# Let $f_{k+1}(x)=f_{k}(\cos x)$ and $f_{1}(x)=\cos x$ then $\lim_{k\to\infty}f_{k}(x)=0.73905\cdots$ [duplicate]

Let $$f_{k+1}(x)=f_{k}(\cos x)$$ and $$f_{1}(x)=\cos x$$ then $$\lim_{k\to\infty}f_{k}(x)=0.73905\cdots$$

I was just piddling around with the calculator one day. I don't know what happened but I just happened to take the cosine of a single number (in radians) repeatedly. It converged to a single value $$0.739085133\dots$$ It converged to this same thing for every number I tried. Like for example, the cosine of the cosine of the cosine of the cosine$$\dots$$ of any arbitrary value is equal to that.

Please tell me if I have made a new observation, or if it's just a false alarm.

• So basically you have been dabbling with the sequence $$x_n = \cos{(x_{n-1})}$$ Which of course converges to a value of $x = \eta$ for which $$\eta = \cos \eta$$ Apr 28, 2020 at 11:54
• It;s the unique fixed point of the cosine function, and it's an attracting fixed point. Apr 28, 2020 at 11:55
• Apr 28, 2020 at 12:04
• the number doesn't need to be in radians , in can be in martian banana unit and the answer should be the same. Apr 29, 2020 at 8:57

The function $$f(x)=\cos (x))$$ has just one fixed point ($$x_0$$ is a fixed point of $$f(x_0)$$ if and only if $$f(x_0)=x_0$$). That fixed point is as you found $$x_0\approx 0.739085133$$.

Then you can see that $$|f'(x_0)|=|-\sin (x_0)|<1$$.

So $$x_0\approx 0.739085133$$ is an attractive fixed point of $$f(x)=\cos (x)$$, and that's why you got that result using your calculator.

• And that's why you get that point, no matter where you start. Apr 28, 2020 at 12:43

I know others have sufficiently answered your question, but I wanted to add a neat graphical representation. You can start from some particular $$x$$ value, go vertically to the $$y=\cos(x)$$ curve, then horizontally to the $$y=x$$ line, and repeat this process to zero in on the fixed point of $$f(x)=\cos(x)$$.

Here I'm starting with an initial $$x$$ value of $$x_0=0.2$$. Or alternatively, you can plot the function $$f(x)=\cos\cos\cos...x$$ for a large number of iterations and notice it is a nearly constant function. (I'll let you do this yourself.) As a final note this number is in fact transcendental and has a special name, the Dottie Number.

• Thanks a lot bro for help it really worked Apr 28, 2020 at 14:40
• btw i did plot it u know Apr 28, 2020 at 14:46
• Picture worth 1,000 words. Apr 29, 2020 at 3:08