# How to prove that the logarithm is a transcendental function

Let's consider the function $$\log x$$; how can I prove that it is a transcendental function on the function field of rational functions, i.e. that a polynomial in two variables $$p(x,y)$$ such that $$p(x,\log x)=0$$ identically does not exist?

I have been trying different approaches: seeing the logarithm on the real numbers, like a formal series or like a holomorphic function on an open in the complex plane, but I was not able to do this. I have also tried looking for the differences with $$\sqrt{(1+x)}$$ which is algebraic on the rational function, and can be defined on a subset of the reals, with a formal series or on an open subset of the complex plane.

• I do not think that this is true: the coefficients of the polynomial $p$ could be themselves transcendental May 23 at 20:57
• An algebraic function can have branch points of at most finite order, whereas for $\log x$ the branch point at $0$ has infinite order. May 23 at 20:58
• Yeah, realized that after my comment and already deleted. May 23 at 20:58
• @GEdgar could you give me some more details about this argument? What do you intend with algebraic function? What are the relations between branching points and algebraic functions May 23 at 21:01

A direct way to do this is to assume there is such a $$p(x,\log x)=0, x\in \mathbb R_+$$ Since $$\log$$ takes infinitely many values, $$p$$ cannot be a polynomial in $$\log x$$ only, so let $$n \ge 1$$ the highest power of $$x$$ with non zero coefficients and write the equation as:

$$a_n(\log x)x^n+a_{n-1}(\log x)x^{n-1}+..a_0(\log x)=0$$, where $$a_0,..a_n$$ are polynomials in one variable.

Dividing by $$x^n$$ and noting that $$\frac{a_k(\log x)}{x^{n-k}} \to 0, x \to \infty$$ for all $$0 \le k \le n-1$$, we get that $$a_n(\log x) \to 0, x \to \infty$$.

But now if $$a_n$$ is non constant, obviously $$a_n(\log x) \to \pm \infty, x \to \infty$$ depending on the sign of its leading coefficient and that is a contradiction. Hence $$a_n(\log x)$$ is constant and then it must be zero, contradicting the original assumption that the coefficient of $$x^n$$ in $$p$$ is non-zero, so we are done!

HINT:

Let us show that we cannot have an equality

$$\sum_{k=0}^m P_k(x) \exp ( \lambda_k x) = 0$$ where $$P_k(x)$$ are non-zero polynomials and $$\lambda_k$$ are distinct complex numbers. Otherwise we would have

$$P_0(x) \exp (\lambda_0 x) = -\sum_{k=1}^m P_k(x) \exp (\lambda_k x)$$

Now the right hand side is annihilated by a differential operator $$\prod_{k=1}^m (D - \lambda_k\cdot I)^{d_k+1}$$ while the LHS by $$(D-\lambda_0)^{d_0+1}$$. But the $$\gcd$$ of the polynomials $$\prod_{k=1}^m(T- \lambda_k)^{d_k+1}$$ and $$(T-\lambda_0)^{d_0+1}$$ is $$1$$. It follows that both sides are annihilated by the identity differential operator, so both are $$0$$, contradiction.