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

Question

I know that $\pi \approx \sqrt{10}$, but that only gives one decimal place correct. I also found an algebraic number approximation that gives ten places but it's so cumbersome it's just much easier to just memorize those ten places.

What's a good approximation to $\pi$ as an irrational algebraic number (or algebraic integer if possible) that is easier to memorize than the number of places it gives correct?

EDIT: Algebraic number preferably of low degree, such as $2$ or $3$ (quadratic or cubic).

Answer 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.

Answer 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:

Answer 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

Answer 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 $4$.

Answer 5

I'm hardly the first to think of this, but I might be the first to say it in this thread: $\sqrt{10} \approx \pi$ suggests that we look at the powers of $\pi$ and see which come closest to integers. Then do floor or ceiling on $\pi^n$ and that gives you an approximation as an irrational algebraic integer of degree $n$.

Hence $\root 3 \of 31$ (already mentioned by Zach), $\root 5 \of 306$, etc.

Answer 6

An interesting simple one is $$ \pi\approx \frac{3(3+\sqrt{5})}{5} = \frac{6\varphi^2}{5} \approx 3.1416407 $$ where $\varphi$ is the golden ratio.

If you don't mind the algebraic numbers expressed in terms of their polynomials, here are some polynomials that have a root very close to $\pi$:

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$$ x^3+120x^2+164x-\frac{86529}{50}=0 $$ for which all solutions are real, and one solution is $x\approx 3.14159265359006$

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$$ x^4-48x^3-12x^2-33x+1613=0 $$ for which one of the two real solutions is $x\approx 3.14159265358842$

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$$ x^5+2x^4-4x^3+76x^2+149x-1595=0 $$ which has exactly one real solution, $x\approx 3.14159265358998831$

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Also, here's an interesting way to express one of the other famous ways to approximate $\pi$, particularly $\sqrt[4]{2143/22}$: $$ \pi\approx \sqrt{\sqrt{\frac12+\frac{3!+4}{5+6}+7+89}} $$ Note that the numbers 1 through 9 appear once each, in order.

Answer 7

Dalzell's integral is related to the rational approximation $\pi\approx \frac{22}{7}$.

$$\pi=\frac{22}{7}-\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx\approx\frac{22}{7}$$

Similar small integrals are related to simple irrational approximations using $\sqrt{2}$ and $\sqrt{3}$.

$$\pi=\frac{20\sqrt{2}}{9}-\frac{2\sqrt{2}}{3} \int_0^1 \frac{x^4(1-x)^4}{1+x^2+x^4+x^6}dx\approx \frac{20\sqrt{2}}{9}$$

$$\pi=\frac{9\sqrt{3}}{5}+\frac{6\sqrt{3}}{5}\int_0^1\frac{x^3(1-x)^2}{1+x^2+x^4}dx\approx \frac{9\sqrt{3}}{5} $$

Fractions $\frac{20}{9}$ and $\frac{9}{5}$ are convergents of $\frac{\pi}{\sqrt{2}}$ and $\frac{\pi}{\sqrt{3}}$ respectively.

Answer 8

How about $$\pi \simeq \sqrt [3]{\cfrac{31}{1-\cfrac{12}{39^3-40}}}\tag{11 d.p.}$$ Easy to remember because $\pi^3\approx 31$ and $39$ is the number before $40$.

Also, a rather cool approximation is: $$\pi - e\simeq 1-\frac 1{\sqrt 3} + \frac{1}{\sqrt{11^6+13^5+19^4+24^3+33^2+\sqrt 5}}\tag{13 d.p.}$$

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Here are a few approximations given by Ramanujan:

$$\pi\simeq \frac{12}{\sqrt {130}}\log_e \bigg\{\frac{(2+\sqrt 5)(3+\sqrt {13})}{\sqrt 2}\bigg\}\tag{15 d.p.}$$

$$\pi \simeq \frac{24}{\sqrt {142}}\log_e \Bigg\{\sqrt{\frac{10+7\sqrt 2}{4}}+\sqrt{\frac{10+11\sqrt 2}{4}}\Bigg\}\tag{16 d.p.}$$

$$\pi \simeq \frac{12}{\sqrt {190}}\log_e\big\{(2\sqrt 2+10)(3+\sqrt {10})\big\}\tag{18 d.p.}$$

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A Chebyshev-type approximation: $$\pi\simeq \bigg(9^2+\frac{19^2}{22}\bigg)^{1/4}\tag{8 d.p.}$$

Answer 9

Some nice approximations can be produced by exploiting the ideas of Archimedes. The difference between a unit circle and an inscribed regular $n$-agon is made by $n$ circle segments. If we approximate them with parabolic segments and call $$ A_n = \frac{n}{2}\sin\frac{2\pi}{n}=n\sin\frac{\pi}{n}\cos\frac{\pi}{n} $$ the area of the inscribed $n$-agon, we get that $$ \pi \approx \frac{4 A_{2n}-A_n}{3} = \frac{n}{3}\sin\frac{\pi}{n}\left(4-\cos\frac{\pi}{n}\right)$$ where the absolute error behaves like $\frac{C}{n^5}$. Here some approximations derived through this geometric method: $$\begin{array}{l|c|l}\hline n=12 & 4\sqrt{6}-4\sqrt{2}-1 & 3.141104722\\ \hline n=24 & \sqrt{2}-\sqrt{6}+8 \sqrt{8-4 \sqrt{2+\sqrt{3}}}&3.141561971\\ \hline\end{array}$$ This can be further improved. For instance, since the MacLaurin series of $\frac{x}{\sin x}$ and $\frac{1}{15}\left(68+11\cos(x)-64\cos(x/2)\right)$ agree up to the $x^6$ term (the same idea has been exploited here) we have $$ \pi \approx \frac{n}{15}\sin\frac{\pi}{n}\left(68+11\cos\frac{\pi}{n}-15\cos\frac{\pi}{2n}\right) $$ and the following algebraic approximations:

$$\begin{array}{l|c|l}\hline n=6 & \frac{1}{10}\left(136+11\sqrt{3}-64\sqrt{2+\sqrt{3}}\right) & 3.141405312\\ \hline n=12 & \frac{\sqrt{3}-1}{5\sqrt{2}}\left(136+11\sqrt{2+\sqrt{3}}-64\sqrt{2+\sqrt{2+\sqrt{3}}}\right) &3.141589664\\ \hline \end{array}$$

Plenty of other approximations (both accurate and reasonably simple) can be derived by combining some version of the Shafer-Fink inequality and Machin formulas, for instance $$\pi\approx \frac{180}{16 \sqrt{20+6 \sqrt{10}}+6 \sqrt{10}+21}+\frac{90}{8 \sqrt{10+4 \sqrt{5}}+3 \sqrt{5}+7}$$ whose error is $<10^{-6}$, or $$ \pi \approx \frac{360}{7+7\sqrt{2}+6 \sqrt{2 \left(2+\sqrt{2}\right)}+16 \sqrt{2 \left(2+\sqrt{2}\right) \left(\sqrt{2+\sqrt{2}}+2\right)}}$$ whose error is $<4\cdot 10^{-7}$.

Answer 10

It is not a low degree polynomial, but easy to remember for sure.
We know that $$\frac{1}{1+x}=1-x+x^2-x^3+\cdots$$ Substituting $x^2$, $$\frac1{1+x^2}=1-x^2+x^4-x^6+\cdots$$ But we also know that $\int\frac1{1+x^2}dx=\arctan x$.
So let us integrate both sides (from $x=0$ to $x=y$), $$\arctan y=y-\frac{y^3}3+\frac{y^5}{5}-\frac{y^7}{7}+\cdots$$ Substitute $y=1$ and we get $$\pi=4(1-\frac13+\frac15-\frac17+\cdots)$$.

Answer 11

How about $$\frac {3.1415926535}{1}$$

It's fairly easy to memorize, and it's good to10 decimal places.