[考古] Tantrasamgraha

hbghlyj

Wikipedia - Tantrasamgraha   Kerala school of astronomy and mathematics
The Discovery of the Series Formula for π by Leibniz, Gregory and Nilakantha - Roy-pi.pdf
In the Tantrasangraha-vakhya, the series for arctan, sine and cosine are given in verse which, when converted to mathematical symbols may be written as
\begin{gathered}
r\arctan\left(\frac{y}{x}\right) = \frac{1}{1}\cdot\frac{ry}{x} -\frac{1}{3}\cdot\frac{ry^3}{x^3} + \frac{1}{5}\cdot\frac{ry^5}{x^5} - \cdots, \text{ where } \frac y x \leq 1\\
y=r\sin \frac{s}{r} = s - s\cdot\frac{s^2}{(2^2+2)r^2} + s\cdot \frac{s^2}{(2^2+2)r^2}\cdot\frac{s^2}{(4^2+4)r^2} - \cdots \\
r-x=r \left( 1 - \cos \frac{s}{r} \right) = r \frac{s^2}{(2^2-2)r^2} - r \frac{s^2}{(2^2-2)r^2}\cdot \frac{s^2}{(4^2-4)r^2} + \cdots
\end{gathered}

Figure 4

There are also some special features in the Tantrasangraha's treatment of the series for π/4 which were not considered by Leibniz and Gregory. Nilakantha states some rational approximations for the error incurred on taking only the first $n$ terms of the series. The expression for the approximation is then used to transform the series for π/4 into one which converges more rapidly. The errors are given as follow\[
\frac{\pi}{4} \simeq 1-\frac{1}{3}+\frac{1}{5}-\cdots \mp \frac{1}{n} \pm f_i(n+1) \quad i=1,2,3\tag{12}\label{12}
\]
where
\[
f_1(n)=\frac{1}{2 n}, f_2(n)=\frac{n / 2}{n^2+1} \text { and } f_3(n)=\frac{(n / 2)^2+1}{\left(n^2+5\right) n / 2} \text {. }
\]
The transformed series are as follows:
\[
\frac{\pi}{4}=\frac{3}{4}+\frac{1}{3^3-3}-\frac{1}{5^3-5}+\frac{1}{7^3-7}-\cdots\tag{13}\label{13}
\]
and
\[
\frac{\pi}{4}=\frac{4}{1^5+4 \cdot 1}-\frac{4}{3^5+4 \cdot 3}+\frac{4}{5^5+4 \cdot 5}-\cdots .
\]
Leibniz's proof of the formula for $\pi / 4$ was found by the quadrature of a circle. The proof in Jyesthadeva's book is by a direct rectification of an arc of a circle. In the diagram given below, the arc $A C$ is a quarter circle of radius one with center $O$ and $O A B C$ is a square. The side $A B$ is divided into $n$ equal parts of length $\delta$ so that $n \delta=1, P_{r-1} P_r=\delta$. $EF$ and $P_{r-1} D$ are perpendicular to $O P_r$. Now, the triangles $O E F$ and $O P_{r-1} D$ are similar, which gives
\[
\frac{E F}{O E}=\frac{P_{r-1} D}{O P_{r-1}}, \quad \text { that is, } E F=\frac{P_{r-1} D}{O P_{r-1}} .
\]

Figure 5

The similarity of the triangles $P_{r-1} P_r D$ and $O A P_r$ gives
\[
\frac{P_{r-1} P_r}{O P_r}=\frac{P_{r-1} D}{O A} \text { or } P_{r-1} D=\frac{P_{r-1} P_r}{O P_r} .
\]
Thus,
\[
E F=\frac{P_{r-1} P_r}{O P_{r-1} O P_r} \simeq \frac{P_{r-1} P_r}{O P_r^2}=\frac{\delta}{1+A P_r^2}=\frac{\delta}{1+r^2 \delta^2}
\]
Since arc $E G \simeq E F \simeq \frac{\delta}{1+r^2 \delta^2}$, $\frac{1}{8}$ arc of circle is
\[
\frac{\pi}{4}=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{\delta}{1+r^2 \delta^2}\tag{14}\label{14}
\]

kuing

想减少减号两边的距离，改成 r{-}x 即可，无需 r\!-\!x

hbghlyj

Of course, a clear idea of limits did not exist at that time so that the relation was understood in an intuitive sense only. To evaluate the limit, Jyesthadeva uses two lemmas. One is the geometric series$$\frac{1}{1+x}=1-x+x^{2}-x^{3}+\cdots$$Jyesthadeva says that the expansion is obtained on iterating the following procedure:
\[
\frac{1}{1+x}=1-x\left(\frac{1}{1+x}\right)=1-x\left(1-x\left(\frac{1}{1+x}\right)\right) \text {. }
\]
The other result is that
\[\tag{15}\label{15}
S_n^{(p)} \equiv 1^p+2^p+\cdots+n^p \sim \frac{n^{p+1}}{p+1} \text { for large } n .
\]
A sketch of a proof is given by Jyesthadeva. He notes first that
\[\tag{16}\label{16}
n S_n^{(p-1)}=S_n^{(p)}+S_1^{(p-1)}+S_2^{(p-1)}+\cdots+S_{n-1}^{(p-1)} .
\]
This is easy to verify. Relation \eqref{16} is also contained in the work of the tenth century Arab mathematician Alhazen, who gives a geometrical proof in the Greek tradition^{See The Historical Development of the Calculus, (mentioned in footnote 1), p. 84}. He uses it to evaluate $S_n^{(3)}$ and $S_n^{(4)}$ which occur in a problem about the volume of a certain solid of revolution. Yuktibhasa shows that for $p=2,3$
\[\tag{17}\label{17}
S_1^{(p-1)}+S_2^{(p-1)}+\cdots+S_{n-1}^{(p-1)} \sim \frac{S_n^{(p)}}{p},
\]
and then suggests that by induction the result will be true for all values of $p$. Once this is granted, it follows that if
\[
S_n^{(p-1)} \sim \frac{n^p}{p},
\]
then by \eqref{16} and \eqref{17},
\[
n S_n^{(p-1)} \sim S_n^{(p)}+\frac{S_n^{(p)}}{p} \quad \text { or } \quad S_n^{(p)} \sim \frac{n^{p+1}}{p+1} \text {, }
\]
and \eqref{15} is inductively proved.

hbghlyj

We now note that \eqref{14} can be rewritten, after expanding $1 /\left(1+r^2 \delta^2\right)$ into a geometric series, as
\[
\begin{aligned}
\frac{\pi}{4} &=\lim _{n \rightarrow \infty}\left[\delta \sum_{r=1}^n 1-\delta^3 \sum_{r=1}^n r^2+\delta^5 \sum_{r=1}^n r^4-\cdots\right] \\
&=\lim _{n \rightarrow \infty}\left[1-\frac{1}{n^3} \sum_{r=1}^n r^2+\frac{1}{n^5} \sum_{r=1}^n r^4-\cdots\right] \\
&=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots
\end{aligned}
\]
where we have used relation \eqref{15} and the fact that $\delta=1 / n$. Now consider the approximation \eqref{12} and its application to the transformation of series. Suppose that
\[
\sigma_n=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots \pm \frac{1}{n} \mp f(n+1),
\]
where $f(n+1)$ is a rational function of $n$ which will make $\sigma_n$ a better approximation of $\pi / 4$ than the $n$th partial sum $S_n$. Changing $n$ to $n-2$ we get
\[
\sigma_{n-2}=1-\frac{1}{3}+\frac{1}{5}-\cdots \mp \frac{1}{n-2} \pm f(n-1)
\]
Subtracting the second relation from the first,
\[\tag{18}\label{18}
\pm u_n=\sigma_n-\sigma_{n-2}=\pm \frac{1}{n} \mp f(n+1) \mp f(n-1) .
\]
Then
\[
\begin{aligned}
\sigma_n &=\sigma_{n-2} \pm u_n \\
&=\sigma_{n-4} \mp u_{n-2} \pm u_n \\
&=\cdots=\sigma_1-u_3+u_5-u_7+\cdots \pm u_n \\
&=1-f(2)-u_3+u_5-u_7+\cdots u_n
\end{aligned}
\]
It is clear that
\[
\lim _{n \rightarrow \infty} \sigma_n=\frac{\pi}{4}
\]
and therefore
\[\tag{19}\label{19}
\frac{\pi}{4}=1-f(2)-u_3+u_5-u_7+\cdots .
\]
Thus, we have a new series for $\pi / 4$ which depends on how the function $f(n)$ is chosen. Naturally, the aim is to choose $f(n)$ in such a way that \eqref{19} is more rapidly convergent than (1). This is the idea behind the series \eqref{13}. Now equation \eqref{18} implies that
\[\tag{20}\label{20}
f(n+1)+f(n-1)=\frac{1}{n}-u_n
\]
For (19) to be more rapidly convergent than (1), $u_n$ should be $o(1 / n)$, that is, negligible compared to $1 / n$. It is reasonable to assume $f(n+1) \simeq f(n-1) \simeq f(n)$. These observations together with \eqref{20} imply that $f(n)=1 / 2 n$ is a possible rational approximation in equation (12). With this $f(n)$, the value of $u_n$ is given by \eqref{20} to be
\[
u_n=\frac{1}{n}-\frac{1}{2(n+1)}-\frac{1}{2(n-1)}=-\frac{1}{n^3-n} .
\]
Substituting this in (19) gives us (13), which is
\[
\frac{\pi}{4}=1-\frac{1}{4}+\frac{1}{3^3-3}-\frac{1}{5^3-5}+\frac{1}{7^3-7}-\cdots .
\]
The other series
\[
\frac{\pi}{4}=\frac{4}{1^5+4 \cdot 1}-\frac{4}{3^5+4 \cdot 3}+\frac{4}{5^5+4 \cdot 5}-\cdots
\]
is obtained by taking $f(n)=\frac{n / 2}{n^2+1}$ in (19).
It should be mentioned that Newton was aware of the correction $f_1(n)=1 / 2 n$. For in the letter to Oldenburg, referred to earlier, he says, "By the series of Leibniz also if half the term in the last place be added and some other like devices be employed, the computation can be carried to many figures." However, he says nothing about transforming the series by means of this correction.
It appears that Nilakantha was aware of the impossibility of finding a finite series of rational numbers to represent $\pi$. In the Aryabhatiya-bhasya he writes, "If the diameter, measured using some unit of measure, were commensurable with that unit, then the circumference would not likewise allow itself to be measured by means of the same unit; so likewise in the case where the circumference is measurable by some unit, then the diameter cannot be measured using the same unit." 19

The Yuktibhasa contains a proof of the arctan series also and it is obtained in exactly the same way except that one rectifies only a part of the $1 / 8$ circle.

It can be shown that if $\pi / 4=S_n+f(n)$, where $S_n$ is the $n$th partial sum, then $f(n)$ has the continued fraction representation
\[
f(n)=\frac{1}{2}\left[\frac{1}{n+} \frac{1^2}{n+} \frac{2^2}{n+} \frac{3^2}{n+} \cdots\right] .
\]
Moreover, the first three convergents are
\[
f_1(n)=\frac{1}{2 n}, \quad f_2(n)=\frac{n / 2}{n^2+1} \text { and } \quad f_3(n)=\frac{(n / 2)^2+1}{\left(n^2+5\right) n / 2},
\]
which are the values quoted in \eqref{13}. Clearly, Nilakantha was using some procedure which gave the successive convergents of the continued fraction (21) but the text contains no suggestion that (20) was actually known to him. This continued fraction implies that
\[
\frac{2}{4-\pi}=2+\frac{1^2}{2+} \frac{2^2}{2+} \frac{3^2}{2+} \cdots
\]