Famous Theorems of Mathematics/π is transcendental/Proof

The mathematical constant $\pi =3.141592\dots$ is a transcendental number (or inalgebraic).

In other words, it is not a root of any polynomial with rational coefficients.

Proof

Let us assume that $\pi$ is algebraic, so there exists a polynomial

$P(z)=a_0+a_{1}z+a_2{z}^2+\cdots +a_d{z}^d\in \mathbb{Q}[z],(a_0\ne 0)$

such that $P(\pi )=0$.

Part 1

Lemma: If $\pi$ is algebraic, then $\pi i$ is algebraic.

Proof: We get

$\begin{array}{cc}P(\pm iz)& =a_0+a_1(\pm iz)+a_2(\pm iz{)}^2+\cdots +a_d(\pm iz{)}^d\\ & =(a_0-a_2{z}^2+\cdots )\pm (a_{1}z-a_3{z}^3+\cdots )i\end{array}$

hence $\pi i$ is a root of the polynomial

$P(iz)P(-iz)=(a_0-a_2{z}^2+\cdots {)}^2+(a_{1}z-a_3{z}^3+\cdots {)}^2\in \mathbb{Q}[z]$

$◻$

Therefore, there exists a polynomial $P_1\in \mathbb{Q}[z]$ of degree $n$ with roots $z_1,\dots ,z_n$, such that $z_1=\pi i$.

By Euler's identity we have ${\text{e}}^{\pi i}+1=0$. Therefore:

$\begin{array}{cc}0& =({\text{e}}^{z_1}+1)({\text{e}}^{z_2}+1)\cdots ({\text{e}}^{z_n}+1)\\ & =\sum _{i=1}^n{\text{e}}^{z_i}+\sum _{1\le i_1<i_2\le n}{\text{e}}^{z_{i_1}}{\text{e}}^{z_{i_2}}+\sum _{1\le i_1<i_2<i_3\le n}{\text{e}}^{z_{i_1}}{\text{e}}^{z_{i_2}}{\text{e}}^{z_{i_3}}+\cdots +\sum _{1\le i_1<\cdots <i_n\le n}{\text{e}}^{z_{i_1}}\cdots {\text{e}}^{z_{i_n}}+1\\ & =\sum _{i=1}^n{\text{e}}^{z_i}+\sum _{1\le i_1<i_2\le n}{\text{e}}^{z_{i_1}+z_{i_2}}+\sum _{1\le i_1<i_2<i_3\le n}{\text{e}}^{z_{i_1}+z_{i_2}+z_{i_3}}+\cdots +{\text{e}}^{z_1+\cdots +z_n}+{\text{e}}^0\\ & =\sum _{i=1}^{2^n}{\text{e}}^{{\beta }_i}\end{array}$

The exponents are symmetric polynomial in $z_1,\dots ,z_n$, and among them are $1\le m\le 2^n-1$ non-zero sums. That is:

${\text{e}}^{{\beta }_1}+\cdots +{\text{e}}^{{\beta }_m}+2^n-m=0$

As we previously learned, for all $0\le k\le n$ there exists a monic polynomial $P_k\in \mathbb{Q}[z]$ such that its roots are the sums of every $k$ of the roots $z_1,\dots ,z_n$. Therefore:

$\begin{array}{cc}Q(z)& =P_0(z)P_1(z)\cdots P_n(z)\in \mathbb{Q}[z]\\ & =(z-{\beta }_1)\cdots (z-{\beta }_m)\cdots (z-{\beta }_{2^n})\\ & =z^{2^n-m}(z-{\beta }_1)\cdots (z-{\beta }_m)\end{array}$

After reduction we get that:

$(z-{\beta }_1)\cdots (z-{\beta }_m)\in \mathbb{Q}[z]$

Multiplying by the least common multiple $b_m\in \mathbb{Z}$ of the rational coefficients, we get a polynomial of the form

$\begin{array}{cc}B(z)& =b_m(z-{\beta }_1)\cdots (z-{\beta }_m)\\ & =b_m{z}^m+b_{m-1}{z}^{m-1}+\cdots +b_{1}z+b_0\in \mathbb{Z}[z]\end{array}$

Part 2

Let $f(z)$ be a polynomial of degree $d$. Let us define $F(z)=\sum _{j=0}^d{f}^{(j)}(z)$. Taking its derivative yields:

$F^{\prime }(z)=\sum _{j=0}^d{f}^{(j+1)}(z)=\sum _{j=1}^d{f}^{(j)}(z)=F(z)-f(z)$

Let us define $G(z)={\text{e}}^{-z}F(z)$. Taking its derivative yields:

$G^{\prime }(z)={\text{e}}^{-z}{F}^{\prime }(z)-{\text{e}}^{-z}F(z)={\text{e}}^{-z}[F^{\prime }(z)-F(z)]=-{\text{e}}^{-z}f(z)$

By the Fundamental Theorem of Calculus, we get:

$\begin{array}{cc}G(z)-G(0)={\text{e}}^{-z}F(z)-{\text{e}}^{-0}F(0)& =-\underset{0}{\overset{z}{\int }}{\text{e}}^{-w}f(w)dw\\ F(z)-{\text{e}}^{z}F(0)& =-\underset{0}{\overset{z}{\int }}{\text{e}}^{z-w}f(w)dw\end{array}$

Now let:

$F({\beta }_i)-{\text{e}}^{{\beta }_i}F(0)=-\underset{0}{\overset{{\beta }_i}{\int }}{\text{e}}^{{\beta }_i-w}f(w)dw=A_i$

Summing all the terms yields:

$\begin{array}{c}\sum _{i=1}^{m}F({\beta }_i)-\sum _{i=1}^m{\text{e}}^{{\beta }_i}F(0)=\sum _{i=1}^m{A}_i\\ \sum _{i=1}^{m}F({\beta }_i)-F(0)\sum _{i=1}^m{\text{e}}^{{\beta }_i}=\sum _{i=1}^m{A}_i\\ (2^n-m)F(0)+\sum _{i=1}^{m}F({\beta }_i)=\sum _{i=1}^m{A}_i\end{array}$

Part 3

Lemma: Let $f(z)$ be a polynomial with a root $z_0$ of multiplicity $d\ge 1$. Then $f^{(j)}(z_0)=0$ for all $0\le j\le d-1$.

Proof: By strong induction.

Let us write $f(z)=(z-z_0{)}^{d}Q(z)$, with $Q(z)$ a polynomial such that $Q(z_0)\ne 0$.

For $d=1$ we get:

$f^{(0)}(z_0)=f(z_0)=0$

Assume that for all $1\le d\le k$ the claim holds for all $0\le j\le d-1$.
We shall prove that for $d=k+1$ the claim holds for all $0\le j\le k$:

$\begin{array}{cc}f(z)& =(z-z_0{)}^{k+1}Q(z)\\ f^{(1)}(z)& =(k+1)(z-z_0{)}^{k}Q(z)+(z-z_0{)}^{k+1}{Q}^{(1)}(z)\\ & =(z-z_0{)}^k[(k+1)Q(z)+(z-z_0)Q^{(1)}(z)]\\ & =(z-z_0{)}^{k}R(z)\end{array}$

The Template:Font color is of multiplicity $k\ge 1$, with $R(z)$ a polynomial such that $R(z_0)\ne 0$.
Hence their product satisfies the induction hypothesis.

$◻$

Part 4

Let us now define:

$\begin{array}{cc}f(z)& =(b_m{)}^{c}g(z),(c=mp-1)\\ g(z)& =\frac{1}{(p-1)!}{z}^{p-1}[B(z){]}^p\\ & =\frac{(b_0{)}^p}{(p-1)!}{z}^{p-1}+\sum _{k=p}^{p+c}\frac{d_k}{(p-1)!}{z}^k,(d_k\in \mathbb{Z})\end{array}$

and $p$ is a prime number such that $p>\max \{b_0,b_m,2^n-m\}$. We get:

$f^{(j)}(z)=(b_m{)}^c{g}^{(j)}(z)=(b_m{)}^c\sum _{k=j}^{p+c}\frac{j!}{(p-1)!}(\genfrac{}{}{0ex}{}{k}{j})d_k{z}^{k-j},(0\le j\le p+c)$

hence for all $j\ge p$, the function $f^{(j)}(z)$ is a polynomial with integer coefficients all divisible by $p$.

By parts 1 and 3, we get:

$\begin{array}{cc}N& =(2^n-m)F(0)+\sum _{i=1}^{m}F({\beta }_i)\\ & =(2^n-m)\sum _{j=0}^{p+c}{f}^{(j)}(0)+\sum _{i=1}^m\sum _{j=0}^{p+c}{f}^{(j)}({\beta }_i)\\ & =(2^n-m)\sum _{j=p-1}^{p+c}{f}^{(j)}(0)+\sum _{i=1}^m\sum _{j=p}^{p+c}{f}^{(j)}({\beta }_i)\\ & =(2^n-m)[f^{(p-1)}(0)+\sum _{j=p}^{p+c}{f}^{(j)}(0)]+\sum _{j=p}^{p+c}\sum _{i=1}^m{f}^{(j)}({\beta }_i)\\ & =(2^n-m)(b_0{)}^p(b_m{)}^c+(2^n-m)\sum _{j=p}^{p+c}{f}^{(j)}(0)+(b_m{)}^c\sum _{j=p}^{p+c}\sum _{i=1}^m{g}^{(j)}({\beta }_i)\end{array}$

The Template:Font color is an integer not divisible by $p$.
The Template:Font color is an integer divisible by $p$.
The Template:Font color is the most important:
By Vieta's formulae we get

$E_k(\overrightarrow{\beta }{}^m)=(-1{)}^k\frac{b_{m-k}}{b_m}\in \mathbb{Q},(1\le k\le m)$

and the sums are symmetric polynomials in ${\beta }_1,\dots ,{\beta }_m$. Therefore, these can be expressed as polynomials

$\begin{array}{cc}\sum _{i=1}^m{g}^{(j)}({\beta }_i)& =G_j(\overrightarrow{E}{}^m(\overrightarrow{\beta }{}^m))\in 𝔽[\overrightarrow{\beta }{}^m]\\ & =\frac{a_j}{(b_m{)}^{c_j}},(a_j,c_j\in \mathbb{Z},c_j\ge 0)\end{array}$

$(b_m{)}^c\sum _{j=p}^{p+c}\sum _{i=1}^m{g}^{(j)}({\beta }_i)=(b_m{)}^c\sum _{j=p}^{p+c}\frac{a_j}{(b_m{)}^{c_j}}=\sum _{j=p}^{p+c}{a}_j(b_m{)}^{c-c_j}$

In addition, we get:

$\begin{array}{cc}\mathrm{deg}(g^{(j)})\le c& ⟹\mathrm{deg}(G_j)\le c,(p\le j\le p+c)\\ & ⟹0\le c_j\le c\\ & ⟹(b_m{)}^{c-c_j}\in \mathbb{Z}\end{array}$

Hence, the blue part is an integer divisible by $p$.

Conclusion: $N$ is an integer not divisible by $p$, and particularly $N\ne 0$.

Part 5

By part 2, by the triangle inequality for integrals we get:

$\begin{array}{cc}|A_i|& =|\underset{0}{\overset{{\beta }_i}{\int }}{\text{e}}^{{\beta }_i-w}f(w)dw|\\ & \le \underset{0}{\overset{{\beta }_i}{\int }}\left|{\text{e}}^{{\beta }_i-w}\right||f(w)||dw|\\ & =\underset{0}{\overset{{\beta }_i}{\int }}\left|{\text{e}}^{{\beta }_i-w}\right|\left|\frac{(b_m{)}^c}{(p-1)!}{w}^{p-1}\right[B(w){]}^p||dw|\\ & =\underset{0}{\overset{{\beta }_i}{\int }}\frac{\left|{\text{e}}^{{\beta }_i-w}\right|}{|b_m|}\frac{|b_m{|}^{mp}|w{|}^{p-1}|B(w){|}^p}{(p-1)!}|dw|\end{array}$

By the triangle inequality, we get:

$0<|N|=\left|\sum _{i=1}^m{A}_i\right|\le \sum _{i=1}^m|A_i|\le \sum _{i=1}^m\underset{0}{\overset{{\beta }_i}{\int }}\frac{\left|{\text{e}}^{{\beta }_i-w}\right|}{|b_m|}\frac{|b_m{|}^{mp}|w{|}^{p-1}|B(w){|}^p}{(p-1)!}|dw|$

On the other hand, we get

$\underset{p\to \mathrm{\infty }}{\lim }\frac{|b_m{|}^{mp}|w{|}^{p-1}|B(w){|}^p}{(p-1)!}=0,(w\in ℂ)$

hence for sufficiently large $p$ we get $0<|N|<1$. A contradiction.

$◻$

Conclusion: $\pi i$ is transcendental, hence $\pi$ is transcendental.

$◼$

Template:Header