Timeline for What's an example of a transcendental power series?

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May 9, 2010 at 7:57	comment	added	Wadim Zudilin		If a function $f(z)=\sum_{n=0}^\infty a_nz^n$ with all $a_n\in\mathbb Q$ is algebraic, then it is not hard to show that an algebraic relation can be given over $\mathbb Q(z)$ rather than $\mathbb C$. Therefore the above proof also shows that the formal power series in question is transcendental not only over $\mathbb C(z)$ but over any $k(z)$ where the field $k$ containts $\mathbb Q$.
May 9, 2010 at 3:58	comment	added	Qiaochu Yuan		Do you use anywhere in this argument that the base field is C?
Apr 16, 2010 at 23:11	comment	added	Wadim Zudilin		The factorial series is even simpler if you just refer to the fact that its value at $1/2$ is transcendental due to Liouville's 1844 theorem. Note that Liouville's theorem does not cover $\sum_{k\ge0}z^{d^k}$, so Mahler's example is a bit harder and gives some hints about the powerful method in transcendence number theory. It's a matter of personal taste to choose between "Liouville" and "Mahler".
Apr 16, 2010 at 14:58	comment	added	fedja		IMHO, the factorial series gives a much simpler proof. All you need there is that $nN!$ can be represented as a sum of not more than $n$ factorials in just one way and the next $D$ numbers have no such representation at all if $N>n+D$, say. So the coefficient at the highest power just appears explicitly in the Taylor series of $\sum_{k=0}^n p_kf^k$
Apr 16, 2010 at 11:26	history	answered	Wadim Zudilin	CC BY-SA 2.5