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The Transcendence of π\pi

The proof that π\pi is transcendental is not well-known despite the fact that it isn't too difficult for a university mathematics student to follow. The purpose of this paper is to make the proof more widely available. A bonus is that the proof also shows that ee is transcendental as well.

The material in these notes are not mine; it is taken from a supplement issued by Ian Stewart as an adjunct to a Rings and Fields course in 1970 at the University of Warwick.


Definition

A complex number is algebraic over Q\mathbb{Q} if it is a root of a polynomial equation with rational coefficients.

Thus aa is algebraic if there are rational numbers α0,α1,,αn\alpha_{0},\alpha_1,\ldots,\alpha_n not all 0, such that α0an+α1an1++αn1a+αn=0.\alpha_0 a^n+\alpha_1 a^{n-1}+\ldots+\alpha_{n-1}a+\alpha_n=0.

Definition

A complex number is transcendental if it is not algebraic, so it is not the root of any polynomial equation with rational coefficients.

In proving that it is impossible to 'square the circle' by a ruler-and-compass construction we have to appeal to the theorem:

   The real number π\pi is transcendental over Q\mathbb{Q}

The purpose of this supplement is to indicate, for those who may be interested, how this theorem may be proved. It is possible to prove that there exist transcendental real numbers by using infinite cardinals, as was first done by Cantor in 1874. Earlier Liouville (1844) had actually constructed transcendentals, for example n=110n!\sum \limits_{n=1}^{\infty}10^{-n!} is transcendental.1 However, no naturally occurring real number (such as ee or π\pi) was proved transcendental until Hermite (1873) disposed of ee. π\pi held out until 1882 when Lindemann, using methods related to those of Hermite, disposed of that. In 1900 David Hilbert proposed the problem:

   If a,ba,b are real numbers algebraic over Q\mathbb{Q}, if a0a\neq0 or 11 and b is irrational, prove aba^b is transcendental.

This was solved independently in 1934 by the Russian, Gelfond, and a German, Schneider.

Before proving transcendence of π\pi we shall prove a number of similar theorems, using simpler versions of the final method, as an aid to comprehension. The tools needed are first-year analysis.2

Theorem

π\pi is irrational

Proof. Let In(x)=1+1(1x2)ncos(αx)dxI_{n}(x)=\int_{-1}^{+1}\left(1-x^{2}\right)^{n}\cos(\alpha x)\:\text{d}x
Integrating by parts we have
α2In=2n(2n1)In14n(n1)In2(n2)\begin{aligned} \alpha^{2}I_n=2n(2n-1)I_{n-1}-4n(n-1)I_{n-2}\qquad (n\geq 2) \end{aligned}
which implies that
α2n+1In=n!(Pnsin(α)+Qncos(α))()\begin{aligned} \alpha^{2n+1}I_{n}=n!\left( P_n\sin(\alpha)+Q_n\cos(\alpha)\right) \qquad(*) \end{aligned}
where Pn,QnP_n,Q_n are polynomials of degree <2n+1<2n+1 in α\alpha with integer coefficients.

Remark. degPn=n, degQn=n1\deg P_n=n,\ \deg Q_n=n-1

Put α=π2,\alpha=\dfrac{\pi}{2}, and assume π\pi is rational, so that π=ba, a,bZ\pi=\dfrac{b}{a},\ a,b\in\mathbb{Z}
From (*) we deduce that Jn=b2n+1Inn!J_{n}=\dfrac{b^{2n+1}I_{n}}{n!} is an integer. On the other hand Jn0J_n \to 0 as nn\to\infty since bb is fixed and InI_n is bounded by
1+1cos(πx2)dx \int_{-1}^{+1}\cos\left(\frac{\pi x}{2}\right)\:\text{d}x
JnJ_n is an integer, 0.\to 0. Thus Jn=0J_n=0 for some nn. But this integrand is continuous, and is >0>0 in most of the range (1,+1)(-1,+1), so Jn0.J_n\neq 0. Contradiction. \blacksquare



Theorem

π2\pi^{2} is irrational (so π\pi does not lie in any quadratic extension of Q\mathbb{Q})

Proof. Assume π2=ab, a,bZ.\pi^{2}=\dfrac{a}{b},\ a,b\in\mathbb{Z}.
Define
f(x)=xn(1x)nn!,G(x)=bn[π2nf(x)π2n2f(x)++(1)nπ0f(2n)(x)]\begin{aligned} f(x) &=\dfrac{x^n\left(1-x\right)^n}{n!}, \\ G(x) &= b^n\left[\pi^{2n}f(x)-\pi^{2n-2}f^{\prime\prime}(x)+\ldots+(-1)^n\pi^0f^{(2n)}(x)\right] \end{aligned}
(superscripts indicating differentiations). We see that the value of any derivative of ff at 00 or 11 is either 00 or an integer. Also G(0)G(0) and G(1)G(1) are integers. Now
ddx[G(x)sin(πx)πG(x)cos(πx)]=[G(x)+π2G(x)]sin(πx)=bnπ2n+2f(x)sin(πx)since f(2n+2)(x)=0=π2ansin(πx)f(x)\begin{aligned} \frac{\text{d}}{\text{d}x}\left[G^{\prime}(x)\sin(\pi x)-\pi G(x)\cos(\pi x)\right] &=\left[G^{\prime\prime}(x)+\pi^{2}G(x)\right]\sin(\pi x)\\ &=b^n\pi^{2n+2}f(x)\sin(\pi x) \qquad \text{since }f^{(2n+2)}(x)=0\\ &=\pi^2 a^n\sin(\pi x)f(x) \end{aligned}
so that
π01ansin(πx)f(x)dx=[G(x)sin(πx)πG(x)cos(πx)]01=0+G(0)+G(1)=integer.\begin{aligned} \pi\int_0^1 a^n\sin(\pi x)f(x)\:\text{d}x & =\left[\dfrac{G^{\prime}(x)\sin(\pi x)}{\pi}-G(x)\cos(\pi x)\right] _0^1\\ &=0+G(0)+G(1) \\ &=\text{integer}. \end{aligned}
But again the integral is non-zero and 0\to 0 as nn\to\infty. Thus again we have a contradiction. \blacksquare

Getting more involved, now:

Theorem (Hermite)

ee is transcendental over Q\mathbb{Q}

Proof. Suppose amem++a1e+a0=0(aiZ).a_m e^m+\ldots+a_1e+a_0=0\qquad(a_i\in\mathbb{Z}). WLOG a00a_0\neq0

   Define f(x)=xp1(x1)p(x2)p(xm)p(p1)!\qquad f(x)=\dfrac{x^{p-1}(x-1)^p(x-2)^p \ldots(x-m)^p}{(p-1)!}
where for the moment pp is arbitrary and prime.

   Define F(x)=f(x)+f(x)++f(mp+p1)(x).\qquad F(x) = f(x)+f^{\prime}(x)+\ldots+f^{(mp+p-1)}(x).

   Now if 0<x<m0< x< m,
f(x)mp1mmp(p1)!=mmp+p1(p1)!\begin{aligned} |f(x)| & \leq\dfrac{m^{p-1}m^{mp}}{(p-1)!}\\ & =\dfrac{m^{mp+p-1}}{(p-1)!} \end{aligned}
Also\qquadddx(exF(x))=ex[F(x)F(x)]=exf(x)\dfrac{\text{d}}{\text{d}x}\left(e^{-x}F(x)\right) =e^{-x}\left[F^{\prime}(x)-F(x)\right] =-e^{-x}f(x)

   so that
aj0jexf(x)dx=aj[exF(x)]0j=ajF(0)ajejF(j).\begin{aligned} a_{j}\int_0^j e^{-x}f(x)\:\text{d}x & =a_j\left[-e^{-x}F(x)\right]_0^j\\ & =a_jF(0)-a_j e^{-j}F(j). \end{aligned}
Multiplying by eje^j and summing over j=0,1,mj=0,1,\ldots m we get
j=0majej0jexf(x)dx=F(0).0j=0majF(j)=j=0mi=0mp+p1ajf(i)(j).\begin{aligned} \sum_{j=0}^m a_je^j\int_0^j e^{-x}f(x)\:\text{d}x & =F(0).0-\sum_{j=0}^m a_j F(j) \\ & =-\:\sum_{j=0}^m \sum_{i=0}^{mp+p-1}a_j f^{(i)}(j). \end{aligned}
We claim that each f(i)(j)f^{(i)}(j) is an integer, divisible by pp except when j=0j=0 and i=p1.i=p-1. For only non-zero terms arise from terms where the factor (xj)p(x-j)^p has been differentiated pp times, and then p!p! cancels (p1)!(p-1)! and leaves pp, except in the exceptional case. We show that in the exceptional case the term is NOT divisible by pp. Clearly f(p1)(0)=(1)p(m)pf^{(p-1)}(0) = (-1)^p \ldots (-m)^p. We CHOOSE pp larger than mm, when this product cannot have a prime factor pp. Hence the right-hand side of the above equation is an integer 0\neq0. But as pp\to\infty the left-hand side tends to 00, using the above estimate for f(x)|f(x)|. This is a contradiction. \blacksquare



Theorem (Lindemann)

π\pi is transcendental over Q\mathbb{Q}

Proof. If π\pi satisfies an algebraic equation with coefficents in Q\mathbb{Q}, so does iπi\pi (i=1i=\sqrt{-1}). Let this equation be θ1(x)=0,\theta_1 (x)=0, with roots iπ=α1,,αn.i\pi=\alpha_1,\ldots,\alpha_n. Now eiπ+1=0e^{i\pi}+1=0 so
(eα1+1)(eαn+1)=0 \left(e^{\alpha_1}+1\right) \ldots \left(e^{\alpha_n}+1\right)=0
We now construct an algebraic equation with integer coefficients whose roots are the exponents of ee in the expansion of the above product. For example, the exponents in pairs are α1+α2,α1+α3,,αn1+αn\alpha_1+\alpha_2,\,\alpha_1+\alpha_3,\,\ldots,\,\alpha_{n-1}+\alpha_n. The αs\alpha\,{\small s} satisfy a polynomial equation over Q\mathbb{Q} so their elementary symmetric functions are rational. Hence the elementary symmetric functions of the sums of pairs are symmetric functions of the αs\alpha\,{\small s} and are also rational. Thus the pairs are roots of the equation θ2(x)=0\theta_2 (x)=0 with rational coefficients. Similarly sums of 3 αs\alpha\,{\small s} are roots of θ3(x)=0\theta_3 (x)=0, etc. Then the equation
θ1(x)θ2(x)θn(x)=0 \theta_1(x)\theta_2(x) \ldots \theta_n (x)=0
is a polynomial equation over Q\mathbb{Q} whose roots are all sums of αs\alpha\,{\small s}. Deleting zero roots from this, if any, we get
θ(x)=0θ(x)=cxr+c1xr1+cr\begin{aligned} \theta(x) &= 0 \\ \theta(x) &= cx^r+c_1 x^{r-1}+\ldots c_r \end{aligned}
and cr0c_r \neq0 since we have deleted zero roots. The roots of this equation are the non-zero exponents of ee in the product when expanded. Call these β1,βr\beta_1,\ldots\beta_r. The original equation becomes
eβ1+eβr+e0+e0=0 e^{\beta_1}+\ldots e^{\beta_r}+e^0+\ldots e^0=0
ie
eβi+k=0 \sum e^{\beta_i}+k=0
where kk is an integer >0(0>0\quad(\neq0 since the term 111\ldots1 exists)
Now define
f(x)=csxp1[θ(x)]p(p1)!f(x)=c^s x^{p-1}\dfrac{\left[\theta(x)\right]^p}{(p-1)!}
where s=rp1s=rp-1 and pp will be determined later.
Define
F(x)=f(x)+f(x)++f(s+p)(x). F(x)=f(x)+f^{\prime}(x)+\ldots+f^{(s+p)}(x).
ddx[exF(x)]=exf(x) as before. \frac{\text{d}}{\text{d}x}\left[e^{-x}F(x)\right]=-e^{-x}f(x) \text{ as before}.
Hence we have
exF(x)F(0)=0xeyf(y)dy e^{-x}F(x)-F(0)=-\int_0 ^x e^{-y}f(y)\:\text{d} y
Putting y=λxy=\lambda x we get
F(x)exF(0)=x01e(1λ)xf(λx)dλ. F(x)-e^x F(0)=-x\int_0 ^1 e^{(1-\lambda)x}f(\lambda x)\:\text{d}\lambda.
Let xx range over the βi\beta_i and sum. Since eβi+k=0\sum e^{\beta_i}+k=0 we get
j=1rF(βj)+kF(0)=j=1rβj01e(1λ)βjf(λβj)  dλ. \sum_{j=1}^r F\left(\beta_j\right)+kF(0) =-\sum_{j=1}^r \beta_j \int_0 ^1 e^{(1-\lambda)\beta_j}f\left(\lambda\beta_j\right)\;\text{d}\lambda.


CLAIM

For large enough pp the LHS is a non-zero integer.

j=1rf(t)(βj)=0(0<t<p)\sum\limits_{j=1}^r f^{(t)}\left(\beta_j\right)=0\quad (0< t< p) by definition of ff. Each derivative of order pp or more has a factor pp and a factor csc^s, since we must differentiate [θ(x)]p\left[\theta(x)\right]^p enough times to get 0\neq0. And f(t)(βj)f^{(t)}\left(\beta_j\right) is a polynomial in βj\beta_j of degree at most ss. The sum is symmetric, and so is an integer provided each coefficient is divisible by csc^s, which it is. (symmetric functions are polynomials in coefficients = polynomials in cic\dfrac{c_i}{c} of degree s)\leq s). Thus we have
j=1rf(t)(βj)=pktt=p,,p+s. \sum_{j=1}^r f^{(t)}\left(\beta_j\right)=pk_t \quad t=p,\ldots, p+s.
Thus LHS=(integer)+kF(0).LHS = (\text{integer})+kF(0). \quad What is F(0)F(0)?
f(t)(0)=0t=0,,p2.f(p1)(0)=cscrp(cr0)f(t)(0)=p (some integer)t=p,p+1,.\begin{aligned} f^{(t)}(0) &=0 \qquad t=0,\ldots,p-2.\\ f^{(p-1)}(0) & =c^s c_r ^p \qquad \left(c_r\neq0\right) \\ f^{(t)}(0) &=p\text{ \small(some integer)} \qquad t=p,p+1,\ldots. \end{aligned}
So the LHS is an integer multiple of p+cscrpkp+c^s c_r ^p k. This is not divisible by pp if p>k,c,crp>k,c,c_{r}. So it is a non-zero integer. But the RHS 0\to 0 as pp\to\infty and we get the usual contradiction. \blacksquare

[1] A proof can be found at Liouville's Constant is Transcendental
[2] This was true in 1970. Is it still true today?