Chapter 2. Seki Takakazu

Column Why does π/4=arctangent (1)=4 arctangent (1/5) - arctangent (1/239) hold true? (Level 3)

With respect to angle α, assume

tanα=1/5 {arctangent(1/5)=α}.

Then, we obtain

tan2α = 2 tanα/(1- Tan squared(α)={2×(1/5)}/{1-(1/5)^2} = 10/24 = 5/12.

By repeating this, we obtain

tan4α = 2 tan(2α)/(1 - tan squared (2α)) ={2×(5/12)}/{1-(5/12)^2} = 120/119 = 1+1/119 = tan(π/4)+1/119.

And now, as

tan(4α-π/4) = {tan(4α) - tan(π/4)} / {1 + tan(4α)×tan(π/4)} = (1/119) / (1+120/119) = 1/239

holds true due to

tan(α+β)=(tanα+tanβ)/(1-tanαtanβ),

the Addition Theorem with respect to tan,

Arctangent(1/239)= 4α-π/4= 4 arctangent(1/5)-π/4

or

π/4 = 4 arctangent (1/5) - arctangent (1/239)

is derived.

J. Machin, an astronomer from London, discovered that

tan4α= tan(π/4)+1/119

held true when

tanα=1/5

, and he calculated Pi to 100 decimal points in 1706 by using this relationship and

Arctangent (x)=x-x^3/3+x^5/5-x^7/7+ ….

When we actually calculate

π=16 arctangent (1/5)-4 arctangent (1/239)≒16(1/5-1/(3×5^3)+1/(5×5^5)-1/(7×5^7))-4(1/239-1/(3×239^3))

with an 8-digit calculator, we obtain a figure of 3.1415918.

Subsequently, relational expressions were discovered including the following:

π=24 arctangent (1/8)+8 arctangent (1/57)+4 arctangent (1/239), π=48 arctangent (1/18)+32 arctangent (1/57)-20 arctangent (1/239), π=32 arctangent (1/10)-4 arctangent (1/239)-16 arctangent (1/515)

Using such expressions, calculations to evaluate Pi to still further decimal places were attempted.

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