pi is actually an infinite product

Infinite product for sin(z):

15 percent off everything using the code MATHS505 on Advanced MathWear:

Complex analysis lectures:

If you like the videos and would like to support the channel:

You can follow me on Instagram for write ups that come in handy for my videos:

My LinkedIn:

[gesucristo]

With the double factorial it’s even more good looking

[markbrightwell]

The final expression for pi/2 includes an infinity over infinity form, i.e., inside the parentheses. An infinite product can be defined by pairing the terms in the numerator with corresponding terms in the denominator. However, with the pairing given by (2/1)(4/3)(6/5)(8/7)…, the first term is already too big to give the desired result. With the alternative pairing given by (2/3)(4/5)(6/7)…, all the terms are less than one. Thus, the result of the infinite product would be less than one.

[MikeB-qv]

Always cool. It's a great way to spend my lunchtime.

[idjles]

I lik Euler's intuition of the zeros of the sine function. It's how you see the Wallis function as really a view of the zeros of the sin function. Same goes for all orders of the Basel problem - they are jus the zeros of the sin function.

[GreenMeansGOF]

I think the final answer is a little deceptive. You use associativity of multiplication on an infinite product which may not be valid.

[illumexhisoka]

I remembered a very useful theorem that will proof that this product conv
if the sum -1+a_n conv
Then the product a_n cov to a non zero value if a_n≠0
but it doesn't necessarily go the other way between the sum and the product

[Anmol_Sinha]

Oh, this is the walli's formula and this derivation is inspired by eulers take on the basels problem

[daniellosh]

((2x4x6x . . . x2n)/(1x3x5x . . . x(2n-1))^2 = (2n+1)xpi/2 which diverges as n tends infinity

[akifbaysal]

I have doubt that this formulation as written in the final result is true. Consider this, single out 2:1, the all other factors 4/3, 6/5 etc is bigger than 1, consider for a moment they are 1 to make a low estimate.. Then

pi low estimate ~ 2 (2)^2 = 8.

I think there is some shift involved in ordering numerator vs denominator as some people pointing..

[mikecaetano]

Camille Paglia? "Look on my Works, ye Mighty, and despair!"

[kennethwilliams]

There some people who go nuts when you talk about this stuff, don’t know why…

[saddenn]

how wonderful, my time at 2am well spend

[aiyinliu]

As many people have commented here. This alternative form of the Wallis formula, while much more attractive, is not kosher in a mathematical sense. To prove the convergence you have to ‘massage’ it back to the original form (2k)^2/(2k-1)/(2k+1). But still, this form is cooler. So f**k mathematical rigor.😂

[UnknownGhost]

This is too easy to understand for engineering students

[illumexhisoka]

Great video
but I don't like the way you write it
It looks like it's going to diverge
Keeping multiplying numbers that are larger than 1

[Ghaith]

"Clear as say to him" they were built different back in the days

[TheBest]

somehow pi, infinity, and 0 are all equal. I know it, just can’t prove it, yet

[kravitzn]

The equation in the thumbnail is false because the product of 2k/(2k-1) from k=1 to infinity diverges.