Yesterday was 50th partial sum, today is 100th partial sum

Friday, March 13, 2015
 

On Friday, the date, in month/day/year notation, reflects the first five significant figures of 3.131592904, the 100th partial sum of the Leibniz formula for pi. The Leibniz formula for pi, named after Gottfried Wilhelm Leibniz, is an alternating series:

$${\displaystyle {\frac {\pi }{4}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}},}$$

The series converges very slowly; it has logarithmic convergence, with ${\displaystyle n}$ correct digits requiring somewhere on the order of 10ⁿ terms. This means it takes hundreds of terms to calculate two digits after the decimal point; it takes 628 terms to get an approximation better than “3.14,” the version of pi that you may have used in your math classes in 7th or 8th grade. It takes 791 terms to get an approximation better than ${\displaystyle {\frac {22}{7}}}$, or 3.142857, another popular and simple approximation of pi. In case that wasn’t bad enough, it gets a lot worse. It takes over one million (10⁶) for six digits, one billion (10⁹) for nine digits, one trillion (10¹²) terms for 12 digits, and so on. This means that for high-precision calculations of pi, the formula should not be used directly; however, numerous “convergence acceleration” methods are available to use this formula to calculate hundreds or even thousands of digits. In 1992, Jonathan Borwein and Mark Limber used the first 1000 Euler numbers to calculate π to 5263 decimal places with the Leibniz formula.

Anyway, today’s date is 3/13/15, and after 100 terms of the Leibniz formula for pi, the sum is 3.13159 29035 58552. Notice how the first five significant figures of this sum (3.1315) have the same five digits, in the same order, as today’s date. Yesterday, on 3/12/15, the date matched the first five figures of the 50th partial sum. Note that 50 is half of 100. The 50th partial sum is 3.12159 46525 91010. Here’s the cool part: although both partial sums are incorrect on the second digit after the decimal points, both sums are correct on the third, fourth, and fifth digits, both being correct on the “159” part. This is not a coincidence. These are generated by the Euler sequence, the first four being 1, −1, 5, and 61. The asymptotic formula for the n-th partial sum is:

$${\displaystyle {\frac {\pi }{2}}-2\sum _{k=1}^{\frac {N}{2}}{\frac {(-1)^{k-1}}{2k-1}}\sim \sum _{m=0}^{\infty }{\frac {E_{2m}}{N^{2m+1}}}}$$

From this formula, we can derive a simple “convergence acceleration.” If ${\displaystyle n}$ is odd, the partial sum is slightly more than pi, and we can subtract ${\displaystyle {\frac {1}{n}}}$ to get a much closer approximation of pi. For even integers ${\displaystyle n}$, we add ${\displaystyle {\frac {1}{n}}}$ instead.

For ${\displaystyle n=50}$, we get 3.12159 46525 + 0.02 = 3.14159 46525.

For ${\displaystyle n=100}$, we get 3.13159 29035 + 0.01 = 3.14159 29035.

Not only is the latter entry correct to six digits (3.141592), but also, it is slightly better than Milü, or ${\displaystyle {\frac {355}{113}}}$, which is, by far, the best fractional approximation using a denominator under 1000.

${\displaystyle {\frac {355}{113}}}$ = 3.14159 29203, which agrees with the first six digits of pi (3.141592).

This weekend, get ready to celebrate a very special Pi Day. This is celebrated annually on March 14, since it is written as 3/14, which has the same digits as “3.14,” and this Saturday, March 14th, it will be 3/14/15, agreeing with FOUR digits after the decimal point (3.1415). Of course, with the Leibniz formula for pi, the sequence “3.141592” does not appear until after 1,530,012 terms. You can figure this out without having to add all those terms. It’s very simple. With Windows Calculator, here is how you can figure out the number of terms in under ten seconds:

pi – ${\displaystyle {\frac {1}{n}}}$ > 3.141592

${\displaystyle {\frac {1}{n}}}$ −7

${\displaystyle n}$ > 1,530,011.653, rounded up to the next integer, 1,530,012.

Now, as a follow-up exercise, how many terms do you need for 9 digits (3.141592653)?

pi – ${\displaystyle {\frac {1}{n}}}$ > 3.141592653

${\displaystyle n}$ > 1,695,509,434.130, rounded up to the next even integer, 1,695,509,436.

The upper partial sums must also begin with “3.141592653,” so:

pi + ${\displaystyle {\frac {1}{n}}}$

${\displaystyle n}$ 2,437,795,019.

Take the larger of the two results. The answer is 2,437,795,019, with a denominator of 2,437,795,019 × 2 − 1 = 4,875,590,037.

More Trivia: This year’s number, 2015, is also the denominator of the 1,008th term, and that is also when the lower partial sum reaches 3.1406 for the first time. “3.1415” first appears after 10,794 terms (denominator 21,587). Here are the numbers of terms for the intervening March 14ths between 2006 and 2015.

• 3.1407 (March 14, 2007): 1,122 terms (denominator 2,243)
• 3.1408 (March 14, 2008): 1,262 terms (denominator 2,523)
• 3.1409 (March 14, 2009): 1,444 terms (denominator 2,887)
• 3.1410 (March 14, 2010): 1,688 terms (denominator 3,375)
• 3.1411 (March 14, 2011): 2,030 terms (denominator 4,059)
• 3.1412 (March 14, 2012): 2,548 terms (denominator 5,095)
• 3.1413 (March 14, 2013): 3,418 terms (denominator 6,835)
• 3.1414 (March 14, 2014): 5,192 terms (denominator 10,383)

For the upper partial sums, the last time “3.1416” appears happens at 136,119 terms, with a denominator of 272,237. From that point onwards, all partial sums start with “3.1415,” and “3.14159” first appears on the 136,121st partial sum (denominator 272,241).