# Is there a natural scale associated with polynomials?

This question is related to a previous question asked here.

Power laws are scale invariant. They don't have a built-in or characteristic scale associated with them. Exponentials such as $e^{-x/\xi}$ are not scale-invariant. They have a characteristic scale $\xi$. What is the matter with polynomials such as $f(x)=ax^2+bx^3$ (where $x,a,b$ are all dimensional parameters with appropriate dimension)? Like exponentials, they too are not scale-invariant. But is there a natural scale associated with it? If yes, how does one find that hidden scale?

• I guess the natural scale may be hidden in constants like $h$, for example. - as implied by answer just now by Steven and the other answer – tom Mar 21 '18 at 16:42
• @tom The question is how to identify the hidden scale. – SRS Mar 21 '18 at 16:49
• This seems to be a question about mathematics, not physics. – sammy gerbil Mar 21 '18 at 17:31
• @sammygerbil It's important to understand if one wishes to understand scale invariance in physics. Moreover, mathematicians (as far as I know) is not used to the language of "length scales", "characteristic scales" etc. And It's, of course, related to physics. See my question here physics.stackexchange.com/questions/394732/… – SRS Mar 21 '18 at 17:34
• So I looked at your Landau free energy scale question and I think I understand more about what you are asking. I will comment there. – tom Mar 21 '18 at 17:43

Write your polynomial as $f(x) = ax^2 [1+ x/(a/b)]$. You see that $a/b$ is the scale at which the $x^3$ term takes over.

Take a look at the log-log plot of $f(x)$ for different values of $l=a/b$. The black lines are there to guide the eye. The dashed line corresponds $x^2$ and the dotted one to $x^3$.

Take a look at the red curve. It scales as $x^2$ for small arguments and as $x^3$ when $x$ is large. The cross-over scale is clearly visible in the middle.

• Do you mean $a/b$ is the natural scale? I don't understand why. Does your argument hold good for polynomials of the form $f(x)=ax^2+bx^3+cx^4$ or of higher orders? You think there will be a single characteristic scale? @StevenMathey – SRS Mar 21 '18 at 16:47
• Yes. For $x\ll l$, the $x^3$ term can be neglected and $f \sim x^2$. For $x \gg l$, the $x^3$ term completely overwhelms the other and $f \sim x^3$. For higher order polynomials, you have more scales in the problem, $f(x) = a x^2 [1 + x/l_1 + (x/l_2)^2]$, with $l_1 = a/b$ and $l_2 = \sqrt{a/c}$. – Steven Mathey Mar 21 '18 at 17:21
• Can you slightly edit your answer? I've accidentally downvoted it, and cannot undo the downvote unless you edit it a bit. @StevemMathey – SRS Mar 24 '18 at 17:29

There's a problem with $$f(x)= x^2+x^3$$ in that, if $x$ is not dimensionless, then various powers in your polynomials will have different dimensions and so cannot strictly be added. If of course you make $x$ dimensionless by dividing $x$ by some characteristic scale $a$, then you're back to the same argument as the exponential examples.

• I've been careless in asking the question. Now I have modified the question. Is there a way I can identify some natural scale with $f(x)$? @ZeroTheHero – SRS Mar 21 '18 at 16:20