I found on wikipedia () that a degree of a general function can be computed as $$\deg f(x) = \lim_{x\to\infty}\frac{\log |f(x)|}{\log x}$$ or $$\deg f(x) = \lim_{x\to\infty}\frac{x f^\prime(x)}{f(x)}$$ So e.g. in case $f(x)=\log x$, $\deg f(x) = 0$. In case of polynomials, it is the classical polynomial degree, i.e., $\deg (x^3+1) = 3$.
Unfortunately, any reference to some relevant literature is missing. Can you please point me at some literature where these kinds of degrees are analyzed? Thanks.
In case like $f(x) = x+\log x$ the degree (by this definition) is $1$. But does it say also something about possible number of real roots? Similarly like when we have a "pure" polynomial (there is at most "$\deg f(x)$" roots)?
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$\begingroup$This may not be exactly the type of reference you're seeking, but G. H. Hardy's book Orders of Infinity may be of interest.
Wikipedia's definitions of "degree" are defined by a function's behavior near infinity, and therefore do not bound the number of real roots for an arbitrary function. If $b > 0$ and $c$ are real numbers, the function $$ f(x) = c + b(x + \sqrt{1 + x^2}) + \sin x $$ has degree $1$. However, the number of positive real roots can be an arbitrary non-negative integer by choosing $b$ and $c$ suitably. (The graph of the middle term is one branch of a hyperbola, with asymptotes the negative $x$-axis and the line $y = 2bx$. Taking $b$ sufficiently small ensures $f$ has an arbitrarily large number of positive roots.) Arranging for infinitely many positive roots is left as an exercise. (Suggestion: Let $f(x) = x^2 \sin(1/x)$.)
Philosophically, to assume a function is polynomial (or rational) is to make a rigid, global hypothesis, which is why the degree of a polynomial bounds the number of roots.
Wikipedia's definitions of degree don't appear to capture any behavior that's not already captured by comparison with powers of $x$. Specifically, if $f = O(x^\alpha)$, then $\deg f \leq \alpha$ (using $\deg$ to denote the Wikipedia definition). Conversely, if $$ \alpha = \deg f = \lim_{x \to \infty} \frac{\log |f(x)|}{\log x}, $$ then $$ \lim_{x \to \infty} \frac{\log |f(x)|}{\log x^\alpha} = 1. $$ That is, for all $\varepsilon > 0$, there exists a real number $R$ such that $x^{\alpha(1-\varepsilon)} < |f(x)| < x^{\alpha(1+\varepsilon)}$ for $x > R$.
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