The famous Chinese estimate $\pi\approx\frac{355}{113}$ is good. I think that is too good. As a continued fraction,
$$\pi=[3; 7,15,1,292,\ldots].$$
That $292$ is a bit too big. Is there a reason for such a good approximation that Chinese mathematics found, or were they just lucky?
[Math] Chinese estimate for $\pi$. Were they lucky
continued-fractionsirrational-numbersmath-historypisoft-question
Related Solutions
This isn't exceptionally good compared to the partial convergents of the continued fraction expansion. Terminating the continued fraction for $\pi$ right before the $292$ gives $\frac{355}{113}= \textbf{3.141592}9203\ldots$, which gets $6$ digits after the decimal place right, while your fraction only gets $3$ digits after the decimal place correct even though it has larger numerator and denominator.
The answer is yes. The method is detailed in the paper by S. K. Lucas. The approximate fractions are obtained from truncating the exact continued fraction of the respected numbers, so the signs are alternating. We first give a few examples.
Results
For $\pi$
The continued fractions are $3, 22/7, 333/106, 355/113, 103993/33102, \dots$.
\begin{align} \pi - \frac{333}{106} &= \int_0^1 \frac{x^4 \, (1-x)^5 \, \left(74 \, x^2-53 \, x+21\right)}{106 \left(x^2+1\right)} \, dx. \\ \frac{355}{113} - \pi &= \int_0^1 \frac{x^{10} \, (1-x)^8 \, \left(886+95\,x^2\right)}{3164 \left(x^2+1\right)} \, dx. \end{align}
For $\pi^2$
The truncated continued fractions are $9, 10, 69/7, 79/8, 227/23, 10748/1089, \dots$,
\begin{align} \pi^2-\frac{69}{7} &= \int_0^1 \frac{4 \, x^{4} \, (1 - x)^3 \left(64 x^2 -39 x + 25\right)} {13 \, (1 + x^2) } \log(x^{-1}) \, dx \\ &= \int_0^1 \frac{24 \, x^{6} \, (1 - x)^2 \left(119 - 72 \, x^2\right)} {191 \, (1 + x^2) } \log(x^{-1}) \, dx. \end{align}
\begin{align} \frac{79}{8} - \pi^2 &= \int_0^1 \frac{4 \, x^6 \, (1-x)^3 (49 - 51 x + 100 x^2)} {17 \, (1 + x^2) } \log(x^{-1}) \, dx \\ &= \int_0^1 \frac{4 \, x^3 \, (1-x)^4 (25 + 2254 x^2)} {743 \, (1 + x^2) } \log(x^{-1}) \, dx \\ &= \int_0^1 \frac{ 24 \, x^5 \, (1-x)^2 \left[37 \, (x^2 + 1) - 73 \, x\right] } { 73 \, (1 + x^2) } \log(x^{-1}) \, dx \end{align}
\begin{align} \pi^2-\frac{227}{23} &= \int_0^1 \frac{4 \, x^{19} \, (1 - x)^4 \left(61847 x^2+87524\right)} {8559 \, (1 + x^2) } \log(x^{-1}) \, dx. \end{align}
For $\pi^3$
The truncated continued fractions are $31, 4930/159, 14821/478, \dots$.
\begin{align} \pi^3-31 &= \int_0^1 \frac{8 \, x^5 \, (1-x)^2 \, \left(324889-120736 \, x^2\right)} {445625 \, (1 + x^2) } \log^2 x \, dx\\ \frac{4930}{159}-\pi^3 &= \int_0^1 \frac{4 \, x^{10} \, (1-x)^4 \, \left(695774836+470936528857 \, x^2\right)} {470240754021 \, (1 + x^2) } \log^2 x \, dx. \end{align}
For $\pi^4$
The truncated continued fractions are $97, 195/2, 487/5, 1656/17, 2143/22, \dots$.
\begin{align} \pi^4-97 &= \int_0^1 \frac{240 \, x^{4} \, (1 - x)^{2} \,\left(3522267 x^2+1681375\right) } {3221561 \, (1 + x^2) } \log^3(x^{-1}) \, dx \\ \frac{195}{2}-\pi^4 &= \int_0^1 \frac{192 \, x^{6} \, (1 - x)^{2} \, \left(5657688 x^2+3056473\right) } {3641701 \, (1 + x^2) } \log^3(x^{-1}) \, dx. \\ \pi^4-\frac{487}{5} &= \int_0^1 \frac{15 \, x^{8} \, (1 - x)^{2} \, \left(3293858975 x^2+746556831\right) } {278611172 \, (1 + x^2) } \log^3(x^{-1}) \, dx. \\ \frac{1656}{17}-\pi^4 &= \int_0^1 \frac{480 \, x^{7} \, (1 - x)^{4} \, \left(8555775811 x^2+2883779820\right) } {39703971937 \, (1 + x^2) } \log^3(x^{-1}) \, dx. \\ \pi^4-\frac{2143}{22} &= \int_0^1 \frac{480 \, x^{31} \, (1 - x)^{4} \, \left(4071997316165706379 x^2+175446796437023645180\right) } {1199623593846005571607 \, (1 + x^2) } \log^3(x^{-1}) \, dx. \end{align}
Method
The idea is simple. We basically combine the following identities.
(1) \begin{align} \int_0^1 \log^{s-1}\left( x^{-1} \right) x^k \, dx = \frac{(s-1)!}{(k+1)^s}. \end{align}
For an even $s$
(2a) \begin{align} \int_0^1 \frac{ \log^{s-1}(x^{-1}) \, x }{1+x^2} \, dx &=2^{-s} \int_0^\infty \frac{ t^{s-1} }{ e^t + 1 } \, dt \\ &=2^{-s} \int_0^\infty t^{s-1} \left( e^{-t} - e^{-2\,t} + e^{-3t} - \cdots \right) \, dt \\ &= \frac{ (s-1)! \, (2^s - 2)}{4^s} \zeta(s) \\ &= \frac{ (1-2^{1-s}) \, |B_s| }{2 \, s} \, \pi^s. \end{align} where $\zeta(s)$ is the Riemann zeta function, $B_s$ is the Bernoulli number, which is rational. The last step is well known.
Similarly, for an odd $s$,
(2b) \begin{align} \int_0^1 \frac{ \log^{s-1}(x^{-1}) }{1+x^2} \, dx &=\int_0^\infty \frac{ t^{s-1} \, e^{-t} }{ e^{-2t} + 1 } \, dt \\ &=\int_0^\infty t^{s-1} \left( e^{-t} - e^{-3\,t} + e^{-5\,t} -\cdots \right) \, dt \\ &= (s-1)! \, \left(1-\frac{1}{3^s}+\frac{1}{5^s}-\cdots \right) \\ &= (s-1)! \, \beta(s) = \frac{|E_{s-1}|}{2^{s+1}} \, \pi^s, \end{align} where $E_s$ is the Euler number, which is also rational.
This means \begin{align} \frac{\pi}{4} &= \int_0^1 \frac{ 1 }{1+x^2} \, dx, \\ \frac{\pi^2}{48} &= \int_0^1 \frac{\log\left( x^{-1} \right) x}{1+x^2} \, dx\\ \frac{\pi^3}{16} &= \int_0^1 \frac{\log^2\left( x^{-1} \right) }{1+x^2} \, dx\\ \frac{7\,\pi^4}{192} &= \int_0^1 \frac{\log^3\left( x^{-1} \right) x}{1+x^2} \, dx. \end{align}
Now suppose we have a polynomial $P(x) = Q(x)(1 + x^2) + R(x)$, and we want $$ \int_0^1 \frac{ \log^s(x^{-1}) \, P(x) } { 1 + x^2 } \, dx = \pi^s - A, $$ where $A$ is an approximation of $\pi^s$ (we have assumed the possible sign, the case of negative sign is similar). To satisfy this equation, we demand,
\begin{align} \int_0^1 \frac{ \log^s(x^{-1}) \, R(x) } { 1 + x^2 } \, dx &= \pi^s \\ \int_0^1 \log^s(x^{-1}) \, Q(x) \, dx &= - A, \end{align}
This requires \begin{align} R(x) &= \begin{cases} \dfrac{ 2^{s+1} } { |E_{s-1}|} & \mathrm{for\; odd\;} s \\ \dfrac{ 2 \, s } { (1-2^{1-s}) \, |B_s| } x & \mathrm{for\; even\;} s \end{cases} \\ \sum_{n=0} \frac{ (s-1)! }{(n+1)^s} q_n &= -A, \end{align} where $Q(x) = \sum_{n=0} q_n \, x^n$.
Using these to rules to design $P(x)$, we get the above formulas. Particularly, we studied the form \begin{align} P(x) = x^u \,(1-x)^v (a \, x^2 + b \, x + c). \end{align}
For a particular set of $u$ and $v$, the constraint for $R(x)$ determines two parameters, say $b$ and $c$. The constraint for $A$ determines $a$. We then check if $a \, x^2 + b \, x + c$ is nonnegative definite. We then vary $u$ and $v$ to seek a simple combination of $a$, $b$ and $c$.
Best Answer
Let us assume that a Chinese mathematician can get an accurate approximation to $\pi$ by calculating the perimeters of regular polygons inscribed in and circumscribed around a circle. Then by trial and error or some other method, he can find small-denominator rational approximations to $\pi$. It is not likely nor necessary that Chinese mathematics had some now-unknown special theory for rational approximation.
Yes, there is a good reason for a near approximation: As you know, a convergent $h_n/k_n$ is the best approximation to a real number of all rational fractions with denominator $k_n$ or less. In fact, if $|\pi b - a| < |\pi k_n - h_n|$ for some rational fraction $a/b$, then $b \ge k_{n+1}$ (see Proposition 16). Hence, it is no surprise that 355/113 is a convergent of $\pi$. Furthermore, the inequality $$\left|\pi - \frac{h_n}{k_n}\right| < \frac{1}{k_n k_{n+1}}$$ given by Theorem 5 shows that the approximation $h_n/k_n$ is especially good (near) relative to the size of the denominator $k_n$ if the denominator $k_{n+1}$ of the next convergent is large (Corollary 2 to Theorem 5), which is the case for 355/113 as we see in the first few convergents of $\pi$: 3/1, 22/7, 333/106, 355/113, 103993/33102.
Your observation that the partial quotient $a_4$ = 292 is large is on target because $k_{n+1}$ is related to $a_{n+1}$ by the recursive formula $k_{n+1} = a_{n+1}k_n + k_{n-1}$.
By the above proposition, to get a rational approximation with error less than $|$113$\pi \, -$ 355$|$, we need a five-digit denominator!
Someone who discovers that 355/133 $\approx \pi$ is not lucky; rather he has a good understanding of rational approximation or is facile with calculations. On the other hand, it is probably lucky that 355/113 is a near approximation to $\pi$ because the partial quotients in the simple continued fraction expansion for $\pi$ are apparently random.