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

graph of \cos(\cos(90 \sqrt{x}))

(from here)

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    what is there to explain?2011-10-08
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    Why it has compression , unlike normal cos , it also expands in the spaces between the peaks2011-10-08
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    It's like Longitudinal Waves but without compression2011-10-08
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    Zooming near $0$ should make it look more like our mind picture of what it should look like.2011-10-08
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    @xsari3x: Well, the compression is because you precomposed the cosine with the square root. ;)2011-10-08
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    The change in frequency is due to the square-root.2011-10-08
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    Ehm , what's "precomposition " ?2011-10-08
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    The cosine of $0$ is $1$, so that's at least part of the explanation.2011-10-08
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    so the square root will make the change in the frequency ?? How , I'm intending to generate this signal through hardware :) ?2011-10-08
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    @Rasmus check the http://bit.ly/rbiyJs , u will find the plot2011-10-08
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    @xsari3x: Precomposing a function $g$ with a function $f$ just means to perform the function $f$ before performing $g$.2011-10-08
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    @Rasmus: The square root of even a smallish number like the $0.0025$ that you mention, is kind of big, in the sense that after you multiply by $90$ you are seriously away from $0$. By zoom I mean *really* zoom, like $[0,10^{-10}]$. Then we get about $0.54$, no more wiggling.2011-10-08

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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 $\cos(90 \sqrt{x})$ enter image description here

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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 $0$.2011-10-08
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    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.2011-10-08