Cosine Button Convergence: Degrees vs Radians

[rumborak]

This is of course just a silly exercise on a calculator, but it is intriguing that when operating in either degrees or "grad", hitting the cosine button will converge incredibly fast, whereas in radians it is a lot slower.

Anybody have a good idea why?

 

[andrewkirk]

It's because the result of the cos function is always in the range [-1,1] which, in degrees, is a tiny angle, whose cosine is very close to 1. So in degrees, the value always gets close to 1 in two steps.

In radians, the range [-1,1] covers almost a third of the circle, so that phenomenon does not apply. And the series does not converge to 1 either (for most starting values).

 

[Ibix]

Anybody have a good idea why?

Cos that's how it goes.

 

[Ibix]

And the series does not converge to 1 either (for most starting values).

It converges to about 0.739 in about fifteen iterations for all initial $\theta$, $0\leq\theta\leq 2\pi$, according to a quick spreadsheet I tried it in. In general it has to converge on $\theta=\cos\theta$, in whatever units you are using, doesn't it?

Edit: ...which can be multi-valued if you use very large units, but not with radians.

 

[andrewkirk]

In general it has to converge on θ=cosθ\theta=\cos\theta, in whatever units you are using, doesn't it?

Edit: ...which can be multi-valued if you use very large units, but not with radians.

Yes, it will converge to the root(s) of the equation $kx=\cos x$ where $k$ is the number of units per radian. For degrees, $k\approx 57.3$. The convergence point will differ according to the units. Graphically, it converges to the abscissa of the intersection point between the lines $y=kx$ and $y=\cos x$. The following image shows the intersection point for units of double-radian, radian, half-radian and degrees ($k=0.5,1,2,180/\pi$). The roots are 0.51, 0.74, 0.90 and 0.9998 respectively.

To work out the smallest units at which there will be multiple possible convergence points, we seek the tangent to the cosine curve that passes through the origin, with the point of touching being in the first quadrant. The tangency condition gives us equations $kx=\cos x$ and $k= -\sin x$, with the second of those equating the gradients. From the first we get $k=\frac{\cos x}x$. Substituting that into the second gives $\frac{\cos x}x+\sin x=0$. Numerically seeking a solution for that in the interval $(\pi,2\pi)$ yields the root $r\approx 6.12$. For units of $r$ radians each, there will be two possible convergence points, and I expect the point wto which it converges will depend on the initial value given to the calculator. For units larger than that, there will be at least three possible convergence points.

Here's a picture of the tangent.