I might do 10 videos on this one

Good video the creat enthusiasm for the subject. Absolutely magical for the learner. Totally trivial for the initiated.

[@owlsmath]

thanks!

[@slavinojunepri7648]

The Wallis product came up like magic. Very good one! 👍

[@owlsmath]

Yeah I enjoy that! It looks like you have a very messy solution and then it gets cleaned up 👍👍👍

[@MikeMagTech]

Great video! I look forward to going through all the variations on this one.

[@owlsmath]

Thanks Mike! Still haven’t decided but we’ll see what happens. I see some interesting ways

[@mohandoshi153]

Lovely evaluation. Waiting for the remaining 9 methods :).

[@owlsmath]

Ha! That would be amazing! Currently I think I only have 2 done out on paper so I may have set the bar too high. 🤣

[@waarschijn]

I'm gonna watch all 10

[@owlsmath]

😂thanks!

[@adandap]

I started with u = - ln(x) and then did a series expansion of 1/(1 + e^(-u) ) to get to the same place. I'm wondering if there is a 'direct' way of doing this that doesn't require recognising Wallis' product or some other derived result? Maybe I'll find my answer somewhere in the next nine videos. 😄

[@owlsmath]

😂😂😂 Method 2 is coming soon. It’s probably a little more complicated than this one. I’m not sure if this one will be considered ‘direct’ or derived. We will see

[@owlsmath]

But the next one doesn’t involve Wallis product

[@krisbrandenberger544]

@ 9:53 and onward, the n's should be k's.

[@owlsmath]

Yep exactly. Mistake. I said I was gonna change the variable but only did it under the PI.

[@akirakato1293]

i was able to get to the final ln(wallis product) form with the feyman technique variable a being (x^a-1)/(x+1)ln(x)dx but forgot about wallis product so I was stuck there lol.

[@owlsmath]

Ha! Makes sense. That step can be hard to recognize.