Machin's Formula for Pi

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Theorem

$\dfrac \pi 4 = 4 \arctan \dfrac 1 5 - \arctan \dfrac 1 {239} \approx 0 \cdotp 78539 \, 81633 \, 9744 \ldots$

This sequence is A003881 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


The calculation of $\pi$ (pi) can then proceed using the Gregory Series:

$\arctan \dfrac 1 x = \dfrac 1 x - \dfrac 1 {3 x^3} + \dfrac 1 {5 x^5} - \dfrac 1 {7 x^7} + \dfrac 1 {9 x^9} - \cdots$

which is valid for $x \ge 1$.


Proof 1

Let $\tan \alpha = \dfrac 1 5$.

Then:

\(\ds \tan 2 \alpha\) \(=\) \(\ds \frac {2 \tan \alpha} {1 - \tan^2 \alpha}\) Double Angle Formula for Tangent
\(\ds \) \(=\) \(\ds \frac {2 / 5} {1 - 1 / 25}\)
\(\ds \) \(=\) \(\ds \frac 5 {12}\)
\(\ds \leadsto \ \ \) \(\ds \tan 4 \alpha\) \(=\) \(\ds \frac {2 \tan 2 \alpha} {1 - \tan^2 2 \alpha}\) Double Angle Formula for Tangent
\(\ds \) \(=\) \(\ds \frac {2 \times 5 / 12} {1 - 25 / 144}\)
\(\ds \) \(=\) \(\ds \frac {120} {119}\)
\(\ds \leadsto \ \ \) \(\ds \tan \left({4 \alpha - \frac \pi 4}\right)\) \(=\) \(\ds \frac {\tan 4 \alpha - \tan \frac \pi 4} {1 + \tan 4 \alpha \tan \frac \pi 4}\) Tangent of Difference
\(\ds \) \(=\) \(\ds \frac {\tan 4 \alpha - 1} {1 + \tan 4 \alpha}\) Tangent of $45^\circ$
\(\ds \) \(=\) \(\ds \frac {120 / 119 - 1} {120 / 119 + 1}\)
\(\ds \) \(=\) \(\ds \frac 1 {239}\)
\(\ds \leadsto \ \ \) \(\ds 4 \alpha - \frac \pi 4\) \(=\) \(\ds \arctan \frac 1 {239}\)
\(\ds \leadsto \ \ \) \(\ds \frac \pi 4\) \(=\) \(\ds 4 \arctan \frac 1 5 - \arctan \frac 1 {239}\)

$\blacksquare$


Proof 2

\(\ds \map \arg {\paren {5 + i }^4 \paren {239 - i} }\) \(=\) \(\ds \map \arg {5 + i}^4 + \map \arg {239 - i}\) Argument of Product equals Sum of Arguments
\(\ds \) \(=\) \(\ds 4 \map \arg {5 + i} + \map \arg {239 - i}\) Argument of Product equals Sum of Arguments
\(\ds \) \(=\) \(\ds 4 \arctan \frac 1 5 + \arctan \frac {-1} {239}\)
\(\ds \) \(=\) \(\ds 4 \arctan \frac 1 5 - \arctan \frac 1 {239}\) Inverse Tangent is Odd Function


The validity of the material on this page is questionable.
In particular: We cannot take it for granted that $\map \arg z = \arctan \dfrac {\Im z} {\Re z}$, we have to be sure we know what quadrant we are in. Clearly here $\Re z > 0$ so complications don't directly arise, but we might want to use the cosine/sine definition to be rigorous
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Furthermore, because:

\(\ds \paren {5 + i}^4\) \(=\) \(\ds 5^4 + 4 \times 5^3 i - 6 \times 5^2 - 4 \times 5 i + 1\) Binomial Theorem
\(\ds \) \(=\) \(\ds 476 + 480i\) simplifying

we can write:

\(\ds \paren {5 + i}^4 \paren {239 - i}\) \(=\) \(\ds \paren {476 + 480 i} \paren {239 - i}\)
\(\ds \) \(=\) \(\ds 113764 + 480 + 114720i - 476i\)
\(\ds \) \(=\) \(\ds 114244 + 114244i\)


From there, we substitute:

\(\ds \map \arg {\paren {5 + i }^4 \paren {239 - i} }\) \(=\) \(\ds \map \arg {114244 + 114244 i}\)
\(\ds \) \(=\) \(\ds \arctan \frac {114244} {114244}\)
\(\ds \) \(=\) \(\ds \arctan 1\)
\(\ds \) \(=\) \(\ds \frac \pi 4\)

By transitivity:

$\dfrac \pi 4 = 4 \arctan \dfrac 1 5 - \arctan \dfrac 1 {239}$

$\blacksquare$


Also known as

Some sources give this as Machin's Identity.


Source of Name

This entry was named for John Machin.


Historical Note

John Machin devised his formula for $\pi$ in $1706$.

It allowed him to calculate $\pi$ to over $100$ decimal places.

This greatly surpassed the work of Ludolph van Ceulen.


Sources

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