Approximating Pi, as an example of FEM method

This page replicates the example in chapter 1.1 of in a book about finite element method《有限元法基础》（张雄，高等教育出版社，2023). The book is a classical Soviet-Chinese style textbook that gives info blandly. Here we give full details.

Compute Some Sine Without Pi

Computing the value of $\sin()$ generally requires the value of $\pi$, however, we can still obtain special values by the definition of Sine:

$\sin(0)=0$
$\sin(\pi/2)=1$

These will be our initial condition.

Then we need the Recurrence Relation between $\sin(a)$ and $sin(2a)$.

From Multiple-angle and half-angle formulas, we may get $\sin(2a)=2\sin(a)\cos(a)$, where if $a<=\pi/2$ then $\cos(a)=\sqrt{1-\sin(a)^2}$ as in Pythagorean identities. Thus,

\[ \begin{aligned} \sin(2a)&=2\sin(a)\cos(a) \\\\ \sin(2a)^2 &= 2^2\sin(a)^2\cos(a)^2 \\\\ \sin(2a)^2 &= 2^2\sin(a)^2(1-\sin(a)^2) \\\\ &=(2\sin(a))^2-[2\sin(a)^2]^2\\\\ \end{aligned} \]

Substitute $n=\sin(2a)^2$ and $x=2\sin(a)^2$,

\[ \begin{align} n^2&=2x-x^2 \\\\ x^2-2x+n^2&=0 \\\\ \end{align} \]

Using the Quadratic formula, we can easily get

\[ x=2\sin(a)^2 = \frac{2\pm\sqrt{4-4n^2}}{2}=1\pm\sqrt{1-n^2}=1\pm\sqrt{1-\sin(2a)^2} \]

Substitute the values back,

\[ \sqrt{2}\sin(a)=\sqrt{1\pm\sqrt{1-\sin(2a)^2}}\ . \]

The $\pm$ sign depends on value of $a$. Plot the right side, we may easily know that when $a<\pi/4$, the sign should be minus.

(* Plotting in Mathematica *)
Plot[{Sqrt[1 + Sqrt[1 - Sin[2 x]^2]], Sqrt[1 - Sqrt[1 - Sin[2 x]^2]], Sqrt[2]*Sin[x]}, {x, 0, 10}]

Then we can get a special sequence of value for $\sin(\pi/n)$ where $n$ is power of 2. For example,

$$ \sin(\pi/4)=\frac{\sqrt{1-\sqrt{1-\sin(\pi/2)^2}}}{\sqrt{2}}=0.707107\ . $$ In Mathematica, this can be defined as

f[0] := 0
f[2] := 1
f[n_] := Sqrt[1 - Sqrt[1 - f[n/2]^2]]/Sqrt[2]
Table[{Sin[Pi/Evaluate[2^x]], f[2^x] // N}, {x, 1, 10}] // TableForm

(* Result *)
Sin[\[Pi]/4]    0.707107
Sin[\[Pi]/8]    0.382683
Sin[\[Pi]/16]   0.19509
Sin[\[Pi]/32]   0.0980171
Sin[\[Pi]/64]   0.0490677
Sin[\[Pi]/128]  0.0245412
Sin[\[Pi]/256]  0.0122715
Sin[\[Pi]/512]  0.00613588
Sin[\[Pi]/1024] 0.00306796

Computing the Value Of $Pi$

For a circle that are divided to $n$ parts, the length of each "element line" is $L_e=d\sin(\pi/n)$ and the total length is thus $L=\sum L_e = n d\sin(\pi/n)$.

By definition of $\pi$, we have$L=d\pi$ and thus $\pi=n\sin(\pi/n)$.

Finally,

calcPi[n_] := n*f[n] // N
Table[With[{n = 2^p}, {n, DecimalForm[calcPi[n], 16], 
    DecimalForm[Abs[Pi - calcPi[n]], 5]}], {p, 2, 10}] // TableForm

(* Result: n, Pi, error *)
4   2.82842712474619    0.31317
8   3.061467458920719   0.080125
16  3.121445152258053   0.020148
32  3.136548490545941   0.0050442
64  3.140331156954739   0.0012615
128 3.141277250932757   0.0003154
256 3.141513801144145   0.000078852
512 3.141572940367883   0.000019713
1024    3.141587725279961   0.0000049283

Appendix

% plot the demo
\documentclass[tikz]{standalone}
\usetikzlibrary {angles,quotes,shapes.geometric,arrows.meta} 
\begin{document}
\begin{tikzpicture}
\coordinate (O) at (0,0);
\draw[dashed] (O) circle [radius=4cm];
\node [name=s,regular polygon, regular polygon sides=8, minimum size=8cm,magenta,ultra thick] at (O) {};
\foreach \i in {1,2,3,4,5,6,7,8} {
    \draw [dashed] (s.center)-- (s.corner \i);
    \filldraw (s.corner \i) circle [radius=0.05];
};

\draw (s.corner 2)--(s.corner 3)--(s.corner 4)--(s.corner 5)--(s.corner 6)--(s.corner 7)--(s.corner 8)--(s.corner 1);

\coordinate (L) at (s.corner 1);
\coordinate (R) at (s.corner 2);
\draw [|<->|,magenta,thick]  (L) -- node[below=0cm,black] {$d\sin(\pi/n)$} (R);

\draw pic [draw,angle radius=2cm, "$2\pi/n$"] {angle = L--O--R};
\end{tikzpicture}
\end{document}

Approximating Pi, as an example of FEM method

Compute Some Sine Without Pi

Computing the Value Of \(Pi\)

Appendix