# 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)

-

1
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

1
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

2
@xsari3x: Well, the compression is because you precomposed the cosine with the square root. ;) – Rasmus Oct 8 '11 at 21:12
show 7 more comments

## 2 Answers

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 $\cos(90 \sqrt{x})$

-

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".

-

1
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
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9410967826843262, "perplexity_flag": "head"}