I have seen this notion of a scale-invariant power law curve exhibiting the property that $f(cx) = a(cx)^{-k} = c^{-k}f(x)$, and I am confused about how I should be thinking of this as "scale-invariant"

When I see this term, scale-invariant, I think "one can zoom in or out and the picture looks exactly the same". If I were to plot a power law, $f(x) = ax^{k}$, wouldn't the only way I could get something that looks the same no matter how much I zoom in be when $k=1$? If I were to plot a parabola in the desmos graphing calculator for instance, and I were to zoom out continuously, eventually the plot just looks like a straight line along the y axis. In what sense is this scale invariant?

I know I definitely have the wrong idea about this somewhere, I just need to figure out where I am going wrong.


1 Answer 1


I'm not an expert, but I would say that the function $f(x) = a x^k$ is scale invariant in the following sense:

Consider some interval $(x, b x)$. In this interval, the function $f(x)$ goes from:

\begin{aligned} f(x) \quad &\text{ at }x,\\ f(bx) = b^{k}f(x) \quad &\text{ at }bx. \end{aligned}

Now, imagine that you scale the interval by some value $\lambda$. In other words, let's look at the interval $(\lambda x, \lambda b x)$. If we do the same analysis as before, we can see that the function $f(x)$ over the interval goes from

\begin{aligned} f(\lambda x) = \lambda^{k} f(x) \quad &\text{ at }\lambda x,\\ f(\lambda bx) = \lambda^{k}b^{k}f(x) \quad &\text{ at }\lambda bx. \end{aligned}

So you should be able to see that over the "scaled" interval, the function $f(x)$ behaves exactly like it does over the "unscaled" interval, except for a global multiplicative factor.

What this means is that if you look at a function like $f(x)$ over a range of (appropriately scaled) intervals, it will look identical, modulo the above-mentioned factor. If you'd like to see this more clearly, I've plotted the parabola that you mention below on Mathematica, over the range $(\lambda , \,2\lambda)$, while varying $\lambda$ from 1 to 100. You should be able to see that the form of the function remains the same (the "jiggling" you see is because the successive images are not aligned well):

Manipulate[Plot[x^2, {x, \[Lambda], 2 \[Lambda]}, PlotRange -> {{\[Lambda], 2 \[Lambda] }, {\[Lambda]^2, 4 \[Lambda]^2}}, PlotStyle -> {Black}], {\[Lambda], 1, 100}]

                                enter image description here

On the other hand, if you did the same thing with a function that isn't scale invariant like the exponential (as @GiorgioP suggested in a comment below), you'd see something very different for the same parameter range:

                                enter image description here

  • 4
    $\begingroup$ Nice answer. Maybe adding a similar animated plot with a non-scale-invariant function, like an exponential, could help to grasp even better the meaning of scale-invariance. $\endgroup$ Commented Sep 18, 2021 at 9:19
  • 5
    $\begingroup$ I want to add to this that the meaning in physical sense is that if you take the ratio of your function at a different scale, say $x$=1 cm and 10 cm you get $a(10cm)^k/a(1cm)^k = 10^k$. If you now do the same using meters (100cm) you get $a(1000cm)^k/a(100cm)^k = 10^k$ i.e. relative ratios between parts of the graph do not depend on the scale you look at. $f$ at 10 times of $x$ is always going to be $10^k$ times $f$ at $x$, whatever the units of measure you choose. With an exponential this does not work $e^{a10cm}/e^{a1cm}=e^{a9cm}$ whereas $e^{a1000cm}/e^{a100cm}=e^{a900cm}$. $\endgroup$
    – JalfredP
    Commented Sep 18, 2021 at 9:43
  • 1
    $\begingroup$ @GiorgioP: very good suggestion, done! Let me know if you don't approve. JalfredP that's an important point, thanks! $\endgroup$
    – Philip
    Commented Sep 18, 2021 at 10:28
  • 1
    $\begingroup$ Now the difference is striking. Unfortunately, I cannot add a second upvote :-) $\endgroup$ Commented Sep 18, 2021 at 15:59
  • 1
    $\begingroup$ Thank you for the great answer! It seems like the bit of intuition I was missing was that I needed to be thinking of "ranges" of values looking the same. Your animations make it very clear. $\endgroup$ Commented Sep 19, 2021 at 8:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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