Wallis product formula

Рет қаралды 12,477

6 жыл бұрын

In this video, I derive the Wallis product formula, which says that 2/1 2/3 4/3 4/5 ... = pi/2, where the numerator runs through all the positive even integers counted twice, and the denominator through all the adjacent odd integers. This proof involves only calculus, and is a delight for all the calculus lovers! Enjoy!

Пікірлер: 47

@drpeyam 6 жыл бұрын

At the beginning it’s supposed to be pi/2 instead of pi. Don’t worry, the proof is still correct 🙂

@philipgales4214 6 жыл бұрын

You sound like Neil Patel!!!!

@faith3174 6 жыл бұрын

I had seen 3b1b's proof for this which was amazing, but now I can say seeing this mathematically rigorous proof was just as amazing! Also, congratulations on passing 10k subscribers!! I'm so happy to see you're getting so much love from the math community

@drpeyam 6 жыл бұрын

Thanks so much ❤️❤️❤️

@dominikstepien2000 6 жыл бұрын

Imho proof presented by Dr. Peyam is even better because you can entirely do it by yourself without getting some crazy, ingenious idea

@user-wu8yq1rb9t 2 жыл бұрын

I was looking for Wallis product, and recently I discovered my dear Peyam already did it. Thank you ... قضیه‌ی فشردگی (Sandwich Theory) ,😀😀

@danielfisher587 6 жыл бұрын

Around 16:30, I think you can tighten the argument by stating that you are interested in powers of sin(x) only between 0 and pi, inclusive. In this range, 0

@elieabourjeily8520 6 жыл бұрын

Having seen this video , I feel nostalgic for my university years like 15 years ago. Big RESPECT Dr. Peyam

@tiripoulain 6 жыл бұрын

Great proof! I had never seen this before, and as someone who's freshly done with calc 2, I love it

@dhunt6618 6 жыл бұрын

I don't remember anything like this in my calculus class. Your proofs like these would have made the class much more fun!

@christopherbassett-romo7078 6 жыл бұрын

This is one of my favourite proofs, thank you Dr. Peyam!!

@MAULIKPATELnamste 3 жыл бұрын

Hearing such a soft tone, make me want to do more math with Dr. Peyam.

@albertemc2stein290 6 жыл бұрын

Great method of proving this formula with the recursive integral! Would be great seeing more of those!

@kuzu2104 6 жыл бұрын

Congrats on 10k subscribers! Amazing video as always 😍

@shanmugasundaram9688 6 жыл бұрын

An infinite product of the ratios of positive even integers to positive odd integers each written twice is equal to the area of a semi unit circle.Mathematcs is always beautiful.

@danielortega5211 6 жыл бұрын

Thank you for this beautiful proof! You're inspiring videos are among the reasons why I'm first year math undergraduate now, still waiting for Pi's Transcendence proof :-)

@bandamkaromi 6 жыл бұрын

Good recitation. Thank you, Dr. Peyam.

@tomvanmoer8202 6 жыл бұрын

Classic. I love this formula

@nathanielsharabi 6 жыл бұрын

Great video! Really suprised that this didnt require heavy tools and methods to prove. A high schooler that knows about the sqeeze theorem and a few other stuff can understand this.

@patrickp.6559 6 жыл бұрын

Wallis' integrals are super cool!

@ManiFunctor 6 жыл бұрын

Great video!

@kutuboxbayzan5967 5 жыл бұрын

good job

@ImPresSiveXD 6 жыл бұрын

Very nice! :D

@dr.rahulgupta7573 3 жыл бұрын

Very good Sir .I will attend your lectures till the last digit of Dr 3.14159..................m.

@John75ify 6 жыл бұрын

Nice video but small error at 14:45. Your product should have been (2k - 1)/(2k). It's odd integer over even integer. You fix it later without noticing though so I guess it was fine anyway.

@theoleblanc9761 6 жыл бұрын

At 20:00 you should use the recursive relationship! Because you have found the exact value of I(n) for all n based on the relationship and the you refind the relationship based on the exact value of I(n)...! By the way, quite good video! But I have a question: We clearly see that the expression is the product of all even squares divide by all odd squares but if we write it like that: Product_{n=1}^{infinity} (2n/(2n-1))^2 Or: Product_{n=1}^{infinity} (2n/(2n+1))^2 But (I am not sure) the first goes to infinity and the second goes to 0 (or at least it is less than 1 and so than π/2). Really interesting how changing a little bit the same "thing" change the result.

@TheGamingWattsit 4 жыл бұрын

At 9:40 why does I(2n) product stop at I(0) ?

@gadxxxx 5 жыл бұрын

I'd be interested to know how many terms of the product need to be taken to get pi / 2 correct to 7 decimal places?

@georgecooper7389 6 жыл бұрын

wow!

@fourier07able 4 жыл бұрын

How did Wallis may find this formula in 1656? Just there is one explanation. An alien did help him. Forget Calculus, and Analysis, it were horrible for that epoch. Even so he could study the integral of even and odd powers of sin x. A real feat!

@kutuboxbayzan5967 5 жыл бұрын

Calculate Γ'(n) and Γ'(n+1/2) n is natural number but n≠0 Know: Γ(1)=-γ γ is euler mascheroni constant γ≈0.577

@alejandroduque772 6 жыл бұрын

Min 16.28 how can i prove that the higher the power of sine is, lower the result. I mean if you take a value of sine from π/2 to π the result is negative, and when you raise It to an even power you Will get a positive, and when you raise It to an odd power you Will get a negative.

@drpeyam 6 жыл бұрын

Yes. But on (0,pi), sin is always positive

@alejandroduque772 6 жыл бұрын

@@drpeyam my bad for some reason I confused sine with cosine. Now I get it.

@philipgales4214 6 жыл бұрын

Anyone else think he sounds like and almost carries himself like Neil Patel?

@Koisheep 6 жыл бұрын

Came here just to like this video

@GreenMeansGOF 6 жыл бұрын

Is it incorrect to move the numerators? The producst could be written as π/2 = [(2/3)(4/5)(6/7)(8/9)...]^2 But I get the feeling that this is wrong.

@frede1905 6 жыл бұрын

I think it is wrong. The expression inside the product sign (pi) is (2k/(2k-1))•(2k/(2k+1)), which of course is not (2k/(2k+1))^2, which would be the expression inside the product sign of your sum.

@Aman_iitbh Жыл бұрын

yes its inncoreect as it will diverge we can check its divergence in wolfram alpha