Decoding The Infinite i Power Tower

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BriTheMathGuy

BriTheMathGuy

10 ай бұрын

Explore the intriguing world of complex numbers as we unravel the mystery of the infinite power tower i^i^i^.... This video breaks down the mathematical journey step-by-step, from logarithmic properties, the Lambert W Function, and Newton's method to the surprising final solution. Discover the beauty and elegance of complex analysis in this engaging visual exploration.
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This video was partially created using Manim. To learn more about animating with Manim, check out:manim.community
Corrections:
1:04 Parenthesis are placed incorrectly - should be: i^{i^i^{i^...}}
6:55 The negative (-) sign should be inside the W function, next to ipi/2
9:25 Should have -ipi/2 for these, not just -pi/2
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. Viewers should always verify the information provided in this video by consulting other reliable sources.

Пікірлер: 106
@BriTheMathGuy
@BriTheMathGuy 11 ай бұрын
🎓Become a Math Master With My Intro To Proofs Course! www.udemy.com/course/prove-it-like-a-mathematician/?referralCode=D4A14680C629BCC9D84C
@Melissaposs
@Melissaposs 11 күн бұрын
Hi
@airatvaliullin8420
@airatvaliullin8420 10 ай бұрын
important notation note: when one writes, say, x^x^x^x without any parentheses, by default the computation is performed from the top down, so x^x^x^x = x^(x^(x^x)) and not ((x^x)^x)^x. if you launch the iteration considering the latter (that is, [z_0 = i, z_{n+1} = z_{n} ^ i] instead of the correct [z_0 = i, z_{n+1} = i^z_{n}]) you don't converge to that value you've found in the video, but fall into the 3-cycle "0.20..., 6.12e-17-i, 4.81...)"
@katakana1
@katakana1 10 ай бұрын
This might be why @jaysn1683 seems to find multiple solutions depending on the amount of i's specified in the tower!
@airatvaliullin8420
@airatvaliullin8420 10 ай бұрын
@@katakana1 not sure I got their point about "local maximum of the amount of solutions". the equation i^z = z has infinite amount of solutions as ln(i) = i * pi / 2 + 2*pi*n with n integers. the simulation converges to the principal branch
@uggupuggu
@uggupuggu 10 ай бұрын
At this point change your name to Brilliant the Math Guy
@James2210
@James2210 10 ай бұрын
4:18
@samueljehanno
@samueljehanno 4 ай бұрын
​@@James22104:17
@jaysn1683
@jaysn1683 10 ай бұрын
If you substitute z=i^i^(…) like in the video as follows: i^i^z=z you get multiple solutions. Adding another i gives you even more solutions. Repeating the pattern, you reach a local maximum of the amounts of solutions after which the amount of solutions decreases again. Maybe worth a follow-up video!
@timothy4229
@timothy4229 10 ай бұрын
Holy fuck what should I look up to read about this
@jaysn1683
@jaysn1683 10 ай бұрын
I don’t no, I found out about it a few weeks ago when experimenting with WolframAlpha
@claytoncoe838
@claytoncoe838 10 ай бұрын
But only the solutions in common between the different substitutions are actual solutions to the original question
@jaysn1683
@jaysn1683 10 ай бұрын
Could you please elaborate further or link a post explaining the situation?
@TFWPLSSUB
@TFWPLSSUB 10 ай бұрын
​@@jaysn1683this same channel has a video on i^i itself being a real number
@nathanisbored
@nathanisbored 10 ай бұрын
careful with parentheses when writing power towers! the expression at 1:04 doesnt represent the spiral. its not even a tower, it collapses down with each parathesis. writing these is annoying for this reason, since we dont have a convenient notation for the anti-log operation
@sumongus
@sumongus 10 ай бұрын
i picked up on this too
@TrixieWolf
@TrixieWolf 10 ай бұрын
I came to the comments to say the same thing: the expression at the beginning is not a power tower, it's iterated exponentiation in the wrong direction.
@raydaypinball
@raydaypinball 10 ай бұрын
I was also gonna say this looks like some other caught it
@Vlodya
@Vlodya 10 ай бұрын
What do you mean we don't have a convenient notation, what about: i^i^i^i^i^...^i?
@HeavyMetalMouse
@HeavyMetalMouse 10 ай бұрын
There is a candidate for convenient antilog... 'exp' with a subscript of the base. It's not terribly commonly used, but I have occasionally seen it to represent exponentiation when the important argument is meant to be the exponent. exp₂(x) = 2^x
@DSN.001
@DSN.001 10 ай бұрын
Fortunately imaginary unit is not called z. Otherwise this infinite tower would equal to a sleeping man.
@swordofjustice7444
@swordofjustice7444 21 күн бұрын
But z is usually used for complex number notation, like z = a+bi lol
@pierrelindgren5727
@pierrelindgren5727 10 ай бұрын
An _i_ for an _i_ for an _i_ ... leaves the whole world plotted on an graph.
@Quasarbooster
@Quasarbooster 10 ай бұрын
0:54 minor correction, the parentheses are the wrong way around. Still, great video - I love infinite power towers!
@bunnygod3948
@bunnygod3948 10 ай бұрын
I also noticed that and wasn't sure what he was doing
@djsmeguk
@djsmeguk 10 ай бұрын
The real question: why are there three "branches" in that graph. It looks suspiciously like the tour a point goes on when inside the mandelbrot set. There's probably one lurking somewhere inside here...
@luminica_
@luminica_ 10 ай бұрын
The points that converge do indeed form a fractal.
@ThisAVaporeon_3333
@ThisAVaporeon_3333 10 ай бұрын
Eon?
@Qermaq
@Qermaq 10 ай бұрын
Somewhere in between Wolfram Alpha and tabulating everything on an electric abacus is one of my fall-back methods: Excel. When I don't know how to find a value within a known range, I can get Excel to do the calculations over 10000+ values that are, say, 0.00001 apart. I can have a column check an answer in two columns and report greater than or less than, and I can just look for where that flips. If It's not in the list, I change the starting value to try to get it in range. If I find a flip point, the lower value (assuming I'm counting up from the starting value) before the flip is my new starting value, and I add some post-decimal-point zeros to the increment value. In a short time I find a value to 15 decimal places. But then I always investigate what better, more accepted means of finding that value is.
@samueljehanno
@samueljehanno 4 ай бұрын
There is a better thing called coding
@Qermaq
@Qermaq 4 ай бұрын
@@samueljehannoYeah but I'd have to learn that.
@user-ik1py5ve6t
@user-ik1py5ve6t 10 ай бұрын
9:25 you wrote W(-π/2) and not W(-iπ/2) in the top right corner of the screen
@TheMemesofDestruction
@TheMemesofDestruction 10 ай бұрын
Love it! Thank you!
@marble17
@marble17 10 ай бұрын
Ain't no way bro, my math bro make definition for √-1 and accidentally make a new branch of math 💀
@imaginaryangle
@imaginaryangle 10 ай бұрын
It would be amazing to see the spirals filled in continuously via functional roots of f(x)=i^x. I wonder if those would simplify..?
@Jack_Callcott_AU
@Jack_Callcott_AU 10 ай бұрын
Very nice❕Thanks Bri.
@Ninja20704
@Ninja20704 10 ай бұрын
While i know it’s prob too long and technical for the video, how would we prove that the sequence converges in the first place? Cause people always say we must make sure the infinite recursive sequence/pattern converges before we can do our substitutions, but i have no idea how to show that.
@birefringent2851
@birefringent2851 10 ай бұрын
you can say the result is the limit of a sequence of maps (if it exists) z_{n+1} = f(z_n) where f(z) = i^z with the initial term z_1 = i. If the limit exists then it must be a fixed point of the map f. To show that the limit exists you just have to show that the initial term z_1 = i lies within the basin of attraction for one of the fixed points of the map.
@The_Commandblock
@The_Commandblock 10 ай бұрын
That spiral looks so cool
@pyropulseIXXI
@pyropulseIXXI 10 ай бұрын
TI-36X Pro Engineering/Scientific Calculator: I was able to easily program the algorithm of _Newton's method_ into the calculator itself such that the only thing I had to do was choose an initial value and then hit the [ENTER] key a bunch of times, until 'convergence.' In less than 2 seconds, I could get a zero with the precision of 12 correct decimal places. Everyone else was doing this 'by hand,' so to speak, whilst I busted out the entire *_hw_* assignment in less tim than it took everyone else to do a _single problem;_ the TI-36X Pro has feature that takes the derivative of a function at x=a; it also has an "OP" function, where every time you hit {ENTER}, it would run the function. If one made the 'OP" function to f(a) -> x (store the value of the function evaluated at x=a into the variable x), then one would have a recursive function
@cleats727
@cleats727 5 ай бұрын
Thanks Bri the Guy Math!
@NotBroihon
@NotBroihon 10 ай бұрын
Always annoyed me that there's no closed form for i^i^i^i... Does the sequence of the tetration of i touch the real axis ever again (other than i^i)?
@dukenukem9770
@dukenukem9770 10 ай бұрын
This’ll be a fun problem to give my son. I would really like to see a good proof for why the limit as x goes to zero (from the right) of the nth tetration of x is zero if n is odd and one if n is even…
@saadalikhan9210
@saadalikhan9210 8 ай бұрын
@5:39 you have implemented log(i) = ln|i|+arg(i) over ln(i)... you will need to divide log(i) with log(e) as well to get to ln(i) so ln(i) will actually be equals to (iπ)/(2*log(e)) or you can say ln(z)/z = iπ/(2*log(e))
@jacksonstarky8288
@jacksonstarky8288 10 ай бұрын
Power towers, imaginary numbers, and pi... these are a few of my favourite things.
@KarlFredrik
@KarlFredrik 10 ай бұрын
An equivalent solution is (i*2/pi)*Lambert(-i*pi/2). Numerically the same although I don't really know how to reduce the solutions into each other.
@Tritibellum
@Tritibellum 10 ай бұрын
im kinda interested in knowing what would be i-th tetration of i. (or to be more exactly, i = e^i(pi)/2, since i know there are infinitely many roots for i, but i want the "principal" root) the only problem would be to define what would be the i-th tetration of some number x... maybe in form of integral, series, product, etc etc...
@trifonmag4205
@trifonmag4205 8 ай бұрын
believe it or not, i↑↑i is solvable
@katcubed
@katcubed 10 ай бұрын
Nice video !
@severnkariuki9129
@severnkariuki9129 10 ай бұрын
I have a subtle problem with this. z is not i^z coz the z in the power has one less i. This is my problem with proofs that 0.9 recurring is 1 and harmonic number diverges to infinity rather than log(infinity). Harmonic sum reaches infinity if done upto e^infinity. If infinity is 1/0 then the number line is scalable and probably 1 is a simple infinity with al properties except scale .
@MrNikeNicke
@MrNikeNicke 10 ай бұрын
But one less than infinity is infinity, and log(infinity) is infinity
@kikones34
@kikones34 10 ай бұрын
The reason why you can say z = i^z is not rigorously explained in the video. You are basically trying to find the fixed point of the function f(z) = i^z. If the expression i^i^i^i... converges to a finite value, then it must follow that this value is a fixed point of the function f(z), i.e. applying any more iterations will not change its value. As such, this value must be the solution to the equation f(z) = z, or i^z = z. Your intuition that the harmonic series tends to infinity logarithmically is totally correct, but we have other ways to express it rigorously. log(infinity) is still infinity. However, if you would take the limit as x goes to infinity of the harmonic series divided by log(x), you will see that the result is 1 www.wolframalpha.com/input?i=lim+x+-%3E+inf+of+%28sum+k%3D1+to+x+of+1%2Fk%29+%2F+ln%28x%29, meaning both functions approach infinity at the same rate. This is one way your idea can be formalized, by framing it as a comparison of growth rates between different functions. Regarding 0.99... recurring, there are many, many different ways to prove that no real number exists between that number and 1, as such, that number must be 1, since there are always infinite real numbers between two distinct real numbers. Your notion that 0.99... is lacking some infinitesimal amount to reach 1 can be formalized by using other number systems, such as the surreal numbers. But this notion does not exist within the real numbers.
@notnilc2107
@notnilc2107 10 ай бұрын
for anyone who also got confused at 8:06 about where w*e^w=-i*pi/2 comes from, it comes from the W(z)e^(W(z))=z equation that was on screen before it. If you plug -i*pi/2 into that equation and let w=W(-i*pi/2), you get w*e^w=-i*pi/2.
@edreds2145
@edreds2145 10 ай бұрын
Thx bro! I really appreciate that you posted this
@fac7orcosplay
@fac7orcosplay 10 ай бұрын
If you will use an iterative method it would have been easier to approach directly i^i^i^i^i^i^i... since the very beginning instead of partially calculating it and then iterating
@Matthew_Klepadlo
@Matthew_Klepadlo 10 ай бұрын
I’m bored as heck this summer, since I sit around doing nothing but imaginary and complex math problems in my notebook. I’ll add this to the list of fun stuff to do on my break! I (the reader) will do this exercise first then come back to the video. Then I might hara$$ my Professor about it afterwards LMAO. EDIT: Ok, I tried it, and I believe I found out that it was impossible for me to do given the current tools I have. I’d have to know what the lambert function is and what arg(z) is, and I didn’t, so whomp whomp.
@axog9776
@axog9776 10 ай бұрын
please continue your real analysis playlist
@CallMeJellyy
@CallMeJellyy 7 ай бұрын
Is there any other reasons why we don't use the other values for the complex logarithm other than making it easier to work with?
@mrosskne
@mrosskne 4 ай бұрын
how do you evaluate i^i?
@Xnoob545
@Xnoob545 10 ай бұрын
1:00 youre not doing it correctly The exponent shpuld be evaluated top down
@skylardeslypere9909
@skylardeslypere9909 10 ай бұрын
I believe the simulation is correct but he just wrote it down wrongly. What he wrote down is something like the sequence a[n] = i^(i^n) which has no limit.
@Vlodya
@Vlodya 10 ай бұрын
A friendly reminder that complex exponentiation and logarithm are multi-valued functions And also Newton's method doesn't always work
@kaustubhpandey3259
@kaustubhpandey3259 10 ай бұрын
Hey... At the beginning when you plotted the points i think the parantheses are doing something else than you want them to I think you were plotting i^(i+i+i+...) Because of the product property of exponents This is a common mistake while coding. You can easily fix it by changing your loop statement Ig mathematicians are just more prone because it can't be bug fixed
@humbledb4jesus
@humbledb4jesus 10 ай бұрын
4:20
@ko-prometheus
@ko-prometheus 10 ай бұрын
The great Russian mathematician Lobachevsky said: "Mathematics is gymnastics of the mind" It is certainly useful to periodically do gymnastics of the mind. But still!!!! What practical advice do you have?? Where do these infinite power series apply in practice???? .
@liquidthe_ethiopiannationalist
@liquidthe_ethiopiannationalist 6 ай бұрын
So basically the infinith tetration of i
@emanuellandeholm5657
@emanuellandeholm5657 10 ай бұрын
7:06 You used -W(x) = W(-x), but W() is not an odd function. What's going on here?
@axog9776
@axog9776 10 ай бұрын
Day 2 of please continue your real analysis playlist
@charlesfrancisdelaostialem1537
@charlesfrancisdelaostialem1537 10 ай бұрын
What does i^i mean?
@uggupuggu
@uggupuggu 10 ай бұрын
are you stupid or stupid
@alexterra2626
@alexterra2626 10 ай бұрын
i is the imaginary unit (the square root of -1) and a^b means a multiplied by itself b times So i^i is i multiplied by itself i times, which on the surface makes no sense, but if you do the math, it gives a real number, the result being approximately 0.20
@Matthew_Klepadlo
@Matthew_Klepadlo 10 ай бұрын
@@alexterra2626 The exact number is e^(-pi/2). Remember, e^(ipi)+1=0. Subtract 1 from both sides, square root both sides, then raise each side to i to get e^(-pi/2). Simple! Easy! FUN 🙃
@scronchman0146
@scronchman0146 10 ай бұрын
Letting z=i^i^i^i... and then solving for it presupposes that z actually exists. So you still need to prove that the limit exists!
@sahildas.
@sahildas. 9 ай бұрын
(i^i^i...) is i because i is 1, but just tilted 90 degrees, and (1^1^1...) is 1
@Ostup_Burtik
@Ostup_Burtik 4 ай бұрын
no, it`s not "just rotated 1".
@scottleung9587
@scottleung9587 10 ай бұрын
Yep, i^i=exp(pi/2)!
@TBCN69
@TBCN69 10 ай бұрын
in the first minute you didnt do the infinite i tower, you did (i^i)^i... instead you should have done i^(i^i)...
@Misteribel
@Misteribel 10 ай бұрын
You start with a power tower at 0:21, but then you don’t calculate it as one at 1:11. Power towers are evaluated right-to-left, as you yourself explained in another video. I.e., 3^3^3 is 3^(3^3), not (3^3)^3 as in the vid. Does it matter with i? I don’t know.
@TBN-350
@TBN-350 7 ай бұрын
Now solve it downwards instead of upwards :)
@chipobject
@chipobject 10 ай бұрын
BRILLIANT IS ACTUALLY HELPFUL but i stayed in neon league for 8 days now im raging
@filipsperl
@filipsperl 10 ай бұрын
Bro talks like CGP grey
@benjaminojeda8094
@benjaminojeda8094 10 ай бұрын
0:53 your parenthesis are wrong
@user-ct1iv9dq1b
@user-ct1iv9dq1b 10 ай бұрын
Tetration on ♾️...
@annxu8219
@annxu8219 8 ай бұрын
also written as i{2}ω
@janitorvoniserlohn
@janitorvoniserlohn 8 ай бұрын
Let y_0 = i = exp(i * pi/2) y_1 = (y_0)^i = i^i = (exp(i * pi/2))^i = exp(-pi/2) y_2 = (y_1)^i = i^i^i = (exp(-pi/2))^i = exp(-i * pi/2) = -i y_3 = (y_2)^i = i^i^i^i = (exp(-i * pi/2))^i = exp(pi/2) y_4 = (y_3)^i = i^i^i^i^i = (exp(pi/2))^i = exp(i * pi/2) =y_0 The derivation manifests a cyclic relation for i^i^i^....... Therefore, i^i^i^......is non-convergent.
@MarloTheBlueberry
@MarloTheBlueberry 10 ай бұрын
This is what I feel like when I watch a math video: beginning of the video: Yeah im smart this is easy stuff middle of the video: uhhh yea that make sense.... right? end of the video: *help my brain melted*
@sans1331
@sans1331 10 ай бұрын
lemme guess: -1. edit: nope i guess not.
@marcosrodriguez2496
@marcosrodriguez2496 10 ай бұрын
I don't think all the math you did after deriving z = i^z does anything. The solution to the problem is simply the solution(s) to this equation. You could have just solved z=i^z numerically using your favorite method. It's not becoming more "correct" because you use some fancy functions to derive another more contrived equation that can also only be solved numerically.
@k_meleon
@k_meleon 10 ай бұрын
no
@Ostup_Burtik
@Ostup_Burtik 4 ай бұрын
yes
@stevenrowbottom1346
@stevenrowbottom1346 4 ай бұрын
100th comment
@Eknoma
@Eknoma 10 ай бұрын
Your visualization from 0:45 shows the formula as (((i^i)^i)^i)^... But what we are interested in, is i^(i^(i^(...)))
@kikitaylor-zd8rc
@kikitaylor-zd8rc 10 ай бұрын
Z=i^z take z√ i=√z z=-1
@TheFraz123
@TheFraz123 10 ай бұрын
I always thought your name was pronounced "bree" not "brai" 💀
@annxu8219
@annxu8219 8 ай бұрын
d
@jesusnthedaisychain
@jesusnthedaisychain 10 ай бұрын
(0.4000 + 0.0392) + (0.4000 - 0.0392) * i Just something I noticed.
@MarloTheBlueberry
@MarloTheBlueberry 10 ай бұрын
At this point change your name to Brilliant the Math Guy
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