Calculating pi to ten decimal places with Machin's formula

Since it was pi day and I'm teaching a calculus course, I thought I'd try making a video about actually calculating pi by hand, along with a bit of the history. The approach we follow is Machin's formula from 1706, which he used to calculate pi to 100 decimal places through series representations of inverse trigonometric functions.
I intended this to go up by 3/15, but it turns out it takes time making videos like these! At least it's by rounded-up Ancient Egyptian pi day (3/17, from their approximation of 3 and 13/81).

[@ahuman2371]

Wow, this might be the wildest internet rabbit hole I've ever gone on. I had to do a chemistry lab for school and happened to find a report you did back in 2007, found your website, your youtube channel, and apparently you're also a math nerd‽ (Albeit, a little bit farther along the career than I am) What a small world

[@memsboii2117]

Thank you so much for this Video!!! I really needed it for a School assignment.

[@romin_1122]

Thanks

[@jamesrobertson9149]

I don't do things by half, so I only celebrate 2 pi

[@mr.sindel]

This is an amazing video! I really wish someone had shown me this when I was in high school after I learned enough trig to understand the derivation of Machin's Formula. @9:00 shouldn't the term -4(1/239-1/(3*239^2)) really be 4(1/239-1/(3*239^3))?

[@mathfacts271]

Yes, I got the exponent wrong there, thanks! (And happy pi day!)

[@mr.sindel]

How did you determine the number of decimal places needed at each calculation @9:57? Is there a technical way to do this? I want to push my students for more digits of accuracy and knowing that they don't have to perform each calculation to n decimal places at each step will speed up our computations quite a bit.

[@mathfacts271]

For the first series, it's from looking at how many 4/100's (25's) we're dividing by. I was conservative, and I rounded that to dividing by 10 for this. So, starting with 11 digits, each successive term needs one fewer digit. Surely you can be a bit less conservative, since it's more like log_10(25) digits per term (about 1.39 when rounded down). For the second series, I corrected the estimates around @11:10, and it's from conservatively thinking about dividing by 239 as dividing by 100, so for ten digits of accuracy you need two fewer digits each term (more like 2.37 digits).