I also mentioned 7^7/4^9 = 3.1415.

http://www.wolframalpha.com/input/?i=6-6^%28-EulerGamma%29-FeigenbaumAlpha uses just a few characters in the input interpretation for 3.14159269

I saw this and immediately thought of convergents of continued fractions.
It turns out that the best approximations found by your method are all convergents:
Last[Convergents[Pi, 4]] = 355/113
Last[Convergents[Sqrt[2], 6]] = 99/70
Last[Convergents[E, 23]] = 848456353/312129649

I am not sure if they will always yield the highest efficiency rational approximation but it seems reasonable (though the base could be important?).

Hmm. We can use the natural logarithm and square roots? This will be hard to beat:

(ln -1)/sqrt(-1)

1/989 (4 digits) is the exact value of itself and its decimal representation is 462 repeating digits. Not sure what the score is since I couldn’t locate a Notebook version of the posting to do the calculations.

Interesting post. Using decimals is not exactly a fair comparison to fractions, since there is an implicit denominator (and base) that decimals get for free, and fractions don’t. It’s surprising to me that the results are so close.

This may not be in the desired spirit, but allow me to try to revive the value of 22/7. It is equal to 3.1 [base 7], which agrees with 3.14 [base 10], for a savings of 25%.

Log[I]*2/I + Exp[-4631]

is accurate up to 2011 digits and I only used roughly 5 digits in the formula. ;-)

@Greg and Lubos, I definitely had Log[-1] and Log[I] in mind with the “Pi related values of special functions” and “inverse trig functions” limitations, although Lubos, extra points for throwing in the useless Exp term to confusing things.
@ Prof Farsnworth Pi/1 definitely falls foul of the “no Pi related constants (such as Pi) rule.

@Vadim and Danial are both in the spirit of the competition. Unfortunately, my definition of using Mathematica InputForm, perhaps slightly unfairly, works against both of you as the “7^^3.1” is the base 7 input form at 6 characters, and cube root is “31^(1/3)” at 8 characters.

Sorry notebook was not included, I will try and get that changed. In the meantime, here are two crucial measures…

matchingCharacters[expr_, target_] := Ceiling[-Log10[Abs[expr – target]]] + 1;
characterCount[expr_] := StringLength[StringReplace[ToString[Unevaluated[expr], InputForm], ” ” -> “”]];
SetAttributes[characterCount, HoldAll];

You can actually get twice the digits you put in for any number of digits input. By using Newton’s iteration to find the root of sin(x), you can calculate pi very efficiently. For example, let a=3.14159. Now, we can find more digits of pi than put in by using:
a=a-(sin(a)/cos(a))
returning 3.1415926535897932(446…), 16 digits for 6 digits memorized.

If we double the digits I put in, I get:
a=3.14159265358, a-(sin(a)/cos(a)), returning 3.14159265358979323846264338327950(319…) = 36 digits for 12 input. (Triple!)

Since I can now double the number of digits I memorize, does that mean I win the contest, or does that count as a “pi-related value of a special function”? It does not directly return infinite digits like, say, sqrt(6*Zeta(2)) or log(-1)/sqrt(-1).

You could also logically extend this to higher order by using a cubic or higher order iteration like Halley’s method. If all else fails, lim n-> inf. of x_n+1=x_n-cos(x_n) = pi/2, letting you use the same principle applied to root solving for memorizing pi/2.

Anyway, just ideas. I guess I win if that is allowed; it doubles the memorized digits.

2*(1.5708+cos(1.5708)+cos(1.5708+cos(1.5708))+cos(1.5708+cos(1.5708)+cos(1.5708+cos(1.5708)))+cos(1.5708+cos(1.5708)+cos(1.5708+cos(1.5708))+cos(1.5708+cos(1.5708)+cos(1.5708+cos(1.5708)))))

Hi,
Here’s a possibility that does marginally better than 26%. Not earthshaking but it works although I don’t know if it is still in the spirit. My candidate is below:

36^^stqx16ojy6l0ingb8qsnm98frdc3kphixo5941pejlq7r26iu8gz30nln9075qzd4t6147fdjkciosc6q22u6fhn91s3nbz323hxlwgk2tft9qvfx4n4bbvwkqc5r79pz2vpdk31g6lt28x9qjj74il991mf73yri5lld3ptunmg08le2qxivkcatjohc9eo1z6ss0kylj2zt56xaylpito2ffezsf29e6tdyvofsnv58mqvid2kty91sxyqfe5hakic47rwcjpw0gr4lkettaqei25wsmzgcsol5baytu5g61o2m0k1g0img82t1 /
36^^96bpnsjun85xvchdyf6ljmy3vtyee0lboj79zymed7mw37nzyd76xzm7y2yx6lxs88dw2kqkybshqvl81106hf35k3u7e2225yxpx6h93fxruawzhiqkiqstipr6alv826rtp8i5h8vpm81vn5qxf9mf3dkvoahhbd4cudb9da9llihtx9ayxgp9u7w5ewizoicnzm8v3ku0phbpbzua2ff4qdxnia6gdyquzcozidwnztctaui0moz3sm1cumh2yswfoetvmlesht4pa479t23j205hco03gp8kww14x2i0mjb03ez9qav5vauy6

The main issue in this challenge is that of data compression. It’s clear that higher bases are often more compact than lower bases. I simply recast numerator and denominators of the convergents of pi into a higher base. The highest base I could use automatically in Mathematica is 36 (otherwise we have to extend the letter base 26 letters + 9 digits + 1). This higher base yielded efficiencies of around 35% at convergents of around 1000.

Of course, bases require extending the digit set from 9 digits to more characters. If we’re allowed to extend the character set indefinitely then we could, in principle, represent all countables (eg. the 1000’th convergent of pi) by one character which gives an efficiency of about 999/1000. This of course would come at the expense of (our) memory in identifying enough single symbols and establishing a universally accepted convention for such digits.

Well, if irrational functions are allowed so why not allow transcendental ones ?

ArcCos[-1]

Just on Number to remeber and its the very first natural one.

Cheers Robert

31415926535897932384626433832795028841 is prime.

If you could find pi(31415926535897932384626433832795028841), then the
nth prime number
n=pi(31415926535897932384626433832795028841) is 3141592653…841
since the n-th prime number is the inverse of the pi(x) for x is prime. So take (pi(3141592653…841)-th prime)/10^37

Unfortunately, the largest calculation of pi(x) I know of only
found values up to 4*10^22, and even that was only specific
values (they did not calculate anything other than a constant
times a power of ten). You could still try and calculate pi(x) for
large values with more modern tech. That calculation used the
most powerful home computers in the world as of the early
2000’s. Anyone with powerful computers today could at least try
and find pi(3141592653…841)

If recurrence relationships are not disallowed, then the following method has good 9th degree convergence (i.e. more “bang for the buck” than the 2nd degree convergence of the Newton-Raphson method):

In[1]:=
(* ——————————————————————— *)
mep=$MaxExtraPrecision;
$MaxExtraPrecision=Infinity;

matchingCharacters[expr_,target_]:=Ceiling[-Log10[Abs[expr-target]]]+1;
characterCount[expr_]:=StringLength[StringReplace[ToString[Unevaluated[expr],InputForm],””->””]];
SetAttributes[characterCount,HoldAll];

Timing[
characterCount[
Block[{n=6,p,aP,rP,sP,t,u,v,w,aN,rN,sN},p=3*9^(n-1);aP=N[1/3,p];rP=N[(Sqrt[3]-1)/2,p];sP=N[(1-rP^3)^(1/3),p];Do[t=1+2rP;u=(9rP(1+rP+rP^2))^(1/3);v=t^2+t*u+u^2;w=27(1+sP+sP^2)/v;aN=aP*w+3^(2(i-1)-1)(1-w);sN=(1-rP)^3/((t+2u)v);rN=(1-sN^3)^(1/3);aP=aN;rP=rN;sP=sN;,{i,n}];aN^-1]
]
]
Timing[
matchingCharacters[
Block[{n=6,p,aP,rP,sP,t,u,v,w,aN,rN,sN},p=3*9^(n-1);aP=N[1/3,p];rP=N[(Sqrt[3]-1)/2,p];sP=N[(1-rP^3)^(1/3),p];Do[t=1+2rP;u=(9rP(1+rP+rP^2))^(1/3);v=t^2+t*u+u^2;w=27(1+sP+sP^2)/v;aN=aP*w+3^(2(i-1)-1)(1-w);sN=(1-rP)^3/((t+2u)v);rN=(1-sN^3)^(1/3);aP=aN;rP=rN;sP=sN;,{i,n}];aN^-1]
,
Pi
]
]

$MaxExtraPrecision=mep;
Remove[mep,matchingCharacters,characterCount];
(* ——————————————————————— *)

This gives the following output:

Out[1]= {0.,383}
Out[2]= {2.776,161125}

So for 383 input characters, this matches the first 161,125 characters of Pi without having to wait around very long (2.776 seconds … if you are also using a laptop from 2008 that’s on its “last leg”).

Note: The “n=6” part in the Block controls everything … and “n=6” means that we only did 6 iterations of this method to get 161,125 characters of Pi. If we bump that up, then we’ll get a better approximation at the cost of time. Example:

– For n = 7, the Timing on matchingCharacters returns {34.897,1450165} … so if n = 7, then we’ll match the first 1,450,165 charcters of Pi in about 35 seconds.
– For n = 8, the Timing on matchingCharacters returns {504.694,13051532} … so if n = 8, then we’ll match the first 13,051,532 characters of Pi in around 8 minutes 30 seconds

Method Reference: J.M. Borwein & F.G. Garvan, “Approximations to Pi via the Dedekind eta function”, March 1996, accessed on July 9, 2011 from:
http://oldweb.cecm.sfu.ca/organics/papers/garvan/paper/html/node12.html

Very Respectfully,
Joshua Burkholder

Motivated by the title of your nice post: some rational approximations give better irrationality measures (μ) than others. The upper bound of 7.6063 is due to Salikhov (2008), according to http://mathworld.wolfram.com/IrrationalityMeasure.html
Very Respectfully,
Américo Tavares

Much prefer the toothpick and floorboard approach – lots of physical exercise

((99998! 2)/BernoulliB[99998])^(1/99998)/2
does very well: it’s not a program, and delivers 30,108 characters for 42 memorised (provided Mathematica is to hand, of course!)

And there’s a theorem commonly found in number theory books that states that the best rational approximations to a real number are the convergents of its continued fraction, although I haven’t got an online version to hand.

Since the month has now passed, I will officially draw the competition to a close.

I am going to declare Larry the winner both by the rules that I laid out and for the depth of his contributions. I think the Wolfram marketing budget can stretch to some runners-up for which I choose Nabil for clever use of bases and Richard Chapling (who I suspect broke the rule on special values of special functions, but in clever enough way that I couldn’t find the definition he based it on) and while they didn’t succeed in terms of the challenge, I liked Greg, Ricardo and Americo Tavares’ contributions. So T-shirts of their choice are going to each of these people. (Contact me if you were one of these people and didn’t get mail from me).

Yes indeed, it is useless to mention more than 2 decimals. Practically it is useless to take into consideration more than 3 digits for any measured quantity because the precision ranges between 1/999 and 1/100.
Think about PC vs slide rule.

I must admit that the article was an eye opener, and the provocative title is aptly choosen.
Still there are situations where the decimal approximation is unfit. 22/7 is better for in the head calculations. In using 16 bits micro processors without floating point 355/113 is usable for scaling: multiply into double precision, then divide. Multiplying by 31,416 then divide by 100,000 is not an option because the latter numbers are too big. This is explained in some detail in the classic “Starting Forth”.

Using 37 decimal digits of PI one can calculate the circumference of a circle the size of the visible universe, and only be off by about 1 nanometer. That’s good enough for most applications.

http://pi.lacim.uqam.ca/eng/approximations_en.html <- I think this is what you were looking for

Ramanujan's log(262537412640768744)/sqrt(163) seems best

here…. all you have to do is remember how to spell pie, which by the time you learn 3.14 does not take any extra memory:
http://s3-ec.buzzfed.com/static/imagebuzz/web03/2010/4/12/5/pi-pie-29590-1271063492-58.jpg

…that’s better than remembering 3 digits.

I win.

I think you may have missed the point of the 22/7 approximation. It’s not there to make things easier to remember. It’s there so that e.g. builders, gardeners building things know that if they have ‘integers’ (aka slabs, plant pots) they can arrange them in lines/circles in a 22/7 ratio with relatively few problems that glue can’t solve. The error at that level is manageable and is ‘positive’ e.g. your center line of 7 things fits into your circle of 22 rather than overlapping it.

22/7 is actually less entropic a string than 3.14. And therefore easier to remember (by a tiny margin)

The whole trick with Ramanujan’s approximation is that the 30-digit number he managed to obtain can be written as a 6-digit cube + 744. Hence the great reduction of digit size. He based himself not on fractions, but on powers (of e to a multiple of Pi). The whole idea is to be able to re-write that N-digit numerator, or denominator, or power in a way which drastically decreases its digit-size. Try calculating for instance Round (Pi ^ N), and then try to muster up some sort of digit reducing alternative spelling for the number thus obtained, then write Pi as Radical of order N from that alternatively spelled large number. Though I have to warn that I’ve already tried that, as well as few other tricks, based on other Ramanujan-inspired approximations, and came up short. :-)

Although I know both Pi and E and Phi with over 40 digits each, I think that the best bargain for Pi is 3. 14 16, because it gives you a precision in the range of 10^-6 for only 4 (instead of 6) digits memorized, with 3 + 1 = 4, and 14 & 16 being consecutive even numbers. 355 / 113 is also easy to remember, once you realize that it’s constructed from the first three odd numbers (1, 3, and 5), each appearing twice, starting with the denominator. Likewise for Sqrt (2): it is best to remember that 1 / Sqrt (2) ~= 0.707 (a simple-looking repeating sequence obviously close to 70 / 99), from which the 99 / 70 approximation flows easily.

So, I’m very late to the party (like all of the last few posters), but I really think it’s important to stress what Paul Draper and others have pointed out:

Human memory doesn’t work the same way computer memory does. We remember patterns far more easily than abstract numbers. I’m not a mathematician or an engineer, so I rarely calculate the circumference of a circle, and thus only remembered pie to 3 digits (3.14, obviously). Paul’s “For I know I chose knowledge to obtain life’s goal” has, in the last 5 minutes, taught me another 7 digits almost immediately. It’s easier to remember than 3.14159, let alone 3.141592654.

I think you should re-do all of these calculations based on entropy, rather than raw digits. If you do, you’ll see the rational representations shoot way ahead.

355/113 is only about 10-15 bits of entropy, assuming 3.3 bits per decimal digit (a standard estimate) as you only have to remember 3 digits – 1, 3, and the pattern. That’s actually really high, considering it’s also ignoring the pattern that cuts the entropy down to somewhere around 1 bit per digit. The value of pi that that 10-15 bits of entropy generates is about 23 bits of entropy – a huge savings. Also remember that each number you add is harder to remember than the last, which is kinda the point of the entropy calculation, so that 15 is a whole lot easier than that 23.

The phrase is a lot harder to calculate, because assuming each word is random you get an insane amount of entropy (thousands of bits), yet it is easier to remember than 355/113, so it’s obviously around 15 bits of entropy or less. It’s hard to quantify how much the pattern cuts it down, but it’s obviously a huge amount. 3.141592654 is much harder to remember. Plus, it looks a heck of a lot better as a signature than the number.

I don’t understand why you would post this. The fraction 22/7 is a much better approximation, both in absolute terms and when you consider how good it is with just a one-digit denominator. For comparison, 3.14 = 157/50.
| pi – 22/7 | ≈ 1.264*10^(-3)
| pi – 157/50 | ≈ 1.593*10^(-3).
I think you owe Archimedes an apology.

The rational approximation 22/7 is useful to teach kids without getting into decimals. Not that I am a fan to that ratio, but just an observation why the approximation lives.

That’s not as efficient as you’d think, when you take 0x into account.

0xFF vs 255 is one extra “digit”.
0xFFFF vs 65535 is no gain, add one and you’re behind by a digit.
0xFFFFFFFF is 10, vs 10 digits for 4294967295, and again, just add one to go behind again.