what is the infinite tetration of i

What is i^i^i^... ? This is a super fun math because we will need the Lambert W function.
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#math #complexnumbers #blackpenredpen

[@rawi6917]

i to the i to the i tatatatatatata

[@blackpenredpen]

Calm Down hahahahaha yea

[@aasyjepale5210]

TATATATATATATA YEA

[@JalebJay]

How is this marked 2 weeks ago?

[@hassanakhtar7874]

Wtf black magic "2 weeks ago"

[@freemodegaming4562]

Wtf how is this posted 2 weeks ago

[@blackpenredpen]

Do you like fish?

[@erickherrerapena8981]

Hazlo con ratas por favor.

[@alexdemoura9972]

Attention BPRP Subscribers: International Mathematical Union (IMU) just approved new standards, as follow: 1. "i over i over i ta ta ta..." can be described as this: ᵢ ᳟ ⁱ i "ta-ta-ta" replaces any kind of continuity; 2. Lambert-function-on-Fish (no, it is not a seafood recipe) definition: 🐟 W(🐟 e ) = 🐟 where 🐟 can be a function or variable, real, imaginary, ta-ta-ta. It replaces all recursive functions and variables of any kind; 3. Any color including purple are allowed (why not?), even when there is a restriction of colors, such as "black and red pens only", in all Mathematical works (paper, classes, exams, ta-ta-ta) from now on;

[@PlutoTheSecond]

Why are all of your fish evil? Are you building an army?

[@zavionw.8052]

🐟 >:)

[@blackpenredpen]

PlutoTheSecond Hahahhahaha Sorry I meant Tatatatatatatatata

[@flowerwithamachinegun2692]

* serious voice * *Here is the fish*

[@jeffkevin3]

It is interesting to plot i, i^i, i^i^i, etc. on the complex plane to observe the process of convergence.

[@Waffle_6]

im assuming it just spirals on to whatever that is equal to

[@PROtoss987]

It does so in shrinking and rotating triangles

[@whythosenames]

Do you know that e to the i to the e i 0 is e to the wau to the tau wau wau?

[@blackpenredpen]

I know the reference!!!

[@dhruvvraghu6226]

Waut the fauk are yau taulking abaut?

[@chinkeehaw9527]

Me reading the comment inside: Hmm, that sounds familiar. Where have I heard it? Oh! Vihart!

[@berenjervin]

Need some pow wow chow to grok the tau wau wau!

[@1976kanthi]

Why did I read this in vihart’s voice

[@transire3450]

Now, what is: W(i)

[@rogiertp]

W(i) :)

[@hamdimohammed2541]

This guy is so damn good that he makes me see how far I still have to go to answer any question in math. Huge respect!

[@ryandx8088]

A question that I have always had is: When to put and when not to put: +2n(pi)i When staying with the exponent of the exponential form in complex numbers.

[@VibingMath]

Yay it's the fish power tower hahahaha! It's happy that Chirayu also did it after your unlisted video!

[@whyit487]

I'm loving these i videos!

[@blackpenredpen]

Thanks!!!

[@Tomaplen]

is there a way to calculate the Lambert function? I have no idea what W(-ipi/2) is, nor a single a idea of close to what value it is

[@crawfy48]

You can approximate it numerically: www.wolframalpha.com/input/?i=Lambert+w+function+of+%28-+i*+pi+%2F2%29

[@jcbuchin]

The Lambert W Function can be implemented in your calculator using Newton's Method. To find W(A), the equation to solve is: f(x) = x*exp(x)-A = 0 (1) f”(x) = (x+1)exp(x) And f(x)/f'(x) can be written as: (x-A/exp(x))/(x+1) Finally, the recursive formula for Newtons Method is: X = x-(x-A/exp(x))/(x+1)) From (1) we have x*exp(x) = A or ln(x) + x = ln(A); x = ln(A)-ln(x); x ~ ln(A); ln(A) is a good initial value for the Newton Method when A>1, if this is not the case we use A. If your calculator can handle complex numbers ln(A) is a good initial value too. You can find that W(-i*pi/2) ~ 0.5664173303 - i0.6884532271 and exp(-W(-i*pi/2) ) ~ 0.4382829367 + i0.3605924719 you can compute i^i^i^i... a few times in your calculator and verify that it converges to this number.

[@gabrielvieira3026]

You can use Newton-Rapson

[@poutineausyropderable7108]

@@jcbuchin You wrote f(x)= f''(x)= Its a small typo but it may confused people. (Just saying)

[@accountfantoccio5608]

@@theseeker7194 I'm pretty sure that the infinite tetration of i converges as i is within the region of convergency of the infinite tetration. Also, you are doing it wrong, you are collapsing the entire exponential from the bottom instead of the top, which is why the answer you gave it's not correct. The tetration of i should converge to 0.4...+i*0.5... iirc.

[@maro4412]

I think you haven't found all solutions, because if you replace i^i^i^.... such i^i^z, you will get another. Maybe I don't have right. But I want to know if other solutions are exist.

[@minijo4289]

lambert w function has multiple branches. answer shown is the principle value

[@axbs4863]

It’s ≈ 0.43828 + 0.36059i for anyone wondering

[@synaestheziac]

Hell yeah, ln is purple for me synesthetically

[@sulfursw9286]

Pretty difficult question to think: tetration of i to the i i↑↑i :P If I am correct, complex tetration was generally solved in 2018.

[@angelmendez-rivera351]

Otherkin SW Can you show me a link for complex tetration being solved?

[@angelmendez-rivera351]

As far as I know, it has been shown that there exists a unique piecewise smooth monotonic function satisfying f(a + 1) = b^f(a) for all complex a provided that 1/e^e < b < e^(1/e), consistent with Schröder's equation and Abel's equation, which is agreed to be the standard definition of tetration. However, that such an f exists does not mean there is closed-form expression for it. i^^i may or may not be defined, but if it is, then it, for all practical intents and purposes, cannot be simplified or calculated other than with some numerical algorithm for computers. As far as I am concerned, I would say we are far from solving the problem of tetration.

[@sulfursw9286]

@@angelmendez-rivera351 I guess you are right, the article is far beyond my skills, that's why I wasn't sure that I understood it :0 link.springer.com/article/10.1007/s10444-018-9615-7

[@joshmcdouglas1720]

I did it a slightly different way and got the answer of 2iW(-ipi/2)/pi which ends up being equal to what you got in the video! Pretty cool stuff

[@durgeshnandinijha6054]

These fishes would invade the entire humanity........ #tatatatatatatata army

[@mooncake8320]

No matter how much you try, the trio of e, π, i will always find a way to be introduced in your proof x')

[@alexdemoura9972]

Attention BPRP Subscribers: International Mathematical Union (IMU) just approved new standards, as follow: 1. "i over i over i ta ta ta..." can be described as this: ᵢ ᳟ ⁱ i "ta-ta-ta" replaces any kind of continuity; 2. Lambert-function-on-Fish (no, it is not a seafood recipe) definition: 🐟 W(🐟 e ) = 🐟 where 🐟 can be a function or variable, real, imaginary, ta-ta-ta. It replaces all recursive functions and variables of any kind; 3. Any color including purple are allowed (why not?), even when there is a restriction of colors, such as "black and red pens only", in all Mathematical works (paper, classes, exams, ta-ta-ta) from now on;

[@blackpenredpen]

Alex de Moura Hahahhahahahha! Love this

[@blackpenredpen]

Sorry I mean Tatatatatatatata

[@mohammedayankhan4497]

@@blackpenredpen can you answer my question? I have a genuine doubt. Pls reply yes if you read this.

[@Manuel-pd9kf]

Do i'th root of the i'th root of the i'th root..... of the i'th root of i

[@usuyus]

that's the same thing as i^(1/i)^(1/i)^(1/i)^(1/i)^..., which would be equivalent to saying i^-i^-i^-i^-i^-i^-i^... (as 1/i = -i). So, do the same thing for -i (-i^-i^-i^-i^-i^... = z), find z and calculate i^z.

[@yurenchu]

@Yusuf Onur, Nope, sorry, that's not correct. What he is asking, is not a tetration (a.k.a. "power tower"). Instead, he is just asking about an expression of the form (..((((i^a)^a)^a)^a)...)^a with a being equal to 1/i = -i. This simplifies to i^[ a*a*a*a*...*a ] , which will be non-convergent. i'th root of the i'th root of the i'th root ... of the i'th rooth of i = = ⁱ√( ⁱ√( ⁱ√( ⁱ√( ⁱ√( ... ⁱ√( i )..))))) = (..((( i^[1/i] )^[1/i] )^[1/i] )^[1/i]... )^[1/i] = (..(((((( i^[-i] )^[-i] )^[-i] )^[-i] )^[-i] )^[-i] )...)^[-i] = i^[ (-i)*(-i)*(-i)* ... *(-i) ] which is either i^(-i) , or i^(-1) , or i^(i) , or i^(1) , and hence isn't convergent. In other words: it's the limit of the sequence a[n+1] = (a[n])^(1/i) , starting from a[0] = i . But as it turns out, z = lim{n-->+infinity} a[n] doesn't exist.

[@nostalgiafactor733]

@@usuyus 1/i isn't -i my dude

[@usuyus]

@@nostalgiafactor733 expand the fraction with i, so you'll get this: 1/i = i/(i*i) = i/-1 = -i Hope that clarifies

[@usuyus]

@@yurenchu ah i see... You are right

[@skylermagnificent5422]

Channel named blackpenredpen, 0:40 of course we’ll use a purple pen

[@alexanderskladovski]

that thumbnail made me click :D

[@blackpenredpen]

nice!!!

[@AlgyCuber]

W(x) and ssqrt(x) are closely related so we can write the answer in terms of ssqrt(x) i^z = z i = z^(1/z) 1/i = 1/z^(1/z) -i = (1/z)^(1/z) 1/z = ssqrt(-i) z = 1/ssqrt(-i)

[@jameeztherandomguy5418]

why you put extra s at the beginning of suare root

[@armax6452]

@@jameeztherandomguy5418 ssqrt stand for super square root not square root

[@jameeztherandomguy5418]

@@armax6452 what ??

[@armax6452]

@@jameeztherandomguy5418 you dont know about super square root?

[@jameeztherandomguy5418]

@@armax6452 no and i cant even find anything about it anywhere lmao

[@plasmacrab_7473]

Thanks for the great video! I have a quick question: Since i is imaginary and i^i=e^(-pi/2) and i^i^i = i^(a real number), each iteration of the power tower makes the result alternate between real and non-real. Would the result you got still be valid, or would the limit of the power tower not converge? Thanks again for releasing great videos in your spare time!

[@angelmendez-rivera351]

PlasmaCrab _ The power tower does converge, so his result is valid. However, proving it does converge is quite a pain in the ass, and probably outside the scope of this comment section. You might wanna check Knesser's result, though. If you search that, you might find something satisfactory.

[@mrsaxobeat6891]

That’s not quite true. The series doesn’t alternate, and i^i^i^i isn’t real. In fact, I believe that the only real answer is i^i

[@yurenchu]

@PlasmaCrab, As @MrSaxobeat already remarked, your claim is not true. The imaginary part of i^i^i is not a multiple of pi/2 , hence neither the imaginary part nor the real part of i^i^i^i will be 0.

[@plasmacrab_7473]

@@yurenchu Ah that makes sense, I got too ahead of myself. Thanks for the explanation!

[@yurenchu]

@PlasmaCrab , No problem! I'm glad to have helped.

[@HHUud78TDuj]

Try integrating e^(-x^2), but use 1.9999999 instead of 2 to see if it prevents having to use the error function

[@neilgerace355]

Going by the intro music, I was looking for some Entertainment in this video, and I wasn't disappointed.

[@Mathmagician73]

God of mathematics 😍🔥🔥🔥

[@titina_85]

수렴값 찾으러 온 한국인 손

[@이수호-y5l]

ㄹㅇ ㅋㅋ

[@sergio6357]

ㅋㅋㅋㄱㅣㄱㅋㅋㅋ 너두? 나두ㅋㅋㅋㅋㅋㅋㅋ

[@segubeam]

ㄹㅇㅋㅋ

[@이주호-y7j6t]

ㅋㅋㅋㅋㅋ

[@disabusement]

ㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋ

[@abdellahtvtube8823]

As a person who likes math and do good in my college math as engineering student. . You are very good and talented Big up 👍👍👍

[@poprockssuck87]

"blackpenredpen" ... Nowhere does it say "purple pen!"

[@MathswithMuneer]

Excellent video friend

[@Sg190th]

Can't _imagine_ doing this kind of work

[@erickherrerapena8981]

Bien video plumanegraplumaroja.

[@akshit7445]

Chirayu covered this in his last to last to last video i guess

[@MathswithMuneer]

akshit chodhary who is he ?

[@blackpenredpen]

Yes he did but I uploaded this video (unlisted) on Nov. 9th. That's why you can see the comment from 2 weeks ago.

[@chirayu_jain]

Yup, I did

[@chirayu_jain]

@@MathswithMuneer I am Chirayu Jain, I also have math channel, Please subscribe to it

[@mineyous]

Wouldn't we need to first show that the complex sequence defined by U0=i and Un+1=i^Un for all integer n converges? Otherwise : 2*2*2*...=x => 2*x=x => x=0 And then 0=2*2*2*...=+∞ is not solution If someone know how to show the existence of a solution (or to show that exp(-W(-ipi/2)) effectively works) it would be great :)

[@angelmendez-rivera351]

youssef yjjou The statement 2*2*2*2*••• = +♾ is not correct, as ♾ is not a number.

[@angelmendez-rivera351]

Also, it is a well-known theorem in mathematics that z^^n as n -> ♾ converges if 1/e^e < |z| < e^(1/e). However, proving it is way beyond the scope of this channel, as it requires some really high-level mathematics. Nothing wrong with taking the theorem for granted.

[@mineyous]

@@angelmendez-rivera351 Thanks a lot for your answer, with such a theorem my problem vanishes and it reassures me ^^ (I was refering to lim[n→+∞](2^n) for 2*2*…=+∞ ) I will still search for a proof :)

[@angelmendez-rivera351]

youssef yjjou lim 2^x (x -> +♾) = +♾, this is correct. However, you have to be careful. In some contexts, it may be useful to say 2*2*2*2*••• = lim 2^x (x -> ♾). However, in the theory of sequences, as well the theory of extensions of the real numbers, we tend to talk about those two expressions as different quantities. Think of it this way. If there exists a quantity that is bigger than all real numbers, called M, and if we consider f(x) = x |-> 2^x : R & {M} -> R, then f(M) =/= lim 2^x (x -> M). This is just a formalism, but the point is that the limit alone doesn't allow you to say the expression cannot take on a finite value.

[@jakubzagrodzki6037]

2:34 „here is THE FISH”

[@NonTwinBrothers]

thumbnail goals

[@mokouf3]

I noticed that you always love to draw the fish when mentioning Lambert W function.

[@SebastienPatriote]

That is because he's a good teacher. That's a mnemonic tool to help us remember the function. He could have used x instead but it's not as effective.

[@rrteppo]

When I first started watching your videos I was in high school and didn't understand any of it. Now I am in college and understand most of it.

[@JSSTyger]

Lambert would be proud.

[@MrMatthewliver]

Wouldn't it be easier to introduce an "inverse tetration" (superroot) as a special function instead of Lambert W function?

[@MrDmitrmipt]

You only found partial limits, you're doing bad because people might misunnderstand. Other solutions are i and -i

[@inarmi7169]

you’re a genius

[@Dlmlai7255]

Too great to watch maths at this level you make it easier.

[@yongdalim9010]

So close to a nice answer

[@chaffle7265]

After watching.... Me : uh.....so fish is delicious. Right?

[@mrgeroi_]

0:29 i to the *ЗИЗЬ* power

[@thedoublehelix5661]

but does it converge??? (love the video and the clever algebra)

[@sergioh5515]

I like my fish fried

[@blackpenredpen]

Nice! I like it like that too

[@sabinrawr]

I love this beautiful channel and the beautiful math within. But am I the only one who is left a bit unsatisfied when we start making sense of i^i^i^... And then hear "and that's it" when we reach an even more comprehensible expression?

[@user-er8tk4ts9g]

I put i, i^i , i^i^i , i^i^i^i, i^i^i^i^i... and got on with the next things on the graph with Geogebra to view approximation of i^i^i^i^i^i^....(this video shows) and it looks like a spiral. What does it mean?

[@kieranblazier4058]

Complex powers represent a combination of rotation and scaling in the complex plane. The successive rotations and scalings converge (namely to the value demonstrated in the video), and in order to converge they have to diminish sufficiently fast. Thus they formal a spiral that funnels down toward the point of convergence. This is often what convergent sequences look like in the complex plane.

[@HarshitSharma-jy2xq]

Blackpenredpen uses a purple pen. *Me- Wait that's illegal!!!!!!*

[@purim_sakamoto]

いやーーー　お見事ですねーー W関数便利すぎる　チートやん 高校の授業って、Excelで出来ることをそろばんでやらされてるようなものやな

[@mathmancalc7753]

The i's (ayes?) have it. Now I got it, too!

[@vmarzein]

I love that thumbnail

[@niom-nx7kb]

can you really do the first line to second line step though? I only know the domain and range of the function in the real plane but what about the complex plane?

[@adlad1563]

UNLIMITED POWER

[@moisesbessalle]

can you manipulate an infinite series agebraically like that?? Im not sure unless that infinite series converges but then you would either have to know that beforehand or just make the assumptions and have a potentially incomplete answer at best no?

[@change_profile_n8755]

Looking forward to this sh@t :)

[@ghotifish1838]

Why is it always e and pi when I solve something

[@adealtas]

Hey, I was wondering, since i=e^[i(Pi/2)], could you subsitute the i in the exponential by e^[i(Pi/2)] ? So you'd get: i= e^[e^[i(Pi/2)](Pi/2)] And most importantly, if you repeat this an inifinite amount of times, is it still equal to i ? And since i never appears on this infinite exponential, does it become a real number equal to i? This might be very poorly explained, and Im sure I made some mistakes here and there, feel free to ask me more and correct my errors.

[@angelmendez-rivera351]

Adeal's TASes i = e^(πi/2) = e^[πe^(πi/2)/2] implies that there is a sequence to consider here. Namely, a(n + 1) = πe^(πa(n)/2)/2, with a(0) being the parameter that one need to be careful about. If a(0) = 1, the sequence diverges. In fact, the sequence only converges if |a(0)| < 1, or if a(0) = i or a(0) = -i. Assuming the sequence converges, one can obtain the limit by setting πe^(πz/2)/2 = z, and then solve for z. However, in the theory of divergent sequences, one can find the value of infinite expressions without regard for their convergence. In this case, it would be possible for the expression (e^π/2)^^♾ to be i.

[@o3.hassan386]

Thanks for your video 😍

[@IkkiMitsui]

Something is annoying me... what if you replace i=e^(i*pi/2) for each i in the infinite tower... we'll have something like e^(e^(e^(...*pi/2)*pi/2)*pi/2), and I can see only real values here...

[@tretyakov3112]

Is it correct to say that ln(i)= i pi/2? Complex log has infinit values and if we change what is inside Lambert function value will change too. So i ^i tatata has many different values, hasn’t it?

[@MPEBC.1170]

W가 뭔가요?

[@기부니너무나도아요]

Professor I didn't understand

[@mariomario-ih6mn]

I made a channel and I didn't get any views. What makes a KZbin channel good

[@pendergastj]

I love your videos but there are so many concepts I don't understand. Will you make an introduction to Calculus for beginners. Your videos are of such a high quality and you are a great teacher but for people like me who only got a D in maths but want to learn it is difficult to get started.

[@PROtoss987]

Professor Leonard has good introductory math courses, and MIT OCW (Gilbert Strang etc.) is also good

[@speictreach2249]

This is the power of requiem

[@takyc7883]

Now do i x i CHALLENGING WARNIG

[@lorenzobarbano]

What if... log in base i? Is there something special you can tell us about log with complex bases?

[@samuelsurfboard9887]

LogiX (log x to base i)= Inx/Ini , In i =iπ/2, In x *2/πi, 2Inx/πi, Rationalizing -2i*Inx/π

[@sasoribi1341]

Your name is not only "blackpen & redpen", but also "bluepen" add, please! Blue is Only alone...oh...sad.

[@RyanLewis-Johnson-wq6xs]

w = lambert function

[@Metros23]

What is this W function? I'm taking complex analysis and I haven't seen it yet

[@yurenchu]

Lambert W function - en.wikipedia.org/wiki/Lambert_W_function - mathworld.wolfram.com/LambertW-Function.html This stuff is way more advanced than the basic "complex analysis" course.

[@Xnoob545]

I still dont know what the answer is What does W(iπ/2) equal

[@disasterarea9341]

i love the tibees t-shirt :D

[@jacobmejer5082]

Does he use Z to show it's a complex number? I'm a little new to this :)

[@ferrumarctus]

yes

[@drChoosen]

exp(-W(-(i*PI)/(2))) and Wolfram says "no roots exists". Hmmm, what does it means? Can not be quantified?

[@gustavoespinoza7940]

Pffft no check for convergence

[@VamsiKrishna-hg3yi]

Brother do 100 probability questions

[@ВикторПоплевко-е2т]

1:50 these 2 fish*

[@lesbloches1142]

Is e^e^e^e^e... a solution for ln(x) = x ?

[@idolevi612]

That's an incredible question.

[@MeowMeow-bt8eg]

What do you think about Patrick jmt? We're you inspired by him?

[@blackpenredpen]

lone wolf He is great! As I mentioned in my 100 calc 2 problems video, he and khan were the OG math KZbinrs and of course I was inspired by them.

[@medabidichannel]

When you put Z = i^i^i^... and you did all the mathematical operations you presumed that i^i^i^... exists. You have to prove it.

[@benjaminbrat3922]

Great! Could you simplify the answer even more using W(x)*exp(W(x))=x, such that exp(-W(x))=W(x)/x ? That would give 2i/pi*W(i*pi/2), I think.

[@angelmendez-rivera351]

Benjamin Brat Yes

[@ezio99ez]

Fish ? OMG !!!

[@allaincumming6313]

I also did this one a month ago :o, and went to the same answer, but my interpretation of that is: if well it's a valid answer, in the sense of convergence and algorithm process it's not. I mean, there are infinite numbers equal to i^i^i^i^..., when the process of tetration is just one, there is no limit but a function of double infinite solutions (by iπ/2 which can be generalized, and by W function which has other subfunctions). Very mindblowing.

[@allaincumming6313]

Also, exponentiate i to itself represents a twist in the complex plane, each time you do it. So, doing it infinitely results on an infinite twisting, without a limit but 0 as a possible convergent answer (because 0 is indifferent to the act of twisting).

[@angelmendez-rivera351]

Allain Cumming No, that is incorrect. The sequence (1, i, i^i, i^i^i, ...) converges, and there is a theorem in complex analysis that directly states this.

[@allaincumming6313]

@@angelmendez-rivera351 Really? What theorem is it? I'm interested. PD: Hablas español?

[@allaincumming6313]

@@angelmendez-rivera351 And the convergence is just the principal value of the solution? Why there is different answers by applying different revolutions to ln(i)?

[@angelmendez-rivera351]

Allain Cumming Each element in the sequence is multivalued, so the limit of the sequence, which is the infinite power tower, is also multivalued. i^i itself is already multivalued. Anyhow, I do not know if the theorem has a name, but the theorem states that if 1/e^e < |a| < e^(1/e), then the sequence (1, a, a^a, a^a^a, ...) converges, and the limit, which is equal to a^a^a^•••, is -W(-ln(a))/ln(a).

[@diogr22]

now do infinite pentation!

[@ranjitsarkar3126]

Pls take my problem

[@Wecoc1]

He used a purple pen o_o

[@MrDmitrmipt]

But from the very beginning you could also write Z^i = Z, and this is simply not solvable. In other words, first you need to prove that Z is a nice limit which is finite and could be found. I doubt this is true.

[@valentinoromitti6005]

≈ 1.36 - 1.12 i

[@pineapplef3m0]

can you do i^^i (tetration)

[@TyLee960]

tahtahtahtahtah

[@Dlmlai7255]

I was imagining about this as an IIT Bombay student

[@ranjitsarkar3126]

Come on bro make a video on ( i to the height of i )

[@balazsb2040]

Does this work if you do i^i^z=z? Because technically it's true

[@angelmendez-rivera351]

Balázs Bartha Yes, it is true, but it does not work because the equation i^(i^z) = z is impossible to solve analytically.

[@millemilliardsdedollars6914]

Do you have the right to do i^i^i^i... = Z = i^Z ? It looks like the mistake with the infinite sum of 1 when you can demonstrate that 1 = 0 with S = S +1 etc...

[@IbrahimNakshbndi]

please master give me an a link for your video which show the fish method.. 🐟🐟🐠🐠🐠🐠🐟🐠🐠🐠🐠🐠🐠🐟🐠🐟🐟🐠🐠

[@brattwurst1979]

Yeah I've got a feeling it's gonna be e something 😂