Pi is a wonderful number, π = 3.14159265…. It’s very useful, ratio of the circumference of a circle to its diameter, or the ratio of area of a circle to the square of its radius, but it is irrational: one can show that it can not be described as an exact fraction. When I was in middle school, I thought to calculate Pi by approximations of the circumference or area, but found that, as soon as I got past some simple techniques, I was left with massive sums involving lots of square-roots. Even with a computer, I found this slow, annoying, and aesthetically unpleasing: I was calculating one irrational number from the sum of many other irrational numbers.

At some point, I moved to try solving via the following fractional sum (Gregory and Leibniz).

π/4 = 1/1 -1/3 +1/5 -1/7 …

This was an appealing approach, but I found the series converges amazingly slowly. I tried to make it converge faster by combining terms, but that just made the terms more complex; it didn’t speed convergence. Next to try was Euler’s formula:

π²/6 = 1/1 + 1/4 + 1/9 + ….

This series converges barely faster than the Gregory/Leibniz series, and now I’ve got a square-root to deal with. And that brings us to my latest attempt, one I’m pretty happy with discovering (I’m probably not the first). I start with the Taylor series for sin x. If x is measured in radians: 180° = π radians; 30° = π/6 radians. With the angle x measured in radians, can show that

sin x = x – x³/6 + x⁵/120 – x⁷/5040 + …

Notice that the series is fractional and that the denominators get large fast. That suggests that the series will converge fast (2 to 3 terms?). To speed things up further, I chose to solve the above for sin 30° = 1/2 = sin π/6. Truncating the series to the first term gives us the following approximation for pi.

1/2 = sin (π/6) ≈ π/6.

Rearrange this and you find π ≈ 6/2 = 3.

That’s not bad for a first order solution. The Gregory/ Leibniz series would have gotten me π ≈ 4, and the Euler series π ≈ √6 = 2.45…: I’m ahead of the game already. Now, lets truncate to the second term.

1/2 ≈ π/6 – (π/6)³/6.

In theory, I could solve this via the cubic equation formula, but that would leave me with two square roots, something I’d like to avoid. Instead, and here’s my innovation, I’ll substitute 3 + ∂ for π . I’ll then use the binomial theorem to claim that (π)³ ≈ 27 + 27∂ = 27(1+∂). Put this into the equation above and we find:

1/2 = (3+∂)/6 – 27(1+∂)/1296

Rearranging and solving for ∂, I find that

27/216 = ∂ (1- 27/216) = ∂ (189/216)

∂ = 27/189 = 1/7 = .1428…

If π ≈ 3 + ∂, I’ve just calculated π ≈ 22/7. This is not bad for an approximation based on just the second term in the series.

Where to go from here? One thought was to revisit the second term, and now say that π = 22/7 + ∂, but it seemed wrong to ignore the third term. Instead, I’ll include the 3rd term, and say that π/6 = 11/21 + ∂. Extending the derivative approximations I used above, (π/6)³ ≈ (11/21)³ + 3∂(11/21)², etc., I find:

1/2 ≈ (11/21 + ∂) -(11/21)³/6 – 3∂(11/21)²/6 + (11/21)⁵/120 + 5∂(11/21)⁴/120.

For a while I tried to solve this for ∂ as fraction using long-hand algebra, but I kept making mistakes. Thus, I’ve chosen to use two faster options: decimals or wolfram alpha. Using decimals is simpler, I find: 11/21 ≈ .523810, (11/21)² =  .274376; (11/21)³ = .143721; (11/21)⁴ = .075282, and (11/21)⁵ = .039434.

Put these numbers into the original equation and I find:

1/2 – .52381 +.143721/6 -.039434/120 = ∂ (1-.274376/2 + .075282/24),

∂ = -.000185/.86595 ≈ -.000214. Based on this,

π ≈ 6 (11/21  -.000214) = 3,141573… Not half bad.

Alternately, using Wolfram alpha to reduce the fractions,

1/2 – 11/21+ 11³/6•21³ -11⁵/(120•21⁵) = ∂ (24(21)⁴/24(21)⁴ – 12•11²21²/24•21⁴+ (11)⁴/24•21⁴)

∂ = -90491/424394565 ≈ -.000213618. This is a more exact solution, but it gives a result that’s no more accurate since it is based on a 3 -term approximation of the infinite series.

We find that π/6 ≈ .523596, or, in fractional form, that π ≈ 444422848 / 141464855 = 3.14158.

Either approach seems OK in terms of accuracy: I can’t imagine needing more (I’m just an engineer). I like that I’ve got a fraction, but find the fraction quite ugly, as fractions go. It’s too big. Working with decimals gets me the same accuracy with less work — I avoided needing square roots, and avoided having to resort to Wolfram.

As an experiment, I’ll see if I get a nicer fraction if I drop the last term (11)⁴/24•21⁴: it is a small correction to a small number, ∂. The equation is now:

1/2 – 11/21+ 11³/6•21³ -11⁵/(120•21⁵) = ∂ (24(21)⁴/24(21)⁴ – 12(11²21)²/24•21⁴).

I’ll multiply both sides by 24•21⁴ and then by (5•21) to find that:

12•21⁴ – 24•11•21³+ 4•21•11³ -11⁵/(5•21) = ∂ (24(21)⁴ – 12•11²21²),

60•21⁵ – 120•11•21⁴+ 20•21^2•11³ -11⁵ = ∂ (120(21)⁵ – 60•11²21³).

Solving for π, I now get, 221406169/70476210 = 3.1415731

It’s still an ugly fraction, about as accurate as before. As with the digital version, I got to 5-decimal accuracy without having to deal with square roots, but I still had to go to Wolfram. If I were to go further, I’d start with the pi value above in digital form, π = 3.141573 + ∂; I’d add the 7th power term, and I’d stick to decimals for the solution. I imagine I’d add 4-5 more decimals that way.

Robert Buxbaum, April 2, 2018