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Consider two real-valued real analytic functions $f$ and $g$. I want to prove that there exists a greatest common divisor $d$, which is a real analytic function. By greatest common divisor, I mean the following:

  1. Common divisor: There exist real analytic functions $q_1, q_2$ such that $f = dq_1, g = dq_2$, and
  2. Greateast: If there is any other function $d_1$ that satisfies 1. above, then there exists a real analytic function $q_3$ such that $d = d_1q_3$.

I am guessing that a proof could be derived from the Taylor series expansion, but I am not sure how to proceed.

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The relevant data are the roots (with multiplicity).

A GCD of two analytic functions $f$ and $g$ will be an analytic function $h$ such that for each $a \in \mathbb{R}$ one has $\operatorname{ord}_a (h) = \min \{\operatorname{ord}_a (f), \operatorname{ord}_a (g) \}$ where $\operatorname{ord}_a$ denotes the multiplicity of the root at $a$ (setting it $0$ if there is no root).

Put differently the order is the index where the Taylor series expansion around $a$ actually starts.

Algebraically the ring of real analytic functions is a Bézout domain, see for example Ring of analytic functions, meaning every finitely generated ideal is principal and in particular any two elements admit a GCD (see GCD domain)

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What do you mean by "greatest"? I.e., what would $\max\{\sin x, \cos x\}$ be?

Check out the development leading up to GCD among integers: It starts with the division algorithm, i.e., for all $a, b \in \mathbb{Z}$ there are $q, r \in \mathbb{Z}$ with $0 \le r \le \lvert b \rvert$ such that $a = b q + r$. Compare to the same situation for polynomials, where now you have $a(x) = b(x) q(x) + r(x)$, where the degree of $r$ is less than the degree of $b$. From this fundamental relationship you develop the idea of greatest common divisor (be it in the sense of greatest absolute value for integers or largest degree for polynomials). This assumes you have an integral domain with a valuation function (absolute value, degree) and a division algorithm. Real valued functions with pointwise operations (sum, product) aren't an integral domain (there are zero divisors, functions that aren't zero that multiply to the zero function).

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  • $\begingroup$ In every UFD there is a notion of GCD. There is no need for a valuation and a division algorithm. Even beyond that the notion of GCD makes sense (as even detailed in OP); "greatest" is with respect to divisibility. See for example the notion of GCD domain. $\endgroup$ – quid Dec 31 '15 at 15:23

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