It would have distracted from the main thread of the video, but for those who are curious about how we can get numbers from this formula: we can obtain a power series for arctan by first taking the geometric series with common ratio -x^2 to get a power series for 1/(1 + x^2). Then integrating term by term we get a power series for arctan. We can use partial sums to get the approximation for both of arctan(1/5) and arctan(1/239) using arithmetic operations (addition, subtraction, multiplication, and division).
Excellent comment Min! Thanks! I did want to maybe say something about how both people used the identity to get the approximations, but I wasn't sure the best way :)
Thanks! Yes. It is more "visually-inspired" than a visual proof. Though the "Garfield's trapezoid" technique employed here can be used for lots of trig formulas like this so it is a nice visual technique.
Calculated 100 Mio digts in about 2 days distributed on 3 computers with 10 active kernels, some years ago. Y-Cruncher, some implementation of the Chudnovsky formula would do same in minutes.
3 minutes, 14 seconds long, released at 3:14 AM EDT on March 14th. Nice.
I aim to please :)
i seeked for this comment
And the likes is 413 (314 in reverse)!
It would have distracted from the main thread of the video, but for those who are curious about how we can get numbers from this formula: we can obtain a power series for arctan by first taking the geometric series with common ratio -x^2 to get a power series for 1/(1 + x^2). Then integrating term by term we get a power series for arctan. We can use partial sums to get the approximation for both of arctan(1/5) and arctan(1/239) using arithmetic operations (addition, subtraction, multiplication, and division).
Excellent comment Min! Thanks! I did want to maybe say something about how both people used the identity to get the approximations, but I wasn't sure the best way :)
at this point i am convinced you made this video weeks ago just to prepare for the timing :)
I definitely made this a few weeks ago :)
Nice video! The visualizations and computations are a bit more involved than in the other videos, but this is still great stuff.
Thanks! Yes. It is more "visually-inspired" than a visual proof. Though the "Garfield's trapezoid" technique employed here can be used for lots of trig formulas like this so it is a nice visual technique.
I like how you've made the video's length approximately as pi xD
😀👍
Calculated 100 Mio digts in about 2 days distributed on 3 computers with 10 active kernels, some years ago. Y-Cruncher, some implementation of the Chudnovsky formula would do same in minutes.
Happy pi day y'all!
A great day to celebrate math!
I am the #(3.14 reversed) likes!
👍😀
How did John Machin calculate arctan(1/5) and arctan(1/239) to compute pi?
Used the series approximation.
Neat!
I AM interested in doing a colaboration with You specially in discrete math, I know some manim myself
That was awesome!
Glad you liked it!
ERROR ángulos alfa y beta son distintos, por ende su arco tangente es distinta
? I said alpha is arctan a/b and beta is arctan x/y... so they are different.
you lost me at "= arctan 120/119" (@2:05). It's in no way explained where that comes from...?!
The formula is there and it says to make a and x both 5 and b and y both 12. Did you try plugging those in?
@@MathVisualProofs oh yes, thanks, stupid me, I missed that... my bad, sorry for "disturbing"... !🙏
@@M-F-H no worries! It’s pretty fast. Still working on pacing.
I'm with Zeno: you are wasting your time