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

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).

• Because I know plenty of rational approximations, and because that's the direction of my curiosity, not of any practical application (e.g., landscaping). – David R. Sep 12 '15 at 21:38
• $\dfrac {355}{113}$ is hard to beat but $\sqrt{\sqrt{\dfrac{2143}{22}}}\approx 3.14159265258$ is kind of neat (double exchange of $1234$). – Raymond Manzoni Sep 12 '15 at 22:36
• It seems that I rediscovered a result of Ramanujan. See too Mathworld. – Raymond Manzoni Sep 12 '15 at 22:57
• @Raymond Ah, a root of $22x^4 - 2143$. Very nice. – Robert Soupe Sep 13 '15 at 3:21
• Thanks @Robert! It was obtained using the continued fraction of powers of $\pi$ and stopping before a 'large' term yielding : $$\frac {355}{113},\; \sqrt{\frac{227}{23}},\; \sqrt{31},\;\sqrt{\frac{2143}{22}},\;\sqrt{306},\cdots, \;\sqrt{294204},\cdots$$ Let's conclude with a $5$-digits palindrome for the fractional part of $\pi$ : $\frac 1{\large{\sqrt{17571}}}\approx 0.141592648$ – Raymond Manzoni Sep 13 '15 at 9:28

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 $\pi.$
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: (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 : $$\frac {355}{113},\; \sqrt{\frac{227}{23}},\; \sqrt{31},\;\sqrt{\frac{2143}{22}},\;\sqrt{306},\cdots, \;\sqrt{294204},\cdots$$ 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{\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$\pi$:$\frac 1{\large{\sqrt{17571}}}\approx 0.1415926\color{#808080}{48}$(with two more terms this becomes$\sqrt{\dfrac{296}{5201015}}\approx 0.141592653589\color{#808080}{63}$). Another one :$\;\dfrac 1{\sqrt{6189766}} \approx 0.141592653\color{#808080}{64}$. 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{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{9.5} &\approx 3.14159\color{#808080}{78}\\ 5-\sqrt{22+\frac{1}6} &\approx 3.14159\color{#808080}{62}\\ \sqrt{\frac{1961}2}-19 &\approx 3.1415\color{#808080}{898}\\ 2+\sqrt{\frac{75}{26}} &\approx 3.141592\color{#808080}{19}\\ \frac{355}{113} &\approx 3.141592\color{#808080}{92}\\ \sqrt{294204} &\approx 3.1415926\color{#808080}{36}\\ \left(\sqrt{\frac {1731}{76}}-3\right)^2 &\approx 3.1415926\color{#808080}{65}\\ \sqrt{6}+\sqrt{\frac {61}{184}}&\approx 3.1415926\color{#808080}{45}\\ \sqrt{35}-\sqrt{\frac{6215}{291}} &\approx 3.14159265\color{#808080}{266}\\ \sqrt{\frac{2143}{22}}&\approx 3.14159265\color{#808080}{258}\\ 5-\sqrt{913+\frac 16} &\approx 3.141592653\color{#808080}{37}\\ \sqrt{5}+\sqrt{\frac{2323}{3455}} &\approx 3.141592653\color{#808080}{436}\\ \sqrt{4508-\frac 1{153}}-64 &\approx 3.1415926535\color{#808080}{28}\\ \sqrt{\frac{788453}{95}}-\sqrt{41} &\approx 3.1415926535\color{#808080}{918} \\ \sqrt{\sqrt{\frac{1087906}{63}}-34}&\approx 3.14159265358\color{#808080}{876}\\ \frac{5419351}{1725033}&\approx 3.141592653589\color{#808080}{815}\\ \sqrt{7}+\sqrt{\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{\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. How about, $$\sqrt {31}=3.14138...$$ 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 • Maybe I'm the only one, but I don't get the$\frac{69}{7}$joke. – Robert Soupe Sep 18 '15 at 3:03 • Search Google, and$7$is a "lucky" number. It's funny if you get the$69$reference, it's a bit dirty ;) – Zach466920 Sep 18 '15 at 3:21 • Oh, I see. Kind of like the$rdr^2$joke in that episode of The Simpsons Bart didn't get. Thanks for explaining it. – Robert Soupe Sep 18 '15 at 15:16$\root 10 \of {93648}$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. 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 $$<4\cdot 10^{-7}$$. 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, $$x\approx 3.14159265358998831$$ • +1. The first approximation you gave is by Ramanujan. See details in this answer. – Paramanand Singh Jan 15 at 5:40 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 $$\pi=4(1-\frac13+\frac15-\frac17+\cdots)$$. • That gives rational approximations, though. – Ian Sep 16 '15 at 10:44 • OP: "Or algebraic integer approximations though" – Aditya Agarwal Sep 16 '15 at 10:46 • In context, "irrational algebraic number (...or algebraic integer)". Between that and the title, I think the situation is pretty clear. – Ian Sep 16 '15 at 11:09 • David should have said "or irrational algebraic integer" for extra clarity. – user153918 Sep 17 '15 at 17:52 • 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. – Xoque55 Apr 19 '16 at 22:36
How about $$\frac {3.1415926535}{1}$$