# Can you explain why plotting $\cos(\cos(90 \sqrt{x}))$ looks like this?

Can you explain why plotting $\cos(\cos(90 \sqrt{x}))$ looks like this:

(from here)

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what is there to explain? – user12205 Oct 8 '11 at 21:02
Why it has compression , unlike normal cos , it also expands in the spaces between the peaks – xsari3x Oct 8 '11 at 21:06
It's like Longitudinal Waves but without compression – xsari3x Oct 8 '11 at 21:10
Zooming near $0$ should make it look more like our mind picture of what it should look like. – André Nicolas Oct 8 '11 at 21:10
@xsari3x: Well, the compression is because you precomposed the cosine with the square root. ;) – Rasmus Oct 8 '11 at 21:12

You can't expect it to look like a Cos(x) for example because the argument $\cos(90 \sqrt{x})$ is not linear, and is itself cyclic. Here is the graph of $\cos(90 \sqrt{x})$

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As $x$ changes, $\sqrt{x}$ changes at a varying rate that approaches $\infty$ as $x$ approaches $0$. You can see that by realizing that $y = \pm\sqrt{x}$ is the same as $x=y^2$, and that's a parabola with a vertical tangent at $x=y=0$. Therefore it oscillates very fast when $x$ is near $0$. As $x$ moves away from $0$, then $\sqrt{x}$ changes more slowly as $x$ changes, so this oscillates more slowly.

$\cos(\cos(90\sqrt{0}))= \cos 1 \approx 0.54,$ so it starts at $0.54$, and returns to $0.54$ whenever $90\sqrt{x}$ returns to something whose cosine is $1$. The function returns to $1$ whenever $90\sqrt{x}$ returns to something whose cosine is $0$.

The infinite rate of change at $x=0$ means the graph will have a vertical tangent at $x=0$, but does not mean it oscillates infinitely many times between $0$ and any particular positive value. The reasons for this can be seen with a little thought. In other words, you should be able to actually count the oscillations between any positive argument and $0$. The number of such oscillations is large because of the "90".

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Dear Michael, I think "The infinite rate of change at $x=0$ means the graph will have a vertical tangent at $x=0$" is not true. The reason being that cosine has derivative $0$ at $0$. – Rasmus Oct 8 '11 at 21:36
You're right: I just computed the limit of the derivative at $0$. It's got a numerator and denominator that both approach $0$, and the limit is a finite positive number. So it's increasing pretty fast at $x=0$, but not infinitely fast. – Michael Hardy Oct 8 '11 at 21:50