A Super Exponential Power Tower

🤩 Hello everyone, I'm very excited to bring you a new channel (aplusbi)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermathshorts
/ @aplusbi
❤️ ❤️ ❤️ My Amazon Store: www.amazon.com...
When you purchase something from here, I will make a small percentage of commission that helps me continue making videos for you.
If you are preparing for Math Competitions and Math Olympiads, then this is the page for you!
CHECK IT OUT!!! ❤️ ❤️ ❤️
❤️ A Differential Equation | The Result Will Surprise You! • A Differential Equatio...
❤️ Crux Mathematicorum: cms.math.ca/pu...
❤️ A Problem From ARML-NYSML Math Contests: • A Problem From ARML-NY...
❤️ LOGARITHMIC/RADICAL EQUATION: • LOGARITHMIC/RADICAL EQ...
❤️ Finding cos36·cos72 | A Nice Trick: • Finding cos36·cos72 | ...
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/...
Follow me → / sybermath
Subscribe → www.youtube.co...
⭐ Suggest → forms.gle/A5bG...
If you need to post a picture of your solution or idea:
in...
#radicals #radicalequations #algebra #calculus #differentialequations #polynomials #prealgebra #polynomialequations #numbertheory #diophantineequations #comparingnumbers #trigonometry #trigonometricequations #complexnumbers #math #mathcompetition #olympiad #matholympiad #mathematics #sybermath #aplusbi #shortsofsyber #iit #iitjee #iitjeepreparation #iitjeemaths #exponentialequations #exponents #exponential #exponent #systemsofequations #systems
#functionalequations #functions #function #maths #counting #sequencesandseries
#algebra #numbertheory #geometry #calculus #counting #mathcontests #mathcompetitions
via @KZbin @Apple @Desmos @NotabilityApp @googledocs @canva
PLAYLISTS 🎵 :
Number Theory Problems: • Number Theory Problems
Challenging Math Problems: • Challenging Math Problems
Trigonometry Problems: • Trigonometry Problems
Diophantine Equations and Systems: • Diophantine Equations ...
Calculus: • Calculus

[@GiuseppeAriano]

Mmm. One need first to prove convergence. Try to use as right hand 2 and 4. You'll have the same result i. e. sqrt2. That's impossible.

[@surelyred]

True

[@nikpuc9413]

Задача решается в уме за 30 секунд. Логарифрируются левая и правая части. получается x^x^x^…*lnx=1 но x^x^x^… равно e. e*lnx=1. lnx=1/e следовательно x=e^(1/e)

[@johnlv12]

Let a_1=0^0, a_2=0^0^0, ..., a_n=0^(a_{n-1}) for n>2. Then, a_1=1, a_2=0^1=0, a_3=0^0=1, ... So, a_n alternates between 0 (n even ) and 1 (n odd). Therefore the infinite tower 0^0^0... does not converge.

[@MortezaSabzian-db1sl]

Separate limitation a_n=1 if n=2k-1 k∈N a_n=0 if n=2k k∈N Limitation with Delta Kroniker Or a_n=δ_(n,2k-1) k∈N Restriction in another way a_n=(1-(-1)^n)/2

[@schatten3158]

It is important to note that A^(B^C) ~= (A^B)^C The question is (X^(X^(X^(X^(X^....))))) = e not ((((((...^X)^X)^X)^X)^X)^X) = e

[@black_eagle]

Why do I have a hard time getting my head around the idea that this infinite exponential with an exponent greater than one (e^(1/e) =~ 1.44) converges? How can 1.44^1.44^1.44 ... converge? I must not be thinking about these infinite exponentials correctly.

[@johnlv12]

Let a_1=e^(1/e) =~ 1.44 and for n>1, let a_n=(e^(1/e))^a_{n-1}. Then, you can prove that a_n is always less than e by induction. Assume inductively that a_{n-1}

[@MichaelRothwell1]

​@@johnlv12 Right. And the fact that sequence is increasing can also be shown by induction. It's convenient to start the sequence at a₀=1, which is OK, as aₙ=(e^(1/e))^aₙ₋₁, implies a₁=(e^(1/e))^1=e^(1/e). Then clearly a₀

[@TedHopp]

10:42 "Can something converge to two different values?" is the wrong question. The real question is "Can more than one value of x converge to the same value?" That hasn't been answered. The second method suggests not, but the first method uses the Lambert W function, which has infinitely many branches. That creates three possibilities: (1) the choice of branch doesn't change the end result; (2) all but one branch correspond to spurious solutions; or (3) there are multiple solutions. Now the answer isn't so clear.

[@Gezraf]

but syber, the first method was just a turn over all the way back to what you did in the second method. you declared x^y = y as a formula, but in that scenario y = e

[@SyberMath]

You're right!

[@giuseppemalaguti435]

x^e=e...x=e^(1/e)

[@surelyred]

I recognized this problem from my own experimentation, i knew it was e^(1/e) from thumbnail

[@maxvangulik1988]

e=x^e x=e^(1/e) easy

[@rogerphelps9939]

Absolutely. No need o use tthe W function. He is tying himself in knots.

[@arenje1]

One of my favorite channels

[@SyberMath]

Thank you! 😍 Let's spread the word 😜

[@ericmiller6056]

The initial equation is incoherent: it assumes that that the exponential tower of x's converges (to e), when it doesn't converge at all. At least not for any x > 1, ... so not to e.

[@georgepaidas1132]

Absolutely "crazy"🤣🤣 and great 👍👍👍

[@SyberMath]

Glad you enjoyed it 😄

[@bkkboy-cm3eb]

x^e = e e·lnx = 1 ∴x = e^(1/e)

[@scottleung9587]

I used the second method.

[@tejpalsingh366]

we can out e power also on x^ x^ e= e now solve .. can we put like that r not ??

[@asheredude]

Isn't it what he did in 2nd method

[@MortezaSabzian-db1sl]

x^x^e=e Let x^e=a So x^a=e So we have two equations like this: a=x^e x^a=e Multiply both sides of the equation; a*x^a=e*x^e ln(x)a*e^(ln(x)*a)=ln(x)e*e^(ln(x)*e) Put both sides in the Lambert function a=e x^e=e so x=e^(1/e)

[@asheredude]

@@MortezaSabzian-db1sl Similar to what he did

[@iikebaana]

You need to prove that x^y=y. It works only in specific cases

[@phill3986]

👍😎✌️😁🔥🔥😁✌️😎👍