Why does any function get thinner as $x$ is multiplied by a constant?

Example:

$$\cos(x)$$

$$\cos(8x)$$

"Thinner" might not be the correct term. But I just want to know why does changing $$x$$ to $$8x$$ make it look like that?

• It's simple, what the first function used to do in an interval from $0$ to $1$ the new function does in the interval from $0$ to $1/8$. In particular if cos rounds once from $-\pi/2$ to $\pi/2$ the new function will do that from $-\pi/16$ to $\pi/16$. So it will round 8 times in the interval to $-\pi/2$ to $\pi/2$. Which makes it look thinner. – Yanko Feb 15 at 9:43
• @Yanko Why are you answering in a comment? – Arthur Feb 15 at 9:46
• indeed, this is like applying the function after a rescale in the x direction. I.e. consider the function $r: \mathbb{R} \to \mathbb{R},x \mapsto ax$ then your functions are just $f\circ r = f(r(x))$ and so just rescaled the whole grid. Also, please observe that if the constant is smaller than 1 it actually gets "fatter" and if it is negative it gets mirrored – Enkidu Feb 15 at 9:47
• @Arthur It's too "non-formal" for me to post it as an answer. I don't mind if someone else turns this into an answer. – Yanko Feb 15 at 9:50
• @Yanko There is no requirement here that answers are formal and I see nothing wrong with yours. And while you may not care about the points, getting an actual answer post upvoted and / or accepted will take this question off the unanswered queue and you will have done a little part in tidying up this place. Comments do not help in that regard. Answers do. – Arthur Feb 15 at 9:52

Taking a stab at a non-mathematical answer (well, minimally mathematical I guess). It makes intuitive sense to me, but I can't quite explain it mathematically.

The first thing to notice is that this is unrelated to using goniometric functions. It applies to any function, even ones as simple as $$y = x$$. The only difference is that it's less clear at first sight that $$y = 8.x$$ is a squashed ("thinner") version of $$y = x$$.
Most people perceive the difference between the two graphs as a rotation instead of a horizontal squashing.

However, if you were to color-code the graph (e.g. red-green-blue-red-... for all integer values of $$x$$ (rounded down)), you will see that it is in fact squashed and not rotated.

Think of the x axis as measure of physical distance, let's say kilometers. From your starting point ($$x=0$$), the Eiffel tower is 10 kilometers ahead ($$x=10$$), and Big Ben is another 20 kilometers further ($$x=30$$). Try to visually imagine the monuments on the x axis.

                                                                 ^
A                                        |o|
/-\                                       | |
-------------------------------------------------------------------------------------> (km)
0 1 2 3 4 5 6 7 8 9 10 . . . . . . . . . 20 . . . . . . . . . 30 . . . . . . . .


I apologize for the mediocre artwork.

Now I'm going to introduce a new unit, the Flatermeter, which happens to be exactly equal to 10km. What would our graph now look like if the X axis expresses distance in Flatermeters?

From your starting point ($$x=0$$), the Eiffel tower is 10 kilometers ahead, which is 1 Flatermeter ($$x=1$$), and Big Ben is another 20 kilometers further, which is another 2 Flatermeters ($$x=3$$). Which would look like this:

        ^
A  |o|
/-\ | |
-------------------------------------------------------------------------------------> (Fm)
0 1 2 3 4 5 6 7 8 9 10 . . . . . . . . . 20 . . . . . . . . . 30 . . . . . . . .


Notice how everything bunched up together, and all the distances shrunk by a factor of 10. Also notice that you could replace $$Fm$$ by $$10.km$$ as they are equal values.

The original $$y = f(km)$$ was quite wide. But the $$y = f(Fm)$$, which is the same as $$y = f(10km)$$ has bunched everything up much closer (which is what you're calling "thinner" in your question).

When you take a graph (e.g. $$y = x$$), and then artificially inflate the "step size" (= value of x) by a factor $$k$$ (e.g. $$y = k.x$$), then the graph will run through its shape $$k$$ times faster. Depending on how you visualize the graph, this has one of two (visual) consequences:

• The markings on the x axis move further apart (by a factor of $$k$$) and the graph has the exact same shape, visually speaking.
• The markings on the x axis stay the same and the graph itself horizontally shrinks (by a factor of $$k$$), visually speaking.

Your example deals with the latter scenario.

• Welcome to MSE! – YiFan Feb 15 at 12:16

From the comment by Yanko above:

It's simple, what the first function used to do in an interval from $$0$$ to $$1$$ the new function does in the interval from $$0$$ to $$1/8$$. In particular if cos rounds once from $$−π/2$$ to $$π/2$$ the new function will do that from $$−π/16$$ to $$π/16$$. So it will round $$8$$ times in the interval to $$−π/2$$ to $$π/2$$. Which makes it look thinner.

Multiplying the argument of a trigonometric function by a constant changes its period, which precisely is the distance between two consecutive local maxima or local minima which in either case must be equal.

Consider the general sinusoidal wave $$y=A\sin(ax+b)+C$$. The period of this wave or trigonometric function is given by $$2\pi/a$$.

In your case, define $$A_1:y=\cos x$$ and $$A_2:y=\cos 8x$$. By merely inspecting the expressions one can observe that these two waves must have a difference of periods (because the coefficients of $$x$$ is different in both the cases). Period of $$A_1$$ is $$2\pi$$, however that of $$A_2$$ is $$\pi/4$$. That is why you observe such a change in graphs of these functions.

The plot of a graph is really just a set of points $$S=\{(x,y)\mid y=f(x)\}$$. Let's say you turned $$x$$ into $$ax$$ for a constant $$a$$. Then surely, $$S$$ will not in general remain the same. The new set will instead contain of $$(x/a,y)$$ for every $$(x,y)$$ that used to be in $$S$$, since now, $$y=f(a(x/a))=f(x)$$ which fufills the definition of a point being on the graph of a function. So the action of "making $$f(x)$$ become $$f(ax)$$" takes each point $$(x,y)$$ to $$(x/a,y)$$, hence "compressing" the $$x$$-axis.