Machin (1680-1751) is distinguished as an influential mathematician,
who contributed to Newton's work, and gave Taylor the insight on what
is now known as Taylor's Theorem.
Machin's pi formula is:
The starting point is the Gregory
Series, by which the arctans are calculated:
The class of formulae, of which the Machin is probably the best, is:
Naturally, as there are 4 unknowns, we need to do some trial and error.
a candidate formula
is not the way that Machin discovered his formula. It is a way to
search for candidate formula of this kind, which need to be verified by
working out the formula using vulgar fractions to preserve absolute
Because many of these formulae were to be used with pencil
and paper, the easier numbers are to be preferred. We therefore,
consider that pi/4 is about 0.78, which is about 4·1/5. This means we
can take m=4 and a=1/5. We can also take n=1, for the first try and
then do some trigonometry to solve the equation.
So 2.1 below is a candidate formula:
We need to solve this equation in one unknown (b).
the Second Tangent
Substituting in 2.1 and rearranging:
Take tangents both sides:
Using the tan(A-B) formula,
We note the formula for tan2θ:
From 2.2 we know that tanθ=1/5. Substituting in 2.6, we get:
We note the formula for tan4θ:
And substitute tan2θ in 2.8:
Substituting this value for tan4θ, and tan(π/4)=1 in 2.5
Noting that arctan(-1/239)=-arctan(1/239), and substituting the latter
into our equation
We obtain the Machin Formula:
For 14 Decimal Places
The number of terms required for 4 decimals accuracy with the Machin is
4 (k=3). Compare this with the Gregory
Series with x=1, which is 10,000 (ignoring rounding errors).
The Machin Formula:
composed of two arctans, and we can see that the first will converge
more slowly than the second. We therefore seek the number of terms for
the first to be accurate to 14 decimal places, ignoring rounding
errors. The general term for the first arctan is:
maximum error in the sum of an alternativing and converging series,
such as the Gregory Series is no greater than the value of the first
omitted term in the series.
Writing k+1=p, for convenience, and ignoring the sign, because it does
not affect value of the error, we seek a term such that:
This will be the first term we can omit. We require 14 decimals, so
need to be accurate to 0.000000000000005 or 1/(2·1014)
Multiplying throughout by p:
Substituting x=1/5, log(16)=1.204, log(0.2)=.301 in 3.4:
Multiplying throughout by -1:
Ignoring logp, as being small compared with p, p≅22 (22.182).
2k+1≅22, k=10.5, say 11 (because it must be an integer)
As this is the first omitted terms, then the last k we need to compute
is k=10, which means we need 11 terms, approximately (The correct
answer is k=9, and we need 10 terms).
We can be more accurate by transforming 3.7 to :
When p=22, 3.8 is 1.22, which means it might be too big. Just choosing
p=20, f(20)=-0.22, which means it is too small. Writing this in the
table below, and taking the differences between the values:
Using simply proportion x:2::0.22:1.44, or x=0.31, say 1. So we take
2k+1=21, k=10, and this is the first omitted term, so our k=9, and the
number of terms is 10. We note that this is the maximum error!