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Boris Gourévitch
The world of Pi - V2.57
modif. 13/04/2013



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John Machin
(1680 - 1751)



Remember !

Slice of life

John Machin is a not-so-well-known mathematician. He was born in 1680 was a profesor in astonomy in London. He discover in 1706 the formula =16arctan(1/5)-4arctan(1/239), which, thanks to the development of the power serie of arctan known since Grégory, allowed him to get the above formula... But for the experience historian of Pi, Machin played an important role, firstly because he was the first to calculate 100 decimal using his formula, but mostly because he opened the door of arctan formula research....

Around

The arctan formula are a simple and fast method to work out the decimals of Pi. Knowing the expansion arctan(x)= for x between -1 and 1, then all we need is to find the combinations of arctan giving /4=arctan(1). The smaller the term inside the arctan brackets is the faster the series associated with it converges. Today still, we sometime check the calculation of the decimals of Pi with this type of formula... It is true that the extration of square roots is still tedious and a rational sequence very useful...
One of the last record to date (51 billion decimals) was checked with Gauss's formula.
The other famous arctan formula whose combination is equal to à (see historic) :
(convention x.arctan(1/y)->x*y )

16*5-4*239 so, as we memorised, is from Machin 1706
20*7+8arctan(3/79) Euler 1755
4*2+4*3 Euler or Hutton 1776 no one agrees...
16*5-4*70+4*99 Euler, him again ! 1764
4*2+4*5+4*8 L. von Strassnitzky
8*3+4*7 Charles Hutton 1776 then Euler 1779
8*2-4*7 Hermann
12*4+4*20+4*1985 S. Loney 1893 then Störmer 1896
32*10-4*239-16*515 S Klingenstierna 1730
48*18+32*57-20*239 by the great Gauss himself !
48*38+80*57+28*239+96*268 Gauss yet again...
24*8+8*57+4*239 Störmer 1896
An by order of usefulness... in easiness of Calculation
44*57+7*239-12*682 85,67%
22*28+2*443-5*1393-10*11018 88,28%
17*23+8*182+10*5118+5*6072 92,41%
88*172+51*239+32*682+44*5357+68*12943 93,56%
100*73+54*239-12*2072-52*2943-24*16432 96,38%
12*18+8*57-5*239 96,51%
8*10-1*239-4*515 96,65%
44*53-20*443-5*1393+22*4443-10*11018 97,09%
17*22+3*172-2*682-7*5357 97,95%
16*20-1*239-4*515-8*4030 99,13%
61*38-14*557-3*1068-17*3458-34*27493 99,14%
227*255-100*682+44*2072+51*2943-27*12943+88*16432 99,32%
24*53+20*57-5*239+12*4443 99,61%
127*241+100*437+44*2072+24*2943-12*16432+27*28800 99,92%
4*5-1*239 100%

We measure the calculation cost of a formula such as Machin's by 1/log(5)+1/log(239).
That's the meaning of the percentage above...

I myself had fun in looking for a few formula and found with others...

128*107+128*122+28*239+96*268+48arctan(19/2167) and

732*530+732*563+128arctan(3/2611)+332arctan(27/64589)+48arctan(53/55479)+
+64arctan(6/15617)+28arctan(6/15617)+28*9703+100*14633

who have heavy calculation cost but with very fast convergence.

The most complete site on arctan (and who studies the usefulness of those formula...) is www.ccsf.caltech.edu/~roy/pi.formulas.html

Precision

Let's try to estimate roughly the number of terms that we need to calculate in the serie so to obtain d correct decimals of Pi. We can observe from the expansion of the arctan in series that we need to estimate n such that ,which after simplification comes down to n>. Given that, in the combination of arctans, this is the term where b is the smallest that predominates, for Machin's formula, we have then n>0,72d, which is a little well respected according to the trials...

Proof

It would be tedious - and to be honest useless! - to completly proof Machin's formula when it is the principle that is mostly important.... We just need to know a few results so to glimpse the full proof or the method to find similar formula.... Here they are:

1) arctan(a)+arctan(b)= (by composition of tan and by avoiding ab=1...)

2) (it follows from the above formula (1) )

3) +arctan(x) (still obvious from (1) )

4) with integers ai ,bi ,k
if and only if (a1+ib1)(a2+ib2)...(an+ibn) has an zero imaginary part.
(we notice that this is the case for Machin's formula with ai=5, bi=1 for i=1,...,16 and ai=239, bi=-1 for i=17,...20 since (5+i)16(239-i)4=-681386607803576157184)

The formula comes from the fact that natural logarithm is defined in complex by
ln(a+ib)= where p is relatif integer, and the property ln(ab)=ln(a)+ln(b) does the job...

5) from the same style : with integers k, a, b
if and only if (1-i)k(a+i)m(b+i)n is real.

Trials

n replace infinity in the above series...

n=2 : we get 3,14182 (3)
n=10 : 3,141592653589793294 (16)
n=50 : 72 correct decimals

we have therefore a convergence of roughly 1.4n (close to 1/0.72=1,388... founded above)

Acceleration of the convergence

Weirdly, if the good old Delta2 of'Aitken functioned well on Leibniz's  series of the same type, it seems that the terms of power (2k+1) deorientate a bit the Delta2. It's use hence is a little bit less useful than the calculation of a higher rank in the serie.


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