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How can I calculate $\alpha$, without using a calculator?
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$enter image description here

I know $x = -\frac{1}{4} \implies y= \frac{\sqrt{15}}{4}, $ now how can I calculate $$\arccos\left(-\frac{1}{4}\right) = \alpha,\quad \arcsin\left(\frac{\sqrt{15}}{4}\right) = (180° - \alpha),$$ without using a calculator? How did the Greeks to calculate the angle?

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    $\begingroup$ Both arccos and arcsin have a taylor series expansion, so if it is only about approximating $\alpha$, we could use that. For the question about how the Greeks did that, I have absolutely no clue. Maybe searching around constructible numbers could give some hints about that? $\endgroup$ – N.Bach Nov 19 '17 at 22:25
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    $\begingroup$ Why do you think the Greeks could do that? $\endgroup$ – Christian Blatter Nov 23 '17 at 8:01
  • $\begingroup$ Do you suppose you have the value of $\pi$ or $pi/180$ available to convert degrees to radians? This would enable you to use the Taylor series for final refinements. $\endgroup$ – LutzL Nov 23 '17 at 12:24
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    $\begingroup$ For example, a sextant could do this without trigonometry. While Eratosthenes did not have a sextant, one could imagine using something as simple as a (large) protractor to estimate the angle. $\endgroup$ – rogerl Nov 23 '17 at 13:46
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    $\begingroup$ Relevant YouTube video. $\endgroup$ – ja72 Nov 23 '17 at 18:12

13 Answers 13

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You can read the binary digits of $\arccos(x)/\pi$ off the signs of $2\cos(2^kx)$, which is an easy to compute sequence defined recursively with $x_{n+1} = x_n^2-2$.

More precisely, you put a $1$ digit when the product of the signs so far is negative, and a $0$ otherwise :

$\begin{matrix}x_0 & -1/2 & - & - \\ x_1 &-7/4 & - & + \\ x_2 & 17/16 & + & + \\ x_3 & -223/256 & - & - \end{matrix}$

Now this starts getting hard because squareing $3$ digits number is a lot of hard work, so let me roughly approximate the fractions with $2$ digit numerators and denominators.

$\begin{matrix} -23/25 & & & \le x_3 \le & & & -11/13 & - & - \\ -11/8 & \le & -217/169 & \le x_4 \le & -721/625 & \le& -8/7 & - & + \\ -34/49 & \le & -34/49 & \le x_5 \le & -7/64 & \le & -7/64 & - & - \\ -2 & \le & -8143/4096 & \le x_6 \le & -3646/2401 & \le & -36/25 & - & + \\ 4/63 & \le & 46/625 & \le x_7 \le & 2 & \le & 2 & + & + \\ \end{matrix}$

And now this is too imprecise to continue.

So far I got the cumulative sign sequence $(-,+,+,-,+,-,+,+)$ and so the angle is between $(2^{-1}+2^{-4}+2^{-6})\pi$ and $(2^{-1}+2^{-4}+2^{-6}+2^{-8})\pi$

In degrees you replace $\pi$ with $180$, so those are $104.06\ldots$ and $104.77\ldots$


The recurrence follows from the addition formula : $2\cos(2x) = 2\cos^2(x)-2\sin^2(x) = 4\cos^2(x)-2 = (2\cos(x))^2-2$

Suppose you call $a_n \in [0 ; \pi]$ the angle whose cosine is $2x_n$. If $x_n\ge 0$ then $a_n \in [0 ; \pi/2] $ and then $a_{n+1} = 2a_n$, so the binary digits of $a_n/\pi$ are $.0$ followed with the binary digits of $a_{n+1}/\pi$
If $x_n \le 0$ then $a_n \in [\pi/2 ; \pi]$ and then $a_{n+1} = 2\pi-2a_n$, so the binary digits of $a_n/\pi$ are $.1$ followed with the inverted binary digits of $a_{n+1}/\pi$

Thus $a_{n+1} = \pm 2 a_n \mod {2\pi}$, and by induction, $a_n = \pm 2^n a_0 \pmod {2\pi}$ where the sign depends on the parity of the number of negative $x_k$ encountered for $0 \le k < n$. The $n$th digit is $0$ if and only if $2^n a_0 \in [0 ; \pi] \pmod {2\pi}$, which means $\pm a_n \in [0;\pi] \pmod {2\pi}$ with the same sign. But since $a_n \in [0;\pi]$, the digit is $0$ if the sign was $+$ and it is $1$ is the sign was $-$.

And so the $n$th binary digit correspond to the parity of the number of negative cosines encountered for $0 \le k < n$.

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    $\begingroup$ please follow me in my movement to replace $\cos$ with $\operatorname{ttcos}$ (short for two times cos) in all the textbooks everywhere. $\endgroup$ – mercio Nov 23 '17 at 13:53
  • $\begingroup$ +1 for the approach with the binary, but surely you are not serious about this ttcos? $\endgroup$ – Nick Pavlov Nov 23 '17 at 19:52
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    $\begingroup$ I like this answer a lot and just upvoted it. But, then, I'm fairly comfortable with the idea of dynamical conjugacy, which I think you're using here to relate $2\cos(2x)$ and $x^2-2$. I wonder if a little elaboration might help? $\endgroup$ – Mark McClure Nov 24 '17 at 1:38
  • $\begingroup$ @mark Since mercio did not feel obliged to add any details, why don't you provide an answer highlighting the theory. I get the gist of the argument, but I'm really out in the cold as far as learning anything. $\endgroup$ – CopyPasteIt Dec 2 '17 at 1:14
  • $\begingroup$ @MikeMathMan I actually think a small edit to this answer would probably do it. I'd totally do it, if I wasn't busy packing to take the kids to Disney World right now. :) If mercio doesn't flesh it out a bit, perhaps I will in a week or so. $\endgroup$ – Mark McClure Dec 2 '17 at 1:22
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Now using a 'linear-crunch' interpolation method with the 'Hipparchus Trig Table Simulation'. Also - how about just interpolating on your favorite $30°$ right triangle!

The ancient Greeks did not use the concept of a function, nor did they use Cartesian Coordinates for the Euclidean Plane, but that did not stop them from making great advances in science.

Hipparchus was a Greek astronomer, and is now called the "the father of trigonometry"; he was the first to tabulate the corresponding values of arc and chord for a series of angles. Searching the web, we can describe his contribution with this high-level (modern) summary:

Hipparchus created trigonometry tables by inscribing a 48 sided regular polygon into the unit circle, allowing scientists to use linear interpolation to approximate angles.

enter image description here


The 48-gon has central angles of $7.5°$ ($\frac{30°}{4}$) with $1,080$ diagonals (chords); see Wolfram.

So, given the OP's question Hipparchus would look as the data in his table to find the corresponding diagonals and then approximate the angle.

His works/tables have been lost so we can't use them here.


I thought it would be fun to create a 48-gon '$\mathbb R \times \mathbb R$ coordinate table' and to solve the OP's problem using linear interpolation. I used google sheets to create a (really accurate) table and looked at the data to find that the angle $\alpha$ was between 97.5° and 105°. I approximated the angle using "lazy $x$ only" interpolation and got $\alpha \approx 104.4844°$.

Here is a screenshot:

enter image description here


Here we interpolated using just the x-coordinate. This approach gave a pretty good approximation since the 'meat of the movement' was in the $x$ coordinate.

The coordinate $(-.25, \frac{\sqrt{15}}{4})$ lies between two points of the 48-gon and you can calculate the lengths of the two corresponding 'division' segments. In the 'linear $x$' interpolation, the 'interpolation factor' was $0.9312585$, but when you combine/crunch both the $x$ and $y$ coordinates (see below) you get a factor of $0.9302951$, and that gives you a very precise approximation of

$\alpha \approx 104.4772°$

compared to the true value of $\alpha = 104.4775\ldots °$.

If you want to see the (new and improved) spreadsheet interpolation formulas in action, click on this; here is the linear-crunch interpolation math:

Let $C_0$, $C_1$ be adjacent vertices of the 48-gon inscribed in the unit circle of $\mathbb R^2$, with corresponding angles of $\alpha_0$ and $\alpha_1$ with $\alpha_1 - \alpha_0 = 7.5°$ and let $(x,y)$ be on the unit circle and between $C_0$ and $C_1$.

Let $S_0$ be the length of the chord line segment joining $C_0$ to $(x,y)$.
Let $S_1$ be the length of the chord line segment joining $C_1$ to $(x,y)$.
Let $S = S_0 + S_1$
Let $R_0 = \frac{S_0}{S}$
Let $R_1 = \frac{S_1}{S}$

Then approximate the angle $\alpha$ for $(x,y)$ with this interpolation:

$\alpha \approx R_1 \, \alpha_0 + R_0 \, \alpha_1$

It should not be surprising that to calculate $arccos$ ($x$ coordinate) using a trig table, better results are obtained by looking for $arcsin$ ($y$ coordinate) at the same time, using circle geometry.

Notice that this 'linear crunch' interpolation gives the exact answer when $S_0 = S_1$. But the angle we are looking for is around $15°$ away from the axis - or around $1/2$ of $30°$. So we apply this general math to the first quadrant:

$C_0 = (1,0) \text{ marking } 0°$

$C_1 = (\frac{\sqrt 3}{2},\frac{1}{2}) \text{ marking } 30°$

$(x,y) = (\frac{\sqrt15}{4},\frac{1}{4})$

Here are the calcs:

$S_0 = 0.2520086$
$S_1 = 0.2700908$
$S = 0.5220993$
$R_0 = 0.4826832$
$R_1 = 0.5173168$

$\alpha \approx R_1 \, 0° + R_0 \, 30° = 14.4804962°$

I wonder if ancient Greek astronomers, like Ptolemy, used similar linear-crunch interpolation techniques, calculating square roots in their angle approximation algorithm. Most likely they found 'good enough' approximations to match with their observations and make predictions.


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I have an idea. it is known that the ancient Greeks were able to calculate the bisect an angle with compass and straightedge or ruler. Through this method you can find an optical approximation to Punt $-\frac{1}{4}$ by interpolating:

enter image description here

In this way you can easily calculate by hand a good approximation to the angle $\beta$

\begin{align}\beta&\approx 90\left[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + \frac{1}{32} - \frac{1}{64}\right]\\ &= 90\left[\left(\frac12+\frac14+\frac18 +\frac1{32}\right)-\left(\frac1{16} + \frac1{64}\right)\right]\\ &=90\left[\frac{29}{32} - \frac{5}{64} \right]\\ &=90\left[\frac{53}{64}\right] \\ &=\quad\frac{4770}{64}\\ &=\quad 74,53^° \end{align}

$$\alpha \approx (180^° - 74,53^°) = 105.46^°$$


I do not think that this simple method has been unknown to the Greeks.

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We present a method utilizing elementary trigonometric identities, small polynomials, geometric series, and the Archimedean property. Of course, Babylonian representation of numbers and square root algorithm will be used for computation. From the angle sum formulas we can derive $$\sin3x=3\sin x-4\sin^3x \label{a}\tag{1}$$ Or equivalently $$\sin x=3\sin\frac x3-4\sin^3\frac x3 \label{b}\tag{2}$$ From equation $(\ref{b})$ we derive $$\begin{align}\sin x+\frac16\sin^3x & =3\sin\frac x3-4\sin^3\frac x3+\frac16\left(3\sin\frac x3-4\sin^3\frac x3\right)^3 \label{c}\tag{3}\\ & =3\left(\sin\frac x3+\frac16\sin^3\frac x3\right)-\frac16\left[\left(9-8\sin^2\frac x3\right)^2+27\right]\sin^5\frac x3\end{align}$$ From equation $(\ref{b})$, given that $x$ is a first-quadrant angle so $0\le x\le\frac{\pi}2$, $$2\sin\frac x3=\sin\frac x3\left(3-4\sin^2\frac{\pi}6\right)\le\sin\frac x3\left(3-4\sin^2\frac x3\right)=\sin x\le3\sin\frac x3 \label{d}\tag{4}$$ It follows easily by mathematical induction that $$\frac1{3^n}\sin x\le\sin\frac x{3^n}\le\frac1{2^n}\sin x \label{e}\tag{5}$$ Also having completed the square, $$28\le\left(9-8\sin^2\frac x3\right)^2+27\le108 \label{f}\tag{6}$$ Substituting $\frac x{3^n}$ for $x$ in equation $(\ref{c})$ and multiplying by $3^n$ and using inequalities $(\ref{e})$ and $(\ref{f})$, we arrive at $$\begin{align}-18\frac{3^n}{2^{5n+5}}\sin^5 x & \le-18\cdot3^n\sin^5\frac x{3^{n+1}} \label{g}\tag{7}\\ & \le3^n\left(\sin\frac x{3^n}+\frac16\sin^3\frac x{3^n}-\frac x{3^n}\right)-3^{n+1}\left(\sin\frac x{3^{n+1}}+\frac16\sin^3\frac x{3^{n+1}}-\frac x{3^{n+1}}\right) \\ & \le-\frac{14}33^n\sin^5\frac x{3^{n+1}}\le-14\frac{3^n}{3^{5n+6}}\sin^5x\end{align}$$ Summing inequalities $(\ref{g})$ from $n=0$ to $n=N-1$, noting that the outer expressions are geometric series and the middle expression is a telescoping series, $$\begin{align}-\frac{18}{29}\left[1-\left(\frac{3}{32}\right)^{N}\right]\sin^5 x &=-\frac9{16}\left[\frac{1-\left(\frac3{32}\right)^{N}}{1-\frac3{32}}\right]\sin^5x \label{h}\tag{8}\\ & \le\sin x+\frac16\sin^3x-x-3^N\left(\sin\frac x{3^N}+\frac16\sin^3\frac x{3^N}-\frac x{3^N}\right) \\ & \le-\frac{14}{729}\left[\frac{1-\left(\frac1{81}\right)^N}{1-\frac1{81}}\right]\sin^5x=-\frac7{360}\left[1-\left(\frac1{81}\right)^N\right]\sin^5x\end{align}$$ Rearrange a bit and we have $$\begin{align}-\frac{18}{29}\sin^5 x & +\frac{18}{29}\left(\frac{3}{32}\right)^{N}\sin^5 x+3^N\left(\sin\frac x{3^N}+\frac16\sin^3\frac x{3^N}-\frac x{3^N}\right) \label{i}\tag{9}\\ & \le\sin x+\frac16\sin^3x-x \\ & \le-\frac7{360}\sin^5x+\frac7{360}\left(\frac1{81}\right)^N\sin^5x+3^N\left(\sin\frac x{3^N}+\frac16\sin^3\frac x{3^N}-\frac x{3^N}\right)\end{align}$$ Recall that a simple construction shows that $$\cos\theta\le\frac{\sin\theta}{\theta}\le1 \label{j}\tag{10}$$ Actually in the above link I had arrived at $\sin\theta\cos\theta<\theta$, but on multiplying by $2$ we get $\sin2\theta<2\theta$ so that $\sin\theta<\theta$, which is the second of inequalities $(\ref{j})$. From this we can see that $$\begin{align}3^N\left(\sin\frac x{3^N}-\frac x{3^N}\right) & =x\left(\frac{\sin\frac x{3^N}}{\frac x{3^N}}-1\right)\ge x\left(\cos\frac x{3^N}-1\right) \label{k}\tag{11}\\ & =-2x\sin^2\frac x{2\cdot3^N}\ge-\frac{2x}{2^{2N}}\sin^2\frac x2\end{align}$$ Where we have used $\cos\theta=1-2\sin^2\frac{\theta}2$ and inequality $(\ref{e})$. This simplifies the first of inequalities $(\ref{i})$ to $$-\frac{18}{29}\sin^5 x-\frac2{2^{2N}}\sin^2\frac x2\le\sin x+\frac16\sin^3x-x \label{l}\tag{12}$$ Meanwhile, working from inequalities $(\ref{j})$ again, $$3^N\left(\sin\frac x{3^N}-\frac x{3^N}\right)=x\left(\frac{\sin\frac x{3^N}}{\frac x{3^N}}-1\right)\le0 \label{m}\tag{13}$$ This, along with inequality $(\ref{e})$ simplifies the second of inequalities $(\ref{i})$ to $$\sin x+\frac16\sin^3x-x\le-\frac7{360}\sin^5x+\frac7{360}\left(\frac1{81}\right)^N\sin^5x+\frac1{6\cdot3^{2N}}\sin^3x \label{n}\tag{14}$$ We can make more progress via the Archimedean property: if for some $0\le x\le\frac{\pi}2$ $$\sin x+\frac16\sin^3x-x<-\frac{18}{29}\sin^5 x \label{o}\tag{15}$$ Then $$u=-\frac{18}{29}\sin^5 x-\left(\sin x+\frac16\sin^3x-x\right)>0 \label{p}\tag{16}$$ From the Archimedean property of the real numbers there exists some positive integer $M$ such that $uM>2\sin^2\frac x2$. Since $4^M>M$ for any positive integer $M$, $2^{2M}u>2\sin^2\frac x2$ so $$u=-\frac{18}{29}\sin^5 x-\left(\sin x+\frac16\sin^3x-x\right)>\frac2{2^{2M}}\sin^2\frac x2 \label{q}\tag{17}$$ Then $$-\frac{18}{29}\sin^5 x-\frac2{2^{2M}}\sin^2\frac x2>\sin x+\frac16\sin^3x-x \label{r}\tag{18}$$ A contradiction to inequality $(\ref{l})$, so there is no $0\le x\le\frac{\pi}2$ satisfying $(\ref{o})$ so $$-\frac{18}{29}\sin^5 x\le\sin x+\frac16\sin^3x-x \label{s}\tag{19}$$ Similarly we can simplify inequality $(\ref{n})$ to $$\sin x+\frac16\sin^3x-x\le-\frac7{360}\sin^5x \label{t}\tag{20}$$ From inequalities $(\ref{s})$ and $(\ref{t})$ we finally get what we were aiming at: $$\sin x+\frac16\sin^3x+\frac7{360}\sin^5x\le x\le\sin x+\frac16\sin^3x+\frac{18}{29}\sin^5 x \label{u}\tag{21}$$ EDIT: There is some room for improvement here in that, if the original angle $x=\frac{\theta}{2^n}$ were assumed $0<x<\frac{\pi}8$, then the right of inequalities $(\ref{e})$ would read $$\sin\frac x{3^n} \le \left(\frac2{2+\sqrt2+\sqrt6}\right)^n\sin x$$ This would improve the right of inequalities $(\ref{u})$ to $$x\le\sin x+\frac16\sin^3x+\frac{18}{\left(\frac{2+\sqrt2+\sqrt6}2\right)^5-3}\sin^5 x$$ For a factor of better than $7$ improvement in the error estimate for the upper bound. Also this assumption would improve the left of inequalities $(\ref{f})$ to $$\left(5+\sqrt2+\sqrt6\right)^2+27\le\left(9-8\sin^2\frac x3\right)^2+27$$ Which improves the left of inequalities $(\ref{u})$ to $$\sin x+\frac16\sin^3x+\frac{\left(5+\sqrt2+\sqrt6\right)^2+27}{1440}\sin^5x\le x$$ That takes the fifth-order term from about $0.019$ to about $0.073$, much closer to the arcsin series value of $\frac3{40}=0.075$.
EDIT: Argh, these improvements mean I can save a bisection and still get as good an answer. They also limit the exponential growth in error terms as more terms of the arcsin series are added. Since $$\frac3{41}<\frac{\left(5+\sqrt2+\sqrt6\right)^2+27}{1440}$$ and $$\frac{18}{\left(\frac{2+\sqrt2+\sqrt6}2\right)^5-3} < \frac7{83}$$ We can modify inequalities $(\ref{u})$ to $$\sin x+\frac16\sin^3x+\frac3{41}\sin^5x\le x\le\sin x+\frac16\sin^3x+\frac{7}{83}\sin^5 x \label{au}\tag{21a}$$ With this formula in hand, which might be in the ballpark of what Archimedes could derive, we can follow his computation of $\pi$ by bisecting angles: $$2\sin\frac{\theta}2=\frac{\sin\theta}{\cos\frac{\theta}2} \label{v}\tag{22}$$ So it follows by mathematical induction that $$2^n\sin\frac{\theta}{2^n}=\frac{\sin\theta}{\prod\limits_{k=1}^n\cos\frac{\theta}{2^k}} \label{w}\tag{23}$$ Also we have $$\cos\frac{\theta}2=\sqrt{\frac{1+\cos\theta}2} \label{x}\tag{24}$$ So we can find the cosines we need for equation $(\ref{w})$ from equation $(\ref{x})$. When we get $2^n\sin\frac x{2^n}$ we can use inequalities $(\ref{au})$ in the form $$2^n\sin\frac x{2^n}+\frac162^n\sin^3\frac x{2^n}+\frac3{41}2^n\sin^5\frac x{2^n}\le x\le2^n\sin\frac x{2^n}+\frac162^n\sin^3\frac x{2^n}+\frac{7}{83}2^n\sin^5\frac x{2^n} \label{y}\tag{25}$$ This enables us to use angle bisection to get approximations of $\sin^{-1}x$ with fourth order error, compared to the second order (in $\theta/2^n$) error of the original Archimedes $\pi$ computation.
The Babylonians seem to have been aware of the iteration $$x_{n+1}=\frac{x_n^2+D}{2x_n} \label{z}\tag{26}$$ to compute the $\sqrt{D}$, but this requires a long division at every iteration. Less laborious is the similar iteration for $\frac1{\sqrt D}$: $$x_{n+1}=x_n\left(\frac32-\frac12Dx_n^2\right) \label{aa}\tag{27}$$ This is another Newton-Raphson iteration, but the Babylonians could probably have proved it by letting $x_n=\frac1{\sqrt D}+\epsilon$ and showing that $$x_{n+1}=\frac1{\sqrt D}-\frac32\epsilon^2\sqrt D-\frac12\epsilon^3D \label{ab}\tag{28}$$ To get a flavor for how the Babylonians might have calculated, let's try to get $C_1=\cos\frac {\theta}{2^0}$ from $C_0=\cos \frac{\theta}{2^0}$ in the table below. First we compute (in sexagesimal of course) $$\begin{align}\frac1{x_0^2} & =\frac{1+C_0}2=\frac{\text{[01]}+\text{[00].[58][05][41][06][02][30]}}{\text{[02]}} \\ & =\frac{\text{[01].[58][05][41][06][02][30]}}{\text{[02]}}=\text{[00].[59][02][50][33][01][15]}\end{align}$$ rounding down as necessary. Then we might calculate the reciprocal of the first two digits of this to get $$x_0^2=\frac{\text{[01]}}{\text{[00].[59][02]}}=\text{[01].[00][59]}$$ This is a little big, so let's divide by $4$ to get $$\frac{x_0^2}4=\frac{\text{[01].[00][59]}}{\text{[04]}}=\text{[00].[15][15]}$$ Now we look this up in a table of squares: $\text{[30]}^2=\text{[15][00]}$, $\text{[31]}^2=\text{[16][01]}$ so $\frac{x_0}2=\text{[00].[30]}$ is the closest, and $x_0=\text{[02]}\times\text{[00].[30]}=\text{[01].[00]}$. We also need $$\frac12D=\frac{\text{[00].[59][02][50][33][01][15]}}{[02]}=\text{[00].[29][31][25][16][30][37]}$$ Then $$\frac12Dx_0=\text{[00].[29][31]}\times\text{[01].[00]}=\text{[00].[29][31]}$$ $$\frac12Dx_0^2=\text{[00].[29][31]}\times\text{[01].[00]}=\text{[00].[29][31]}$$ $$\frac32-\frac12Dx_0^2=\text{[01].[30]}-\text{[00].[29][31]}={\text[01].[00][29]}$$ $$x_1=x_0\left(\frac32-\frac12Dx_0^2\right)=\text{[01].[00]}\times\text[01].[00][29]=\text[01].[00][29]$$ $$\frac12Dx_1=\text{[00].[29][31][25][16]}\times\text{[01].[00][29]}=\text{[00].[29][45][41][27]}$$ $$\frac12Dx_1^2=\text{[00].[29][45][41][27]}\times\text{[01].[00][29]}=\text{[00].[30][00][04][32]}$$ $$\frac32-\frac12Dx_1^2=\text{[01].[30]}-\text{[00].[30][00][04][32]}=\text{[00].[59][59][55][28]}$$ $$x_2=x_1\left(\frac32-\frac12Dx_1^2\right)=\text{[01].[00][29]}\times\text{[00].[59][59][55][28]}=\text{[01].[00][28][55][25]}$$ $$\frac12Dx_2=\text{[00].[29][31][25][16][30][37]}\times\text{[01].[00][28][55][25]}=\text{[00].[29][45][39][12][24][34][57]}$$ $$\frac12Dx_2^2=\text{[00].[29][45][39][12][24][34][57]}\times\text{[01].[00][28][55][25]}=\text{[00].[29][59][59][59][57][20][08]}$$ $$\frac32-\frac12Dx_2^2=\text{[01].[30]}-\text{[00].[29][59][59][59][57][20][08]}=\text{[01].[00][00][00][00][02][39][52]}$$ $$\begin{align}\frac12C_1 & =\frac12Dx_2\left(\frac32-\frac12Dx_2^2\right) \\ & =\text{[00].[29][45][39][12][24][34][57]}\times\text{[01].[00][00][00][00][02][39][52]}\\ & =\text{[00].[29][45][39][12][25][54][14]}\end{align}$$ $$C_1=2\frac12C_1=\text{[02]}\times\text{[00].[29][45][39][12][25][54][14]}=\text{[00].[59][31][18][24][51][48]}$$ $$\begin{array}{c|c|c} \text{Symbol} & \text{Low} & \text{High} \\ \hline S_{0} & \text{[00].[15][00][00][00][00][00]} & \text{[00].[15][00][00][00][00][00]}\\ C_{0} & \text{[00].[58][05][41][06][02][30]} & \text{[00].[58][05][41][06][02][31]}\\ C_{1} & \text{[00].[59][31][18][24][51][48]} & \text{[00].[59][31][18][24][51][50]}\\ C_{2} & \text{[00].[59][52][49][10][26][11]} & \text{[00].[59][52][49][10][26][12]}\\ C_{1}C_{2} & \text{[00].[59][24][11][01][19][42]} & \text{[00].[59][24][11][01][19][46]}\\ C_{3} & \text{[00].[59][58][12][15][59][49]} & \text{[00].[59][58][12][15][59][50]}\\ C_{4} & \text{[00].[59][59][33][03][53][54]} & \text{[00].[59][59][33][03][53][55]}\\ C_{3}C_{4} & \text{[00].[59][57][45][20][42][04]} & \text{[00].[59][57][45][20][42][07]}\\ C_{1}C_{2}C_{3}C_{4} & \text{[00].[59][21][57][42][24][36]} & \text{[00].[59][21][57][42][24][44]}\\ C_{5} & \text{[00].[59][59][53][15][58][05]} & \text{[00].[59][59][53][15][58][07]}\\ C_{1}C_{2}C_{3}C_{4}C_{5} & \text{[00].[59][21][51][02][38][49]} & \text{[00].[59][21][51][02][39][00]}\\ 2^{5}S_{5} & \text{[00].[15][09][38][22][04][33]} & \text{[00].[15][09][38][22][04][37]}\\ 2^{10}S_{5}^2 & \text{[00].[03][49][50][43][57][25]} & \text{[00].[03][49][50][43][57][28]}\\ 2^{20}S_{5}^4 & \text{[00].[00][14][40][28][58][26]} & \text{[00].[00][14][40][28][58][27]}\\ \frac162^{5}S_{5}^3 & \text{[00].[00][00][00][34][01][45]} & \text{[00].[00][00][00][34][01][46]}\\ \frac3{41}2^{5}S_{5}^5 & \text{[00].[00][00][00][00][00][03]} & \\ \frac{7}{83}2^{5}S_{5}^5 & & \text{[00].[00][00][00][00][00][04]}\\ \theta(\text{rad}) & \text{[00].[15][09][38][56][06][21]} & \text{[00].[15][09][38][56][06][27]}\\ \theta(°) & \text{[14].[28][39][02][37][52][13]} & \text{[14].[28][39][02][37][58][53]} \end{array}$$ Some explanation of the above table: $S_n=\sin\frac{\theta}{2^n}$, $\\C_n=\cos\frac{\theta}{2^n}$, $C_1C_2$ is the product of cosines, the Low column is the lower bound of the computed value (everything rounded down) and the High column is the upper bound. This is to conform with the Archimedes $\pi$ computation where he proved lower and upper bounds for $\pi$. Here is the program that calculates $\sin^{-1}\left(-\frac14\right)$ in $6$-digit fixed point sexagesimal arithmetic:

! archpi2.f90 -- Computes asin(1/4) using Archimedean method of bisection
module funcs
! Need 64-bit integers for 10 digit accuracy
   use ISO_FORTRAN_ENV, only: wp => REAL64, ik => INT64
   implicit none
integer, parameter :: ikd = 16
   contains
! Quick and dirty function to create sexagesimal strings from numbers.
! Input is 60**ndigits*number, 0 <= number < 60
! Output is sexagesimal string
      function printf(n,digits)
         integer(ik) n
         integer digits
         character(4*(digits+1)+1) printf
         integer i
         write(printf,'(*(a,i0.2,a))') &
           ('[',mod(n/60_ik**i,60_ik),']'//trim(merge('.',' ',i==digits)),i=digits,0,-1)
      end function printf
end module funcs

program archpi2
   use funcs
   implicit none
! Values suitable for asin(0.25)
   integer, parameter :: digits = 6      ! Sexagesimal digits working precision
   integer, parameter :: depth = 5       ! Number of bisections
   real(wp), parameter :: S0 = 1.0_wp/4  ! sin(theta)
! Values suitable for pi/6
!   integer, parameter :: digits = 7
!   integer, parameter :: depth = 6
!   real(wp), parameter :: S0 = 1.0_wp/2
! Scale factor for fixed-point sexagesimal arithmetic
   integer(ik), parameter :: factor = 60_ik**digits
! Holds info for cosines
   type node
      character(:), allocatable :: label ! What is stored
      integer(ik) vL,vH                  ! Low and high fixed-point approximations
   end type node
   type(node) data(depth)                ! Cos and cosine product data
   integer last                          ! Last element of data active
   integer current                       ! Current element of data
   integer i
! Intermediated data in fixed-point sexagesimal format
   integer(ik) S0L, S0H            ! sin(theta), lower and upper bounds
   integer(ik) C0L, C0H            ! cos(theta), lower and upper bounds
   integer(ik) SnL, SnH            ! 2**depth*sin(theta/2**depth),
                                   ! lower and upper bounds
   integer(ik) SnL2, SnH2          ! SnL**2 and SnH**2
   integer(ik) SnL3, SnH3          ! 2**depth*sin(theta/2***depth)**3/6,
                                   ! lower and upper bounds
   integer(ik) SnL4, SnH4          ! Snl**4 and SnH**4
   integer(ik) SnL5, SnH5          ! 7/360*2**depth*sin(theta/2**depth), lower and
                                   ! 18/29*2**depth*sin(theta/2**depth), upper
   integer(ik) thetaL, thetaH      ! Theta(rad), lower and upper bounds
   integer(ik) thetaLD, thetaHD    ! Theta(°), lower and upper bounds

   write(*,'(a)') '$$\begin{array}{c|c|c}'
   write(*,'(a)') '\text{Symbol} & \text{Low} & \text{High} \\'
! Convert sin(theta) to sexagesimal fixed point format
   S0L = floor(S0*factor,ik)
   S0H = ceiling(S0*factor,ik)
   write(*,'(*(g0))') '\hline S_{0} & \text{',printf(S0L,digits),'} & \text{',printf(S0H,digits),'}\\'
! Get cos(theta)
   C0L = floor(sqrt((1.0_wp*factor)*factor-(1.0_wp*S0H)*S0H),ik)
   C0H = ceiling(sqrt((1.0_wp*factor)*factor-(1.0_wp*S0L)*S0H),ik)
   write(*,'(*(g0))') 'C_{0} & \text{',printf(C0L,digits),'} & \text{',printf(C0H,digits),'}\\'
! Compute cos(theta/2**i) and their products
   last = 0
   do i = 1, depth
      last = last+1
      C0L = (factor+C0L)/2
      C0H = (factor+C0H+1)/2
      C0L = floor(sqrt((1.0_wp*C0L)*factor),ik)
      C0H = ceiling(sqrt((1.0_wp*C0H)*factor),ik)
      allocate(character(7) :: data(last)%label)
      write(data(last)%label,'(*(g0))') 'C_{',i,'}'
      data(last)%label = trim(data(last)%label)
      data(last)%vL = C0L
      data(last)%vH = C0H
      write(*,'(*(g0))') data(last)%label,' & \text{',printf(data(last)%vL,digits), &
         '} & \text{',printf(data(last)%vH,digits),'}\\'
      current = i
! Consolidate tree of cosine products
      do while(mod(current,2)==0)
         last = last-1
         current = current/2
         data(last)%label = data(last)%label // data(last+1)%label
         deallocate(data(last+1)%label)
         data(last)%vL = floor((1.0_wp*data(last)%vL)*data(last+1)%vL/factor,ik)
         data(last)%vH = ceiling((1.0_wp*data(last)%vH)*data(last+1)%vH/factor,ik)
         write(*,'(*(g0))') data(last)%label,' & \text{',printf(data(last)%vL,digits), &
            '} & \text{',printf(data(last)%vH,digits),'}\\'
      end do
   end do
! Compute product of all cosines
   do last = last-1 , 1, -1
      data(last)%label = data(last)%label // data(last+1)%label
      deallocate(data(last+1)%label)
      data(last)%vL = floor((1.0_wp*data(last)%vL)*data(last+1)%vL/factor,ik)
      data(last)%vH = ceiling((1.0_wp*data(last)%vH)*data(last+1)%vH/factor,ik)
      write(*,'(*(g0))') data(last)%label,' & \text{',printf(data(last)%vL,digits), &
         '} & \text{',printf(data(last)%vH,digits),'}\\'
   end do
! Compute 2**depth*sin(theta/2**depth)
! Must divide by upper bound of cosine product to get lower bound of sin!
   SnL = floor((1.0_wp*S0L)*factor/data(1)%vH,ik)
   SnH = ceiling((1.0_wp*S0H)*factor/data(1)%vL,ik)
   write(*,'(*(g0))') '2^{',depth,'}S_{',depth,'} & \text{',printf(SnL,digits), &
      '} & \text{',printf(SnH,digits),'}\\'
! Get terms of the formula
   SnL2 = floor((1.0_wp*SnL)*SnL/factor,ik)
   SnH2 = ceiling((1.0_wp*SnH)*SnH/factor,ik)
   write(*,'(*(g0))') '2^{',2*depth,'}S_{',depth,'}^2 & \text{',printf(SnL2,digits), &
      '} & \text{',printf(SnH2,digits),'}\\'
   SnL3 = floor((1.0_wp*SnL2)*SnL/factor/6/2_ik**(2*depth),ik)
   SnH3 = ceiling((1.0_wp*SnH2)*SnH/factor/6/2_ik**(2*depth),ik)
   SnL4 = floor((1.0_wp*SnL2)*SnL2/factor,ik)
   SnH4 = ceiling((1.0_wp*SnH2)*SnH2/factor,ik)
   write(*,'(*(g0))') '2^{',4*depth,'}S_{',depth,'}^4 & \text{',printf(SnL4,digits), &
      '} & \text{',printf(SnH4,digits),'}\\'
   write(*,'(*(g0))') '\frac162^{',depth,'}S_{',depth,'}^3 & \text{',printf(SnL3,digits), &
      '} & \text{',printf(SnH3,digits),'}\\'
   SnL5 = floor((1.0_wp*SnL4)*SnL/factor*3/41/2_ik**(4*depth),ik)
   SnH5 = ceiling((1.0_wp*SnH4)*SnH/factor*7/83/2_ik**(4*depth),ik)
   write(*,'(*(g0))') '\frac3{41}2^{',depth,'}S_{',depth,'}^5 & \text{', &
      printf(SnL5,digits), '} & \\'
   write(*,'(*(g0))') '\frac{7}{83}2^{',depth,'}S_{',depth,'}^5 & ', &
      ' & \text{',printf(SnH5,digits),'}\\'
! Add up terms and print results
   thetaL = SnL+SnL3+SnL5
   thetaH = SnH+SnH3+SnH5
   write(*,'(*(g0))') '\theta(\text{rad}) & \text{',printf(thetaL,digits), &
      '} & \text{',printf(thetaH,digits),'}\\'
   thetaLD = floor((1.0_wp*thetaL)*factor/24429024476_ik*30,ik)
   thetaHD = ceiling((1.0_wp*thetaH)*factor/24429024474_ik*30,ik)
   write(*,'(*(g0))') '\theta(°) & \text{',printf(thetaLD,digits), &
      '} & \text{',printf(thetaHD,digits),'}'
   write(*,'(a)') '\end{array}$$'
end program archpi2

Oh, so how did I get $\theta$ in degrees? I had to compute $\frac{\pi}6=\sin^{-1}\frac12$ and then calculate $$\theta(°)=\theta(\text{rad})\frac{180°}{\pi\text{ rad}}=\frac{30\theta(\text{rad})}{\frac{\pi}6}$$. $$\begin{array}{c|c|c} \text{Symbol} & \text{Low} & \text{High} \\ \hline S_{0} & \text{[00].[30][00][00][00][00][00][00]} & \text{[00].[30][00][00][00][00][00][00]}\\ C_{0} & \text{[00].[51][57][41][29][13][58][58]} & \text{[00].[51][57][41][29][13][58][59]}\\ C_{1} & \text{[00].[57][57][19][58][42][31][20]} & \text{[00].[57][57][19][58][42][31][22]}\\ C_{2} & \text{[00].[59][29][12][05][24][12][15]} & \text{[00].[59][29][12][05][24][12][16]}\\ C_{1}C_{2} & \text{[00].[57][27][35][02][04][20][32]} & \text{[00].[57][27][35][02][04][20][36]}\\ C_{3} & \text{[00].[59][52][17][31][38][42][36]} & \text{[00].[59][52][17][31][38][42][38]}\\ C_{4} & \text{[00].[59][58][04][21][03][13][17]} & \text{[00].[59][58][04][21][03][13][19]}\\ C_{3}C_{4} & \text{[00].[59][50][22][07][33][20][25]} & \text{[00].[59][50][22][07][33][20][30]}\\ C_{1}C_{2}C_{3}C_{4} & \text{[00].[57][18][21][37][34][59][22]} & \text{[00].[57][18][21][37][34][59][32]}\\ C_{5} & \text{[00].[59][59][31][05][08][50][18]} & \text{[00].[59][59][31][05][08][50][20]}\\ C_{6} & \text{[00].[59][59][52][46][16][46][26]} & \text{[00].[59][59][52][46][16][46][28]}\\ C_{5}C_{6} & \text{[00].[59][59][23][51][29][05][44]} & \text{[00].[59][59][23][51][29][05][49]}\\ C_{1}C_{2}C_{3}C_{4}C_{5}C_{6} & \text{[00].[57][17][47][06][26][02][55]} & \text{[00].[57][17][47][06][26][03][11]}\\ 2^{6}S_{6} & \text{[00].[31][24][56][04][25][59][02]} & \text{[00].[31][24][56][04][25][59][12]}\\ 2^{12}S_{6}^2 & \text{[00].[16][26][56][18][18][48][04]} & \text{[00].[16][26][56][18][18][48][15]}\\ 2^{24}S_{6}^4 & \text{[00].[04][30][34][07][26][44][12]} & \text{[00].[04][30][34][07][26][44][19]}\\ \frac162^{6}S_{6}^3 & \text{[00].[00][00][01][15][41][46][59]} & \text{[00].[00][00][01][15][41][47][00]}\\ \frac3{41}2^{6}S_{6}^5 & \text{[00].[00][00][00][00][00][08][00]} & \\ \frac{7}{83}2^{6}S_{6}^5 & & \text{[00].[00][00][00][00][00][09][14]}\\ \frac{\pi}6 & \text{[00].[31][24][57][20][07][54][01]} & \text{[00].[31][24][57][20][07][55][26]} \end{array}$$ In decimals I get $104.4775121813486°\le\cos^{-1}\left(-\frac14\right)=\sin^{-1}\left(\frac14\right)+90°<104.4775121899220°$. Had I not gone to all the trouble of getting a third-order approximation to $\sin^{-1}x$, I would have had to perform many more bisections.

$\endgroup$
4
$\begingroup$

So you found that the angle is the argument of $-1+\sqrt{15}i$. As it is in the II. quadrant, you can remove $90°$ from it and get a point in the first quadrant, $$ α=90°+\arg((-1+\sqrt{15}+i)(-i))=\arg(\sqrt{15}+i). $$ Then by squaring (and removing positive common factors) the angle is doubled, $$ α=90°+\frac12\arg(7+\sqrt{15}i) $$ Now $7^2$ is close to $3⋅15$ which means this angle is close to $30°=\arg(\sqrt3+i)$. Computing the difference to that angle gives \begin{align} α&=105°+\frac12\arg((7+\sqrt{15}i)(\sqrt3-i))=105°+\frac12\arg(\sqrt{3}(\sqrt5+7)-(7-3\sqrt{5})i)\\ &\underline{=105°-\frac12\arg(\sqrt{3}(7\sqrt5+16)+i)}. \end{align}


As $7\sqrt5=\sqrt{16^2-11}\approx 16-\frac{11}{32}$ this means that at least the 32th power, most surely an earlier one, of $\sqrt{3}(7\sqrt5+16)+i$ crosses over the $30°$ line. The reduction from $105°$ is thus a little larger than $\frac{15°}{32}$, giving the angle $α$ around tending a bit below $$\underline{α⪅104.5°}.$$


To be a bit more precise, one would have to compute the exact powers. Using $\epsilon=(\sqrt{3}(7\sqrt5+16))^{-1}$, one gets \begin{array}{r|l} k&\frac12(\sqrt3-i)(1+ϵi)^k\\\hline 27& 1.00401311706-0.0313008181359i\\ 28& 1.00458405300-0.0129873299181i\\ 29& 1.00482094563+0.0053365723355i\\ 30& 1.00472360502+0.0236647955789i\\ 31& 1.00429195233+0.0419912433015i\\ 32& 1.00352601997+0.0603098175549i\\ \end{array} which gives the zero crossing interpolated at about $k=28+\frac23$ and the angle estimate as $$ \underline{α\simeq 105°-\frac{15°}{28+\frac23}=104.4767...°} $$

$\endgroup$
4
$\begingroup$

This answer makes use of the Small Angle Approximation for $\sin$ and radian measures for angles. The latter was not known to the Greeks. I'm unsure if the small angle approximation was known to them in some way.

Our approximation method has the advantage that it can very easily be done without a calculator, though the error term would not have been known until the classical period of Indian mathematics and astronomy

We seek an $\alpha$ such that $\cos(\alpha) = -\frac{1}{4}$. A quick investigation shows us that $\alpha$ lies in the second quadrant (i.e. $\frac{\pi}{2} < \alpha < \pi$). Let $\alpha = \frac{\pi}{2} + \beta$. Let's use the identity $\cos(\frac{\pi}{2} + \beta) = -\sin(\beta)$.

We then seek a $\beta$ such that $\sin(\beta) = \frac{1}{4}$. Making use of the linear approximation for $\sin$, we find $\beta \approx \frac{1}{4}$. Thus $\alpha \approx \frac{\pi}{2} + \frac{1}{4}$.

Converting to degrees gives $\alpha \approx 90^{\circ} + \frac{45}{\pi}^{\circ}$. We can use the approximation $\pi \approx \frac{22}{7}$ and long division to get

$$\frac{45}{\pi} \approx \frac{315}{22} = 14.3\overline{18}$$

Giving us the approximation $$\alpha \approx 104.32^{\circ}$$ This is not far from the true value

$$\alpha = 104.4775\ldots^{\circ}$$

$\endgroup$
3
$\begingroup$

Eratosthenes could have used his friend Archimedes's work on the quadrature of parabolic segments to obtain upper and lower bounds on the value of $\alpha - \tfrac{\pi}{2}$, or indeed any acute angle $\eta$, given its sine:

$$ \frac{8\sin\tfrac{1}{2}\eta - \sin\eta}{3} < \eta < \frac{4\tan\tfrac{1}{2}\eta + \sin\eta}{3} \quad \left(0 < \eta \leqslant \frac{\pi}{2}\right). $$

A modern proof is short, so we get it out of the way first. The result has the form $f(\eta) < 3\eta < g(\eta)$, where $f(0) = g(0) = 0$, so it is enough to prove $f'(\eta) < 3 < g'(\eta)$, for all $\eta$ such that $0 < \eta < \tfrac{\pi}{2}$. Taking the upper bound first, and writing $\cos\eta = y$, we have $0 \leqslant y < 1$, and: $$ g'(\eta) - 3 = \cos\eta + 2\sec^2\tfrac{1}{2}\eta - 3 = y + \frac{4}{1 + y} - 3 = \frac{1 - 2y + y^2}{1 + y} = \frac{(1 - y)^2}{1 + y} > 0. $$ As for the lower bound: $$ 3 - f'(\eta) = 3 - 4\cos\tfrac{1}{2}\eta + \cos\eta = 3 - 4\cos\tfrac{1}{2}\eta + 2\cos^2\tfrac{1}{2}\eta - 1 = 2\left(1 - \cos\tfrac{1}{2}\eta\right)^2 > 0, $$ and as this is true for all $\eta \ne 4n\pi$, we even have $f(\eta) < 3\eta$ for all $\eta > 0$, needing no upper bound. $\square$

Recalling the trigonometric identities \begin{align*} \tan\left(\tfrac{1}{2}\sin^{-1}x\right) & = \left(1 - \sqrt{1 - x^2}\right)/x && (|x| \leqslant 1, \ x \ne 0), \\ \sin\left(\tfrac{1}{2}\sin^{-1}x\right) & = \tfrac{1}{2}\left(\sqrt{1 + x} - \sqrt{1 - x}\right) && (|x| \leqslant 1), \end{align*} we can rewrite the theorem just proved as:

$$ \frac{4\left(\sqrt{1 + x} - \sqrt{1 - x}\right) - x}{3} < \sin^{-1}x < \frac{4\left(1 - \sqrt{1 - x^2}\right) + x^2}{3x} \quad (0 < x \leqslant 1). $$

In the case of present interest:

$$ \frac{8\sqrt{5} - 8\sqrt{3} - 1}{12} < \sin^{-1}\left(\frac{1}{4}\right) < \frac{64 - 16\sqrt{15} + 1}{12}. $$

In radians, this is $0.2526781133 < \eta < 0.2526888717$, or $\eta = 0.2526835 \pm .0000054$.

Converting to degrees, and adding $90^\circ$, we get $104.477383947^\circ < \alpha < 104.47800588^\circ$, or:

$$\alpha = 104.47770^\circ \pm 0.00031^\circ.$$

Converting to degrees, minutes and seconds of arc, we get $104^\circ28'38.6'' < \alpha < 104^\circ28'40.8''$, or:

$$\alpha = 104^\circ28'39.7'' \pm 0^\circ00'01.1''.$$

The conversion factor $180/\pi$ needed here is obtained by long division by an estimate of $\pi$, which can be obtained by exactly the same method. Rather than perform the laborious computation needed to obtain once and for all a sufficiently accurate value of $\pi$, we shall just do enough to give a general idea.

Even the crudest estimate using the theorem, given by substitution in the equation $\pi = 2\sin^{-1}1$, isn't bad: $(8\sqrt{2} - 2)/3 < \pi < 10/3$, or $\pi = 3.22 \pm 0.11$, but we do significantly better than this by substituting in $\pi = 6\sin^{-1}\tfrac{1}{2}$, using the half-angle identities recalled above, as well as this one: \begin{gather*} \cos\left(\tfrac{1}{2}\sin^{-1}x\right) = \tfrac{1}{2}\left(\sqrt{1 + x} + \sqrt{1 - x}\right) \quad (|x| \leqslant 1). \\ \tan\frac{\pi}{12} = 2 - \sqrt{3}, \quad \sin\frac{\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}, \quad \cos\frac{\pi}{12} = \frac{\sqrt{6} + \sqrt{2}}{4}, \\ 4\sqrt{6} - 4\sqrt{2} - 1 < \pi < 17 - 8\sqrt{3}, \\ \text{i.e. } 3.141104722 < \pi < 3.143593539, \\ \pi = 3.1423 \pm 0.0012. \end{gather*} Explicit algebraic Euclidean-constructible bounds remain tractable for hand computation even when we apply the half-angle formulae again, substituting in $\pi = 12\sin^{-1}\tfrac{\sqrt{6} - \sqrt{2}}{4}$: \begin{gather*} \tan\frac{\pi}{24} = \frac{1 - \cos\frac{\pi}{12}}{\sin\frac{\pi}{12}} = \frac{4\cos\frac{\pi}{12} - 2\cos\frac{\pi}{6} - 2} {2\sin\frac{\pi}{6}} = \sqrt{6} + \sqrt{2} - \sqrt{3} - 2, \\ \sin\frac{\pi}{24} = \frac{\tan\frac{\pi}{24}}{\sqrt{1 + \tan^2\frac{\pi}{24}}}, \\ \frac{\sqrt{6} + \sqrt{2} - \sqrt{3} - 2} {\sqrt{\sqrt{16} + 8\sqrt{3} - 10\sqrt{2} - 6\sqrt{6}}} - \sqrt{6} + \sqrt{4} < \pi < 17\sqrt{6} + 15\sqrt{2} - 16\sqrt{3} - 32, \\ \text{i.e. } 3.141561971 < \pi < 3.141716142, \\ \pi = 3.141639 \pm 0.000077. \end{gather*} This is the estimate of $\pi$ obtained by inscribing a regular icositetragon in a circle, and erecting major and minor parabolic segments on each of its $24$ sides. It considerably improves the estimate made by Archimedes using inscribed and circumscribed regular enneacontahexagons ($96$-sided polygons), while involving less computation; but it is still not as good as the approximation $\pi \bumpeq \tfrac{355}{113}$.

In this accelerated variant of Archimedes's construction (and also of Viète's formula, except that if we follow Archimedes we start with a hexagon instead of a square - although a square is also acceptable), we start with a regular $m$-gon inscribed in the unit circle in $\mathbb{C}$, with one vertex being $1$, and the next in anticlockwise order being $z_0 = x_0 + iy_0$, where the value of $x_0 = \cos(2\pi/m)$ is supposed to be already known. We have chosen $m = 6$, so $x_0 = \cos(\pi/3) = 1/2$. For $n \geqslant 0$, recursively create a $2^nm$-gon, by bisecting the arcs between the vertices of the preceding $2^{n-1}m$-gon when $n \ne 0$.

The usual method of computation, associated with Viète's formula, which is indeed well suited to the use of a calculator (see e.g. here), uses the recursive formula $2x_{n+1}=\sqrt{2 + 2x_n}$, and the area of the inscribed polygon, $A_n = 2^{n-1}my_n$. This is of course a strict lower bound for $\pi$. One can also get a strict upper bound for $\pi$ from the area of the circumscribed polygon, which is $A_n/x_n^2 = A_{n+1}/x_{n+1}$. What I noticed when working on the other question was that an arithmetic mean of these two areas,

$$ \frac{A_n + 2A_n/x_n^2}{3} = \frac{A_n}{3}\left(1 + \frac{4}{1 + x_n}\right), $$

gives a much better approximation to $\pi$ than either $A_n$ or $A_n/x_n^2$ separately, and it always seems to be an overestimate. I had already thought of using parabolic arcs for the present question (but had no idea how it would work out in detail), and I began to suspect a serendipitous connection between the two questions!

Because the upper bound given by the theorem is much easier to work with than the lower bound (although it is not quite as precise), my suggested method for computing $\pi$ is to follow the method for Viète's formula as usual, but instead of using $A_n$ as the estimate for $\pi$, use the formula highlighted above, which as we shall see is the same as the upper bound obtained from the theorem.

It took me embarrassingly long to see how simple it is to incorporate Archimedes's own statement of his quadrature result, completely unmodified, into the computational procedure for estimating $\pi$ (or, more generally, for estimating the area of a circular sector), thus computing the lower bound, as well as the upper one. See Addendum 1 for an example.

Although the square roots occurring in the formulae are tractable for hand computation by any of the usual methods, one might consider evaluating them using the binomial series: \begin{gather*} 1 - \sqrt{1 - x} = \tfrac{1}{2}\sum_{k=0}^\infty a_kx^{k+1} \quad (|x| < 1), \qquad a_k = \frac{1\cdot3\cdots(2k-1)}{4\cdot6\cdots(2k+2)}, \\ a_0 = 1, \ a_1 = \tfrac{1}{4}, \ a_2 = \tfrac{1}{8}, \ a_3 = \tfrac{5}{64}, \ a_4 = \tfrac{7}{128}, \ a_5 = \tfrac{21}{512}, \ a_6 = \tfrac{33}{1024}, \ a_7 = \tfrac{429}{16384}. \end{gather*} Because $a_{k+1} < a_k$, we can estimate the remainder using the geometric series: $$ 0 < 1 - \sqrt{1 - x} - \tfrac{1}{2}\sum_{k=0}^{p-1} a_kx^{k+1} < \frac{a_px^{p+1}}{2(1 - x)} \quad (p \geqslant 0, |x| < 1). $$ I don't know if this suggestion is very practical in general, but in the case of a rational sine with a small denominator, such as our $\eta = \sin^{-1}\tfrac{1}{4}$, it can yield estimates very well suited to hand calculation, such as this one (I'll skip the details of its derivation): \begin{align*} \eta & > \frac{1}{2^2} + \frac{1}{3\cdot2^7} + \frac{1}{3\cdot2^{13}} + \frac{1}{3\cdot2^{14}} + \frac{1}{3\cdot2^{15}}, \\ \eta & < \frac{1}{2^2} + \frac{1}{3\cdot2^7} + \frac{1}{3\cdot2^{13}} + \frac{1}{3\cdot2^{14}} + \frac{1}{3\cdot2^{15}} + \frac{1}{3^2\cdot2^{13}}. \end{align*} All that is then required, apart from the "seed" value of $180/\pi$, is a series of divisions by $2$, a couple of divisions by $3$, and a few additions. One could begin by tabulating the terms to be added, thus: $$ \begin{array}{|c||c|c|c|} \hline & \text{radians} & \text{degrees} & \text{deg}^{\circ}\text{min}'\text{sec}'' \\ \hline \hline 1 & 1.00000000 & 57.29577951 & 57^\circ17'44.8062'' \\ 2^{-1} & 0.50000000 & 28.64788976 & 28^\circ38'52.4031'' \\ 2^{-2} & 0.25000000 & 14.32394488 & 14^\circ19'26.2016'' \\ 2^{-3} & 0.12500000 & \text{etc.} & \text{etc.} \\ \hline \end{array} $$

Such dubious practicalities aside, it is interesting to use the binomial series to compare the upper and lower bounds from the theorem with the Maclaurin series for $\sin^{-1}x$: $$ x + \sum_{j=1}^\infty \frac{4a_{2j}x^{2j+1}}{3} < x + \sum_{j=1}^\infty \frac{(j+1)a_jx^{2j+1}}{2j+1} < x + \sum_{j=1}^\infty \frac{2a_jx^{2j+1}}{3}. $$ Comparing the three coefficients of $x^{2j+1}$: $$ \begin{array}{|c||c|c|c|} \hline j & < \sin^{-1}x & \sin^{-1}x & > \sin^{-1}x \\ \hline \hline 0 & 1 & 1 & 1 \\ % j > 0 & 4a_{2j}/3 & a_j(k+1)/(2k+1) & 2a_j/3 \\ 1 & 1/6 & 1/6 & 1/6 \\ 2 & 7/96 & 3/40 & 1/12 \\ 3 & 11/256 & 5/112 & 5/96 \\ \hline \end{array} $$ There is a puzzle here. While it is immediately obvious by inspecting the coefficients that the series on the right exceeds $\sin^{-1}x$, the same is not at all true of the series on the left. (This is another instance of the sharper lower bound also being harder to work with.) We require the inequality:

$$ \frac{2a_{2j}}{a_j} \leqslant \frac{3j+3}{4j+2} \quad (j \geqslant 1), $$

with equality only for $j = 1$. One way to prove this is to use the inequality between arithmetic and harmonic means, in conjunction with the Weierstrass product inequality. Thus, for $j > 1$: \begin{gather*} \frac{2a_{2j}}{a_j} = \frac{(2j+3)\cdots(4j-1)}{(2j+4)\cdots(4j)} = \left(1 - \frac{1}{2j+4}\right)\cdots\left(1 - \frac{1}{4j}\right) < \frac{1}{1 + \frac{j-1}{3j+2}} = \frac{3j+2}{4j+1}, \end{gather*} which is slightly better than was required. $\square$

We end by giving a proof that Archimedes or Eratosthenes could have given (see Addendum 2):

Using two parabolic segments and a triangle to bound a circular sector.

We wish to obtain upper and lower bounds for the area $\theta$ of the sector $OPRQ$ of a circle, centre $O$, whose radius is taken as the unit of length. The angle at the vertex of the sector is $\eta = 2\theta$, and we are given the length of the straight line segment $QV = \sin\eta$.

It is shown below (simply, using modern methods) that: (i) there exists a unique parabola passing through $P$ and $Q$ and touching the circle at the midpoint $R$ of the circular arc $PRQ$, and the parabolic arc $PRQ$ lies inside the circular segment $PRQ$ except for the three points mentioned; and (ii) there exists a unique parabola touching the circle at $P$ and $Q$, and it lies outside the circle except at these two points. It is a well-known theorem (but it is also proved below) that if the tangents meet the axis $OR$ at the point $Z$, then the vertex of the parabola is the midpoint of $SZ$.

By Archimedes's famous quadrature result, the area of the minor parabolic segment (as I shall call it, while I call the other the "major" parabolic segment) is $\tfrac{4}{3}$ the area of the triangle $\triangle QPR$, which is also $\tfrac{4}{3}$ the area of the rectangle $QSRX$ (or equivalently, $RSPY$). The area of the major parabolic segment is $\tfrac{2}{3}$ of the area of the trapezium $PTUQ$ (this was clear from trigonometry, but see also Addendum 2 for a more natural proof by synthetic Euclidean methods), just as the area of the minor parabolic segment is $\tfrac{2}{3}$ of the area of the rectangle $PYXQ$. Therefore, the difference between the two bounds is given by $\tfrac{2}{3}$ of the combined area of the two little triangles $\triangle QXU$, $\triangle TYP$.

When there is time, I will edit the diagram to show the point $Z$, lying on the other side of $R$ from $S$, and the two tangents $PZ$ and $QZ$, both of them being common to the circle and the "major" parabola.

(The diagram has now been edited suitably.)

Meanwhile, I hope it is easy to "see" that because $PZ$ and $QZ$ touch the circle, the angles $\angle OPZ$ and $\angle OQZ$ are right angles, and therefore $OZ = \sec\theta$. Given that (as is proved below) the vertex of the major parabola lies midway between $S$ and $Z$, and of course $OS = \cos\theta$, it follows that the "height" (horizontal in the diagram!) of the triangle that is inscribed in the major parabolic sector and that shares its base $PQ$ and its vertex (which has to remain nameless, because it is far too close to $R$ to be shown in the diagram!), is $\tfrac{1}{2}(\sec\theta - \cos\theta)$. Therefore, because $PQ = 2\sin\theta$, the area of the major inscribed triangle is $\tfrac{1}{2}\sin\theta(\sec\theta - \cos\theta) = \tfrac{1}{2}\tan\theta - \tfrac{1}{4}\sin2\theta$. Therefore, by Archimedes's theorem on the quadrature of a parabolic segment (we only need the simplest case, in which the base of the segment is perpendicular to the axis), the area of the major parabolic segment is $\tfrac{2}{3}\tan\theta - \tfrac{1}{3}\sin2\theta$. Therefore, the area of the circular segment $PRQ$, which (as is shown below) is contained within the major parabolic segment $PRQ$, is strictly less than this. But the area of $\triangle OPQ$ is of course $\tfrac{1}{2}\sin2\theta$, therefore the area $\theta$ of the circular sector $OPRQ$ is strictly less than $\tfrac{2}{3}\tan\theta + \tfrac{1}{6}\sin2\theta$, as claimed. (Of course we need to double the value, as shown, if we want to estimate $\eta$ rather than $\theta$.) $\square$

That proves the upper bound. The proof of the lower bound is simpler (and easier to "see", because it does not involve any points or lines not shown in the diagram!). The area of $\triangle PRQ$ inscribed in the minor parabolic segment $PRQ$ is $QS{\cdot}RS = \sin\theta(1 - \cos\theta) = \sin\theta - \tfrac{1}{2}\sin2\theta$. By Archimedes's theorem again, the area of the minor parabolic segment $PRQ$ is $\tfrac{4}{3}$ of this value. It is shown below that the minor parabolic segment $PRQ$ is contained within the circular segment $PRQ$. Therefore, the area of the circular segment $PRQ$ is strictly greater than $\tfrac{4}{3}(\sin\theta - \tfrac{1}{2}\sin2\theta)$. Adding the area $\tfrac{1}{2}\sin2\theta$ of $\triangle OPQ$ again, it follows that the area of the circular sector is strictly greater than $\tfrac{4}{3}\sin\theta - \tfrac{1}{6}\sin2\theta$. (Again, double this value to estimate $\eta$ rather than $\theta$.) $\square$

It remains only to prove the two lemmas assumed in the above proof, as well as the "well-known" [translation: I found it in an old schoolbook from 50 years ago!] theorem that the tangent to a parabola at a point $P$ intersects the axis of the parabola at a point whose distance from the vertex of the parabola is the same as that of the foot of the perpendicular from $P$ to the axis. Fortunately, all three of these results are virtually trivial to prove using good old coordinate geometry and school calculus, which I can almost remember:

If the vertex of a parabola is taken as the origin of a Cartesian coordinate system, and the axis of the parabola is taken as the positive $y$-axis, then the unit of length can be chosen so that the equation of the parabola is $y = x^2$. That makes everything really easy.

First, of course, is Archimedes's great quadrature result, which nowadays any schoolchild (standing on the shoulders of giants!) can prove. For all $a > 0$, we have $\int_{-a}^ax^2\,dx = \tfrac{2}{3}a^3$, therefore the area of the parabolic segment below the line $y = a^2$ is $2a^3 - \tfrac{2}{3}a^3 = \tfrac{4}{3}a^3$, which is $\tfrac{4}{3}$ the area of the triangle inscribed in the segment. $\square$

Equally easy (after centuries of progress!) is the theorem about the tangent: the slope of the tangent at $(a, a^2)$ is $2a$, so it intersects the axis of the parabola, i.e. the $y$-axis, at the point $(0, -a^2)$. $\square$

For the lemma of the minor parabolic segment, suppose that a circle with centre inside the parabola and on its axis passes through the vertex of the parabola and also some other point of the parabola, which by symmetry we may take to be $(a, a^2)$ for some $a > 0$. If the radius of the circle is $r > 0$, then its centre is $(0, r)$, and we have $a^2 + (a^2 - r)^2 = r^2$, which simplifies to $a^2 = 2r - 1$. A point $(x, x^2)$ of the parabola other than $(0, 0)$ lies inside the circle if and only if $x^2 + (x^2 - r)^2 < r^2$, which simplifies to $x^2 < 2r - 1 = a^2$, i.e. to $|x| < |a|$. Therefore the arc of the parabola between $(a, a^2)$ and $(-a, a^2)$ lies inside the circle, apart from its point of contact at the origin, i.e. the vertex. $\square$

For the lemma of the major parabolic segment, observe that the slope of the normal to the parabola at $(a, a^2)$ is $-\tfrac{1}{2a}$, so it meets the $y$-axis at $(0, a^2 + \tfrac{1}{2})$, which must therefore be the centre of the circle touching the parabola at $(\pm a, a^2)$. The radius of the circle is $\sqrt{a^2 + \tfrac{1}{4}}$. For all $x$, we have \begin{align*} x^2 + \left(x^2 - a^2 - \tfrac{1}{2}\right)^2 - a^2 - \tfrac{1}{4} & = x^2 + x^4 - (2a^2 + 1)x^2 + a^4 + a^2 + \tfrac{1}{4} - a^2 - \tfrac{1}{4} \\ & = x^4 - 2a^2x^2 + a^4 \\ & = (x^2 - a^2)^2, \end{align*} and this is strictly positive except when $x = \pm a$, showing that apart from the two given points of contact, every other point of the parabola lies outside the circle. $\square$

Addendum 1

Here is a computation of $\pi$ to 9 decimal places, using (in 8 iterations, which more than sufficed): 8 square root extractions, 16 long divisions, 17 divisions by single-digit numbers, 25 multiplications by single-digit numbers, and 24 additions or subtractions. It would be perfectly feasible to do this all literally "by hand". I'm not that masochistic, but I did stubbornly stick to using only a cheap non-programmable calculator (Casio fx-85GT+). Here, at the risk of making myself look very silly indeed, is the informal "program" that I have just "executed":

 1 Mentally (and on paper) set n = 0
 2 1 --> X
 3 3sqrt(3)/2 --> A
 4 Begin loop:
 5 Write X in  third column of table
 6 Write A in fourth column of table
 7 If n > 0, write   (4A - M)/3 in fifth column of table
 8 If n > 0, write (4A/X + M)/3 in sixth column of table
 9 A --> M
10 sqrt(2 + X) --> X
11 2A/X --> A
12 Terminate or pause here if desired (memories aren't volatile);
   otherwise, mentally (and on paper) increment n, and return to
   the beginning of the loop

The above instructions represent the recursive formulae \begin{gather*} 2x_0 = 1, \quad A_0 = \frac{3\sqrt{3}}{2}; \\ 2x_n = \sqrt{2 + 2x_{n-1}}, \quad A_n = A_{n-1}/x_n \quad (n > 0); \\ B_n = \frac{4A_n -A_{n-1}}{3}, \quad C_n = \frac{2A_n/x_n + A_{n-1}}{3} \quad (n > 0). \end{gather*}

$$ \begin{array}{|r||r||c|c|c|c|} \hline n & 3\cdot2^{n+1} & 2x_n & A_n & B_n & C_n \\ \hline \hline 0 & 6 & 1.000000000 & 2.598076211 & - & - \\ 1 & 12 & 1.732050808 & 3.000000000 & 3.133974596 & 3.175426481 \\ 2 & 24 & 1.931851653 & 3.105828541 & 3.141104722 & 3.143593539 \\ 3 & 48 & 1.982889723 & 3.132628613 & 3.141561971 & 3.141716142 \\ 4 & 96 & 1.995717846 & 3.139350203 & 3.141590733 & 3.141600348 \\ 5 & 192 & 1.998929175 & 3.141031951 & 3.141592534 & 3.141593134 \\ 6 & 384 & 1.999732276 & 3.141452472 & 3.141592646 & 3.141592684 \\ 7 & 768 & 1.999933068 & 3.141557608 & 3.141592653 & 3.141592655 \\ 8 & 1536 & 1.999983267 & 3.141583892 & 3.141592654 & 3.141592654 \\ 9 & 3072 & 1.999995817 & 3.141590463 & 3.141592654 & 3.141592654 \\ 10 & 6144 & 1.999998954 & 3.141592106 & \bumpeq \pi & \bumpeq \pi \\ \hline \end{array} $$

Although $B_n$ and $C_n$ are the bounds calculated by erecting parabolic arcs on the sides of a $3\cdot2^n$-gon, their computation effectively requires the construction of a $3\cdot2^{n+1}$-gon, which is why they appear in row $n$ rather than row $n-1$ of the table.

The entries "$\bumpeq \pi$" in the last row represent where the calculator reached the limit of its internal precision, threw up its hands, and declared that both numbers were exactly $\pi$, poor thing.

Addendum 2

I stated that the area of the major parabolic segment is $\tfrac{2}{3}$ of the area of the trapezium $PTUQ$, but I gave no argument for this, other than an anachronistically "modern" trigonometric calculation, leaving it unclear what kind of proof Archimedes or Eratosthenes might have come up with.

Here is such a proof:

Using parabolic segments and a triangle to bound a circular sector.

The area of the triangle $\triangle QSZ$ is equal to that of the quadrilateral $QSRU$. This follows from the congruence of the triangles $\triangle OZQ, \triangle OUR$, upon subtracting the area of the triangle $\triangle OSQ$.

Because $N$ is the midpoint of $SZ$, the area of $\triangle PNQ$ is half that of $\triangle PZQ$, therefore the area of the major parabolic segment $PRQ$ is $\tfrac{2}{3}$ that of $\triangle PZQ$, therefore it is $\tfrac{4}{3}$ that of $\triangle QSZ$, therefore it is $\tfrac{4}{3}$ that of quadrilateral $QSRU$, therefore it is $\tfrac{2}{3}$ that of quadrilateral $PTUQ$. $\square$

I also only proved by trigonometry (and didn't even mention) that $QR$ bisects $\angle ZQS$ in $\triangle ZQS$. Archimedes and his colleagues could prove this by drawing a line from $Q$ to the other end of the diameter from $R$. See the new figure. $\square$

Knowing the ratio $QV/OQ$ (in this case $\tfrac{1}{4}$), our ancient Greek geometers would have been able to calculate $QS$, and then $RS$, by twice bisecting the given angle $\angle POQ$. And they could even more easily calculate $RU$. Thus the area of the circular segment $PRQ$, or rather $\tfrac{3}{2}$ of the area, has been trapped between the two known polygonal areas $PYXQ$ and $PTUQ$.

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  • $\begingroup$ I have been watching your exposition with interest because it's of a similar spirit to mine. If you change $\eta$ to $2\eta$ in your first equation, to order $\sin^5\eta$ it reads $$\sin\eta+\frac16\sin^3\eta+\frac1{24}\sin^5\eta<\eta<\sin\eta+\frac16\sin^3\eta+\frac5{24}\sin^5\eta$$ although you could improve the upper bound by taking the tangent points of your outer parabola closer to the center of the arc. Also you could compute $$\cos(x/2^n)\prod_{i=1}^n\cos(x/2^n)$$ and so only do 1 long division to get $\tan(x/2^n)$ and multiplications for the sines. $\endgroup$ – user5713492 Nov 30 '17 at 8:13
  • $\begingroup$ About the upper bound: I see you have $$\theta=\sin\theta\cos\theta+\int_{-\sin\theta}^{\sin\theta}\left(\sqrt{1-x^2}-\cos\theta\right)dx \\ =\sin\theta\cos\theta+\int_{-1}^1\left(\sqrt{1-u^2\sin^2\theta}-\cos\theta\right)\sin\theta du$$ Simpson's rule = lower bound. Upper bound is $$\int_{-1}^1f(x)dx\approx f(-r)-\frac12\left(\frac1{3r}-r\right)f^{\prime}(-r)+\frac12\left(\frac1{3r}-r\right)f^{\prime}(r)+f(r)$$ exact for cubics, min error for quartics is $r=1/\sqrt3$. You have $r=1$. Both are superior parabola. $\endgroup$ – user5713492 Nov 30 '17 at 11:26
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    $\begingroup$ @user5713492 Thank you for your very interesting comments (and answer), which I will endeavour to attend to properly when I am less sleep-deprived. (This question has really had me burning the midnight oil!) $\endgroup$ – Calum Gilhooley Nov 30 '17 at 15:15
  • $\begingroup$ I'm just starting to try to grips with this. (Excuse the delay, but I'm still having sleepless nights. It's been ridiculous!) I'll delete my comments as they're answered, and compile any results agreed between us into another addendum, with acknowledgement to you, if that's OK. (I'm also planning an addendum of my own, but putting it off for the moment.) My notation is $\eta=2\theta$. I proved $8\sin\theta-\sin2\theta<6\theta<4\tan\theta+\sin2\theta$. Using the $a_k$ series expansion, LHS $=8\sin\theta-2\sin\theta\sqrt{1-\sin^2\theta}=6\sin\theta+\sum_{k=0}^\infty a_{k}\sin^{2k+3}\theta$. Yes? $\endgroup$ – Calum Gilhooley Dec 3 '17 at 1:19
  • $\begingroup$ @user5713492 Continuing: By binomial series, or differentiation, $1/\sqrt{1-x}=\sum_{k=0}^\infty(k+1)a_kx^k$. RHS $=[4\sin\theta+2\sin\theta(1-\sin^2\theta)]/\sqrt{1-\sin^2\theta}=(6\sin\theta-2\sin^3\theta)\sum_{k=0}^\infty(k+1)a_k\sin^{2k}\theta$. The first 3 terms of LHS, RHS agree with the inequalities you wrote. (I'll stop there for the moment, try to get some sleep!) $\endgroup$ – Calum Gilhooley Dec 3 '17 at 1:32
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If I were a disciple of Archimedes, having learnt his Exhaustion Method and his computation of $\pi$, I would have probably approached the problem with that technique, thus arriving to use the same algorithm developed by the chinese mathematician Liu Hui some 5 centuries later.

enter image description here

Let's start with a general scheme:
given the point $P_0$ on the unit circle, we know one of its coordinates (e.g. $P_x$) and want to determine the angle $\beta$ defined with the $x$ axis.
Of course, at the times of Archimedes we would have spoken about the triangle $O-P_0-P_x$, etc., but let's use now a more agile Cartesian representation.

At that time I was supposed to know also the Pythagorean theorem and so I would have no difficulty in computing $$ \bbox[lightyellow] { \eqalign{ & y_{\,0} = \overline {OP_{\,y} } = \sqrt {1 - \overline {OP_{\,x} } ^{\,2} } = \sqrt {1 - x_{\,0} ^{\,2} } \cr & Q_{\,0} = \left( {1,\;y_{\,0} /x_{\,0} } \right) \cr & h_{\,0} = y_{\,0} /x_{\,0} \cr & s_{\,0} = \overline {UP_{\,0} } = \sqrt {\left( {1 - x_{\,0} } \right)^{\,2} + y_{\,0} ^{\,2} } \cr & M_{\,1} = \left( {\left( {1 + x_{\,0} } \right)/2,\;y_{\,0} /2} \right) \cr & m_{\,1} = \overline {OM_{\,1} } = {1 \over 2}\sqrt {\left( {1 + x_{\,0} } \right)^{\,2} + y_{\,0} ^{\,2} } \cr & \cr} }$$

then apply proportions to get $$ \bbox[lightyellow] { {1 \over {m_{\,1} }} = {{x_{\,1} } \over {\left( {1 + x_{\,0} } \right)/2}} = {{y_{\,1} } \over {y_{\,0} /2}} }$$ i.e. $$ \bbox[lightyellow] { \left\{ \matrix{ x_{\,1} = \left( {1 + x_{\,0} } \right)\;/\;\sqrt {\left( {1 + x_{\,0} } \right)^{\,2} + y_{\,0} ^{\,2} } \hfill \cr y_{\,1} = y_{\,0} \;/\;\sqrt {\left( {1 + x_{\,0} } \right)^{\,2} + y_{\,0} ^{\,2} } \hfill \cr h_{\,1} = y_{\,1} /x_{\,1} = y_{\,0} /\left( {1 + x_{\,0} } \right) \hfill \cr s_{\,1} = \overline {UP_{\,1} } = \sqrt {\left( {1 - x_{\,1} } \right)^{\,2} + y_{\,1} ^{\,2} } \hfill \cr} \right. }$$

Then I can iterate the recurrence until obtaining that the difference between $y_n$ and $h_n$ is acceptably small (also considering the number of significant digits that my avatar could manage).

From my master I know infact that the length of the arc is comprised among the sides of the inscribed and circumscribed polygon, so I know that $$ \bbox[lightyellow] { y_{\,n} < s_{\,n} < {\beta \over {2^{\,n} }} < h_{\,n} }$$

In the example you proposed we get, just after $4$ iterations $$ \bbox[lightyellow] { y_4=0.082289 \quad s_4=0.082359 \quad h_4= 0.082569 }$$ that is $$ \bbox[lightyellow] { 1.31774\approx 75.5^\circ \; <\; \beta \; < \; 1.3211\approx 75.7^\circ }$$ and actually it is $\beta= \arccos(1/4) \approx 1.318 \approx 75.52^\circ$

Concerning the comment from Fractional Inquirer, note that the recursion above can be reverted to give $$ \bbox[lightyellow] { \eqalign{ & x_{\,n} = \left( {1 + x_{\,n - 1} } \right)\;/\;\sqrt {\left( {1 + x_{\,n - 1} } \right)^{\,2} + 1 - x_{\,n - 1} ^{\,2} } \cr & \quad \quad \Downarrow \cr & 2\left( {1 + x_{\,n - 1} } \right)x_{\,n} ^{\,2} = \left( {1 + x_{\,n - 1} } \right)^{\,2} \cr & \quad \quad \Downarrow \cr & x_{\,n - 1} = 2x_{\,n} ^{\,2} - 1\quad \Leftrightarrow \quad \cos \left( {2\alpha } \right) = 2\cos ^{\,2} \alpha - 1 \cr} }$$ and similarly for $y_n$, $h_n$, that is the Duplication formulas were essentially known at that time (and in fact those given for the direct recurrence ahead are just the half-angle wrt full angle formulas ).

So, after halving the angle a sufficient number of times to assume it to be approximately equal to the sine (tangent), I could have go back through the duplication formula to compute the cosine (or sine or tangent) of the original angle.

That means that, as to avoid having to repeatedly take the square root, I could have proceeded as follows $$ \bbox[lightyellow] { \eqalign{ & \beta \; \to \;\beta /2^{\,n} \approx \sin \left( {\beta /2^{\,n} } \right) \cr & \cos ^2 \left( {\beta /2^{\,n} } \right) \approx 1 - \left( {\beta /2^{\,n} } \right)^2 \cr & \cos \left( {\beta /2^{\,n - 1} } \right) = 2\cos ^2 \left( {\beta /2^{\,n} } \right) - 1 = 1 - 2\left( {\beta /2^{\,n} } \right)^2 \cr & \cos \left( {\beta /2^{\,n - 2} } \right) = 2\left( {1 - 2\left( {\beta /2^{\,n} } \right)^2 } \right)^2 - 1 = 2\left( {2\left( {\beta /2^{\,n} } \right)^2 - 1} \right)^2 - 1 \cr & f(x) = 2x^2 - 1\quad \to \quad \cos \beta = f^{\left( n \right)} (\beta /2^{\,n} ) \cr} }$$

But $$ \bbox[lightyellow] { T_{2n} (x) = 2T_n ^2 (x) - 1 }$$ is one of the relations obeyed by the Chebishev Polynomials of 1st kind, so that the iteration in the recurrence above, interestingly, translates into $$ \bbox[lightyellow] { \cos \beta = T_{2^{\,n} } \left( {\beta /2^{\,n} } \right) }$$

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  • $\begingroup$ Do you know if there exist some similar geometric approach to compute $\cos x$? $\endgroup$ – user346361 Nov 29 '17 at 19:43
  • $\begingroup$ @FractionalInquirer : do you mean given $x$ to compute $\cos x$ ? If so, a) you can always reduce to $x</pi/2$. b) you can apply the method above to compute the sine, and from that the cosine. $\endgroup$ – G Cab Nov 29 '17 at 23:40
  • $\begingroup$ But how exactly the above procedure allows one to compute $\sin x$? $\endgroup$ – user346361 Nov 30 '17 at 8:40
  • $\begingroup$ @FractionalInquirer: I added to my answer the way to revert the recurrence. $\endgroup$ – G Cab Nov 30 '17 at 16:37
  • $\begingroup$ Thanks for your addendum. So, this iteration is the same as computing $\sin(\pi/2^n) \approx \pi/2^n$, for large $n$, and so using the double angle formula to compute the original sine. Is that right, or is this iteration different? $\endgroup$ – user346361 Dec 1 '17 at 3:48
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Assuming we know the addition formulas

$$\sin(x\pm y) = \sin(x)\cos(y)\pm \sin(y)\cos(x) $$ $$\cos(x\pm y) = \cos(x)\cos(y) \mp \sin(x)\sin(y)$$

we are able to also calculate $\sin(\frac{x}{2})$ and $\sin(nx)$ for $\in \mathbb N$ assuming that $\sin(x)$ is known. (Same with $\cos$)

Consequently we can find an approximate solution to $\cos(\phi) = a$ for any $a\in[-1,+1]$ by choosing some big $N$ and then finding $n\in \mathbb N$ such that:

$$ \cos\bigg(\frac{n \pi}{2^N}\bigg) \approx a$$

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HINT.- Getting into the skin of an ancient Greek geometry (no Archimedes of course), I would proceed as follows:

1) Construct the angle $\beta$ whose cosinus is equal to $\dfrac14$ (clearly with a right triangle having $1$ and $\sqrt{15}$ as legs and $4$ as hypotenuse).The searched angle $\alpha$ obviously satisfies $$\alpha=\dfrac{\pi}{2}+\beta$$ 2) I notice that $\cos15^{\circ}=\sqrt{\dfrac{1-\cos 30^{\circ}}{2}}=\dfrac{\sqrt{2-\sqrt3}}{2}\approx0.2588190\approx\dfrac14$

Since I am not Archimede by hypothesis, I take for $\beta$ what seems to me a good approximation so$$\beta\approx\frac{\pi}{12}\Rightarrow \alpha\approx \frac{\pi}{2}+\frac{\pi}{12}=\frac{7\pi}{12}$$ $$\color{red}{\alpha\approx\frac{7\pi}{12}}$$

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If I were an ancient Greek, I would carefully draw a large circle $K$ in the sand with a fixed stake at the center $O$ connected to a fixed length of string with a marker at the other end. I would make it as large as I needed it to be. Perhaps a few cubits in diameter would be enough for my purposes. I would draw a diameter of the circle $K$ from a fixed point $A$ through $O$ to the opposite side, and, using bisection twice, mark the point $B$ one quarter of the way from $O$ to the opposite side. I now use a simple geometric construction with the fixed length of string to find two points that determine the line perpendicular to the diameter and passing through point $B$. I now draw that line and find where it intersects the circle $K$ at a point $\Gamma$.

I am now interested in the arc of the circle $K$ from point $A$ to point $\Gamma$ and the central angle $\Theta$ subtended by this arc from the center $O$. The measure of the distance from $A$ to $\Gamma$, which is the chord corresponding to the arc and central angle, is one possible measure of the angle $\Theta$ I am intereseted in. Alternatively, depending on my need for $\Theta$, I could lay a length of string along the arc of the circle $K$ from point $A$ to point $\Gamma$ and measure its length. Now using 360 degrees for the length of the circumference of the circle $K$, I can convert the arc length to degrees and minutes of arc for $\Theta$. This practical procedure would work as well today as it would in those ancient times.

Admittedly, this is a solution by geometric construction and measuring, and not by calculating with numbers (except perhaps for converting to degree measure), but for all practical purposes, and also considering the technology of those times, this method has a few advantages. Measuring was just a relatively simple and well known procedure. Doing all but simple calculations with Greek numerals and operations like taking square roots was harder and much less well known. Alternatively, If I were somebody like Archimedes, I might use other methods like his approximate calculation of $\pi$.

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  • $\begingroup$ Your approach looks more "babylonian" or "egyptian" or even "roman". The greeks brought the invaluable revolution of the speculative approach over the practical collection of rules and tables: they brought mathematics into SCIENCE ! thus you would have received a downvote (if not an ostrakon :( ) from greek Scholars ! $\endgroup$ – G Cab Nov 30 '17 at 0:04
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We (as did the Greeks too) can use trig to express the required angles as $ \alpha_1=(\pi-\tan^{-1} 4) ,\,\alpha_2 = (\pi+\tan^{-1} 4),$ when terminator radius vectors lie in quadrants $ 2,3$.

When a chord perpendicular to a diameter is cut inside an eccentric circle radius $4$, the geometric mean of unequal segments so formed is the length of bisected segment = $\sqrt{15} $ for unequal segments $3,5$ in the present case. Such circle construction is well known to Greeks, and is one of Euclid's theorems. That circle not drawn as the sides $(\sqrt 15 ,1,4) $ also make a more obvious Pythagorean triplet.

 arcsin-1/4

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The equation of the unity circle is $$x^2+y^2=1$$ then you have a function $f(x)=\sqrt{1-x^2}$ defined on $[-1/4,1]$ the length of a curve by defintion is $$ \int_{-1/4}^1\sqrt{1+(f'(x))^2}dx$$ We have $f'(x)=\frac{-x}{\sqrt{1-x^2}}$, then $$\alpha=\int_{-1/4}^1\sqrt{1+\frac{x^2}{1-x^2}}dx$$ After simplification: $$\alpha=\int_{-1/4}^1\frac{1}{\sqrt{1-x^2}}dx=\arcsin(1)-\arcsin(-1/4)$$ Hence $$\alpha=\frac{\pi}{2}-\arcsin(-1/4)$$ The problematic consists in how to approximate $\arcsin(-1/4)$;

we can use a Taylor series: decomposition of $\arcsin(x)$, one has $$\arcsin(x)= x+ 1/2 (x^3/3) + (1/2)(3/4)(x^5/5) + (1/2)(3/4)(5/6)(x^7/7) ...$$ $$\arcsin(-1/4)= -1/4+ 1/2 ((-1/4)^3/3) + (1/2)(3/4)((-1/4)^5/5) + (1/2)(3/4)(5/6)((-1/4)^7/7)...= - 0.252$$ Then $$\alpha= 1.822 rad$$

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  • $\begingroup$ And now please demonstrate how you get a good approximation by manual computation. $\endgroup$ – LutzL Nov 23 '17 at 13:04
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    $\begingroup$ This was a lot of effort to come to $\arccos(x)=\frac\pi2+\arcsin(-x)$ (modulo multiples of $2\pi$) which follows directly from $\cos α=\sin(\fracπ2-α)$. $\endgroup$ – LutzL Nov 23 '17 at 15:16
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    $\begingroup$ @LutzL In fact the greeks does not have better than that or even similar methods to calculate angles but i did my best. As for my effort, i was following the calculation that i could effect. $\endgroup$ – AlphaXY Nov 23 '17 at 15:33
  • $\begingroup$ I doubt that the Greeks knew about the arcus sine series. However the relation stemming from reflection on the diagonal is easy enough. In principle you could have started your answer with the line $α=\frac π2−\arcsin(−\frac14)$ as that is a well-known trigonometric fact. $\endgroup$ – LutzL Nov 23 '17 at 16:40

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