# Monthly Archives: March 2013

## An irrationality criterion

Recall Corollary 4 from this post:

Corollary 4. If $S$ is a subgroup of $(\mathbb R,+)$, then the following are equivalent:

(i) $S$ is well-ordered;

(ii) $S$ is not dense;

(iii) $S$ is cyclic.

Let $\alpha$ be a non-zero real number and take $S=\langle 1,\alpha\rangle$ to be the subgroup of $(\mathbb R,+)$ generated by $1$ and $\alpha$. This is a cyclic group if and only if $\exists \beta, \langle 1,\alpha\rangle=\langle \beta\rangle$
$\Leftrightarrow 1=m\beta', \alpha=n\beta'$ for some $m,n\in\mathbb Z$ and $\beta'$
$\Leftrightarrow \alpha/1=n/m$, i.e. $\alpha$ is rational.

Now by the above corollary, $S$ is dense iff it is not well-ordered, i.e. iff

$(*)\qquad\forall\varepsilon>0,\exists m,n\in\mathbb Z, 0.

So we have the following criterion for irrationality:

Criterion 1. $\alpha$ is irrational if and only if $(*)$ holds.

Note that $S$ is dense in $\mathbb R$ iff $S/\mathbb Z=\{\{n\alpha\}:=n\alpha-\lfloor n\alpha\rfloor : n\in\mathbb Z\}$ is dense in $(0,1)$. So we deduce:

Criterion 2. $\alpha$ is irrational $\Leftrightarrow\forall\varepsilon>0,\exists n\in\mathbb Z, 0<\{n\alpha\}<\varepsilon$.

We demonstrate how this may be useful by proving that certain types of numbers are irrational.

Proposition 1. Let $a_1,a_2,\dots$ be a sequence of non-zero integers such that

$(\dagger)\qquad\displaystyle S=\frac{1}{a_1}+\frac{1}{a_1a_2}+\frac{1}{a_1a_2a_3}+\dots$ exists, and

$(\ddagger)\qquad\displaystyle\frac{1}{a_{n+1}}+\frac{1}{a_{n+1}a_{n+2}}+\frac{1}{a_{n+1}a_{n+2}a_{n+3}}+\dots\to 0$ as $n\to\infty$.

Then $S$ is irrational.

Proof. We have

$\displaystyle \left\{a_1\cdots a_nS\right\}=\left\{\frac{1}{a_{n+1}}+\frac{1}{a_{n+1}a_{n+2}}+\frac{1}{a_{n+1}a_{n+2}a_{n+3}}+\dots\right\}$.

Multiplying by $-1$ if necessary, we can take the expression in brackets on the right to be positive and it tends to $0$ as $n\to \infty$. Moreover, if

$\displaystyle\frac{1}{a_{n+1}}+\frac{1}{a_{n+1}a_{n+2}}+\frac{1}{a_{n+1}a_{n+2}a_{n+3}}+\dots=0$

then

$\displaystyle\frac{1}{a_{n+2}}+\frac{1}{a_{n+2}a_{n+3}}+\frac{1}{a_{n+2}a_{n+3}a_{n+4}}\dots=-1$

which cannot happen infinitely often as the left hand side tends to zero. So the conclusion follows by Criterion 2. $\square$

Proposition 2. If $a_1,a_2,\dots$ is a sequence of non-zero integers such that $|a_1|\le |a_2|\le\dots$ and $\displaystyle\lim_{n\to\infty}|a_n|=\infty$, then

$\displaystyle S=\frac{1}{a_1}+\frac{1}{a_1a_2}+\frac{1}{a_1a_2a_3}+\dots$

exists and is irrational.

Proof. It suffices to show that $(\dagger)$ and $(\ddagger)$ above hold.

Convergence follows easily from the ratio test, so $(\dagger)$ holds. Now

$\displaystyle\left|\frac{1}{a_{n+1}}+\frac{1}{a_{n+1}a_{n+2}}+\frac{1}{a_{n+1}a_{n+2}a_{n+3}}+\dots\right|$

$\displaystyle\le\frac{1}{|a_{n+1}|}+\frac{1}{|a_{n+1}a_{n+2}|}+\frac{1}{|a_{n+1}a_{n+2}a_{n+3}|}+\dots$

$\displaystyle\le\frac{1}{|a_{n+1}|}+\frac{1}{|a_{n+1}|^2}+\frac{1}{|a_{n+1}|^3}+\dots$

$\displaystyle =\frac{1}{|a_{n+1}|-1}\to 0$ as $n\to \infty$,

i.e. $(\ddagger)$ holds. $\square$

Some special cases of Proposition 2 are particularly interesting:

Corollary 1. $\displaystyle \sum_{n=0}^\infty\frac{1}{(n!)^k}$ is irrational for all positive integers $k$.

Proof. Take $a_n=n^k$. $\square$.

Corollary 2. $e$ is irrational.

Proof. Take $k=1$ in Corollary 1. $\square$

Corollary 3. $\sin 1$ and $\cos 1$ are irrational.

Proof. Take $a_1=1$ and for $n>1$, $a_n=-(2n-2)(2n-1)$ for sine, $a_n=-(2n-3)(2n-2)$ for cosine. $\square$

Corollary 4. $I_0(2)$ and $I_1(2)$ are irrational, where $I_\alpha(x)$ is the modified Bessel function of the first kind.

Proof. Taking $k=2$ in Corollary 1 shows that $I_0(2)$ is irrational. Taking $a_n=n(n+1)$ shows that $I_1(2)$ is irrational. $\square$

Corollary 5. $e^{\sqrt{2}}$ is irrational.

Proof. If it is rational, then so is $e^{-\sqrt{2}}$, and so is

$\displaystyle \cosh(\sqrt 2)=\frac{e^{\sqrt{2}}+e^{-\sqrt 2}}{2}=\sum_{n=0}^\infty\frac{2^n}{(2n)!}$.

Taking $a_n=n(2n-1)$ in the above shows that this is false. $\square$

Corollary 6. Let $k>1$ be an integer and $F_0,F_1,F_2,\dots$ the Fibonacci sequence. Then

(i) $\displaystyle \sum_{n=0}^\infty\frac{1}{k^{F_n}}$ is irrational.

(ii) $\displaystyle \sum_{n=0}^\infty\frac{1}{F_{k^n}}$ is irrational.

Proof. (i) Take $a_n=k^{F_n}$ and use $F_0+F_1+\dots+F_n=F_{n+2}-1$.

(ii) Take $a_n=F_{k^n}/F_{k^{n-1}}$. $\square$

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Filed under Analysis, Number theory

## Trivial zeros of the Riemann zeta function

The Riemann hypothesis is perhaps the most famous unsolved problem in mathematics. It says that all non-trivial zeros of the Riemann zeta function have real part $1/2$. What is the Riemann zeta function? A common misconception is that it is defined as

$\displaystyle\zeta (z)=\sum_{k=1}^\infty k^{-z}=\frac{1}{1^z}+\frac{1}{2^z}+\frac{1}{3^z}+\dots,\quad z\in\mathbb C$.

This is false! Because the series on the right only converges for $\Re(z)>1$. Also, it is a well-established fact that the trivial zeros of the Riemann zeta function occur at the negative even numbers $-2,-4,-6\dots$. If we substitute $z=-2$ in the above equation we get $1^2+2^2+3^2+\dots$ which is far from being zero. So even though these are called ‘trivial’ zeros, it seems highly non-trivial why these are actually zeros. What’s going on?

The problem lies in the above definition. This equation does NOT define the Riemann zeta function, which is a meromorphic function (see below) on the whole complex plane. The above definition only works for $\Re(z)>1$. Then what about the other values of $z$? Before proceeding further, let us familiarise ourselves with some terminology from the theory of complex analysis.

Complex analysis is one of the most beautiful branches of mathematics. It deals with functions $f:\mathbb C\to \mathbb C$ in the complex plane that are differentiable in some open subset $D\subseteq \mathbb C$. (Open sets are like open intervals; open intervals are line segments excluding the endpoints, open sets in the complex plane can be thought of as discs excluding their boundaries.) Some basic definitions are:

• $f:D\to\mathbb C$ is said to be analytic (or holomorphic) at $z_0$ if it is differentiable in a neighbourhood of $z_0$, i.e. at all points $z$ such that $|z-z_0|<\varepsilon$ for some $\varepsilon >0$.
• $f$ above is called entire if it is analytic at every $z\in\mathbb C$.
• Every analytic function has a Laurent series, which is like a Taylor series, but negative powers are allowed.
• If $\displaystyle f(z)=\sum_{k=-n}^\infty a_k(z-z_0)^k$ is a Laurent series in some neighbourhood of $z_0$ then $(z-z_0)^nf(z)$ is analytic, and $z_0$ is called a pole of $f$ of order $n$. The coefficient $a_{-1}$ of $(z-z_0)^{-1}$ is the residue of $f$ at $z_0$, which we shall denote as $\text{res}(f,z_0)$.
• If $f:D\to\mathbb C$ is analytic except for a set of isolated poles, it is called meromorphic.

Unlike real analysis, amazing things happen when analysis is done in the complex plane. Some important results in complex analysis are:

• Cauchy’s integral theorem, that says if $f$ is analytic on some open set $D\subseteq\mathbb C$ and $\gamma$ is any closed curve in $D$, then

$\displaystyle\int_\gamma f(z)dz=0$.

• Cauchy’s integral formula: If $\gamma$ above is a circle oriented counter-clockwise and $w$ is any interior point, then

$\displaystyle f(w)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-w}dz$.

• Cauchy’s residue theorem: If $A$ is the set of poles of $f$ inside the closed curve $\gamma$ and $f$ is analytic inside $\gamma$ except for these poles, then

$\displaystyle \int_\gamma f(z)dz=2\pi i\sum_{a\in A}\text{res}(f,a)$,

• Liouville’s theorem: Any entire function that is bounded must be constant.
• Identity theorem: If $D'\subseteq D\subseteq\mathbb C$ are (connected) open sets with $f,g$ analytic on $D$, such that $f=g$ on $D'$, then $f=g$ on $D$.

The identity theorem gives a useful method of extending a function analytically. Suppose $D'\subseteq D\subseteq\mathbb C$ are open sets and $f:D'\to\mathbb C$ and $g:D\to\mathbb C$ are analytic. If $f=g$ on $D'$, and we were to define $f$ on $D$ preserving analyticity, then the identity theorem says $f=g$ on $D$. $g$ in this case is called the (unique) analytic continuation of $f$ to $D$. Now let’s go back to the Riemann zeta function. It’s proper definition is the following:

$\zeta(z)=\begin{cases}&\displaystyle\sum_{k=1}^\infty k^{-z},\;\Re(z)>1,\\ &\text{analytic continuation elsewhere.}\end{cases}$

Recall the gamma function

$\displaystyle\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}dt,\; \Re(z)>0$.

It can be shown using integration by parts that $\Gamma(z+1)=z\Gamma(z)$. The function $\Gamma(z+1)$ is analytic for $\Re(z)>-1$, so this equation provides the analytic continuation of $\Gamma(z)$ to $\Re(z)>-1$. By repeating this procedure we obtain the analytic continuation of $\Gamma$ to the whole complex plane (except for the non-positive integers, which are in fact the poles of $\Gamma$).

It is an exercise to show that

$\displaystyle \zeta(z)=\frac{1}{\Gamma(z)}\int_0^\infty\frac{t^{z-1}}{e^t-1} dt,\; \Re(z)>1$.

It is another exercise to show that

$\displaystyle \frac{1}{\Gamma(z)}\int_0^\infty\frac{t^{z-1}}{e^t-1} dt=\frac{\Gamma(1-z)}{2\pi i}\int_{\gamma}\frac{t^{z-1}}{e^{-t}-1}dt$

where $\gamma$ is the Hankel contour:

Now the right side of the above equation is analytic for $\Re(z)<1$. Hence this equation provides the analytic continuation of $\zeta(z)$ to $\mathbb C\backslash\{1\}$ (in fact, there is a pole at $z=1$.)

The Bernoulli numbers $B_n$ are defined by the equation

$\displaystyle \frac{1}{e^t-1}=\sum_{m=0}^\infty B_m\frac{t^{m-1}}{m!}$,

and $B_0=1, B_1=1/2, B_{2m+1}=0$ for $m=1,2,\dots$. Substituting this into the equation above, setting $z=-n$ and using $\Gamma(n+1)=n!$ we obtain

$\displaystyle \zeta(-n)=\frac{n!}{2\pi i}\int_{\gamma}\frac{t^{-n-1}}{e^{-t}-1}dt =\frac{n!}{2\pi i}\sum_{m=0}^\infty (-1)^{m-1}\frac{B_m}{m!}\int_\gamma t^{m-n-2}dt$

By Cauchy’s residue theorem,

$\displaystyle\int_\gamma t^{m-n-2} dt=\begin{cases} &2\pi i\;\text{ if }m=n+1\\ & 0 \; \text{ otherwise}.\end{cases}$

Thus $\displaystyle\zeta(-n)=(-1)^n\frac{B_{n+1}}{n+1}$, which is zero if $n>1$ is even.

Filed under Complex analysis

## What Dirichlet had to do with the pigeonhole principle

Suppose you have six pigeons. You want to put them into five holes. Then at least one of the holes must contain more than one pigeon.

This simple idea is a very important mathematical principle known as the pigeonhole principle, or sometimes Dirichlet principle, Dirichlet’s box principle etc. If you have come across this you have probably wondered (like me) why it is named after Dirichlet, or how exactly did he use it. Dirichlet was one of the most prominent mathematicians of all time, having many important things named after him. E.g.

1. Dirichlet’s theorem on arithmetic progressions that says any arithmetic progression $a, a+d, a+2d,\dots$, where $a$ and $d$ are integers sharing no common factor greater than $1$, must contain infinitely many prime numbers.
2. Dirichlet’s unit theorem, a result about the units in the ring of integers of an algebraic number field.
3. Dirichlet L-function, a generalisation of the Riemann Zeta function.
4. The Dirichlet conditions for a Fourier series.
5. And many, many more.

When I first came to know that the pigeonhole principle was credited to Dirichlet, I was curious as to what Dirichlet had to do with it. A number theory lecture from last term happened to shed some light on it. First we need to know some basic facts about rational approximations.

We know that any real number has arbitrarily good rational approximations: just take the decimal expansion and truncate it anywhere. The more digits, the better the approximation. However, there is a different notion of a ‘good rational approximation’ in number theory.

Let $\alpha$ be a real number. A way of measuring how good a rational number $p/q$ approximates $\alpha$ is to check how small $|q\alpha-p|$ is. We say that $p/q$ is a best approximation to $\alpha$ if for any integers $p'$ and $0 we have $|q'\alpha-p'|>|q\alpha-p|$.

Dirichlet proved the following theorem using his principle:

Theorem. (Dirichlet) Let $\alpha$ be a real number. Then for any integer $N>1$ there exist integers $p,q$ with $0 such that

$\displaystyle|q\alpha-p|<\frac 1N$.

Proof. By $\{x\}$ and $\lfloor x\rfloor$ we respectively denote the fractional part and integer part of the real number $x$. E.g.
$\{20/3\}=2/3$,
$\lfloor 20/3\rfloor=6$,
$\{\pi\}=\pi-3=0.1415926535\dots$,
$\lfloor \pi\rfloor=3$ etc.

Note that $x=\lfloor x\rfloor+\{x\}$.

Consider the numbers $0,1,\{\alpha\}, \{2\alpha\},\dots,\{(N-1)\alpha\}$. These are $N+1$ numbers all lying between $0$ and $1$ (inclusive). Divide the interval $[0,1]$ into the $N$ intervals $[0,1/N], [1/N, 2/N],\dots,[(N-1)/N, 1]$. Since we have $N$ intervals and $N+1$ numbers each lying in one of these, the pigeonhole principle says that there must be two numbers lying in the same interval. Then the difference of these two numbers must be less than $1/N$, the length of the interval.

If one of the two numbers is $1$, then $1/N>|\{m\alpha\}-1|=|m\alpha-(\lfloor m\alpha\rfloor+1)|$ for some $m$, so take $p=\lfloor m\alpha\rfloor+1$ and $q=m$. Otherwise, let the two numbers be $\{m\alpha\},\{n\alpha\}$ with $N-1\ge m>n\ge 0$. Then

$\displaystyle \quad |\{m\alpha\}-\{n\alpha\}|<\frac 1N$
$\displaystyle \Rightarrow |(m\alpha-\lfloor m\alpha\rfloor)-(n\alpha-\lfloor n\alpha\rfloor)|<\frac 1N$
$\displaystyle \Rightarrow |q\alpha-p|<\frac 1N$

where $p=\lfloor m\alpha\rfloor-\lfloor n\alpha\rfloor$ and $q=m-n$, with $0, as required. $\square$