I'm always amazed when somebody discovers something this obscure. Why would anybody ever think to look for this?
@MasterChakra7 Жыл бұрын
It's the same old answer for Maths and Physics : cocaine
@donach9 Жыл бұрын
@@MasterChakra7amphetamine, I reckon
@plislegalineu3005 Жыл бұрын
@@donach9 Walter Weiss, the competitor of Leonard Euler
@logician1234 Жыл бұрын
What a crossover
@BikeArea Жыл бұрын
How far fetched can an identity be? Michael: Yes.
@looney1023 Жыл бұрын
I love e. My favorite is this stochastic definition: Let X_i ~ Uniform(0,1) for integers i. If N = min { n | X_1+X_2+...+X_n > 1 }, then E[N] = e. In words, if you repeatedly draw uniform random numbers between 0 and 1 and keep a running total of the sum, then on average, it will take e draws for the sum to exceed 1.
@Rócherz Жыл бұрын
Unrelated, but the video made me realize this: If Φ = (1+√5)/2> 1 is the positive root of x² -x -1 = 0, then Φ² = Φ +1, so that 1 = Φ² -Φ = Φ(Φ-1), but then, 1 = 1/Φ(Φ-1) = 1/Φ² × 1/(1-1/Φ). Since Φ > 1, we have 0 < 1/Φ < 1, thus *1 = 1/Φ² +1/Φ³ +1/Φ⁴ +1/Φ⁵ +1/Φ⁶ +…*
@cycklist Жыл бұрын
I love the magical tapping on the board.
@johntse5770 Жыл бұрын
U reli LOVE using Dominated Convergence Theorem to switch the order of summation and integration in proving summation identities. u've already shot a video on when Feynman's trick doesn't work. When will u shoot a video when this Theorem doesn't work? 🤩🤩
@gm-123-0 Жыл бұрын
10:04 is this wizardry?
@ilyafoskin Жыл бұрын
Michael is the math teacher at Hogwarts
@donach9 Жыл бұрын
I love the magic knock
@DanielGomes-sw2fd Жыл бұрын
8:30 The terms are all positive so you can always exchange the two summations.
@wesleydeng71 Жыл бұрын
Interesting. You could plug in other values than 1/ψ. For example, for x=1/2 one gets e^0.5 = replacing ψ with 2 in RHS.
@ingobojak5666 Жыл бұрын
Thanks for that! Indeed, for any 0
@sunritpal1037 Жыл бұрын
What the hell is up with that amazing thumbnail 😳
@mikecaetano Жыл бұрын
Sweet! Decomposition of unity meets up with friends...
@goodplacetostop2973 Жыл бұрын
16:50
@debussy_69 Жыл бұрын
Haven't checked out Michael in a while, glad you're still here
@aarong2374 Жыл бұрын
Didn't realize R. Schneider is in the band "the Apples in Stereo!"
@CM63_France Жыл бұрын
Hi, I suppose you mean : "our favorit identity involving e", but also some way : "defining" e . So mine is : lim_{n->infty} (1 + 1/n)^n . But if you mean only "involving", my prefered indentity is e^ix = cos x + i sin x .
@2manypeople1 Жыл бұрын
There is a yt-video titled "Unveiling Connections between Mathematical Constants: The Conservative Matrix Field". That would be a nice topic in this context.
@Anonymous-zp4hb Жыл бұрын
Damn, that's beautiful.
@minwithoutintroduction Жыл бұрын
رائع جدا كالعادة. الطريق شاقة و الوصول ممتع
@d-nize Жыл бұрын
Why should numbertheoretical functions like these two (especially my(n)) combined with an geometric (classic sense, ratio of certain lengths) defined number be in any relation to a trancendental number? This is mathemaGics. :D Would be nice to see following scenario. Use this expression as a Definition for exp(1). Extend the definition to exp(x) for real x and assuming exp(ix) = cos(x)+isin(x) find similar expressions for sin(x) and cos(x).
@hcgreier6037 Жыл бұрын
Woouh! That's hard to grasp, but very nice!🤣
@ceebongo Жыл бұрын
What. In. The. Actual. Phi.
@wenkoibital4779 Жыл бұрын
e is everywhere /\ e is golden ---> means : gold is everywhere.😀
@Alan-zf2tt Жыл бұрын
Amazing - it has a poetic beauty all of its own. And I suppose it emphasizes: all of math is really a human construct or does it? 🙂
@abdoshaat3304 Жыл бұрын
This topic under what branch in mathematics
@synaestheziac Жыл бұрын
Almost e hundred thousand subs!
@gp-ht7ug Жыл бұрын
Very interesting identity! I am a simple man and my favorite identity is Euler’s identity e=cosθ+isinθ
@lucid_ Жыл бұрын
Thats e^(iθ) but yeah i see your point
@ruffifuffler8711 Жыл бұрын
? Something about pincers and prongs ?
@ahmedgg8867 Жыл бұрын
Nice
@charleyhoward4594 Жыл бұрын
too abstract ....
@mariochavez3834 Жыл бұрын
Indeed m8 But you can't deny the coolness of this identity
@michaelgolub2019 Жыл бұрын
It seems that there was a mistyping in #\delta$ definition: $\mu e0$ is to be replaced by $\mu e1$.