1-p^{-s}\right)^{-1}. \end{split}


Now that we have the Euler Product formula, we can use it to prove other results. One of Euler's most famous theorems is the proof that the sum of the reciprocals of the prime numbers diverges. In particular, this shows that there are infinitely many prime numbers.

Theorem: The sum $\displaystyle \sum_{p} \frac{1}{p}$ diverges.

Proof: Recall the series for $\log(1-x)$:

\begin{align} \log(1-x) = - \sum_{r=1}^\infty \frac{x^r}{r}. \end{align}

This lets us compute the logarithm of the zeta function:

\begin{align} \log(\zeta(s)) = \log\left(\prod_{p}(1-p^{-s})^{-1}\right) = -\sum_p \log(1-p^{-s}) = \sum_p \sum_{r=1}^\infty \frac{1}{rp^{rs}}. \end{align}

Now we'll split this into two terms:

\begin{align} \log(\zeta(s)) =\sum_p \sum_{r=1}^\infty \frac{1}{rp^{rs}} = \underbrace{\sum_p \frac{1}{p}}_{T_1} + \underbrace{\sum_p \sum_{r=2}^\infty \frac{1}{rp^{rs}}}_{T_2}. \end{align}

Of course we want to show that $T_1$ is infinite.

To do this, we first notice that we can bound the second term. In fact, notice that since $s>1$ and $r \geq 1$, we have $\frac{1}{rp^{rs}} \leq \frac{1}{p^r}$. Hence we have

\begin{align} T_2 = \sum_p \sum_{r=2}^\infty \frac{1}{rp^{rs}} \leq \sum_p \sum_{r=2}^{\infty} \frac{1}{p^r}. \end{align}

The latter term is a geometric series which we can sum using the standard geometric series formula

\begin{align} \sum_{k=i}^\infty x^k = \frac{x^i}{1-x}. \end{align}

Substituting this formula back into Equation (13) (with $x = \frac{1}{p}$) then gives

\begin{align} T_2 \leq \sum_p \sum_{r=2}^{\infty} \frac{1}{p^r} = \sum_p \frac{(\frac{1}{p})^2}{1-\frac{1}{p}} = \sum_p \frac{1}{p(p-1)}. \end{align}

Notice that this term is finite, since you can compare it to the convergent sequence $\sum \frac{1}{n^2}$. Hence we can conclude that

\begin{align} \log(\zeta(s)) = \sum_{p}\frac{1}{p} + \mbox{some finite quantity}. \end{align}

To finish off the proof, note that $\lim_{s \to 1^+} \zeta(s) = \infty$, since the harmonic series diverges. Hence we know that $\lim_{s \to 1^+} \log(\zeta(s))$ diverges as well. Considering our formula for $\log(\zeta(s))$ above, the only way that it can diverge is if

\begin{align} \sum_p \frac{1}{p} = \lim_{s \to 1^+} \sum_p \frac{1}{p} = \infty. \end{align}

This is the result we wanted to prove. $\square$

As a final note before moving onto the next topic, we point out that the Riemann Zeta function is connected in some intimate ways to prime numbers. For instance, the function $\pi(x)$ — the prime counting function — shows up in the evaluation of $\log(\zeta(s))$; one can prove that

\begin{align} \log(\zeta(s)) = s\int_2^\infty \frac{\pi(x)}{x(x^s-1)}~dx. \end{align}

The proof isn't all that hard, but we didn't have time to talk about it in class.

Twisted Riemann Zeta Functions

In class I failed to mention that the proof we gave for the Euler Product formula for the Riemann Zeta function could be extended in the following way:

Theorem: If f is a multiplicative function, then

$\displaystyle \sum_{n=1}^\infty \frac{f(n)}{n^s} = \prod_p (1+\frac{f(p)}{p^s} + \frac{f(p^2)}{p^{2s}} + \frac{f(p^3)}{p^{3s}} + \cdots )$.


Let's use this theorem to compute $\sum_{n=1}^\infty \frac{\sigma(n)}{n^s}$. According to our formula above, we get

\begin{split} \sum_{n=1}^\infty \frac{\sigma(n)}{n^s} &= \prod_p \left(1+\frac{\sigma(p)}{p^s} + \frac{\sigma(p^2)}{p^{2s}} + \frac{\sigma(p^3)}{p^{3s}} + \cdots \\ &= \prod_p (1+\frac{p+1}{p^s} + \frac{p^2+p+1}{p^{2s}} + \frac{p^3+p^2+p+1}{p^{3s}} + \cdots ) \\ &= \prod_p (1+\frac{1}{p^{s-1}} + \frac{1}{p^s} + \frac{1}{p^{2s-2}} + \frac{1}{p^{2s-1}} + \frac{1}{p^{2s}} + \frac{1}{p^{3s-3}} + \frac{1}{p^{3s-2}} + \frac{1}{p^{3s-1}} + \frac{1}{p^{3s}} + \cdots). \end{split}

If you stare at the above expansion long enough, you'll recognize that it can also be written as

\begin{align} \prod_p (1+\frac{1}{p^{s-1}}+\frac{1}{p^{2(s-1)}} + \frac{1}{p^{3(s-1)}} + \cdots ) \prod_{p} (1+\frac{1}{p^s} + \frac{1}{p^{2s}} + \frac{1}{p^{3s}} + \cdots ), \end{align}

which is the same as $\zeta(s-1)\zeta(s)$. $\square$

Extending the Riemann Zeta Function

We've been studying the Riemann Zeta function as a function of a real variable, meaning that we've been thinking of it as a function which we evaluate at some real number s. One of the big ideas in this branch of number theory is to recognize that we can also place complex numbers into this function. In doing so, it turns out that $\zeta(s)$ is a "really nice" complex function whenever the real part of the input s is greater than 1; the actual lingo is "analytic" on the domain $\mathfrak{R}(s)>1$, which basically means that it is a super smooth complex function. (The notation $\mathfrak{R}(s)>1$ means "the real part of s is greater than 1".) This "analytic-ness" of the $\zeta$ function has lots of important consequences. One of those consequences is something called the "analytic continuation" of the $\zeta$ function. Basically, the result is that there is one (and only one) way to extend the Riemann Zeta function to a domain larger than $\mathfrak{R}(s)>1$. In fact, one can define $\zeta(s)$ for every complex number except $s=1$. One way to do this is to use the so-called functional equation of $\zeta$:

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

where the $\Gamma$ function is

\begin{align} \Gamma(s) = \int_0^\infty e^{-t}t^{s-1}~dt \quad \quad (\mbox{ when }s>0). \end{align}

In particular, we can use (21) to evaluate $\zeta(s)$ when the real part of s is smaller than 1: in this case, the real part of 1-s is bigger than one, and so we can evaluate the left-hand side (when the definition of $\zeta(s)$ is otherwise unknown) by evaluating the right hand side (where evaluating $\zeta(1-s)$ is defined).

The big question about the Riemann Zeta function is the following

Riemann Hypothesis: The "nontrivial" values of s for which $\zeta(s) = 0$ are all found on the line $\mathfrak{R}(s) = \frac{1}{2}$.

Believe it or not, this is the "holy grail" of number theory.

Some Analytic Number Theory Results

To finish the class, we pointed out a few results which are born from an analytic approach to number theory. These results were "average value" results, stating the "average value" of a given arithmetic function. The average value of an arithmetic function f is computed as

\begin{align} \lim_{N \to \infty}\frac{\sum_{n \leq N}f(n)}{N}. \end{align}

Can you see why this value earns the name "average value"?

Here are some cool results:

\begin{split} \lim_{N \to \infty} \frac{\sum_{n \leq N}\nu(n)}{N} &\approx \log(N)\\ \lim_{N \to \infty} \frac{\sum_{n \leq N}\phi(n)}{N} &\approx \frac{3N}{\pi^2} \end{split}

These results says that if you pick up a random number n, you should expect that

\begin{align} \nu(n) \approx \log(n) \quad \mbox{ and } \quad \phi(n) \approx \frac{3n}{\pi^2}. \end{align}
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