# Is $f(\sqrt{x}) + f(-\sqrt{x})$ always rational?

Suppose that $$f$$ is a rational function over $$\mathbb{R}$$ (or $$\mathbb{C}$$, or whatever field makes this easier to answer). Is $$\frac{f(\sqrt{x}) + f(-\sqrt{x})}{2}$$ always a rational function of $$x$$?

This question is inspired by a sequence question. Given a sequence $$a_n$$ with a rational generating function, is it true that, say, $$a_{2n}$$ also has a rational generating function? If $$f$$ is the generating function for $$a_n$$, then $$(f(\sqrt{x}) + f(-\sqrt{x})) / 2$$ is the generating function for $$a_{2n}$$.

As an example, take $$f(x) = 1 / (1 - x)$$, which corresponds to the all-ones sequence $$a_n = 1$$. Then $$\frac{f(\sqrt{x}) + f(-\sqrt{x})}{2} = \frac{1}{1 - x},$$ as we would expect.

This specific case might be "easy" to answer (not for me), but when I first heard about this someone mumbled something about Galois theory and field extensions that I didn't understand. I would like to know more about this if it's relevant, since this question generalizes to seemingly harder cases.

• I think Galois theory is the way to go here. $\mathbb R(\sqrt x)$ is a quadratic extension of $\mathbb R(x)$, so the Galois group is cyclic of order $2$. The only non-trivial element is $\sigma$ which changes the sign of $\sqrt x$. Since your expression is fixed under $\sigma$ it must be in the base field. Does that answer your question?
– lulu
Jan 14, 2020 at 20:35
• @lulu That sounds exactly like what the person said. I don't know anything about Galois theory - is this the type of argument I could learn with any introductory text? Jan 14, 2020 at 20:38
• In a nutshell: the answer is yes. This can be justified by noting that switching $\sqrt{x}$ with $-\sqrt{x}$ defines a "field automorphism", and the only elements that are fixed by this field automorphism are the rational functions over $x$. Jan 14, 2020 at 20:38
• this question provides some basic references in Galois Theory. I'd say it's worth looking up one (or more) of these. It's a great and powerful topic, well worth learning.
– lulu
Jan 14, 2020 at 20:43

Here is a brutal-force solution: Write $$f$$ in the form
$$f(x) = \frac{A_0(x^2) + x A_1(x^2)}{B_0(x^2) + x B_1(x^2)}$$
where $$A_0, A_1, B_0, B_1$$ are polynomials. Then
$$\frac{f(\sqrt{x}) + f(-\sqrt{x})}{2} = \frac{A_0(x)B_0(x) - x A_1(x)B_1(x)}{B_0(x)^2 - x B_1(x)^2}.$$