What are better approximations to $\pi$ as algebraic though irrational number?

Solution 1:

(expanding my comments)

Let's start with the fraction $\;\dfrac{355}{113}\,$ easy to remember with something like :
"doubling the odds to be near the pi" (whatever this may mean...).

It is easy to find starting with the continued fraction of $\pi$ and stopping just before the (relatively) large term $292$ : \begin{align} \pi&=[3; 7, 15, 1\color{#00ff00}{, 292, 1, 1, 1, 2, 1, 3, 1, 14,\cdots}]\\ \pi&\approx \frac{355}{113}\approx 3.141592\color{#808080}{035}\\ \end{align}

My next step will be to compute the continued fractions of the first powers of $\pi\,$ and stop the expansion before the first large term (as previously) to get :

$$\frac {355}{113},\; \sqrt{\frac{227}{23}},\; \sqrt[3]{31},\;\sqrt[4]{\frac{2143}{22}},\;\sqrt[5]{306},\cdots, \;\sqrt[11]{294204},\cdots$$

After the first power the most interesting term was the fourth : \begin{align} \pi^4&=[97; 2, 2, 3, 1\color{#00ff00}{, 16539, 1, 6, 7, 6,\cdots}]\\ \pi^4&\approx \frac{2143}{22}\\ \pi&\approx \sqrt[4]{\frac{2143}{22}}\approx 3.14159265\color{#808080}{258}\\ \end{align}

Mnemonic : think at "three ways to reverse two two" that I'll note $\pi\approx \sqrt{+ \negthickspace+/}$ :

1. Power way: two $\sqrt{}\;$ to reverse the double squaring $^2$ $^2$.
2. Incremental way : reverse two times two terms of the $2\times 2$ terms $\;\underbrace{12}\underbrace{34}$
3. Divide by $\,22$.

This solution is interesting because of the large (omitted) $16539$. Should we incorporate this term in the c.f. then the next numerator and denominator would have around $4$ additional digits (since $\log_{10}(16539)\approx 4.2\;$ and from the method to obtain the next fraction in the first link).
The precision will be better with this supplementary term (say $4.3$ digits more) but we needed $4+4$ more digits for this. Without this term we used $4+2=6$ digits for a result of $10$ digits (excellent), with this term we have $8+6=14$ digits for a result of $14$ digits (average for a c.f.).

Searching the largest terms at the beginning of a c.f. (excluding the first non-zero term) should thus be rather interesting! Unfortunately c.f. coefficients as large as $16539$ are rather uncommon.

This result was found by Ramanujan and is given too by Mathworld with many others. $$-$$ Some additional results :

A palindrome for the fractional part of $\pi$ : $\frac 1{\large{\sqrt[5]{17571}}}\approx 0.1415926\color{#808080}{48}$ (with two more terms this becomes $\sqrt[5]{\dfrac{296}{5201015}}\approx 0.141592653589\color{#808080}{63}$). Another one : $\;\dfrac 1{\sqrt[8]{6189766}} \approx 0.141592653\color{#808080}{64}$.

We may too search continued fractions $\dfrac{\log\pi}{\log n}\,$ to obtain : \begin{align} 7^{10/17}&\approx 3.141\color{#808080}{35}\\ 6^{23/36}&\approx 3.1416\color{#808080}{09}\\ 7^{58701/99785}&\approx 3.1415926535\color{#808080}{9651}\\ \end{align}

Other random solutions perhaps nearer to OP's question (with some usual c.f. for reference) : \begin{align} \frac{22}7 &\approx 3.14\color{#808080}{2857}\\ \frac{8.5^2}{23} &\approx 3.141\color{#808080}{30}\\ \sqrt[3]{31}&\approx 3.141\color{#808080}{38}\\ \sqrt{51}-4 &\approx 3.141\color{#808080}{428}\\ \sqrt{4508}-64 &\approx 3.141\color{#808080}{64}\\ 4-\sqrt{\frac {14}{19}} &\approx 3.141\color{#505050}{60}\color{#808080}{49}\\ 7-\left(\frac{55}{28}\right)^2 &\approx 3.1415\color{#808080}{816}\\ 1+\left(\frac{60}{41}\right)^2 &\approx 3.1415\color{#808080}{82}\\ \sqrt{14434}-117 &\approx 3.1415\color{#808080}{83}\\ 2+\sqrt[17]{9.5} &\approx 3.14159\color{#808080}{78}\\ 5-\sqrt[5]{22+\frac{1}6} &\approx 3.14159\color{#808080}{62}\\ \sqrt{\frac{1961}2}-19 &\approx 3.1415\color{#808080}{898}\\ 2+\sqrt[8]{\frac{75}{26}} &\approx 3.141592\color{#808080}{19}\\ \frac{355}{113} &\approx 3.141592\color{#808080}{92}\\ \sqrt[11]{294204} &\approx 3.1415926\color{#808080}{36}\\ \left(\sqrt{\frac {1731}{76}}-3\right)^2 &\approx 3.1415926\color{#808080}{65}\\ \sqrt{6}+\sqrt[3]{\frac {61}{184}}&\approx 3.1415926\color{#808080}{45}\\ \sqrt{35}-\sqrt[3]{\frac{6215}{291}} &\approx 3.14159265\color{#808080}{266}\\ \sqrt[4]{\frac{2143}{22}}&\approx 3.14159265\color{#808080}{258}\\ 5-\sqrt[11]{913+\frac 16} &\approx 3.141592653\color{#808080}{37}\\ \sqrt{5}+\sqrt[4]{\frac{2323}{3455}} &\approx 3.141592653\color{#808080}{436}\\ \sqrt{4508-\frac 1{153}}-64 &\approx 3.1415926535\color{#808080}{28}\\ \sqrt[4]{\frac{788453}{95}}-\sqrt{41} &\approx 3.1415926535\color{#808080}{918} \\ \sqrt[4]{\sqrt{\frac{1087906}{63}}-34}&\approx 3.14159265358\color{#808080}{876}\\ \frac{5419351}{1725033}&\approx 3.141592653589\color{#808080}{815}\\ \sqrt{7}+\sqrt[8]{\frac{94680}{25912921}} &\approx 3.141592653589793\color{#808080}{309}\\ \sqrt{\sqrt{\frac{10521363651}{311209}}-174} &\approx 3.141592653589793238\color{#808080}{01}\\ \frac{21053343141}{6701487259}&\approx 3.141592653589793238462\color{#808080}{38}\\ \sqrt{\sqrt{\frac{20448668456155}{3958899937}}-62} &\approx 3.14159265358979323846264338\color{#808080}{5}\\ \sqrt{12}-\sqrt[3]{\frac{626510899334}{18676834489131}} &\approx 3.1415926535897932384626433832\color{#505050}{80}\color{#808080}{4} \end{align}

We could too use the integer relation algorithms as in Will Jagy's answer or this one but this seems more cumbersome for this problem.

Solution 2:

If you want to stay with degree two or three but no larger, find an implementation of PSLQ and feed it the quadruple (at incredible decimal accuracy) $$ \left(\pi^3, \; \pi^2, \; \pi, \; 1 \right) $$ so as to ask for integer relations, that is integers $a_3, a_2, a_1, a_0$ of not terribly large absolute value, so that $$ a_3 \pi^3 + a_2 \pi^2 + a_1 \pi + a_0 $$ is very close to zero. Then the relevant root of $a_3 x^3 + a_2 x^2 + a_1 x + a_0$ is a good approximation for $\pi.$

The others appear to be getting good results with degree four, you might try that, no more difficult once you have the code for the general cubic correct.

jagy@phobeusjunior:~$ gp
Reading GPRC: /etc/gprc ...Done.

                               GP/PARI CALCULATOR Version 2.5.5 (released)
                        i686 running linux (ix86/GMP-5.1.2 kernel) 32-bit version
                    compiled: Sep 30 2013, gcc-4.8.1 (Ubuntu/Linaro 4.8.1-10ubuntu4) 
                 (readline v6.3 enabled [was v6.2 in Configure], extended help enabled)

                                 Copyright (C) 2000-2013 The PARI Group

PARI/GP is free software, 
? Pi
%6 = 3.141592653589793238462643383
? q = algdep(Pi,4)
%7 = 5871*x^4 - 22872*x^3 - 7585*x^2 + 60199*x + 23027
? polroots(q)
%8 = [-1.311564323926921157096862611 + 0.E-28*I, -0.3879438664397374306161177256 + 0.E-28*I, 
2.453674351288873525029590438 + 0.E-28*I, 
3.141592653589793238462643859 + 0.E-28*I]~
? 

degrees five to ten

?  algdep(Pi,5)
%19 = 909*x^5 - 3060*x^4 + 1814*x^3 - 3389*x^2 - 723*x - 626
?  algdep(Pi,6)
%20 = 820*x^6 - 2340*x^5 - 565*x^4 + 67*x^3 - 1782*x^2 - 1008*x + 1460
?  algdep(Pi,7)
%21 = 306*x^7 - 1189*x^6 + 532*x^5 + 224*x^4 + 899*x^3 + 474*x^2 + 389*x + 485
?  algdep(Pi,8)
%22 = 27*x^8 + 46*x^7 - 256*x^6 - 564*x^5 + 43*x^4 + 672*x^3 - 104*x^2 - 201*x + 220
?  algdep(Pi,9)
%23 = 20*x^9 - 53*x^8 + 32*x^7 - 178*x^6 - 86*x^5 - 11*x^4 + 142*x^3 + 410*x^2 + 34*x + 21
?  algdep(Pi,10)
%24 = 2*x^10 - 5*x^9 - 17*x^8 + 47*x^7 - 64*x^6 + 146*x^5 - 58*x^4 + 79*x^3 + 110*x^2 + 23*x - 7
? 

degree three:

    ?  r = algdep(Pi,3)
    %26 = 91273*x^3 + 8437*x^2 - 960500*x + 104194
    ? polroots(r)
    %27 = [-3.342734408288101386537745201 + 0.E-28*I, 0.1087047799083921816885401406 + 0.E-28*I, 
3.141592653589793238462650438 + 0.E-28*I]~
    ? 
    ? 

degree two:

?  s = algdep(Pi,2)
%28 = 12610705*x^2 - 51111434*x + 36108636
? polroots(s)
%29 = [0.9114269040003652816200798826 + 0.E-28*I, 3.141592653589793238462659346 + 0.E-28*I]~

repeating degree ten, I like how the coefficients are small and begin with 2, I have not found any of these monic (beginning with $1$)

?  t = algdep(Pi,10)
%30 =   2*x^10 - 5*x^9 - 17*x^8 + 47*x^7 - 64*x^6 + 
       146*x^5 - 58*x^4 + 79*x^3 + 110*x^2 + 23*x - 7
? polroots(t)
%31 = [-3.416642530754670637725737702 + 0.E-28*I,
        0.1631777144832237629669559802 + 0.E-28*I, 
        2.659776825745310085407479343 + 0.E-28*I, 
        3.141592653589793238462643332 + 0.E-28*I,
       -0.4285725799568636122958113382 - 0.1971284716837764691749795140*I,
       -0.4285725799568636122958113382 + 0.1971284716837764691749795140*I,
        0.6277749736794889930752953905 - 1.073388946479318133923381580*I, 
        0.6277749736794889930752953905 + 1.073388946479318133923381580*I, 
       -0.2231547252544536053351545286 - 1.460683263806221846450712438*I, 
       -0.2231547252544536053351545286 + 1.460683263806221846450712438*I]~
? 

pretty graph:

Solution 3:

How about,

$$ \sqrt[3] {31}=3.14138...$$

Where, $31$ is the length of a month.

If you want memorable, you could always use,

$$\pi \sim \sqrt{{{69} \over {7}}}=3.139...$$

Do I really need to explain this one?

You could also use,

$$\sqrt{{69 \cdot 1001} \over {7 \cdot 1000}}=3.14117...$$

Where, $1001$ refers to the book 1001 Arabian Nights

Solution 4:

$\root 10 \of {93648}$ is marginally better than $\sqrt{10}$.

But one of the comments has a much better answer, with degree of just $4$.