Existence of some sort of 'infinite algebraicity' of transcendental numbers
Given an arbitrary number, say, $\alpha \in \mathbb{C}$, can anyone supply either
(a) a reason for the existence (or general non-existence) of, or
(b) the reverse engineering of
a (convergent) power series $f(z) = \sum a_n z^n$ with rational coefficients for which $f(\alpha) = 0$? (This problem is trivial when the number in question is algebraic, so feel free to assume transcendentality.) I'm looking for a push in the right direction; any input is welcome.
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prototypical example: $\alpha = \pi$ with $f(z) = \sum \frac{1}{(2n+1)!} z^{2n+1}$ = sin z
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I imagine that this question could easily be recast in several different directions. Please add any appropriate tags if you suspect this too.
Solution 1:
In fact you can do better. Suppose $\alpha_n$ is a sequence of distinct nonzero complex numbers with $|\alpha_n| \to \infty$ as $n \to \infty$, and $\beta_n$ is any sequence of complex numbers. Of course a power series with real coefficients satisfies $f(\overline{z}) = \overline{f(z)}$, so I'll require that if $\alpha_n = \overline{\alpha_m}$ then $\beta_n = \overline{\beta_m}$. Then there is an entire function $f(z)$, given by a power series $f(z) = \sum_{j=0}^\infty c_j z^j$ with all $c_j$ rational, such that $f(\alpha_n) = \beta_n$ for all $n$.
Note that by a consequence of Weierstrass's and Mittag-Lefler's theorems, for any such sequence $\alpha_n$ and any $N$, there is an entire function taking prescribed values at all $\alpha_n$ and also prescribed values for the first $N$ coefficients of its Maclaurin series.
Start with an entire function (guaranteed by the cited theorem) $f_0$ such that $f_0(\alpha_n) = \beta_n$ and
$f_0(\overline{\alpha_n}) = \overline{\beta_n}$ for all $n$. Since $(f_0(z) + \overline{f_0(\overline{z}})/2$ satisfies the same conditions, we may assume $f_0(\overline{z}) = \overline{f_0(z)}$.
Suppose $f_n$ is such a function where in addition the coefficients of $c_0, \ldots, c_{n-1}$ of its Maclaurin series are rational.
Consider $f_{n+1}(z) = f_n(z) + t g_n(z)$ where all $g_n(\alpha_k) = 0$,
$g_n(\overline{z}) = \overline{g_n(z)}$, and the Maclaurin series of $g_n$ starts
with $z^n$. There is a dense set of $t \in \mathbb R$ for which the coefficient of $z^n$ in $f_{n+1}(z)$ will be rational. I will choose such a $t$ small enough that
$|t| \max \{|g_n(z)|: |z| \le n\} < 1/n^2$. Then the sequence $f_n(z)$ converges uniformly on compact sets to an entire function $f$, which has the required properties.
Solution 2:
This is straightforward for arbitrary real $r$. First, let $a_0 = 1$ WLOG. Choose $a_1 \in \mathbb{Q}$ so that $|a_0 + a_1 r| < \frac{1}{2}$ (a sufficiently good rational approximation to $- \frac{a_0}{r}$ suffices). For any $n$, if $a_0, ... a_{n-1}$ have been chosen, it is always possible to choose in a similar fashion $a_n \in \mathbb{Q}$ so that $$\left| a_0 + a_1 r + ... + a_n r^n \right| < \frac{1}{2^n}.$$
The conclusion follows. Of course you can replace $\frac{1}{2^n}$ by a sequence decaying arbitrarily quickly to $0$.
For complex $r$ it's trickier; note that for purely imaginary $r$ the first step already fails. If some integer power of $r$ is real then we can proceed as above; otherwise, the argument is messier. We should consider an increasing sequence $t_n$ of positive integers such that $r^{t_n}$ has very small imaginary part relative to its real part (but not too small or else we won't be able to cancel out the imaginary part of the previous terms) and only tweak the corresponding terms $a_{t_n}$. The desired result is true in this case but the details are a bit annoying (the key observation is that the sequence $\frac{r^n}{|r|^n}$ is dense in the unit circle).