Hi, I am going to make a catalog of all these constants or functions defined by series, because I do not know them. "Terribly sorry about that" : 0:07 , 3:58 , 4:03 , 4:11 , 5:39 , 12:49 , 14:41 , 14:45 , 16:37 , 17:51 , 22:27 , "ok, cool" : 4:11 , 7:11 , 9:12 , 14:55 , 15:36 , 17:04 , 18:53 , 20:23 .
@maths_5058 ай бұрын
Wow there were alot in this video
@CM63_France8 ай бұрын
@@maths_505 Yes, and you can notice that at 4:11 I spotted them both, the reason for this is that, at this time, you said something about my counting, unfortunately I don't understand what, but it ends with "thank you very much", or some thing like that, so I'm not that worried 😃. It's a pleasure for me. By the way could you make a video about the alternate ways(1) of prooving that zeta(2)=pi^2 / 6 . There is one that starts from int_0^infty { int_0^infty { dx dy / (1-xy) }} that is not that easy, because it involves a couple of variable substitutions a bit tricky. (1) otherwise than the famous Euler proof.
@Tosi314158 ай бұрын
the result happens to be extremely close to the euler mascheroni constant aswell
@ericthegreat78058 ай бұрын
Oily-Macaroni
@xloppyschannel48818 ай бұрын
23 minute video has cost me my entire 8 hours of sleep, thanks.
@johnanderson2908 ай бұрын
Wow Kamaal, this was a beautiful beast! Most impressive! 👏
@thomasrichard70548 ай бұрын
Very good! The result is remarkably close to sqrt(3)/3, so numerical integration may fool us. 😁
@emanuellandeholm56578 ай бұрын
I believe the term "King's Property" is an expression used by speakers of Indian English. At least I've never seen it being used by a native US/UK English speaker.
@sanamite7 ай бұрын
Interestingly enough, I've just read it in a french instagram post too ! What term do US/UK native speakers usually use to name it?
@emanuellandeholm56577 ай бұрын
@@sanamite I guess they call it a "change of variable". It's interesting to me that the idiom exists in French. "King's property" sounds like royal real estate to me. :D
@juniorcyans29888 ай бұрын
This is a cool problem worth trying out. Thanks a lot!
@hewhomustnotbenamed59128 ай бұрын
8:08 We can take comfort in the fact that you're speaking from experience.
@txikitofandango8 ай бұрын
I understand how you did the integration by parts, but I can't imagine myself ever escaping from uv - integral of v du paradigm
@keyaanmatin48048 ай бұрын
4:37 is so relatable
@Mathematician61248 ай бұрын
It's very easy using fourier series. I got it 😊. Dear Friend
@mcalkis57718 ай бұрын
Look at that, the gang's all here. I guess e and γ didn't get the invite though. And I suppose Ω isn't really friends with anyone these days. Although at this point I fully trust that you would be able to find an integral whose value contains everything. π, e, γ, G, ζ(3), ln2
@romanvolotov8 ай бұрын
the amount of missing du's in this one is insane lmao
@christophercalvaire20148 ай бұрын
from ln(1+tan u),we could expand to series with bernoulli number at once
@christophercalvaire20148 ай бұрын
or I =int x ln (1+ tan au ) dx,using feymann trick, but it seems that I got an much simpler solution through this ode, I am not sure if I am right.dI/da=C-I*2/a
@MrWael19708 ай бұрын
There are some simple mistakes, but overall is very smart solution. Thank you indeed.
@PritamDas-zu9im8 ай бұрын
sir i think there is a mistake in 9:17 .you forgot to write cosine of log. its ln(cos{pi/4-u})
@uggupuggu4 ай бұрын
dw he still wrote cos in the next line
@txikitofandango8 ай бұрын
Didn't think a math video could be so salty :-p Result is very close to (error less than 0.1% of) the Euler-Mascheroni constant as well as the others mentioned
@mihaipuiu62318 ай бұрын
Sir math 505,...I try to understand a very nice integral from "a deceivingly difficult integral" and I met a formula, which I couldn't find proof. If you are kind to help me find this demonstration, because I can't continue to watch your integral.the formula is...ln(cosx) in terms of cos(2kx). Thank you very much for your help.If you agree to help the beginners like me to locate on the internet formula that you use without proof. This will be a huge help for people like me.I'm engineer and for me math is a HOBBY. Again thanks.
@maths_5058 ай бұрын
kzbin.info/www/bejne/o6Kzhal7ga-Dg5I
@malino05278 ай бұрын
Could you make a video about the integral from 0 to infinity of 1/(x^ln(x))? The result is the fourth root of e times the square root of pi, which I think is really beautiful. It seems that it uses the error function, if you plug it into wolframalpha you'll see.
@sarahakkak4087 ай бұрын
Great solution, but you was can use king rule in the first step
@Shindashi5 ай бұрын
Your pronunciation of Catalan's constant 18:44 threw me off, almost like Caplan.
@Shindashi5 ай бұрын
7:47 lmao thanks for that. F.
@yoav6138 ай бұрын
Very nice!
@Anonymous-Indian..20038 ай бұрын
I guess i solved that question some weeks ago.🤔 But i didn't remember the book in which I've found that. Btw, your solution is also amazing. Love❤ you bro !(No HOMO)
@Anonymous-Indian..20038 ай бұрын
I remembered now, I solved by simplifying. Integral Term is: Int(0 - π/2) (x²/2) { tan(x/2) + tan(π/4 - x/2) }dx = Int(0 - π/2) (1/2) { x² + (π/2 - x)² } tan(x/2) dx Now it'll be easy, Substitute, (1/2)tan(x/2) = sinx - sin2x + sin3x - sin4x + ......... And booooom.
@Aditya_1968 ай бұрын
😂 I understand the nomenclature being kings rule like it's something great but in reality and significance it's just simple , idk myself why they say it maybe something like chess related where u swap King and took just moving around, it helps a lot in problem solving despite being so simple
@kingzenoiii8 ай бұрын
F, definitely watching cuz i have no friends 😭
@maths_5058 ай бұрын
(with musical effects) you've got a friend in me
@kingzenoiii8 ай бұрын
@@maths_505 🤩
@maxvangulik19888 ай бұрын
now factor the result in terms of pi
@DestroManiak8 ай бұрын
How is it deceivingly difficult lol, it definitely looks difficult.
@maths_5058 ай бұрын
It honestly looks kinda innocent....throw in an x² and a couple trig functions....and then you question all your life decisions leading up to that point 💀