Famous Theorems of Mathematics/π is transcendental/Proof

The mathematical constant $\pi =3.141592\ldots$ 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

$P(z)=a_{0}+a_{1}z+a_{2}z^{2}+\cdots +a_{d}z^{d}\in \mathbb {Q} [z],\quad (a_{0}\neq 0)$

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

Part 1

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

Proof: We get

${\begin{aligned}P(\pm iz)&=a_{0}+a_{1}(\pm iz)+a_{2}(\pm iz)^{2}+\cdots +a_{d}(\pm iz)^{d}\\[5pt]&=(a_{0}-a_{2}z^{2}+\cdots \,)\pm (a_{1}z-a_{3}z^{3}+\cdots \,)\,i\end{aligned}}$

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

$P(iz)P(-iz)=(a_{0}-a_{2}z^{2}+\cdots \,)^{2}+(a_{1}z-a_{3}z^{3}+\cdots \,)^{2}\in \mathbb {Q} [z]$

$\square$

Therefore, there exists a polynomial $P_{1}\in \mathbb {Q} [z]$ of degree $n$ with roots $z_{1},\ldots ,z_{n}$, such that $z_{1}=\pi i$.

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

${\begin{aligned}0&=({\text{e}}^{z_{1}}\!+1)({\text{e}}^{z_{2}}\!+1)\cdots ({\text{e}}^{z_{n}}\!+1)\\[3pt]&=\sum _{i\,=\,1}^{n}{\text{e}}^{z_{i}}+\!\!\!\!\sum _{1\leq i_{1}<i_{2}\leq n}\!\!\!\!\!{\text{e}}^{z_{i_{1}}}{\text{e}}^{z_{i_{2}}}+\!\!\!\!\!\!\!\!\sum _{1\leq i_{1}<i_{2}<i_{3}\leq n}\!\!\!\!\!\!\!\!{\text{e}}^{z_{i_{1}}}{\text{e}}^{z_{i_{2}}}{\text{e}}^{z_{i_{3}}}\!+\cdots +\!\!\!\!\!\!\!\!\sum _{1\leq i_{1}<\cdots <i_{n}\leq n}\!\!\!\!\!\!\!\!\!{\text{e}}^{z_{i_{1}}}\!\cdots {\text{e}}^{z_{i_{n}}}\!+1\\[3pt]&=\sum _{i\,=\,1}^{n}{\text{e}}^{z_{i}}+\!\!\!\!\sum _{1\leq i_{1}<i_{2}\leq n}\!\!\!\!\!{\text{e}}^{z_{i_{1}}+\,z_{i_{2}}}+\!\!\!\!\!\!\!\!\sum _{1\leq i_{1}<i_{2}<i_{3}\leq n}\!\!\!\!\!\!\!\!{\text{e}}^{z_{i_{1}}+\,z_{i_{2}}+\,z_{i_{3}}}\!+\cdots +{\text{e}}^{z_{1}+\,\cdots \,+\,z_{n}}\!+{\text{e}}^{0}\\[3pt]&=\sum _{i\,=\,1}^{2^{n}}{\text{e}}^{\beta _{i}}\end{aligned}}$

The exponents are symmetric polynomial in $z_{1},\ldots ,z_{n}$, and among them are $1\leq m\leq 2^{n}-1$ non-zero sums. That is:

${\text{e}}^{\beta _{1}}\!+{\text{e}}^{\beta _{2}}\!+\cdots +{\text{e}}^{\beta _{m}}\!+2^{n}\!-m=0$

As we previously learned, for all $0\leq k\leq n$ there exists a monic polynomial $P_{k}\in \mathbb {Q} [z]$ of degree ${\tbinom {n}{k}}$ such that its roots are the sums of every $k$ of the root $z_{1},\ldots ,z_{n}$. Therefore:

${\begin{aligned}Q(z)&=P_{0}(z)\,P_{1}(z)\cdots P_{n}(z)\in \mathbb {Q} [z]\\[5pt]&=(z-\beta _{1})(z-\beta _{2})\cdots (z-\beta _{m})\cdots (z-\beta _{2^{n}})\\[5pt]&=z^{2^{n}-\,m}(z-\beta _{1})(z-\beta _{2})\cdots (z-\beta _{m})\end{aligned}}$

After reduction we get that:

$(z-\beta _{1})(z-\beta _{2})\cdots (z-\beta _{m})\in \mathbb {Q} [z]$

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

${\begin{aligned}B(z)&=b_{m}(z-\beta _{1})(z-\beta _{2})\cdots (z-\beta _{m})\\[5pt]&=b_{m}z^{m}+b_{m-1}z^{m-1}+\cdots +b_{1}z+b_{0}\in \mathbb {Z} [z]\end{aligned}}$

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:

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

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

$G'\!(z)={\text{e}}^{-z}F'\!(z)-{\text{e}}^{-z}F(z)={\text{e}}^{-z}{\bigl [}F'\!(z)-F(z){\bigr ]}=-{\text{e}}^{-z}f(z)$

By fundamental theorem of calculus, we get:

${\begin{aligned}G(z)-G(0)={\text{e}}^{-z}F(z)-{\text{e}}^{-0}F(0)&=-\!\int \limits _{0}^{z}{\text{e}}^{-w}f(w)dw\\F(z)-{\text{e}}^{z}F(0)&=-\!\int \limits _{0}^{z}{\text{e}}^{z-w}f(w)dw\end{aligned}}$

Now let:

$F(\beta _{i})-{\text{e}}^{\beta _{i}}F(0)=-\!\int \limits _{0}^{\beta _{i}}{\text{e}}^{\beta _{i}-w}f(w)dw=A_{i}$

Summing all the terms yields:

${\begin{aligned}\sum _{i\,=\,1}^{m}F(\beta _{i})-\sum _{i\,=\,1}^{m}{\text{e}}^{\beta _{i}}F(0)=\sum _{i\,=\,1}^{m}A_{i}\\[5pt]\sum _{i\,=\,1}^{m}F(\beta _{i})-F(0)\sum _{i\,=\,1}^{m}{\text{e}}^{\beta _{i}}=\sum _{i\,=\,1}^{m}A_{i}\\[5pt](2^{n}\!-m)F(0)+\sum _{i\,=\,1}^{m}F(\beta _{i})=\sum _{i\,=\,1}^{m}A_{i}\end{aligned}}$

Part 3

Let $f(z)$ be a polynomial with a root $z_{0}$ of multiplicity $d$. We will show that for all $0\leq j\leq d-1$ we get $f^{(j)}\!(z_{0})=0$.

Let us write $f(z)=(z-z_{0})^{d}Q_{0}(z)$, such that $Q_{0}(z)$ is a polynomial with $Q_{0}(z_{0})\neq 0$.

${\begin{aligned}f^{(1)}\!(z)&=d(z-z_{0})^{d-1}Q_{0}(z)+(z-z_{0})^{d}Q_{0}'(z)\\[5pt]&=(z-z_{0})^{d-1}{\Big [}d\,Q_{0}(z)+(z-z_{0})Q_{0}'(z){\Big ]}\\[5pt]&=(z-z_{0})^{d-1}Q_{1}(z)\\[5pt]Q_{1}(z_{0})&=d\,Q_{0}(z_{0})\neq 0\\\\[5pt]f^{(2)}\!(z)&=(d-1)(z-z_{0})^{d-2}Q_{1}(z)+(z-z_{0})^{d-1}Q_{1}'(z)\\[5pt]&=(z-z_{0})^{d-2}{\Big [}(d-1)Q_{1}(z)+(z-z_{0})Q_{1}'(z){\Big ]}\\[5pt]&=(z-z_{0})^{d-2}Q_{2}(z)\\[5pt]Q_{2}(z_{0})&=(d-1)Q_{1}(z_{0})\neq 0\\[5pt]&\,\,\,\vdots \\[5pt]f^{(d-1)}\!(z)&=2(z-z_{0})Q_{d-2}(z)+(z-z_{0})^{2}Q_{d-2}'(z)\\[5pt]&=(z-z_{0}){\Big [}2Q_{d-2}(z)+(z-z_{0})Q_{d-2}'(z){\Big ]}\\[5pt]&=(z-z_{0})Q_{d-1}(z)\\[5pt]Q_{d-1}(z_{0})&=2Q_{d-2}(z_{0})\neq 0\end{aligned}}$

with $Q_{0}'(z),Q_{1}'(z),\ldots ,Q_{d-2}'(z)$ all polynomials.

Part 4

Let us now define:

${\begin{aligned}f(z)&=(b_{m})^{c}g(z),\quad (c=mp-1)\\[5pt]g(z)&={\frac {1}{(p-1)!}}\,z^{p-1}{\bigl [}B(z){\bigr ]}^{p}\\[5pt]&={\frac {(b_{0})^{p}}{(p-1)!}}z^{p-1}+\sum _{k\,=\,p}^{p\,+\,c}{\frac {d_{k}}{(p-1)!}}z^{k},\quad (d_{k}\in \mathbb {Z} )\end{aligned}}$

and $p$ is a prime number such that $p>\max \!{\bigl \{}b_{0},b_{m},2^{n}\!-m{\bigr \}}$. We get:

$f^{(j)}\!(z)=(b_{m})^{c}g^{(j)}\!(z)=(b_{m})^{c}\sum _{k\,=\,j}^{p\,+\,c}{\frac {j!}{(p-1)!}}{\binom {k}{j}}d_{k}z^{k-j},\quad (0\leq j\leq p+c)$

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

By parts 1 and 3, we get:

${\begin{aligned}N&=(2^{n}\!-m)F(0)+\sum _{i\,=\,1}^{m}F(\beta _{i})\\[2pt]&=(2^{n}\!-m)\sum _{j\,=\,0}^{p\,+\,c}f^{(j)}\!(0)+\sum _{i\,=\,1}^{m}\sum _{j\,=\,0}^{p\,+\,c}f^{(j)}\!(\beta _{i})\\[2pt]&=(2^{n}\!-m)\!\!\sum _{j\,=\,p-1}^{p\,+\,c}\!\!\!f^{(j)}\!(0)+\sum _{i\,=\,1}^{m}\sum _{j\,=\,p}^{p\,+\,c}f^{(j)}\!(\beta _{i})\\[2pt]&=(2^{n}\!-m)\,{\bigg [}\,f^{(p-1)}\!(0)+\sum _{j\,=\,p}^{p\,+\,c}f^{(j)}\!(0)\,{\bigg ]}+\sum _{j\,=\,p}^{p\,+\,c}\sum _{i\,=\,1}^{m}f^{(j)}\!(\beta _{i})\\[2pt]&={\color {red}(2^{n}\!-m)(b_{0})^{p}(b_{m})^{c}}\!+{\color {green}(2^{n}\!-m)\sum _{j\,=\,p}^{p\,+\,c}f^{(j)}\!(0)}+{\color {blue}(b_{m})^{c}\sum _{j\,=\,p}^{p\,+\,c}\sum _{i\,=\,1}^{m}g^{(j)}\!(\beta _{i})}\end{aligned}}$

The red part is an integer not divisible by $p$.
The green part is an integer divisible by $p$.
The blue part is the most important:
The sums are symmetric polynomials in $\beta _{1},\ldots ,\beta _{m}$. Therefore we get:

${\begin{aligned}&\alpha _{k}=E_{k}(\beta _{1},\ldots ,\beta _{m})=(-1)^{k}{\frac {b_{m-k}}{b_{m}}}\in \mathbb {Q} ,\quad (1\leq k\leq m)\\[5pt]&\sum _{i\,=\,1}^{m}g^{(j)}\!(\beta _{i})=G_{j}(\alpha _{1},\ldots ,\alpha _{m})={\frac {a_{j}}{(b_{m})^{c_{j}}}},\quad (a_{j},c_{j}\in \mathbb {Z} ,\,c_{j}\geq 0)\\[5pt]&{\color {blue}(b_{m})^{c}\sum _{j\,=\,p}^{p\,+\,c}\sum _{i\,=\,1}^{m}g^{(j)}\!(\beta _{i})}=(b_{m})^{c}\sum _{j\,=\,p}^{p\,+\,c}{\frac {a_{j}}{(b_{m})^{c_{j}}}}=\sum _{j\,=\,p}^{p\,+\,c}a_{j}(b_{m})^{c-c_{j}}\end{aligned}}$

In addition, we get:

${\begin{aligned}\deg(g^{(j)})\leq c&\implies \deg(G_{j})\leq c,\quad (p\leq j\leq p+c)\\[3pt]&\implies 0\leq c_{j}\leq c\\[3pt]&\implies (b_{m})^{c-c_{j}}\in \mathbb {Z} \end{aligned}}$

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

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

Part 5

By part 2, we get:

$A_{i}=-\!\int \limits _{0}^{\beta _{i}}{\text{e}}^{\beta _{i}-w}f(w)dw$

By the triangle inequality for integrals, we get:

${\begin{aligned}|A_{i}|&={\Bigg |}\int \limits _{0}^{\beta _{i}}{\text{e}}^{\beta _{i}-w}f(w)dw\,{\Bigg |}\\[2pt]&\leq \int \limits _{0}^{\beta _{i}}\,{\bigl |}{\text{e}}^{\beta _{i}-w}{\bigr |}\,{\bigr |}f(w){\bigl |}\,|dw|\\[2pt]&=\int \limits _{0}^{\beta _{i}}{\bigl |}{\text{e}}^{\beta _{i}-w}{\bigr |}\left|{\frac {(b_{m})^{c}}{(p-1)!}}\,w^{p-1}{\bigl [}B(w){\bigr ]}^{p}\right||dw|\\[2pt]&=\int \limits _{0}^{\beta _{i}}{\frac {{\bigl |}{\text{e}}^{\beta _{i}-w}{\bigr |}}{|b_{m}|}}\,{\frac {|b_{m}|^{mp}|w|^{p-1}|B(w)|^{p}}{(p-1)!}}\,|dw|\end{aligned}}$

By the triangle inequality, we get:

$0<|N|=\left|\,\sum _{i\,=\,1}^{m}A_{i}\right|\leq \sum _{i\,=\,1}^{m}|A_{i}|\leq \sum _{i\,=\,1}^{m}\,\int \limits _{0}^{\beta _{i}}{\frac {{\bigl |}{\text{e}}^{\beta _{i}-w}{\bigr |}}{|b_{m}|}}\,{\frac {|b_{m}|^{mp}|w|^{p-1}|B(w)|^{p}}{(p-1)!}}\,|dw|$

On the other hand, we get

$\lim _{p\to \infty }{\frac {|b_{m}|^{mp}|w|^{p-1}|B(w)|^{p}}{(p-1)!}}=0,\quad (w\in \mathbb {C} )$

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

$\square$

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

$\blacksquare$

←Fundamental theorem of symmetric polynomials

Proof