The World's Most Beautiful Formula For Pi
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- เผยแพร่เมื่อ 22 พ.ค. 2022
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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
#math #brithemathguy #wallisproduct
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🎓Become a Math Master With My Intro To Proofs Course!
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When I click the link, it just directs me to the homepage of brilliant. Does this mean 200 people limit has already been exceeded?
“It’s great that you like to learn by video, but the best way to learn is to do it yourself”
One of the few quotes that lasted all this time
Hey Bri, I truly appreciate your endeavors to show us all that fascinating math stuff in such a straightforward way. Keep up the good work! :D
For the curious, I wrote this python ten-liner to compute pi using the Wallis product:
a = 2.
n = 1
while n < 6543210:
if n % 2 == 1:
x = float(n + 1) / float(n)
else:
x = float(n) / float(n + 1)
a = a * x
n = n + 1
print a
a = 1
for n in range(2, 13086420, 2):
a = a * n * n / (n ** 2 - 1)
print(a)
Also the same thing
I think this might be the best video you've done yet! Super interesting, sufficiently rigorous, and extremely well-presented.
Good stuff! This is how I showed my boy how to derive the Wallis product. We used this result to derive Sterling’s approximation for the factorial.
This is the clearest version of the
proof on youtube! Thank you!!
Whoa, Nice to see u again buddy; was afraid as there had been no videos so far
these videos are really great i tell you you keep up your good work , and i will keep watching
Cool! Great content, your channel is exceptionally useful!!!
One of the best videos that i have ever seen about math
thanks for the concept.
bro could you please tell form which software do make this kind of video❤
Excellent video Thank-you very much. Never knew that pi can be expressed as an infinite product of some sequences of rational numbers.
Riemann's zeta function can be expressed as an infinite product of rationals (proof from Euler) and zeta(2) = pi² / 6. So that was at least true for pi². For pi, it was indeed not obvious.
Thank you
Great Video!
That was actually a beautiful proof!
Another way
In your video of the Bessel problem
sinx/x=(1-x²/π²)(1-x²/4π²)…
Now plug in x=π/2
2/π=(1-1/2²)(1-1/4²)…=1/2·3/2·3/4·5/4…
π/2=2/1·2/3·4/3·4/5…
Great observation. I think you mean the Basel problem.
@@MichaelRothwell1 Oh thank you
I kinda "discovered" this while playing on my calculator with fractions lol
It's a Beautiful Formula for Pi. Pi can be expressed as an infinite product of some ordered sequence of rational numbers. It raised me in doubt that "Is Every Irrational number can be expressed as an infinite product of some ordered sequence of rational numbers ?"
Hi
I love math
my brain hurting can't force me to not thanking you because of this awesome content
Very neat! 💯
Thanks a lot!
Welcome!
U are a genius
I always wanted to know that 👍
Was this animated with manim?
“The most beautiful expression for π" that’s what I called it when I first learnt about the Wallis Product!!!
meanwhile me, someone who doesn't know math other than the basics, watching this pretending it makes sense to me
Bro I don't know how to become a member of your Channel. Please tell me how can I become your channel member. I already subscribed your channel. And I also love your videos so much because of your problem solving skills. Please tell me
Hit the blue Join button under a video :)
Great video, but you made a mistake on the left side of the equation. It should be tau over four. Other than that, spot on!
😏
Nice video. The starting formula for pi does not converge well at all as after 300 terms you do not even get the first correct digit if 3 for Pi.
What about (2/1)*(3/4)*(6/5)*(7/8)*(10/9)*(11/12)*...?
Me who has no clue what that equation is: I N T E R E S T I N G
Is there a way to use lower knowledge?
Numbers
@@BambinaSaldana numbers what?
@@anhaomaithi1273 You do things to numbers with other numbers then you get a new different unexpected number.
@@BambinaSaldana Oui, oui bambina.
But...you can't use L'Hopital's rule for the limit because n is in the Natural set, not the real one, therefore you can't take any derivative since there is no infinitesimal variation
Actually, when taking n to infinity yes you can apply L’H (assuming it’s in an indeterminate form to start with). We had to prove this in my real analysis class. You’d have issues if you were approaching an actual number, like n approaching 0 for example. Also, you don’t have to use L’H for that limit, you can actually prove it using standard limit techniques. First you show it’s bounded (I’ll leave that to you to do), then you say lim (2n+1)/(2n)=lim (1+(1/2n)) =1+lim(1/2n)=1+½lim(1/n) and clearly lim(1/n) approaches zero, so the original limit must approach 1
@@andrewkarsten5268 oooh, i'd live to see the proof of it in the natural set (also yes, there is always another way to do the proof without usine L'hôpital's rule, which is often the one my teachers do prefer).
@@calebgatty5773 well generally you should only use L’H when none of the other standard tools work. As far as L’H on the natural set, I think we need to be clear on a few things first. The functions 2n and 2n+1 are defined for all real numbers n. Therefore, if we consider using them as continuous functions L’H has no issues. There’s a criterion for limits which states if lim f(c) exists, then for all sequences a_n such that a_n approaches c, lim f(a_n) must approach the same limit as lim f(c). Therefor, if L’H works on a continuous function as n approaches infinity, then the limit as n approaches infinity on the natural set must be the same. This is kind of an understood thing, that in order to apply L’H on the natural set as n approaches infinity, the functions need to be continuous and differentiable on [1,∞). Thing is, for basically every function of the naturals, there exists a unique analytic continuation for that function that is continuous and infinitely differentiable, and you can apply L’H to that analytic continuation (think of the gamma function, it is the analytic continuation of (n+1)!). With this in mind, it’s well defined to say take the analytic continuation, and apply L’H to that continuation, and this continuation will not contradict the true limit in any way because it will give the same result.
Niiiiice
kuliah kalkulus matematika formula
forgot *e at the end. ;
W0W!
why write the product like that, why not (2/3)^2(4/5)^2(6/7)^2...
Cool
This integration was asked in JEE.
Hey Bri, I came up with a new math theorem. This theorem is used to find some pythagorean triples, so thanks to my theorem, there’s no need to memorize them. Let the shortest side of the right triangle have an odd length equal to n, where n is > or =3. Then square n and divide by 2. Then add and subtract this result by 0.5. These are the other 2 side lengths. For example, let 17 be the shortest side. 17^2/2 = 144.5. Add and subtract 0.5 to get 144 and 145. And there you go, 17-144-145 is a pythagorean triple. And you can multiply and divide these numbers by anything you want, as long as all side lengths have all been divided/multiplied by the same amount, and it’s also a pythagorean triple because the resulting triangle is similar.
Now what it the shortest side, n, was even? Well, given n> or =4, then (n/2)^2 +/- 1 would give the lengths of the other 2 sides. For example, if 20 was the shortest side, then (20/2)^2= 100, +/- 1 = 99 and 101. And there you go, 20-99-101 is a pythagorean triple. In the 3 4 5 right triangle, 4 isn’t the shortest side, but still, (4/2)^2 +/-1= 3 and 5.
Doesn’t work for a lot of numbers, for example 13, the other triple with 20: 20,21,29 etc.
@@Hesselaer You’re right, but my formula can still find many pythagorean triples.
@Paul L First of all, Nice work 👏 But sadly it's already done by someone else. The formula is already there. For n is the odd shorter side, (n^2-1)/2 is longer side & (n^2+1)/2 is hypotenuse. For n is the even shorter side, (n/2)^2-1 is longer side & (n/2)^2+1 is hypotenuse. Therefore, Pythagorean triplets are if n is odd, { n, (n^2-1)/2, (n^2+1)/2 } , if n is even, { n, (n/2)^2-1, (n/2)^2+1 }. In fact, this formula doesn't make every Pythagorean triplets.
This only finds A Pythagorean triple, it does not find EVERY Pythagorean triple
The Ancient Greeks came up with the general formula for Primitive Pythagorean triples (i.e. those with no common factor). It is m²-n², 2mn, m²+n², with m, n coprime and of opposite parity with m>n. Your formulae are partícular cases:
For odd numbers k=2p+1, your formula gives the triple k=2p+1, (k²-1)/2=2p²+2p, (k²+1)/2=2p²+2p+1, e.g. 5,12,13 in the case p=2, k=5, which corresponds to the general formula with m=p+1, n=p.
For even numbers k=2p, your formula gives the triple k=2p, (k²-4)/4=p²-1, (k²+4)/4=p²+1, e.g. 8,15,17 in the case p=4, k=8, which corresponds to the general formula with m=p, n=1.
I am going to sleep
you had to ruin it by using stupid notation
third
Hi bri..
Please start a precalculas and calculas courses for all of us don't know calculas.I understood nothing in this video.I love the way You teach so please...🤎💜
3blue1brown does a great course on calculus called “the essence of calculus” that’s how I learned. Pre calc is just trig
then learn calc, it's easy lol.
@@andrewkarsten5268 You didn't learn calc. You might have gained a bit of intuition, but you're not capable of doing proper calculus.
@@ffc1a28c7 actually I did learn calc. I’m a junior in college as a math major, and have already taken through calc III. I’ll give you the course doesn’t give you the rigorous “how to” knowledge of calculus, but the original comment was “I understood nothing” meaning he needs the conceptual understanding of what calculus does. Arguably, the conceptual is more important than the specific details on which to learn first, so it’s a great introduction for calculus. But don’t tell me I don’t know calculus, that’s beyond condescending, and you clearly know nothing about me so to assume I’m not capable of passing a calc exam is really low
@@ffc1a28c7 also as far as this video is concerned, 3blue1brown does teach everything you need to know to understand the “calc” done here