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In Radian mode, continually pressing the $\cos$ function of a number causes the result to converge to $x=0.739085133$, a fixed point of $\cos(x)$. Repeating this behavior with the $\sin$ function causes the result to converge to $x=0$, a fixed point of $\sin(x)$.

What happens if this is done using the tan function? It seems that $x=0$ is a repelling fixed point with no convergence in this case.

What would be the best way to explain these results mathematically?

stevetronix
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  • https://en.wikipedia.org/wiki/Fixed_point_(mathematics)#Attractive_fixed_points – Wojowu Oct 15 '15 at 20:14
  • The reason is that $(tan)'(x)\ge 1$ for all $x$ for which it is defined. – Bernard Oct 15 '15 at 20:16
  • It's like moving a ball at the bottom of a valley or at the top of a hill. In the valley, if you are a little off-center, the forces acting on the ball move it back to center, so it is stable. but at the top of the hill, the forces acting on the ball move it even further away, so it is unstable. In this case, the "force" is effectively the absolute value of the slope. When it is greater than $1$, points move farther away. When it is less than $1$, they move closer. – Paul Sinclair Oct 15 '15 at 20:25

2 Answers2

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I think what you do is to find the invariant point function of $\sin x$, $\cos x$ and $\tan x$. In other words, $x=0$ is the root for $\sin x=x$ and $x=0.739085$ is the root for $\cos x=x$. And that is why you can do it iteratively.

Martin Sleziak
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Esther Jacob
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    I think OP's question is more about why it works for some functions ($\sin,\cos$), and why it doesn't work for others ($\tan$). – Wojowu Oct 15 '15 at 20:19
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    Since $tanx$ is unbounded and cyclic, the root of $tanx=x$ is not unique. So, what you get depends on the first value you plug in. @Wojowu – Esther Jacob Oct 15 '15 at 20:34
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This is explained by stability theory.

The fixed point of $cos$ is the so called Dottie number. The derivative of $cos$ at this number is strictly less than $1$, so the Dottie number is a stable point. With $sin$ and $tan$ the derivative at $0$ equals $1$, so you need to do a more in-depth analysis. The fixed point of $sin$ turns out to be an attractor as well, whereas the dynamics can get quite wild with $tan$. To see what is happening here, it is best you plot a graph of these functions, together with the line $y = x$.

darko
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  • tan(x) seems to have an infinite number of fixed points when graphed with y = x. All fixed points seem to be repelling. Is this right? – stevetronix Oct 15 '15 at 21:23
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    Yes. All fixed points (except for $0$) are repelling, because the derivative there is greater than $1$. With $0$, you have to be a bit more careful. – darko Oct 15 '15 at 21:37
  • Selecting a point close to x=0, say x=0.25, and repeatedly computing the tan of that doesn't seem to get to 0. It doesn't seem like x=0 is converging. Is it neutral? – stevetronix Oct 15 '15 at 21:47
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    No, $0$ is also repelling. The reason for this is because $0$ is a saddle point for $tan$. As I wrote, a picture is worth a thousand words here. – darko Oct 15 '15 at 21:50
  • @darko You say "whereas the dynamics can get quite wild with tan"...can you point to a reference? I'm curious, but not sure where to look. – ThatsRightJack Jan 04 '20 at 10:09