what is the infinite tetration of i

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  • Опубліковано 28 лис 2019
  • What is i^i^i^... ? This is a super fun math because we will need the Lambert W function.
    T-shirts: teespring.com/stores/blackpen...
    Patreon: / blackpenredpen
    #math #complexnumbers #blackpenredpen

КОМЕНТАРІ • 362

  • @rawi6917
    @rawi6917 4 роки тому +850

    i to the i to the i tatatatatatata

  • @flowerwithamachinegun2692
    @flowerwithamachinegun2692 4 роки тому +182

    * serious voice *
    *Here is the fish*

  • @whythosenames
    @whythosenames 4 роки тому +241

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

    • @blackpenredpen
      @blackpenredpen  4 роки тому +74

      I know the reference!!!

    • @dhruvvraghu6226
      @dhruvvraghu6226 4 роки тому +5

      Waut the fauk are yau taulking abaut?

    • @chinkeehaw9527
      @chinkeehaw9527 4 роки тому +1

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

    • @berenjervin
      @berenjervin 4 роки тому

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

    • @1976kanthi
      @1976kanthi 3 роки тому +1

      Why did I read this in vihart’s voice

  • @jeffkevin3
    @jeffkevin3 Рік тому +25

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

    • @Waffle_6
      @Waffle_6 2 місяці тому

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

    • @PROtoss987
      @PROtoss987 Місяць тому

      It does so in shrinking and rotating triangles

  • @blackpenredpen
    @blackpenredpen  4 роки тому +508

    Do you like fish?

    • @erickherrerapena8981
      @erickherrerapena8981 4 роки тому +6

      Hazlo con ratas por favor.

    • @alexdemoura9972
      @alexdemoura9972 4 роки тому +44

      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
      "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
      @PlutoTheSecond 4 роки тому +14

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

    • @zavionw.8052
      @zavionw.8052 4 роки тому +2

      🐟 >:)

    • @blackpenredpen
      @blackpenredpen  4 роки тому +33

      PlutoTheSecond
      Hahahhahaha
      Sorry I meant
      Tatatatatatatatata

  • @transire3450
    @transire3450 4 роки тому +51

    Now, what is: W(i)

  • @ryandx8088
    @ryandx8088 Рік тому +5

    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.

  • @hamdimohammed2541
    @hamdimohammed2541 3 роки тому +27

    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!

  • @Tomaplen
    @Tomaplen 4 роки тому +134

    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
      @crawfy48 4 роки тому +12

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

    • @jcbuchin
      @jcbuchin 4 роки тому +68

      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
      @gabrielvieira3026 4 роки тому +6

      You can use Newton-Rapson

    • @poutineausyropderable7108
      @poutineausyropderable7108 4 роки тому +2

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

    • @accountfantoccio5608
      @accountfantoccio5608 4 роки тому +2

      @@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.

  • @whyit487
    @whyit487 4 роки тому +20

    I'm loving these i videos!

  • @VibingMath
    @VibingMath 4 роки тому +9

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

  • @Manuel-pd9kf
    @Manuel-pd9kf 4 роки тому +33

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

    • @usuyus
      @usuyus 4 роки тому +6

      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
      @yurenchu 4 роки тому +3

      @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
      @nostalgiafactor733 4 роки тому

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

    • @usuyus
      @usuyus 4 роки тому +2

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

    • @usuyus
      @usuyus 4 роки тому +1

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

  • @maro4412
    @maro4412 4 роки тому +11

    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
      @minijo4289 3 роки тому +1

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

  • @joshmcdouglas1720
    @joshmcdouglas1720 3 роки тому +3

    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

  • @mooncake8320
    @mooncake8320 4 роки тому +6

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

  • @synaestheziac
    @synaestheziac Рік тому +2

    Hell yeah, ln is purple for me synesthetically

  • @sulfursw9286
    @sulfursw9286 4 роки тому +45

    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
      @angelmendez-rivera351 4 роки тому +6

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

    • @angelmendez-rivera351
      @angelmendez-rivera351 4 роки тому +9

      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
      @sulfursw9286 4 роки тому

      @@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

  • @abdellahtvtube8823
    @abdellahtvtube8823 4 роки тому

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

  • @plasmacrab_7473
    @plasmacrab_7473 4 роки тому +10

    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
      @angelmendez-rivera351 4 роки тому +3

      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
      @mrsaxobeat6891 4 роки тому +3

      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
      @yurenchu 4 роки тому +2

      @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
      @plasmacrab_7473 4 роки тому

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

    • @yurenchu
      @yurenchu 4 роки тому

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

  • @alexdemoura9972
    @alexdemoura9972 4 роки тому +53

    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
    "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
      @blackpenredpen  4 роки тому +8

      Alex de Moura Hahahhahahahha! Love this

    • @blackpenredpen
      @blackpenredpen  4 роки тому +10

      Sorry I mean
      Tatatatatatatata

    • @mohammedayankhan4497
      @mohammedayankhan4497 3 роки тому +1

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

  • @tretyakov3112
    @tretyakov3112 4 роки тому

    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?

  • @o3.hassan386
    @o3.hassan386 4 роки тому

    Thanks for your video 😍

  • @AlgyCuber
    @AlgyCuber 4 роки тому +12

    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
      @jameeztherandomguy5418 Рік тому

      why you put extra s at the beginning of suare root

    • @armax6452
      @armax6452 Рік тому

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

    • @jameeztherandomguy5418
      @jameeztherandomguy5418 Рік тому

      @@armax6452 what ??

    • @armax6452
      @armax6452 Рік тому

      @@jameeztherandomguy5418 you dont know about super square root?

    • @jameeztherandomguy5418
      @jameeztherandomguy5418 Рік тому

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

  • @moisesbessalle
    @moisesbessalle 9 місяців тому

    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?

  • @Sg190th
    @Sg190th 4 роки тому +1

    Can't _imagine_ doing this kind of work

  • @HHUud78TDuj
    @HHUud78TDuj Місяць тому

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

  • @neilgerace355
    @neilgerace355 2 роки тому

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

  • @Dlmlai7255
    @Dlmlai7255 4 роки тому

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

  • @cake2843
    @cake2843 4 роки тому

    What classes will you be teaching next semester?

  • @skylermagnificent5422
    @skylermagnificent5422 4 роки тому +2

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

  • @projetoedsf3572
    @projetoedsf3572 3 роки тому

    MESTER, COULD YOU TALK ABOUT LA CONSTANT DE INTEGRACION POR LO METODO DE SUSTITUCION?

  • @rrteppo
    @rrteppo 4 роки тому

    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.

  • @durgeshnandinijha6054
    @durgeshnandinijha6054 4 роки тому +9

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

  • @MrMatthewliver
    @MrMatthewliver 8 місяців тому

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

  • @mihirtejaswi5080
    @mihirtejaswi5080 3 роки тому +1

    How do we compute the exact(or approx) value of Lambert w function of a constant without wolfram alpha?

    • @sehr.geheim
      @sehr.geheim Рік тому +1

      by using newtons method:
      suppose you want to calculate this value, which is W(-iπ/2)≈ W(-1.57i)
      The Newton method for reverse engineering functions goes as follows. You make a guess on what the answer should be (any guess will do, but a better guess will take you fewer steps)
      my first guess g will just be 1, which will be used to get a better value g1, which we will use to get g2 ...
      because I am using a calculator the initial guess doesn't have to be good at all, but if you want to do it on paper, then maybe take some value inside the complex unit circle, that's usually a good place to start when dealing with i
      For newtons method we need the derivative of the function, in this case xe^x, which is xe^x + e^x
      g1 = g - (ge^g - iπ/2)÷(ge^g + e^g)
      If you are confused, generally newtons method looks like this:
      g1 = g - (function(g) - inputValue)÷(derivative(g))
      If your calculator has an Ans button, you can do the following: Enter 1 to save it as an answer
      Then input Ans - (Ans • e^Ans - iπ/2)÷(Ans • e^Ans + e^Ans)
      I got the following values for g0=1
      g1 = 0.5 + 0.289i
      g2 = 0.43 + 0.719i
      g3 = 0.58 + 0.679i
      g4 = 0.566 + 0.688i
      g5 = 0.566 + 0.688i
      ...
      Ok, after writing all this, I noticed your comment is 2 years old, so sorry for being in your inbox if you already know how to do this, and if not, you're welcome

  • @Mathmagician73
    @Mathmagician73 4 роки тому +2

    God of mathematics 😍🔥🔥🔥

  • @inarmi7169
    @inarmi7169 4 роки тому

    you’re a genius

  • @JustAnthon
    @JustAnthon 4 роки тому

    Man, I wish I could handle equations like that...
    I was trying to figure out how to figure out the parameters of the binomial distribution formula (like when you have P(X>k), n and p and need to figure out a k) and failed miserably...

  • @MathswithMuneer
    @MathswithMuneer 4 роки тому +1

    Excellent video friend

  • @ghotifish1838
    @ghotifish1838 4 роки тому +2

    Why is it always e and pi when I solve something

  • @johannesh7610
    @johannesh7610 4 роки тому

    I tried really long to find a set where x -> i^x is a contraction, so it must be a mapping onto itself and the imaginary part must be greater than ln(pi/2)/(pi/2) (where |f'|

    • @johannesh7610
      @johannesh7610 4 роки тому

      It is possible to find that, if you also restrict re(x), but then i is not in it, so you have to find a i^i^... ^i thats in there...

  • @user-li1gb8nn5t
    @user-li1gb8nn5t 2 роки тому +12

    수렴값 찾으러 온 한국인 손

  • @purim_sakamoto
    @purim_sakamoto 3 роки тому

    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

  • @Metros23
    @Metros23 4 роки тому +2

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

    • @yurenchu
      @yurenchu 4 роки тому

      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.

  • @alexanderskladovski
    @alexanderskladovski 4 роки тому +21

    that thumbnail made me click :D

  • @thedoublehelix5661
    @thedoublehelix5661 4 роки тому +1

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

  • @adealtas
    @adealtas 4 роки тому

    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
      @angelmendez-rivera351 4 роки тому

      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.

  • @edsanville
    @edsanville 4 роки тому

    What in blazes is the intuitive meaning of putting something to the power of an imaginary number?

  • @allaincumming6313
    @allaincumming6313 4 роки тому

    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
      @allaincumming6313 4 роки тому

      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
      @angelmendez-rivera351 4 роки тому +1

      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
      @allaincumming6313 4 роки тому

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

    • @allaincumming6313
      @allaincumming6313 4 роки тому

      @@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
      @angelmendez-rivera351 4 роки тому +1

      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).

  • @change_profile_n8755
    @change_profile_n8755 4 роки тому +1

    Looking forward to this sh@t :)

  • @erickherrerapena8981
    @erickherrerapena8981 4 роки тому +1

    Bien video plumanegraplumaroja.

  • @pendergastj
    @pendergastj 4 роки тому

    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
      @PROtoss987 Місяць тому

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

  • @disasterarea9341
    @disasterarea9341 4 роки тому

    i love the tibees t-shirt :D

  • @axbs4863
    @axbs4863 2 роки тому

    It’s ≈ 0.43828 + 0.36059i for anyone wondering

  • @NonTwinBrothers
    @NonTwinBrothers 4 роки тому

    thumbnail goals

  • @user-er8tk4ts9g
    @user-er8tk4ts9g 4 роки тому +1

    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
      @kieranblazier4058 4 роки тому

      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.

  • @any13th
    @any13th 4 роки тому

    What happens if you use this lambert w function solution for a power tower of a positive number like 2 or 3?
    My logic says you should get z=infinity for a number like 2 or 3, but wolfram alpha gives me something imaginary instead...
    Is infinity imaginary...

    • @angelmendez-rivera351
      @angelmendez-rivera351 4 роки тому +1

      any13th No. The problem is that those sequences are divergent. If you assign any actual value to the infinite expression, then that value cannot be equal to the limit of the sequences. In other words, the sequence is discontinuous at infinity. They become maps from R U {+♾, -♾} -> R.

  • @vmarzein
    @vmarzein 4 роки тому

    I love that thumbnail

  • @o11k
    @o11k 4 роки тому

    How do you know it converges in the first place?

  • @yongdalim9010
    @yongdalim9010 4 роки тому

    So close to a nice answer

  • @poprockssuck87
    @poprockssuck87 4 роки тому +3

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

  • @Xnoob545
    @Xnoob545 3 роки тому +1

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

  • @sabinrawr
    @sabinrawr 4 роки тому

    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?

  • @JSSTyger
    @JSSTyger 4 роки тому

    Lambert would be proud.

  • @Dlmlai7255
    @Dlmlai7255 4 роки тому

    I was imagining about this as an IIT Bombay student

  • @mathmancalc7753
    @mathmancalc7753 4 роки тому

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

  • @OOKIEDOKIE
    @OOKIEDOKIE 4 роки тому

    When I tried this on my own I did Z^i = Z then when I took the ln of both sides got ilnZ=lnZ which got i=1. What went wrong?

  • @drChoosen
    @drChoosen 4 роки тому

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

  • @IbrahimNakshbndi
    @IbrahimNakshbndi 4 роки тому

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

  • @FunctionalIntegral
    @FunctionalIntegral 4 роки тому

    Another equivalent representation would be : z= W(-alpha)/(- alpha) ; where alpha= i pi/2

  • @MeowMeow-bt8eg
    @MeowMeow-bt8eg 4 роки тому +1

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

    • @blackpenredpen
      @blackpenredpen  4 роки тому

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

  • @lorenzobarbano8022
    @lorenzobarbano8022 4 роки тому +1

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

    • @samuelsurfboard9887
      @samuelsurfboard9887 2 роки тому

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

  • @drizling1
    @drizling1 4 роки тому

    Its 3 am in india ..what is the time there sir??? Right now??

  • @akshit7445
    @akshit7445 4 роки тому +6

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

    • @MathswithMuneer
      @MathswithMuneer 4 роки тому +1

      akshit chodhary who is he ?

    • @blackpenredpen
      @blackpenredpen  4 роки тому

      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
      @chirayu_jain 4 роки тому +1

      Yup, I did

    • @chirayu_jain
      @chirayu_jain 4 роки тому +1

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

  • @KipSupernova
    @KipSupernova 4 роки тому

    Very interesting! Greetings from Germany

  • @rafciopranks3570
    @rafciopranks3570 4 роки тому

    I have no idea how to aproximate it

  • @IkkiMitsui
    @IkkiMitsui 3 роки тому

    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...

  • @arisetotell5484
    @arisetotell5484 4 роки тому

    Just wondering, is x ↑↑i = π, possible to solve?

  • @HTJ5685
    @HTJ5685 2 роки тому

    Thank you a lot!!( from Korea

  • @speictreach2249
    @speictreach2249 4 роки тому

    This is the power of requiem

  • @chaffle7265
    @chaffle7265 2 роки тому +1

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

  • @shakob2
    @shakob2 4 роки тому

    can you do x-ln(x)=ln(45)
    has two solutions but i dont know how to get them :(

  • @mineyous
    @mineyous 4 роки тому +7

    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
      @angelmendez-rivera351 4 роки тому

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

    • @angelmendez-rivera351
      @angelmendez-rivera351 4 роки тому +2

      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
      @mineyous 4 роки тому

      @@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
      @angelmendez-rivera351 4 роки тому +1

      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.

  • @supercool1312
    @supercool1312 4 роки тому

    -1?
    guessed because sqrt(2)^sqrt(2) infinitely is 2 so i guessed it would work the same with other numbers although i have no idea what i’m on about

  • @lesbloches1142
    @lesbloches1142 4 роки тому

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

    • @idolevi612
      @idolevi612 4 роки тому

      That's an incredible question.

  • @cristiannavarro2890
    @cristiannavarro2890 4 роки тому

    What's is a decimal answer?

  • @poo2uhaha
    @poo2uhaha 4 роки тому

    I just was wondering about this question not 2 hours ago and I find this video!? What luck!

  • @MrDmitrmipt
    @MrDmitrmipt 4 роки тому

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

  • @nIaboratory
    @nIaboratory 4 роки тому

    How do you know that i^i^i.... doesn't diverge?

  • @GreenMeansGOF
    @GreenMeansGOF 4 роки тому

    blackpenredpen yay!

  • @jakubzagrodzki6037
    @jakubzagrodzki6037 3 роки тому

    2:34 „here is THE FISH”

  • @HarshitSharma-jy2xq
    @HarshitSharma-jy2xq 4 роки тому

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

  • @balduran2003
    @balduran2003 4 роки тому

    Ok, can you do it without the W function in the answer?

  • @johnny_eth
    @johnny_eth 4 роки тому

    Interesting.
    For which values of X given an infinite power tower x^x^x^... does it converge??

    • @yurenchu
      @yurenchu 4 роки тому

      If we regard the positive real number line, it would be for x between (and including) 1/e^e = 0.0659880358... and e^(1/e) = 1.44466786... . The corresponding values of the infinite power tower at these boundary values would be respectively y = 1/e and y = e .
      See also the article by R. Arthur Knoebel, which is referenced (with a clickable link to an online copy of the article) at ua-cam.com/video/DmP3sFIZ0XE/v-deo.html
      I don't know about convergence outside the positive real number line. (Apparently there is also convergence when x = -1 .)

  • @ezio99ez
    @ezio99ez 3 роки тому

    So the result is approx 0.44 + 0.36i. Wow. But what does it mean ?

  • @gustavoespinoza7940
    @gustavoespinoza7940 3 роки тому

    Pffft no check for convergence

  • @freemodegaming4562
    @freemodegaming4562 4 роки тому +2

    Please somebody tell me how some comments were posted 1 or 2 weeks ago😂😂

    • @blackpenredpen
      @blackpenredpen  4 роки тому +1

      I uploaded this on nov 9th, kept it unlisted in my playlist.

  • @iimuller1240
    @iimuller1240 4 роки тому

    nice video

  • @saharhaimyaccov4977
    @saharhaimyaccov4977 4 роки тому

    What is w fanction?

  • @rdoskoch
    @rdoskoch 4 роки тому

    What is the w function?