9.2 Calculation of $\pi$

The proof of the next lemma depends on the following assumption, which is explicitly stated by Archimedes [2, page 3]. This assumption involves the ideas of curve with given endpoints and length of curve (which I will leave undiscussed).

9.32   Assumption. Let $\mbox{{\bf a}}$ and $\mbox{{\bf b}}$ be points in $\mbox{{\bf R}}^2$. Then of all curves with endpoints $\mbox{{\bf a}}$ and $\mbox{{\bf b}}$, the segment $[\mbox{{\bf a}}\mbox{{\bf b}}]$ is the shortest.

9.33   Lemma.

$$\sin (x)<x \mbox{ for all } x \in \mbox{${\mbox{{\bf R}}}^{+}$},$$

and

$$\vert\sin (x)\vert\leq \vert x\vert \mbox{ for all } x\in\mbox{{\bf R}}.$$

Proof:

Case 1: Suppose $${0<x<{\pi\over 2}}$$.

Then (see the figure) the length of the arc joining $W(-x)$ to $W(x)$ in the first and fourth quadrants is $x+x=2x$. (This follows from the definition of $W$.) The length of the segment $[W(x)W(-x)]$ is $2\sin (x)$. By our assumption, $2\sin(x)\leq 2x$, i.e., $\sin(x)\leq x$. Since both $x$ and $\sin (x)$ are positive when ${0<x<{\pi\over 2}}$, we also have $\vert\sin (x)\vert\leq \vert x\vert$.

Case 2: Suppose $${x\geq {\pi\over 2}}$$. Then

$$\sin (x)\leq \vert\sin (x)\vert\leq 1<{\pi\over 2}\leq x=\vert x\vert$$

so $\sin(x)\leq x$ and $\vert\sin (x)\vert\leq \vert x\vert$ in this case also. This proves the first assertion of lemma 9.33. If $x<0$, then $-x>0$, so

$$\vert\sin (x)\vert=\vert\sin (-x)\vert\leq \vert-x\vert=\vert x\vert.$$

Thus

$$\vert\sin (x)\vert\leq \vert x\vert \mbox{ for all } x\in\mbox{{\bf R}}\setminus \{0\},$$

and since the relation clearly holds when $x=0$ the lemma is proved. $\diamondsuit$

9.34   Lemma (Limits of sine and cosine.) Let $a\in\mbox{{\bf R}}$. Let $\{a_n\}$ be a sequence in $\mbox{{\bf R}}$ such that $\{a_n\}\to a$. Then

$$\{\cos (a_n)\}\to\cos (a) \mbox{ and }\{\sin (a_n)\}\to\sin (a).$$

Proof: By (9.25) we have

$$\cos (a_n)-\cos (a)=-2\sin\Big( {{a_n+a}\over 2}\Big)\sin\Big( {{a_n-a}\over 2}\Big),$$

so

$$\begin{eqnarray*} 0 \leq \vert\cos (a_n)-\cos (a)\vert &\leq& 2\vert \sin \Big( ... ... 2}\Big)\vert\leq 2\vert {{a_n-a}\over 2}\vert=\vert a_n-a\vert. \end{eqnarray*}$$

If $\{a_n\}\to a$, then $\{ \vert a_n-a\vert\} \to 0$, so by the squeezing rule,

$$\{\vert\cos(a_n) - \cos(a)\vert\}\to 0.$$

This means that $\{ \cos (a_n) \} \to \cos(a)$.

The proof that $\{\sin (a_n)\}\to\sin (a)$ is similar. $\diamondsuit$

The proof of the next lemma involves another new assumption.

9.35   Assumption. Suppose $${0<x<{\pi\over 2}}$$. Let the tangent to the unit circle at $W(x)$ intersect the $x$ axis at $\mbox{{\bf p}}$, and let $\mbox{{\bf q}}=(1,0)$.

Then the circular arc joining $W(x)$ to $W(-x)$ (and passing through $\mbox{{\bf q}}$) is shorter than the curve made of the two segments $[W(x)\mbox{{\bf p}}]$ and $[\mbox{{\bf p}}W(-x)]$ (see the figure).

Remark: Archimedes makes a general assumption about curves that are concave in the same direction [2, pages 2-4] which allows him to prove our assumption.

9.36   Lemma. If $${0<x<{\pi\over 2}}$$, then

$$x \leq {\sin(x)\over \cos(x)}.$$

Proof: Suppose $${0<x<{\pi\over 2}}$$. Draw the tangents to the unit circle at $W(x)$ and $W(-x)$ and let the point at which they intersect the $x$-axis be $\mbox{{\bf p}}$. (By symmetry both tangents intersect the $x$-axis at the same point.) Let $\mbox{{\bf b}}$ be the point where the segment $[W(x)W(-x)]$ intersects the $x$-axis, and let $\mbox{{\bf r}}=W(x)$. Triangles $\mbox{{\bf0}} \mbox{{\bf b}}\mbox{{\bf r}}$ and $\mbox{{\bf0}} \mbox{{\bf r}}\mbox{{\bf p}}$ are similar since they are right triangles with a common acute angle.

Hence

$${{\mbox{distance}(\mbox{{\bf r}},\mbox{{\bf b}})}\over {\mbox... ...\bf r}})}\over {\mbox{distance}(\mbox{{\bf0}},\mbox{{\bf r}})}}$$

i.e.,

$${{\sin (x)}\over {\cos (x)}}= {{\mbox{distance}(\mbox{{\bf p}},\mbox{{\bf r}})}\over 1}.$$

Now the length of the arc joining $W(x)$ to $W(-x)$ is $2x$, and the length of the broken line from $\mbox{{\bf r}}$ to $\mbox{{\bf p}}$ to $W(-x)$ is $${2\Big(\mbox{distance}(\mbox{{\bf p}},\mbox{{\bf r}})\Big)=2{{\sin (x)}\over {\cos(x)}}}$$, so by assumption 9.35,

$$2x\leq 2{{\sin(x)}\over {\cos x}}$$

i.e.,

$$x \leq {\sin(x)\over \cos(x)}.$$

This proves our lemma. $\mbox{ $\diamondsuit$}$

9.37   Theorem. Let $\{x_n\}$ be any sequence such that $x_n\neq 0$ for all $n$, and $\{x_n\} \to 0$. Then

$$\Big\{ {{\sin (x_n)}\over {x_n}}\Big\}\to 1.$$ (9.38)

Hence if $\sin(x_n)\neq 0$ for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ we also have

$$\Big\{ {{x_n}\over {\sin(x_n)}}\Big\}\to 1.$$

Proof: If $x \in (0, {\pi\over 2})$, then it follows from lemma(9.36) that $${\cos(x) \leq {\sin(x) \over x} }$$. Since

$$\cos(-x) = \cos(x) \mbox{ and }{\sin(-x) \over -x} = {\sin(x) \over x},$$

it follows that

$$\cos(x) \leq {\sin(x) \over x} \mbox{ whenever } 0 < \vert x\vert < {\pi \over 2}.$$

Hence by lemma 9.33 we have

$$\cos(x) \leq {\sin(x) \over x} \leq 1 \mbox{ whenever } 0 < \vert x\vert < {\pi \over 2}.$$ (9.39)

Let $\{x_n\}$ be a sequence for which $x_n\neq 0$ for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ and $\{x_n\} \to 0$. Then we can find a number $N\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that for all $${n\in\mbox{{\bf Z}}_{\geq N} (\vert x_n\vert<{\pi\over 2}})$$. By (9.39)

$$n\in\mbox{{\bf Z}}_{\geq N}\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\cos(x_n)\leq {{\sin(x_n)}\over {x_n}}\leq 1.$$

By lemma 9.34, we know that $\{\cos (x_n)\}\to 1$, so by the squeezing rule $${ \Big\{ {{\sin(x_n)}\over {x_n}} \Big\}\to 1}$$. $\diamondsuit$

9.40   Example (Calculation of $\pi$.) Since $${ \Big\{ {\pi\over n}\Big\} \to 0 }$$, it follows from (9.38) that

$$\lim\Big\{ {{\sin \Big( {\pi\over n}\Big)}\over { {\pi\over n} }}\Big\}=1$$

and hence that

$$\lim\Big\{n\sin\Big( {\pi\over n}\Big)\Big\}=\pi.$$

This result can be used to find a good approximation to $\pi$. By the half-angle formula, we have

$$\sin^2 \Big( {t\over 2}\Big) ={{1-\cos t}\over 2}={1\over 2}\Big(1-\sqrt{1-\sin^2t}\Big)$$

for $0\leq t\leq \displaystyle { {\pi\over 2}}$. Here I have used the fact that $\cos t\geq 0$ for $0\leq t\leq \displaystyle { {\pi\over 2}}$. Also $${ \sin ({\pi\over 2})=1}$$ so

$$\begin{eqnarray*} \sin^2({\pi\over 4})&=&{1\over 2}\Big(1-\sqrt{1-\sin^2{\pi\ove... ...sqrt{1-{1\over 2}}\Big)={1\over 2}\Big(1-\sqrt{{1\over 2}}\Big). \end{eqnarray*}$$

By repeated applications of this process I can find $${\sin^2\Big( {\pi\over {2^n}}\Big)}$$ for arbitrary $n$, and then find

$$2^n\sin \Big( {\pi\over {2^n}}\Big)$$

which will be a good approximation to $\pi$.

I wrote a set of Maple routines to do the calculations above. The procedure sinsq(n) calculates $${\sin^2\Big( {\pi\over {2^n}}\Big)}$$ and the procedure mypi(m) calculates $${ 2^m\sin\Big( {\pi\over {2^m}}\Big)}$$. The `` fi" (which is `` if" spelled backwards) is Maple's way of ending an `` if" statement. `` Digits $:= 20$" indicates that all calculations are done to $20$ decimal digits accuracy. The command `` evalf(Pi)" requests the decimal approximation to $\pi$ to be printed.

> sinsq := 
>    n-> if n=1 then 1;
>        else .5*(1-sqrt(1 - sinsq(n-1)));
>        fi;

sinsq := proc(n) options operator,arrow; if n = 1 then 1 else .5 -.5*sqrt(1-sinsq(n-1)) fi end

> mypi := m -> 2^m*sqrt(sinsq(m));

$${\it mypi} := {m} \rightarrow 2^{{m}} {\rm sqrt}( {\rm sinsq}( {m} ) )$$

> Digits := 20;

$${\it Digits} := 20$$

> mypi(4);

$$3.1214451522580522853$$

> mypi(8);

$$3.1415138011443010542$$

> mypi(12);

$$3.1415923455701030907$$

> mypi(16);

$$3.1415926523835057093$$

> mypi(20);

$$3.1415926533473327481$$

> mypi(24);

$$3.1415922701132732445$$

> mypi(28);

$$3.1414977446171452114$$

> mypi(32);

$$3.1267833885746006944$$

> mypi(36);

$$0$$

> mypi(40);

$$0$$

> evalf(Pi);

$$3.1415926535897932385$$

9.41   Exercise. Examine the output of the program above. It appears that $\pi=0$. This certainly is not right. What can I conclude about $\pi$ from my computer program?

9.42   Exercise. Show that the number $${ n\sin\Big( {\pi\over n}\Big)}$$ is the area of a regular $2n$-gon inscribed in the unit circle. Make any reasonable geometric assumptions, but explain your ideas clearly.