# Intuition behind power-law scale invariance

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.

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}]


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:

• 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. Sep 18 at 9:19
• 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 $e^{a1000cm}/e^{a100cm}=e^{a900cm}$. Sep 18 at 9:43
• @GiorgioP: very good suggestion, done! Let me know if you don't approve. JalfredP that's an important point, thanks! Sep 18 at 10:28
• Now the difference is striking. Unfortunately, I cannot add a second upvote :-) Sep 18 at 15:59
• 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. Sep 19 at 8:37