Proof that $e^x$ is a transcendental function of $x$?

Question

Let a function $f(x)$ be algebraic if it satisfies an equation of the form $$c_n(x)(f(x))^n + c_{n-1}(x)(f(x))^{n-1} + \cdots + c_0(x)=0,$$ for $c_k(x)$ rational functions of $x$, and let $f$ be called transcendental if it is not algebraic. Is it possible to use this definition directly to show that $e^x$ is transcendental?

One way I have been considering for any complex number is this:

Let $x_0\in\mathbb{C}$ and $x_n=x_0+2\pi i n$, where $n\in\mathbb{Z}$. Hence $x_n\neq x_m$ for all $n\neq m$, but we do have $e^{x_n}=e^{x_m}$ for all $n,m\in\mathbb{Z}$ (since $e^{2\pi i n} = 1$ for all $n\in\mathbb{Z}$). But since the Implicit Function Theorem suggests there exists an exact algebraic formula for $x$ using the above definition of an algebraic function, then $e^x$ can not be algebraic since there are an infinite number of representations $x_n$ of $x$.

Answer 1

One could use the growth at infinity of the function $f:x\mapsto\mathrm e^x$.

Assume that $f$ is algebraic and choose a real number $x\geqslant0$. Then $|f(x)|\geqslant1$ and $$ |c_n(x)|\,|f(x)|^n\leqslant b(x)|f(x)|^{n-1},\qquad b(x)=\sum\limits_{k=0}^{n-1}|c_k(x)|. $$ Hence, for every real number $x\geqslant0$ such that $c_n(x)\ne0$, $|f(x)|\leqslant b(x)/|c_n(x)|$. But indeed, $c_n(x)\ne0$ for every real number $x$ large enough and $b(x)/|c_n(x)|$ can grow at most polynomially when the real number $x$ goes to $+\infty$ while $|f(x)|=\mathrm e^x$ grows... well, exponentially. This is a contradiction.

Answer 2

Here is a purely algebraic proof using only the fact that if $y=e^x$, then $y'=y$, but no analytic facts about the exponential function. In other words it should hold in any differential field.

Let $K$ be a differential field, such that $K=k(x)$ where $k$ is the field of constants of $K$ (ie those numbers whose derivative is 0) and with $x'=1$ (think field of rational functions in $x$). And $K(y)$ is a differential extension with $y'=y.$

First observe that $y\notin K$. For elements of $K$ are of the form $c p/q$ with $p,q$ monic polynomials in $x$ over $k$ and $c$ a constant. If $(p/q)'=p/q,$ then $p'q-pq'=pq.$ But the polynomial on the right has degree $\deg p+\deg q$, while the polynomial on the left has degree at most $\deg p+\deg q-1,$ and the right-hand side is monic, so they are not equal.

Now suppose $y$ were algebraic over $K$, then let the minimal polynomial of $y$ be $y^n+a_{n-1}y^{n-1}+\dotsb+a_0=0.$ Taking the derivative, we have $ny^n+(n-1)a_{n-1}y^{n-1}+a_{n-1}'y^{n-1}+\dotsb+a_0'=0.$ Subtracting from $n$ times the first equation gives $(a_{n-1}-a_{n-1}')y^{n-1}+\dotsb+na_0-a_0'=0.$ We already observed that $y'=y$ has no non-zero solutions in $K$, hence $a_{n-1}-a_{n-1}'\neq 0,$ so we have a monic polynomial of degree $n-1$ annihilating $y$, which contradicts the minimality of our polynomial.

Answer 3

Let's use your idea: assume that we have $F(y,x) = \sum_{k=0}^N A_k(y) x^k$ such that $F(f(x), x) = 0$. Since $f(\cdot)$ is periodic with period $d$. we have

$$\sum_{k=0}^N A_k(f(x)) t^k = 0$$ for all $t = x+ m d$, $m \in \mathbb{Z}$, and so the polynomial in $t$

$$\sum_{k=0}^N A_k(f(x)) t^k =0$$

is the zero polynomial, that is $A_k(f(x)) = 0$ for all $x$. Say one of the $A_k(\cdot)$ is analytic and non-zero. Then the set of zeros of $A_k$ is discrete. Moreover, it contains the image of $f$. We conclude $f$ is constant.