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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?

SRS
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2 Answers2

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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.

[See also this answer.]

ZeroTheHero
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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 $x^3$.

enter image description here

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.

Steven Mathey
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