6
$\begingroup$

Example:

$$\cos(x)$$

Graph of y=cos(x)

$$\cos(8x)$$

Graph of y=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?

$\endgroup$
6
  • 2
    $\begingroup$ 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. $\endgroup$
    – Yanko
    Feb 15, 2019 at 9:43
  • $\begingroup$ @Yanko Why are you answering in a comment? $\endgroup$
    – Arthur
    Feb 15, 2019 at 9:46
  • 2
    $\begingroup$ 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 $\endgroup$
    – Felix
    Feb 15, 2019 at 9:47
  • $\begingroup$ @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. $\endgroup$
    – Yanko
    Feb 15, 2019 at 9:50
  • $\begingroup$ @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. $\endgroup$
    – Arthur
    Feb 15, 2019 at 9:52

4 Answers 4

6
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Welcome to MSE! $\endgroup$
    – YiFan Tey
    Feb 15, 2019 at 12:16
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .