Hey! Yep I put method 1 link in the description and here it is as well: ua-cam.com/video/EFwzcOaXRVI/v-deo.html Let me know if I missed any other links.
you dont need to use digamma function, can just integrate the sum wrt a and when you plug in initial condition of f(0)=0 you get the -sum (-1)^n ln(n+1) as the constant C so youre all set to use wallis prouct. Although this way also has the same convergence issue as the digamma way since the series C by itself is not convergent
Yep. I think I had a pop up message in the middle of the video somewhere but the only reason I didn't go back to Wallis product is that its too similar to the other video (first method).
This is mind blowing highly cerebral
thanks!
This is fantastic. Thanks for sharing and keep up the great work.
Hi Slavino. Thanks! Appreciate it 😊
very nice approach
Thanks Nico!
Wow, that was...insane! ; )
Yep I thought so too! 🤣🤣🤣
Could you please link the other methods.
Hey! Yep I put method 1 link in the description and here it is as well: ua-cam.com/video/EFwzcOaXRVI/v-deo.html
Let me know if I missed any other links.
made it to 10 already (in binary)
Quite true!!! 😂 And good because I’m out of ideas! 🤣
you dont need to use digamma function, can just integrate the sum wrt a and when you plug in initial condition of f(0)=0 you get the -sum (-1)^n ln(n+1) as the constant C so youre all set to use wallis prouct. Although this way also has the same convergence issue as the digamma way since the series C by itself is not convergent
Yep. I think I had a pop up message in the middle of the video somewhere but the only reason I didn't go back to Wallis product is that its too similar to the other video (first method).