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).

  • 7
    $\begingroup$ Because I know plenty of rational approximations, and because that's the direction of my curiosity, not of any practical application (e.g., landscaping). $\endgroup$
    – David R.
    Sep 12, 2015 at 21:38
  • 12
    $\begingroup$ $\dfrac {355}{113}$ is hard to beat but $\sqrt{\sqrt{\dfrac{2143}{22}}}\approx 3.14159265258$ is kind of neat (double exchange of $1234$). $\endgroup$ Sep 12, 2015 at 22:36
  • 4
    $\begingroup$ It seems that I rediscovered a result of Ramanujan. See too Mathworld. $\endgroup$ Sep 12, 2015 at 22:57
  • 3
    $\begingroup$ @Raymond Ah, a root of $22x^4 - 2143$. Very nice. $\endgroup$ Sep 13, 2015 at 3:21
  • 4
    $\begingroup$ Thanks @Robert! It was obtained using the continued fraction of powers of $\pi$ and stopping before a 'large' term yielding : $$\frac {355}{113},\; \sqrt{\frac{227}{23}},\; \sqrt[3]{31},\;\sqrt[4]{\frac{2143}{22}},\;\sqrt[5]{306},\cdots, \;\sqrt[11]{294204},\cdots$$ Let's conclude with a $5$-digits palindrome for the fractional part of $\pi$ : $\frac 1{\large{\sqrt[5]{17571}}}\approx 0.141592648$ $\endgroup$ Sep 13, 2015 at 9:28

11 Answers 11


(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.


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:

enter image description here


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

  • $\begingroup$ Maybe I'm the only one, but I don't get the $\frac{69}{7}$ joke. $\endgroup$ Sep 18, 2015 at 3:03
  • 3
    $\begingroup$ Search Google, and $7$ is a "lucky" number. It's funny if you get the $69$ reference, it's a bit dirty ;) $\endgroup$
    – Zach466920
    Sep 18, 2015 at 3:21
  • 1
    $\begingroup$ Oh, I see. Kind of like the $rdr^2$ joke in that episode of The Simpsons Bart didn't get. Thanks for explaining it. $\endgroup$ Sep 18, 2015 at 15:16
  • $\begingroup$ @RobertSoupe hardy har har :) $\endgroup$
    – Mr Pie
    Mar 9, 2020 at 22:54

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

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


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.


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 $\pi$:

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

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

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

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.

  • 1
    $\begingroup$ +1. The first approximation you gave is by Ramanujan. See details in this answer. $\endgroup$
    – Paramanand Singh
    Jan 15, 2020 at 5:40

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.


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 $<4\cdot 10^{-7}$.


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

Here are a few approximations given by Ramanujan:

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

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

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

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


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 $$\pi=4(1-\frac13+\frac15-\frac17+\cdots)$$.

  • $\begingroup$ That gives rational approximations, though. $\endgroup$
    – Ian
    Sep 16, 2015 at 10:44
  • 1
    $\begingroup$ OP: "Or algebraic integer approximations though" $\endgroup$ Sep 16, 2015 at 10:46
  • $\begingroup$ In context, "irrational algebraic number (...or algebraic integer)". Between that and the title, I think the situation is pretty clear. $\endgroup$
    – Ian
    Sep 16, 2015 at 11:09
  • 1
    $\begingroup$ David should have said "or irrational algebraic integer" for extra clarity. $\endgroup$
    – user153918
    Sep 17, 2015 at 17:52
  • $\begingroup$ I love this series and props for @AdityaAgarwal for explaining it so clearly! However, users should be warned that this series converges very slowly as it bounces up and down above the true value of $\pi$ infinitely often. With that said, taking averages of successive approximations is a little better. $\endgroup$
    – Xoque55
    Apr 19, 2016 at 22:36

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

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

  • 3
    $\begingroup$ This is good only because I neglected to say "irrational algebraic integer" like Alonzo suggested. So no vote either up or down from me. $\endgroup$
    – David R.
    Sep 18, 2015 at 21:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .