Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

(from here)

share|improve this question
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

2 Answers 2

up vote 1 down vote accepted

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

share|improve this answer

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".

share|improve this answer
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.