I'd like to refer the following answer:

@robjohn claims that:


$\pi\over 4$ is $a_1$ but where does $1\over \sqrt(2)$ came from?

My actual question is:
Given $a_1 = {\pi \over 4}$, $a_n = \cos(a_{n-1})$

Why does the range of this recurrence is $\left[\frac1{\sqrt2},\frac\pi4\right]$



$\dfrac{1}{\sqrt{2}}$ is $\cos\left(\dfrac{\pi}{4}\right)$

  • $\begingroup$ So, $a_1 ={ \pi\over 4}, a_n = cos(a_{n-1})$ will always be in that range? $\endgroup$ – AnnieOK May 28 '14 at 7:42
  • $\begingroup$ @AnnieOK: true. $\endgroup$ – DeepSea May 28 '14 at 7:44
  • $\begingroup$ can you explain why? $\endgroup$ – AnnieOK May 28 '14 at 7:51

The statement is that if you take the cosine of any value in $\left[\dfrac{1}{\sqrt{2}},\dfrac{\pi}{4}\right]$ the result will be in $\left[\dfrac{1}{\sqrt{2}},\dfrac{\pi}{4}\right]$.

You should be able to verify this. With regards to your specific question, $\cos\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}.$

Our first interval will be $\left[\dfrac{1}{\sqrt{2}},\dfrac{\pi}{4}\right]$ and then our second interval will be: $$\left[\cos\left(\dfrac{\pi}{4}\right),\cos\left(\dfrac{1}{\sqrt{2}}\right)\right] \approx \left[.7071067,.760244...\right].$$ Consider that our original interval was: $$\left[\dfrac{1}{\sqrt{2}},\dfrac{\pi}{4}\right] \approx [.7071067, .78539...]$$

So the statement makes sense.

  • $\begingroup$ Indeed make sense, but I shall prove it mathematically.. $\endgroup$ – AnnieOK May 28 '14 at 7:52
  • $\begingroup$ @AnnieOK: What part do you want to prove mathematically? Do you want a proof of $\cos\left(\frac{1}{\sqrt{2}}\right) \in \left[\frac{1}{\sqrt{2}}, \frac{\pi}{4}\right]$ or are you OK with the demonstration above using approximate values? $\endgroup$ – Michael Albanese May 28 '14 at 8:40
  • $\begingroup$ Yes, I shall prove it, and not demonstrate it.. $\endgroup$ – AnnieOK May 28 '14 at 16:01

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