Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question
Note that not every function has a degree by this definition, not even if the function is continuous, differentiable, infinitely differentiable, whatever. Note also the degree of $e^x$ is infinite by this definition. As it should be, I suppose, but is that useful? – Gerry Myerson Mar 4 '14 at 12:29
@GerryMyerson I don't know if it is useful. But e.g. when $f(x)=x+\log x$, then there is only one root $xe^x=1$ (Lambert function $W(1) = 0.5671..$). Which may be suggested also by the fact that $\deg f(x)=1$. Maybe if $\deg f(x)=\infty$ there is no real root...I don't know if there is any connection with the "degree of a general function" and the properties of the function. That's why I hoped that I could find some book about this. – pisoir Mar 4 '14 at 12:56
I do not know how available it is, but Prince Askold Khovanski's (of Bernshtein-Kushnirenko-Khovanski or BKK fame) "Fewnomials" deals with functions which are polynomials of few terms or solutions of differential equations of few polynomial terms, and towers thereof. This theory extends the theorems of polynomials like Descartes rule of signs also to this more general function class. – LutzL Mar 4 '14 at 16:38
@LutzL I tried to look into these "fewnomials" but as I see it it deals only with some type of polynomials. I could not find any example where it would handle "mixed" functions like $f(x)=x+\log x$. Can you maybe give me some more hints? Thanks. – pisoir Mar 5 '14 at 18:19
The original book also deals with solutions of linear differential equations where the coefficients are polynomials. Since $xf'(x)=x+1$, this would be covered by this more general idea. But I do not have the book available, and it is more than 10 years ago that I read through it. – LutzL Mar 5 '14 at 18:22

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 $x > R$.

share|cite|improve this answer
Thanks for the answer! I still wonder how did that definition got into wikipedia when it is so hard to find any book (reference) that would use such definition of degree. – pisoir Mar 4 '14 at 12:24
As for your example. If I am not wrong, by the first definition (with logarithms) the degree is really $1$. By the second definition (using the derivative) the degree is $\infty$. Strangely, but it seems that those definitions are not equivalent. – pisoir Mar 4 '14 at 12:38
For $f(x) = c + b(x + \sqrt{1+x^2}) + \sin x$, the second definition of degree is undefined (i.e., $\lim\limits_{x\to\infty}xf'(x)/f(x)$ does not exist), so you're right that the definitions aren't generally equivalent. (If the first definition of $\deg$ formally gives $\infty/\infty$ and the second exists, the two do agree, by l'Hopital.) Given the style and lack of detail in that article, I have a not-entirely-charitable hypothesis about how the definition got into Wikipedia.... – Andrew D. Hwang Mar 4 '14 at 13:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.