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- เผยแพร่เมื่อ 30 ก.ย. 2024
- Full solution development for this ridiculously awesome integral making use of the golden ratio and leading to a beautiful result.
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Im so early there isn't even audio
Weird YT glitch, no worries ;)
Very satisfying integral. The fact that η(2) popped in both sub integrals is nice and makes you think about whether or not it could have been arrived at from that integral before splitting it.
Where do you find such integrals? They're all really cool. Do you have any textbooks you can recommend which have integrals like this?
This is a salivating beauty, and I cannot think of another way of describing it.
This integral possesses beautiful properties indeed. When replacing the golden ratio by a generic power z in C, we obtain the closed form: I(z) = (z^2 +1)/ (12*z) * Pi^2. And this yields the elegant reflexion formula I(z) = I(1/z). Its only zero seems to occur when z = +/- i 🙂
This is nice
This was gorgeous 😍. Thanks for the amazing result.
π²√5/12
Claude 3.5 finds it too from the phi fraction
You could simplify even more at the end since 1/(phi -1) is just phi. It turns into 3*phi-phi^2 = 3*phi - (phi + 1) = 2*phi - 1 = sqrt(5)
Hi,
The final result can be simplified into : sqrt(5) * pi^2/12
"ok, cool" : 2:31 , 2:58 , 11:26 ,
"terribly sorry about that" : 3:55 , 4:23 , 6:05 , 6:08 , 10:10 , 10:13 , 13:11 , 14:18 .
@@CM63_France damn I was terribly sorry a terribly lot this time 😂
**slaps roof of video** this bad boy can fit so much nice cancellation taking place
1/(φ-1)=φ
(3-φ)φ= 3φ-φ-1=2φ-1
@@insouciantFox I know but I just loved that final form 😭
φ + 1/φ is even nicer
Now i am starting a war 😅😅sqrt 5* pi^2 /12 is lot better. Kust kidding any form in maths is as beautiful &satisfactory as the other one
How many years work in integral department (Years of experience)
13:47 gamma(z) should be gamma(s), no? Also at 11:10 shouldn't it be t^(phi-2)?
11:13 shouldn't it be "phi -2" instead of "phi -1"? Cool result nevertheless.
Take it one step further by relating 'phi to kewness of fruit trees, thereby expanding the integral repetoir of Golden Ratios.
15:30 Here you could save a few steps by substituting phi - 1 = 1/phi.
You end up with: (1/phi^2 + 1) / (1/phi) = (phi^2 + 1) / phi = 1/phi + phi = phi - 1 + phi = 2*phi - 1 = 1.
Just your usual dilemma with "how much simplification should I actually do?" 😂
It's cheating to put phi in the intergrand I feel. Not surprising that phi pops out in the result.
It's cheating only if the solution did not make use of the properties of phi. Phi at the end is simply our reward😂
Thank you.
17th
Si arriva facilmente a I=Σ((-1)^k/(k+1))π/sinπ(k+1)Φ..poi,boh..tu hai usato un metodo diverso .io ,invece, ho usato..la serie logaritmica,la funzione beta,e poi la gamma reflection.. poi mi sono bloccato..ah ah...forse ho trovato l'errore:non si può sviluppare in serie logaritmica perché ln(1+x)...x,tra 0 e 1, è maggiore di 1..
Σ author💅
Mashallah! I've said it once, whoever has given you the name Kamal (perfection) has depicted you exactly! Please thank him/her for me.😊
@@trelosyiaellinika you're message has been conveyed to my mother 😂
Nice!
Just out of curiosity, Where do you get these integrals? Like what book/s?
I mostly just make em up or find them on the internet. Math stackexchange is awesome 🔥🔥