how is e^e^x=1 solvable??

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  • เผยแพร่เมื่อ 24 ธ.ค. 2024

ความคิดเห็น • 803

  • @blackpenredpen
    @blackpenredpen  ปีที่แล้ว +248

    Can 1^x=2?
    Solution here: th-cam.com/video/9wJ9YBwHXGI/w-d-xo.htmlsi=dx-XGv_Wf3_0VDH2
    🛍 Euler's Identity e^(iπ)+1=0 t-shirt: amzn.to/427Seae

    • @ISoldßinLadensViagraOnEbayఔ
      @ISoldßinLadensViagraOnEbayఔ ปีที่แล้ว +4

      Can you solve (e^i)+(i^e) pls

    • @MATHMASTERPRO
      @MATHMASTERPRO ปีที่แล้ว +3

      @@ISoldßinLadensViagraOnEbayఔ where the equation?

    • @ISoldßinLadensViagraOnEbayఔ
      @ISoldßinLadensViagraOnEbayఔ ปีที่แล้ว +7

      @@MATHMASTERPRO I am asking for the solution of the answer, this is not an equation, since e and i are numbers

    • @MATHMASTERPRO
      @MATHMASTERPRO ปีที่แล้ว +1

      @@ISoldßinLadensViagraOnEbayఔ (e^i)+(i^e)=cos(1)+cos(i*pi/2)+i(sin(1)+sin(i*pi/2))

    • @xinpingdonohoe3978
      @xinpingdonohoe3978 ปีที่แล้ว

      ​@@ISoldßinLadensViagraOnEbayఔ
      cos(1)+cos(πe/2)+i sin(1)+i sin(πe/2)
      It won't really contract into any nice form.

  • @MathFromAlphaToOmega
    @MathFromAlphaToOmega ปีที่แล้ว +2563

    There's a theorem (called Picard's little theorem) that says that any non-constant holomorphic function on the whole complex plane can miss at most one value. Since e^e^x definitely can't equal 0, it must hit 1 somewhere. :D

    • @pluieuwu
      @pluieuwu ปีที่แล้ว +98

      thats so cool!!!

    • @Isvakk
      @Isvakk ปีที่แล้ว +47

      Are you sure it's holomorphic?

    • @MathFromAlphaToOmega
      @MathFromAlphaToOmega ปีที่แล้ว +130

      Yes, it's infinitely differentiable everywhere.

    • @jakobullmann7586
      @jakobullmann7586 ปีที่แล้ว +189

      @@Isvakk It’s a composition of holomorphic functions, this holomorphic itself.

    • @blackpenredpen
      @blackpenredpen  ปีที่แล้ว +556

      Wow! This is super cool!!

  • @PauloChacal
    @PauloChacal ปีที่แล้ว +2857

    Number “1” will never be the same after your explanation

    • @doodle1726
      @doodle1726 ปีที่แล้ว +54

      Real true man now i can't get it out of my head arrrrghh😱😱🤯

    • @onesecondbaobab
      @onesecondbaobab ปีที่แล้ว +4

      I giggled out loud xD

    • @Hardcore_Remixer
      @Hardcore_Remixer ปีที่แล้ว +22

      You should watch Animation vs Math. The fun begins when it comes to -1 being replaced by e^(i(pi)) which has the same value as -1 😂

    • @threeuniquefingers
      @threeuniquefingers ปีที่แล้ว +3

      Ahh I gave this comment the 1000th like…the transition from 999 to 1k seemed surreal

    • @Questiala124
      @Questiala124 ปีที่แล้ว +5

      Instead of saying “I am number one” say “ I am number e^(2i(Pi)n)”.

  • @EnigmaticChillout
    @EnigmaticChillout ปีที่แล้ว +119

    I graduated six years ago and I used to watch your videos back then, just stumbled across this now and it takes me back. Thank you for making these videos with so much enthusiasm.

  • @DanoshTech
    @DanoshTech 11 หลายเดือนก่อน +163

    I love how passionate he is he makes math seem cool and interesting

    • @alexeynezhdanov2362
      @alexeynezhdanov2362 8 หลายเดือนก่อน +11

      Because it is - cool and interesting.

    • @DanoshTech
      @DanoshTech 8 หลายเดือนก่อน +4

      @@alexeynezhdanov2362 some aspects aren't at the moment I am doing 'math methods' and 'specialist' math' its Australian senior math and essentially it is calculus, geometry, algebra the top senior maths that is done in America and we just covered permutations and combinations and damn they are boring

  • @acykablyatley
    @acykablyatley ปีที่แล้ว +365

    i agree it is very straightforward to see now that e^2ipi = 1 and e^ln(2ipi) = 2ipi so that e^e^ln(2ipi) = 1 is true. but only unintuitive because it is easy to forget that e^ix is periodic in the complex plane...

    • @seja098
      @seja098 ปีที่แล้ว +7

      lol, you definitely would have never guessed this answer, be humble.

    • @_cran
      @_cran 11 หลายเดือนก่อน

      It's not really easy to forget there are so many proofs and techniques over moivre formula 😭 I'm pretty sure you learn it in calc lessons and the start of complex analysis starts with it since you use it to solve everything in that whole class. If anyone reading who doesn't know what's moivre formula is e^ix=cosx+isinx

    • @CertifiedOrc
      @CertifiedOrc 10 หลายเดือนก่อน +38

      ​@@seja098when you're not specified whether the solution is purely real, it is instinctive to check both in the real and complex plane, especially for someone studying higher grade maths, also don't talk shit to people you don't know

    • @acm-gs6bl
      @acm-gs6bl 8 หลายเดือนก่อน +5

      i like your funny words magic man

    • @acykablyatley
      @acykablyatley 8 หลายเดือนก่อน +17

      @@seja098 i did not say that i guessed that answer, and my comment clearly says it was easy to see /after/ watching the video.

  • @isilverboy
    @isilverboy ปีที่แล้ว +40

    @3:00 instead of convert 1, I would prefer change i into e^i(pi/2+2pi c2). In this way you do not have the log of a complex number in the solution.

    • @xinpingdonohoe3978
      @xinpingdonohoe3978 ปีที่แล้ว +20

      If you rewrite a complex number, you still have a complex number, only written differently. Don't be a coward; the log of a complex number is real man business.

    • @isilverboy
      @isilverboy ปีที่แล้ว

      @@xinpingdonohoe3978I have no problems with logs of complex numbers, but imho they still need to be simplified: the log of a complex number can be further simplified by using ln(i) = i(pi/2+2 pi c2).

    • @vadimpetruhanov4150
      @vadimpetruhanov4150 ปีที่แล้ว +3

      Log of a complex number is many-valued function, therefore it is preferable not to use it when it is possible

    • @XJWill1
      @XJWill1 ปีที่แล้ว +8

      The "method" of solving complex-valued equations by randomly converting constants to exp(i*something) is not a reliable way to do it. It may work on some simpler equations, but it will fail on other equations.
      A more reliable way is to use the multi-valued complex natural logarithm function, which is written log() in complex analysis.
      exp(exp(x)) = 1
      log(exp(exp(x)) = log(1)
      exp(x) + i*j*2*pi = 0 + i*k*2*pi where j, k are any integer, this is because log() is multi-valued
      exp(x) = i*m*2*pi where m is any integer
      log(exp(x)) = log(i*m*2*pi)
      x = log(i*m*2*pi) + i*n*2*pi where m and n are any integer

    • @rainerzufall42
      @rainerzufall42 8 หลายเดือนก่อน +1

      Agreed! See above: x = i π (4 c_2 + 1) / 2 + ln(2 π c_1) and c_1 !=0 and c_1 element Z and c_2 element Z
      Clear real and imaginary parts, no complex log, just real ln!

  • @timothyrosenvall1496
    @timothyrosenvall1496 ปีที่แล้ว +203

    I’ve worked with an equation in the past that seemed to reduce to e^x = 0.
    I wondered if x was always just undefined but I have the vaguest memory of reducing it from e^e^x = 1. This is a phenomenal result

  • @dougdimmedome5552
    @dougdimmedome5552 ปีที่แล้ว +86

    The greatest thing about complex analysis is slowly overtime making more insane infinite expressions to approximate 1.

    • @Aramil4
      @Aramil4 17 วันที่ผ่านมา +1

      Not approximate, equal

  • @Sg190th
    @Sg190th ปีที่แล้ว +230

    It's nice seeing the complex world being used more.

  • @camelloy
    @camelloy ปีที่แล้ว +14

    I want to make this very clear this video rescued my desire to learn math. It gave me the first visualization of what e^ipi was instead of rote memorization that drove me up a wall. I actually understand what the complex plain is after being told repeatedly by my professor not to bother looking into it. Can’t wait to dive in further.

    • @csconnor5160
      @csconnor5160 21 วันที่ผ่านมา

      Wow if your professor is telling you not to study the complex plane he's kind of a dick for that

  • @stephanelem822
    @stephanelem822 ปีที่แล้ว +47

    Each time I watch one of your video, I discover one more time, the set of constraints I used to know to solve an equation is largely incomplete. I've no idea to discover without Wolfram I'd be wrong.

  • @leoniekrenzer7716
    @leoniekrenzer7716 ปีที่แล้ว +45

    This is really cool! Also a great way to show why we can't just hit complex functions with the logarithm (since the exponential function is not injective on the complex plane)

    • @Josp101
      @Josp101 ปีที่แล้ว +2

      Wow okay so this is the reason why the simple approach misses solutions!

    • @dethfr491
      @dethfr491 10 หลายเดือนก่อน +2

      That's why there is term called "principal logarithm" of complex numbers .

  • @TFclife
    @TFclife 10 หลายเดือนก่อน +2

    I integrated both sides,
    e^u =e^u
    e^e^x = x
    In both sides
    e^x = ln x
    Find derivative of both sides
    e^x = 1/x
    In both sides
    x = In (1/x)
    Derivative of both sides
    1= - Ln x
    1= - 1/x
    -1= x

    • @thetaomegatheta
      @thetaomegatheta 7 หลายเดือนก่อน +1

      And you got the wrong answer.

  • @dentonyoung4314
    @dentonyoung4314 ปีที่แล้ว +18

    Wow. That was an amazing explanation.

  • @okWishFull
    @okWishFull ปีที่แล้ว +3

    I love your excitement! Loved every moment of this!!!

  •  21 วันที่ผ่านมา +1

    Even now, if we choose any integer for c1 and c2, the value of ln(2*i*pi) will be -infinity. So this solution is no other than of x=-infinity.

  • @dodddsqa
    @dodddsqa ปีที่แล้ว +8

    The way a complex solution appeared out of nowhere is by avoiding the e^nothing=0, and instead to find a certain number that's equivalent to 0 when on the power of e. Which, leads to a fact that e^2cπi = e^0 = 1, works for every integer c.

  • @michaelz2270
    @michaelz2270 ปีที่แล้ว +5

    You can do this systematically. Let w = e^z. Then you are solving e^w = 1, solved by w = 2pi i n for an integer n. So you wish to solve e^z = 2pi i n. For n > 0 one has 2 pi i n = e^(i pi /2 + ln 2pi n). Then e^z = e^(i pi /2 + ln 2pi n) is solved by z = i pi /2 + ln (2pi n) + 2pi i m for integers m. This works for n < 0 too if you replace ln (2pi n) by pi i + ln (2pi |n|).
    Stated in terms of the multivalued logarithm, these are log(log(1)).

    • @TheDoh007
      @TheDoh007 ปีที่แล้ว

      I simply brute-forced my way to e^(e^(0.5*pi*i+1.8378770664093455)) lmao

  • @punpcklbw
    @punpcklbw ปีที่แล้ว +3

    The logarithm cannot be defined for the whole complex plane, as exp(z) = exp(z+2πki) for any integer k. You're basically left with log(0) that is also undefined and approaches negative infinity in the limit.

  • @robsmith9696
    @robsmith9696 ปีที่แล้ว +7

    For anyone missing why the 1 was added back in on the fourth line, it’s because the 1 is still there and multiplied in. When you take the ln() of both sides, ln(1) shows up and gives you the initial issue.

  • @vivianriver6450
    @vivianriver6450 ปีที่แล้ว +13

    If I'm not mistaken, the reason that e^x = 0 has no solution, but e^(e^x) = 1 has a set of complex solutions is because e^x is periodic with period 2*pi*i, but ln(x) is computed using *only one* period.
    It's similar to how x^2 = 1 has *two* solutions, but sqrt(1) is always evaluated to 1.

    • @vibaj16
      @vibaj16 ปีที่แล้ว +1

      I don't think that's the full explanation. sqrt(1) is always evaluated to 1 because that's just how sqrt(x) is defined: the positive number that when squared equals x.

    • @vivianriver6450
      @vivianriver6450 ปีที่แล้ว

      @@vibaj16 x^2 = 1 has two solutions, but *one* of those solutions vanishes when you take the square root of both sides of the equation.
      Likewise, e^(e^x) = 1 has a family of solutions that disappear when you take the log of both sides of the equation because ln(x) evaluates to only *one* value, even tho e^(ln(x) + 2*pi*i) also evaluates to x.

    • @kazedcat
      @kazedcat ปีที่แล้ว +5

      ​@@vivianriver6450It's a mathematical trick called equivalence classes. The solutions do not disappear you are just picking one value that represent and infinite set of solutions. The reason this is needed is because functions by definition must have unique mapping.

    • @alaingamache3908
      @alaingamache3908 16 วันที่ผ่านมา

      @@vivianriver6450no, you’re confusing it with principal square root. x^2=1 => abs(x) = 1 hence why you have two roots

  • @frostboi77
    @frostboi77 14 วันที่ผ่านมา

    i can see his excitement and exasperation growing and its so fun to watch

  • @gitanjalideb454
    @gitanjalideb454 8 หลายเดือนก่อน +3

    Thinking differently- Can't the value of x=-∞?
    Given, e^(e^x)=1
    Or, e^(e^x)=e^0
    Therefore, e^x=0
    If x->∞, then e^∞=∞
    But if x->-∞, then e^(-∞)=1/e^∞=1/∞=0
    Hence, x=-∞.

    • @thetaomegatheta
      @thetaomegatheta 7 หลายเดือนก่อน

      'Thinking differently- Can't the value of x=-∞?'
      The expressions '1/∞', 'e^(-∞)' and 'e^∞' are not defined in this context.
      You have also failed to find all of the roots, which is required to solve an equation.

    • @gitanjalideb454
      @gitanjalideb454 7 หลายเดือนก่อน +2

      @@thetaomegatheta I know everybody seeks for defined solutions, but there is no harm on working with ∞ or 0, I just simply thought in that way, hence I said that. It will absolutely vary how an individual seeks its solution.

    • @thetaomegatheta
      @thetaomegatheta 7 หลายเดือนก่อน

      'I know everybody seeks for defined solutions, but there is no harm on working with ∞ or 0'
      If you define the relevant expressions - sure.
      You are also forgetting that you need to find all roots (in the space where we are looking for solutions). You did not do so.
      Also, your reasoning is bad:
      'Or, e^(e^x)=e^0
      Therefore, e^x=0'
      That is incorrect. Even just watching the video would have clarified that fact for you.

    • @gitanjalideb454
      @gitanjalideb454 7 หลายเดือนก่อน +1

      @@thetaomegatheta Okay, I am sorry about my context. Thank you.
      But I won't stop working (∞ &0) 🗿

    • @dev_time
      @dev_time หลายเดือนก่อน

      This kinda was my first thought, working with limits you can see this one is a singular solution... not sure if there are other apart from the identity shown in the video; kinda hard to tell with such a particular equation 🤔

  • @donwald3436
    @donwald3436 ปีที่แล้ว +29

    It's 2am why am I watching this lol.

  • @SuperDeadparrot
    @SuperDeadparrot ปีที่แล้ว +10

    Ln( 2pi*i*c1 ) = Ln( 2pi * c1 ) + i * pi/2 + i * 2k*pi because ln( i ) = ln( exp( i*pi/2 + 2k*pi*i ) ).
    Also, in complex functions, ln becomes log.

    • @lieman7136
      @lieman7136 ปีที่แล้ว

      ln doesn't become log
      instead ln becomes Ln and log(a)b becomes Log(a)b - capital letters

  • @Ligatmarping
    @Ligatmarping ปีที่แล้ว +4

    Making the mistake of reducing e^(e^x) = 1 => e^x = 0 is like the graduate level version of a^2 = b^2 => a = b hahaha. Nice video!

  • @Jaymac720
    @Jaymac720 ปีที่แล้ว

    In my mental math, it is equal when X equals -infinity.
    Take the natural log of both sides. You get e^x = 0
    Take the natural log again, but you have to apply limits. As X approaches 0 from the left, ln(x) approaches -infinity.
    You can also graph it. Y=1 is an asymptote of the function. It will only reach 1 at -infinity. Math is fun.
    The kinda complication with this though is that the function isn’t symmetrical because you can’t take a log of a negative number. Maybe you can with imaginary numbers, but I don’t know anything about that

    • @thetaomegatheta
      @thetaomegatheta 7 หลายเดือนก่อน

      The expression 'e^x' is undefined in this context.
      Also, functions do not have asymptotes. Curves do.

  • @algorithminc.8850
    @algorithminc.8850 ปีที่แล้ว

    There are many great bits on this channel ... but I really loved this one. Still chuckling ... sincere thanks. Cheers

  • @elemento7351
    @elemento7351 หลายเดือนก่อน +4

    Why do we need to multiply by one at 3:00? That extra 2ipiC2 does nothing in the answer it just goes to 1

    • @user-bz3kd2mt3u
      @user-bz3kd2mt3u 11 วันที่ผ่านมา

      the answer he gives is *A* solution to e^e^x = 1, but choosing any two integers C1, C2 (where C2 != 0) will give you a different equally-valid solution.

  • @ManjulaMathew-wb3zn
    @ManjulaMathew-wb3zn 8 หลายเดือนก่อน

    The answer could be simplified a bit more.
    e^x =(2PIn)i =(2PIn)e^(2PIk+PI/2)i
    Taking ln you get x=ln(2PIn). + i (,2PIk+PI/2) which is in standard a+ib format.

    • @ManjulaMathew-wb3zn
      @ManjulaMathew-wb3zn 8 หลายเดือนก่อน

      Now the back substitution of the general solution.
      e^x=(e^(ln2PIn))*e^(2PIk+PI/2)i
      =(2PIn)*(cos(2PIk+PI/2)+isin(2PIk+PI/2))
      e^x=(2PIn)(0+i)=2PIni
      e^(e^x)=e^2PIni =cos2PIn+isin2PIn = 1+i*0=1

  • @aliariftawfq5354
    @aliariftawfq5354 ปีที่แล้ว +21

    Take integral both side to get (c1,c2) value
    Thank you

    • @bobh6728
      @bobh6728 ปีที่แล้ว +6

      The c’s can be any non-negative integer. How does a integrating find a value?

    • @epikherolol8189
      @epikherolol8189 ปีที่แล้ว +2

      ​@@bobh6728Bros just Messing around lol

  • @1どらごん
    @1どらごん ปีที่แล้ว +2

    オイラーの定理から
    e^x=2iπ  ∴x=ln2iπ
    まではすぐわかりました
    オイラーの定理から
    e^(iπ/2)=i
    ですから
    x=ln2iπ=ln2+lnπ+iπ/2
    となるんですね

  • @Kingpl8.
    @Kingpl8. 10 หลายเดือนก่อน +2

    Limx->-infinity (e^x) =0. => e^e^x=1
    Has anser -infinity

    • @thetaomegatheta
      @thetaomegatheta 7 หลายเดือนก่อน

      Incorrect.
      e^x = 0 has no solutions.
      'e^(-inf)' is an undefined expression in this context.
      Also, google 'continuous function'.

  • @EPMTUNES
    @EPMTUNES ปีที่แล้ว +8

    Hey bprp! I am currently a freshman in college and I came up with this limit while bored in Calc. Give it a shot (and let me know if you think its tricky or cool)
    Lim pi(n-e/pi)
    n * Sin( ___________________ )
    n-> infinity n
    I'll give you a hint: i came across the sequence that this limit describes when iterating in geometry.

    • @tomctutor
      @tomctutor ปีที่แล้ว

      Lim(n->∞): n.sin(π-e/n) = n.sin(π) = n -> ∞
      if I am reading your input correctly!

    • @livinglogically8180
      @livinglogically8180 ปีที่แล้ว +3

      It's e right?

    • @EPMTUNES
      @EPMTUNES ปีที่แล้ว

      @@livinglogically8180 yes

    • @Remon_
      @Remon_ ปีที่แล้ว +1

      Physicist evaluation:
      Series expansion of sin(pi(n-e/pi)/n) so you get n*sin(pi(n-e/pi)/n) = e+O(1/n) and you just slap the limit and it's e.
      Probably some fancier way too but that's the easiest way to evaluate it fast.

    • @vibaj16
      @vibaj16 ปีที่แล้ว +1

      @@tomctutor No.
      lim(n → ∞) nsin(π(n - e/π)/n) = lim(n → ∞) nsin(π - e/n) = lim(n → ∞) nsin(e/n) = lim(n → ∞) esin(e/n)/(e/n)
      lim(n → ∞) e/n = 0
      let x = e/n
      lim(n → ∞) esin(e/n)/(e/n) = lim(x → 0) esin(x)/x = e
      So the answer is e.

  • @James-m3c1k
    @James-m3c1k 9 วันที่ผ่านมา

    Equations like e^(e^x) = 1 used to feel like they were from another planet! It’s amazing how much easier they become when you have tools that guide you step by step. SolutionInn’s AI study tool has been incredibly helpful for me-it breaks down such problems and even provides clear explanations along the way. Definitely recommend it to anyone tackling advanced math concepts like this

  • @jimschneider799
    @jimschneider799 ปีที่แล้ว +2

    Similar to the way that ln(1) = 2*i*pi*C[1], you also have ln(i) = (4*C[3]+1)*i*pi/2, so ln(2*i*pi*C[2]) = ln(2*pi*C[2]) + (4*C[3]+1)*i*pi/2, making the entire solution into x = ln(2*pi*C[2]) + (4*C[1]+4*C[3]+1)*i*pi/2. And, since C[1] and C[3] are just arbitrary integers, that can be further simplified to x = ln(2*pi*C[2]) + (4*D+1)*i*pi/2, for arbitrary integers constants D and C[2], with C[2] != 0.

  • @Black_Hole_Institute
    @Black_Hole_Institute 10 หลายเดือนก่อน

    Ln(2*pi*i*c1)=ln(2*pi*c1)+i*pi/2. Without this your solution is incomplete. You should also plot the distribution of answers on complex plane for wide range of c1 and c2.

  • @ayssinaattori9313
    @ayssinaattori9313 ปีที่แล้ว +8

    Thank you gor this video! I've been thinking about complex exponents recently and this was something really interesting I hadn't thought of before.

  • @lucanina8221
    @lucanina8221 ปีที่แล้ว +1

    x=ln(|2c1pi|) +i*(2pic2 + pi/2*sign(c1) ) where c1 integer different from 0 and c2 integer and ln the natural log (real to real) is the correct solution. Expressing x in terms of complex logarithm is kind of cheating the exercise.

  • @mangouschase
    @mangouschase 18 วันที่ผ่านมา

    maybe its kinda like integral from 1 to infinity of 1/x**p, where if p is any number bigger than 1 the integral converges.
    e**x has not enough power to converge to 1 but e**e**x does

  • @garywalker6216
    @garywalker6216 ปีที่แล้ว +4

    Love your videos!

  • @Rikky-the-Unstoppable
    @Rikky-the-Unstoppable ปีที่แล้ว +1

    e^eˣ = 1
    ⇒eˣ = In(1)/In(e)
    ∵ aˣ = b x = In(b)/In(a)
    ⇒eˣ = 0
    ⇒x = In(0)/In(e)
    ⇒x = In(0)
    [The value of In(0) is undefined]
    ∴ The solution of x is undefined

    • @thetaomegatheta
      @thetaomegatheta 7 หลายเดือนก่อน +1

      'e^eˣ = 1
      ⇒eˣ = In(1)/In(e)'
      No such implication if x is complex.

  • @WhosBean
    @WhosBean 11 หลายเดือนก่อน

    This comes about because when you are using the analytical continuation of the ln function all outputs have a +2pi*n at the end. For example ln(e^2) = 2+2pi*n. So ln(1) = 0 + 2pi*n.

  • @rainbye4291
    @rainbye4291 10 หลายเดือนก่อน +2

    Complex numbers are saviors and sadists at the same time.

  • @Alislaboratory
    @Alislaboratory ปีที่แล้ว +2

    I'm confused. Why do we have to bring the one back at 3:11 instead of taking ln? And what is stopping you from multiplying infinite ones to get different solutions?

    • @shalopo
      @shalopo ปีที่แล้ว

      It confused me as well. I think it was not necessary to multiply like that. However when you perform ln on both sides, THEN you'd get the additional additive term for the solution.
      If you keep multiplying, you'll get an equivalent result. You're just adding arbitrary constants *2i pi (after ln, it's adding), so they are equivalent to adding one constant * 2i pi.

  • @Kallian-corrluz
    @Kallian-corrluz 11 หลายเดือนก่อน +3

    0:50 *a very familiar whistle*

    • @blackholesun4942
      @blackholesun4942 11 หลายเดือนก่อน

      😂 whistle baby

    • @Kallian-corrluz
      @Kallian-corrluz 11 หลายเดือนก่อน

      @@blackholesun4942 here we go
      😙😙😙😙😙

  • @DoxxTheMathGeek
    @DoxxTheMathGeek 9 หลายเดือนก่อน

    It's actually pretty easy because e^x is periodic with 2pi*i. So:
    e^(e^x) = 1
    e^x = ln(1)+2kpi*i = 2kpi*i
    x = ln(2kpi*i)+2jpi*i
    With j and k being whole numbers. And well k can't be 0 cuz otherwise it wouldn't be defined. X3

  • @RyanStewart-j2m
    @RyanStewart-j2m ปีที่แล้ว

    I tried to solve it by raising both sides by e^e^ln(), however, I arrived at the conclusion that if e^e^x =1, then e^e^x = 1. I am truly a mathematical wizard

  • @ScenicFlyer4
    @ScenicFlyer4 ปีที่แล้ว +12

    Ok here's my question, because the answer is very much dependent on the last step. When you have e^e^ln(2ipi), if you cancel the e^ln first, you are left with e^2ipi, which is 1. And for that to be true, 2ipi must be equal to 0, because only being raised to the power of 0 will equal 1. However, if you first evaluate ln(2ipi), you would get undefined, as the natural log of 0 is undefined. Why can you do it the way that you did? I feel like this is a sort of divide by 0 cheat. Order of operations, parenthesis, then exponents. You go from the top down, so I think you are not allowed to cancel the e^ln first. I say you must first evaluate the ln(2ipi). Why? It's at the top. Take this as an example. 3^2^3. This is equal to 6561. First you do 2^3, which is 8, then you have 3^8, which is 6561. You are not allowed to do 3^2 first, as 9^3 is 729, not 6561. Order of exponential operations is extremely important in this scenario. This is like that old way of saying 1=2 by slyly dividing by 0, it's just in an (a-b)/(a-b) form, where a = b.
    Basically all I'm asking is reasoning why you can start evaluating in the middle of the exponents, and not the top. I don't think you can which is why I say e^e^x=1 has no solution. If you have any number, which e is, it's just like pi, if you want to raise it to a power in order to get 1, it MUST be 0. There is no other number where raising something to the power of it results in 0. Therefore e^e^x would have to mean that the second e^x=0. But apparantly it has no solution, therefore e^e^x=1 can't have a solution.
    Also I just had a thought. If 2ipi is zero, which it would have to be to make e^2ipi=1, then what you are doing is e^ln0, and calceling the e^ln to get 0. If you are allowed to do this, the that means that e^x=0 does have a soltion. It would be the natual log of 0. But if you say it does not have a solution, then that means you must start at the top of the exponents, and you would not be allowed to do what you did.
    Idk I'm not the math expert, there's probably an absolute boatload I'm overlooking and don't understand. But my feeble mind would like some answers if possible. Thanks.

    • @bigbaston6329
      @bigbaston6329 ปีที่แล้ว +8

      My friend, this is a good observation, but there is one point that you may be missing.
      If I may quote you : "if you cancel the e^ln first, you are left with e^2ipi, which is 1. And for that to be true, 2ipi must be equal to 0".
      2 i pi doesent have to be zero for e^ 2 i pi to be equal one because of periodicity.
      If we can take an example in real numbers of periodic functions like sin and cos, you have that cos(0) = cos (2pi) = cos (4pi) = ... = 1. However, 0, 2pi and 4pi are not equal. Therefore, if cos (a) = cos (b), it doesn't automatically mean that a = b.
      The same applies to complex exponential functions which can be a representation of complex numbers in the polar form.
      In the cartesian form, you have the complex number z = a + ib, where a is the real component and ib, the imaginary component.
      In the polar form, you can transform a and b (the cartesian coordinates) into a radius r and an angle thetha (ϴ), where r = sqrt(a^2+b^2) and ϴ = tan (b/a). Using trigonometry, you can also rearrange a and b so that you obtain that a = r cos (ϴ) and b = r sin (ϴ).
      Thus, z = r (cos (ϴ) + i sin(ϴ) ), which is also equal to e^irϴ by definition. As you can see, we also have a periodicity in the complex exponential function, notably because it is equal to a sum of periodic function (sin and cos). Therefore, two complex exponential values can be equal while their exponents are different.
      Your reasoning is correct only for real values of exponents, because there is no periodicity in that case.
      You can try it for yourself, i.e. compute e^2pi i and e^4pi i. You find that both are equal to 1 although 2pi i is not equal to 4pi i.
      After that you can see that in his solution, you have C1 (the one in the log) cannot be 0 to respect the definition of a log and by extension to respect the fact that e^x = 0 has no solution.
      I hope it makes it more clear. It is definitely not intuitive!

    • @stevenfallinge7149
      @stevenfallinge7149 ปีที่แล้ว +2

      ln(2iπ) is actually equal to ln(2π) + iπ/2, not undefined.

  • @GMPranav
    @GMPranav ปีที่แล้ว +2

    Nobody:
    The number 1 - "Now I am become death, the destroyers of worlds".

  • @danielmoylan3033
    @danielmoylan3033 ปีที่แล้ว

    Yeah I paused and tried to figure it out myself.
    So if e^(pi*i) = -1, then e^( e^( ln(pi*i) ) ) = -1, thus ( e^( e^( ln(pi*i) ) ) )^2 = 1, thus e^( 2*e^(ln(pi*i) ) ) = 1, thus e^( e^( ln(2*pi*i) ) ) = 1 (what I did is basically cancel the e/ln pair and recombine), and since 1*1 = 1, we can put a c next to the 2 pi i.
    Of course that doesn't explain the non-log version.

  • @mqb3gofjzkko7nzx38
    @mqb3gofjzkko7nzx38 ปีที่แล้ว +44

    Black pen red pen using a blue pen?

    • @hungry-sandwitch1355
      @hungry-sandwitch1355 7 หลายเดือนก่อน +2

      I know how impossible this sounds, but black pen red pen is using a blue marker

    • @hyperbroli6672
      @hyperbroli6672 7 หลายเดือนก่อน

      Calculate the concentration of opium that is in your bloodstream

    • @SimpdePaint
      @SimpdePaint 7 หลายเดือนก่อน +3

      Im waiting for blackpenredpenbluepengreenpenorangepenpinkpen

  • @wabc2336
    @wabc2336 ปีที่แล้ว

    I thought you would also write out what the ln of the imaginary 2iπc_1 equaled.
    ln(2iπc) = ln(2πc) + ln(i)
    ln(i) = ln(e^(iπ/2)) = iπ/2 + 2πc_3 but this is redundant with c_2.
    So our final answer is apparently ln(2πa) + i(π/2 + 2πb) for any integers a,b

  • @OptimusPhillip
    @OptimusPhillip 10 หลายเดือนก่อน

    Took me a second, but I got it. 0 is not the only solution to z=ln(1), any integer multiple of 2πi will also satisfy it. So e^x just needs to be an integer multiple of 2πi. This would make x=ln(2nπi), when n is some non-zero integer, or roughly (1.838+1.571i)+ln(n).

  • @alipourzand6499
    @alipourzand6499 ปีที่แล้ว

    At some point we have:
    e^x = i.2.pi.c1
    i = e^i.(pi/2 + 2.pi.c2)
    e^x = 2.pi.c1.e^i.(pi/2 + 2.pi.c2)
    x = ln(2.pi.c1) + i.(pi/2 + 2.pi.c2)
    Right? Wrong?

    • @lucanina8221
      @lucanina8221 ปีที่แล้ว +1

      almost there, note that c1 cannot be negative inside your logarithm

  • @francescoragghianti6068
    @francescoragghianti6068 17 วันที่ผ่านมา

    As always with complex numbers, it seems a paradox at first, but once you introduce a new dimension it becomes natural to find more solutions. Seems totally legit to me as an explaination: the solution is not in the number one in itself, but the journey you make when you approach it, great :D

  • @rainerzufall42
    @rainerzufall42 8 หลายเดือนก่อน +2

    Sorry Wolfram, I don't like the form of the solution: x = 2 i π c_2 + log(2 i π c_1) and c_1 !=0 and c_1 element Z and c_2 element Z
    It contains the log of a complex argument without any need.
    What about: x = i π (4 c_2 + 1) / 2 + ln(2 π c_1) and c_1 !=0 and c_1 element Z and c_2 element Z?

  • @TauGeneration
    @TauGeneration ปีที่แล้ว

    i like that i was surprised that the "let c1 = c2 = 1" did result into 1. of course that's the solution, you tried to prove that to begin with

  • @gheffz
    @gheffz ปีที่แล้ว +2

    Love it! Thank you.

  • @GSDKXV
    @GSDKXV 10 หลายเดือนก่อน

    Best channel on TH-cam idc

  • @Ytterbium176
    @Ytterbium176 7 หลายเดือนก่อน

    I think I came across another, yet slightly different solution: x = ln(2pi*|m|) + pi/2 *n*i for all integers m,n (and m0). Based on the equation e^x = 2pi*m*i, I supposed that x is a complex number of the form a+bi. This lead me to e^(a+bi) = e^a * e^bi = 2pi*m*i. Therefore, e^a = 2pi*m and e^bi = i, resulting in a = ln(2pi*|m|) and b = pi/2 * n. I'm not an expert on complex numbers, though... Is this a valid approach/result?

  • @mtaur4113
    @mtaur4113 ปีที่แล้ว +1

    ln(2m pi i) is also multivalued and could be broken down into more elementary parts. I suppose the definite integral of 1/z from 1 to 2m pi i works, but it's not something you can plug into a standard scientific calculator, or learned about in most algebra or Calc 1 classes.

  • @NibbaHibba
    @NibbaHibba ปีที่แล้ว +1

    The answer i got when i solved it was
    pi/2*mi + 2pin
    where n is any integer and m is any integer congruent to 1 modulo 4. Is this still the same thing?
    ( i solved for e^x = 2(pi)n(i) and said that angle is pi/2*m and amplitude is 2pi(n) )

  • @daniwalmsley611
    @daniwalmsley611 ปีที่แล้ว +3

    I feel like this should've been taught in schools,
    Like Sqrt(x^2] removes negative solutions, we needed a warning for logarithms too
    I am now slightly scared of how their might be a whole other set of numbers like okaginary but for logs instead of sqrt

    • @davejohnsondeveloper
      @davejohnsondeveloper ปีที่แล้ว

      Reminds me of this video: th-cam.com/video/MP3pO7Ao88o/w-d-xo.htmlsi=T40ITkfSMril8ZGG

    • @Sidnv
      @Sidnv ปีที่แล้ว

      Complex analysis already deals with how to define logarithms for negative numbers (and any nonzero complex number in general). One difference is unlike square root having two values, logarithms are infinitely multi-valued. That is really what this calculation is doing. Any complex number can be represented as re^(i theta) where r is the distance from the origin and theta is the angle the line segment joining 0 to the number makes with the real line. But adding 2 pi to the angle doesn't change the value of the complex number, so it actually has infinitely many possible representations, each separated by 2pi in the angle. When you take the logarithm, you get ln(r) (a positive real number) + i (theta + 2 pi n) where n can be any integer, so you have infinitely many values. The only number for which you cannot definite a logarithm is 0, and there is no way to actually make sense of ln(0), because ln has an "essential" singularity at 0.
      Here's a specific example, supposed you want to take ln(-1). -1 can be represented as e^(i (2n+1)pi) for any integer n. So taking the logarithm, you get the set {i (2n+1) pi: n is any integer}. So any of these values can be considered a logarithm of -1, and all these logs already exist in the complex plane.

    • @mschuhler
      @mschuhler 11 หลายเดือนก่อน

      this was taught in schools.

  • @fetch7312
    @fetch7312 ปีที่แล้ว

    i dont know if you read or take suggestions from the comments, but here's something that stumped me and my calc BC teacher while I was trying to prove the derivative of sin(x):
    I wanted to solve for the sum of sines without using the geometric proof, so I decided to implement euler's formula so I could use properties of exponents and real and imaginary parts to solve for the sum of sines. However, I wanted to first prove the formula, so after solving for the summation representation of powers of e through binomial expansion using lim(n>inf)(1+a/n)^n, i plugged in iz to the sum and separated it into the real and imaginary parts, giving me two taylor series, infsum(n=0)((-1)^n(z^2n)/(2n!)) and i*infsum(n=0)((-1)^n(z^(2n+1))/(2n+1)!). I already knew these series functions would create the equation cos(z)+isin(z) but I wanted to figure out if there was a way to work backwards from the taylor series functions to the original functions assuming that we don't know the equations the taylor series functions correspond to. My efforts were fruitless, so I'm curious if you could take a shot at it.

  • @danencel157
    @danencel157 11 หลายเดือนก่อน +1

    Great video ! What is the goal of putting at 3:13 the e^i2πC2 ?

  • @epicurodesamus
    @epicurodesamus 23 วันที่ผ่านมา

    I didn't understand a thing but I'm glad he was happy to reach that answer

  • @JesusSavedSolomon
    @JesusSavedSolomon 6 หลายเดือนก่อน

    well anything to the 0 power is equal to 1, therefore if e^x=0 then we can rewrite e^e^x as e^0 which equals 1

  • @samuelatienzo4627
    @samuelatienzo4627 ปีที่แล้ว +1

    I love the excitement at 6:30 😂

  • @mathmachine4266
    @mathmachine4266 ปีที่แล้ว

    x=ln(2πN1)+(2N2+1)πi/2, where N1 and N2 are integers

  • @nicolastorres147
    @nicolastorres147 ปีที่แล้ว +2

    For the final answer i don't like to be inside a log

  • @JCCyC
    @JCCyC ปีที่แล้ว

    Wait a sec. The first term, ln(2iπC₁), is just ln(i) + ln(2πC₁). But we know ln(i) is many values, namely iπ/2+2πiC₃ -- so the complete solution is iπ/2+2πiC₃+ln(2πC₁)+2πiC₂ -- which reduces to...
    ln(2πC₁) + i(π/2+2πC₂) -- which I think looks better because it's not expressed in terms of something weird like ln(i).

  • @xlorrix-6320
    @xlorrix-6320 ปีที่แล้ว +2

    why in the third step did you take i2pi*C1 anc multiply it by the polar form of 1 instead of just taking the logarithm?

  • @zhenhuazhao6100
    @zhenhuazhao6100 ปีที่แล้ว +14

    I am not sure if anyone has commented on it already. There are literally boxes and boxes of "blackpenandredpen" under the table. 😂😂😂

  • @greenrocket23
    @greenrocket23 9 หลายเดือนก่อน

    Complex analysis is mind-blowing! I wish I had more time to study that area of mathematics.

  • @faresk3168
    @faresk3168 ปีที่แล้ว +3

    What if the 1 of the 2nd line of your calculation wasn't converted into an exp^(2iPiC_1)? What would that change? in order words, why do we need a solution with 2 integers C_1, C_2? Thanks!

    • @jessewolf7649
      @jessewolf7649 ปีที่แล้ว +1

      Or for that matter, why not a solution with 3 constants or 4 constants or…?

    • @faresk3168
      @faresk3168 ปีที่แล้ว

      @@jessewolf7649 exactly, 1 could be exploded into a product of exp^2piC_n

    • @semiconnerd
      @semiconnerd ปีที่แล้ว

      I didn't get this either. Maybe it was in the solution wolfram alpha gave?

  • @johncirillo9544
    @johncirillo9544 ปีที่แล้ว

    This made me smile! 😊

  • @taito404
    @taito404 ปีที่แล้ว

    2:56 but why is it necessary to multiply it by 1? Can't we just say that the final answer is the natural log of that without the other 1?

  • @nicklanders5178
    @nicklanders5178 ปีที่แล้ว

    Fascinated by the way you hold and switch between markers

  • @1dayofmusic748
    @1dayofmusic748 ปีที่แล้ว +2

    i guess its just what complex jumbers are all about. you can solve more things at the sacrifice of your result being the only one. just the number 1 can be expressed by a lot similar to like the complex roots having more than one solution etc...

  • @PaweMateuszBytner
    @PaweMateuszBytner ปีที่แล้ว +3

    I have no time for checking it for sure, but my guess is that the catch is:
    1. while x = y e^x = e^y is true in Real,
    2. in Complex it is implied only one way: x = y => e^x = e^y, but not necessarily the other way around
    3. in the series of equations we actually need reversed implications to find a solution.
    4. what we've got here is a candidate for a solution, which we checked to not be valid.
    Example of this in Real is x = 2 => x^2 = 4, but x^2 = 4 => x = 2 OR x = -2. The negative solution to quadratic obviously is not a solution for linear.

  • @subzero4190295
    @subzero4190295 หลายเดือนก่อน

    ln(1) = e^x. The only way to have this is with limits, as x approaches -infinity

  • @yagoorviz7573
    @yagoorviz7573 10 หลายเดือนก่อน +2

    Why in the first one can't we do this? e^x=0, e^x=ln(1), e^x=ln(e^(2*pi*c1*i)), e^x=2*pi*c1*i, ln(e^x)=ln(2*pi*c1*i)+ln(e^(2*pi*c2*i)), x=ln(2*pi*c1*i)+2*pi*c2*i. And this leaves us with the same answer. I'd love to know if I'm wrong and where.

  • @ZolarV
    @ZolarV ปีที่แล้ว

    I think its not obvious by the way our math is defined, but is obvious when you expand the concpet of number's to the complex plane.
    E^x =0 doesnt really tell us the whole story, it only has no solutions when you exclude the complex plane, it has many solutions when extended.
    I've long thought that numbers are multidimensional, not just having a single value. When you start thinking about numbers as multidimensional objects, the strong links between types of numbers and fields such as the complex plane become apparent.

  • @danielduranloosli
    @danielduranloosli 11 หลายเดือนก่อน

    I think leaving that "i" in the argument of the natural logarithm makes the solution look a bit ugly despite its simplicity. I used algebraic notation (x=a+i*b) and found a more explicit form for the solution, which I think comes down to x=ln(2*pi*k)+i*(pi/2+pi*n), where k is a natural number and n is an integer... Am I wrong?

  • @mhm6421
    @mhm6421 ปีที่แล้ว +1

    Once you see what 1 actually is, you can't unsee it.

  • @bendeguz_
    @bendeguz_ ปีที่แล้ว

    loved the video!

  • @rubikaz
    @rubikaz 9 หลายเดือนก่อน

    The solution without using complex logarithms is ln(2πn)+(2m-1)πi/2 with n, m integer numbers

  • @Cagri2001
    @Cagri2001 8 หลายเดือนก่อน

    cool video, but the property ln(e^z) = z doesn’t always hold in the complex realm! ln(z1z2) also doesn’t need to equal ln(z1) + ln(z2). An explanation would be nice in the video

  • @-wx-78-
    @-wx-78- ปีที่แล้ว +7

    In complex world, things are… complex? 😉
    Complex numbers are marvelous. Cf. Gauss numbers.

  • @_cran
    @_cran 11 หลายเดือนก่อน

    You can just use the unit circle to find 2pi*i or using moivre formula, is quicker. Your way is a way to approach too but it's kinda longer 😅

  • @cordoba9502
    @cordoba9502 ปีที่แล้ว

    I don't know why the algorithm recommended me this and I can't say I understood much of it but I'm definitely not complaining about it

  • @zeniixlucil3590
    @zeniixlucil3590 2 หลายเดือนก่อน

    my problem with these always been that, why do we do 1=e*2ipi only once, there are countless 1s multiplied always, also given that e*2ipi=1 in the end, this gives x=2ipi which is different from the one we originally calculated

  • @AlessioLapponi
    @AlessioLapponi ปีที่แล้ว

    I do not agree since the logarithm function is not well defined in all the complex plane. You have to put a branch cut so that you cannot go around zero by a phase 2Pi. Once you choose the range of the phase (in your case ln1=i2Pi), you have to be consistent and considering that phase in all the logarithms. In the first step in particular, you have considered the real numbers to have the phase 2Pi c_1 at the right hand side and 0 in the left hand side, which is inconsistent and then wrong

  • @wren.10.
    @wren.10. ปีที่แล้ว +1

    I desperately dream to be as excited to unravel mathematics as he is.

  • @trojanleo123
    @trojanleo123 ปีที่แล้ว +1

    Can I use Euler's Identity for this to conclude that
    e^x = 0 = e^(i*pi) + 1
    Therefore,
    x = ln[e^(i*pi) + 1]
    Would that be an acceptable answer?

    • @wilwdr96
      @wilwdr96 ปีที่แล้ว +2

      well e^(i*pi) + 1 = -1 + 1 = 0 so thats just the same thing as saying x =ln (0) though, which is just as undefinined as e^x = 0, so you have simply written that there is no solution in a more confusing way

  • @et427gamer9
    @et427gamer9 ปีที่แล้ว +1

    I understood very little of this as a high school student but it was very enjoyable

  • @spudhead169
    @spudhead169 ปีที่แล้ว

    if you graph W(x) from 0 to e, also graph ln(x) from 1 to e, both plots intersect at e and they form a "shark's fin" shape. find the area of the shark's fin?

    • @jensraab2902
      @jensraab2902 ปีที่แล้ว

      e-2

    • @spudhead169
      @spudhead169 ปีที่แล้ว

      @@jensraab2902You're right, and it was pretty easy to check that with WFA, but I wanted to see how W(x) could be integrated.

    • @jensraab2902
      @jensraab2902 ปีที่แล้ว

      @@spudhead169Ah, but that's not what you asked! 😁

    • @spudhead169
      @spudhead169 ปีที่แล้ว

      @@jensraab2902 Well no, you're right again, but with any problem it's nice to see the work done to get the answer. I was just trying to encapsulate it into a interesting problem. WFA will give you the anti-derivative of W(x) but it'd be nice to see how that's done.

    • @jensraab2902
      @jensraab2902 ปีที่แล้ว

      @@spudhead169I understand. Unfortunately, I have no idea how it's done, even though I'd be also interested to see the process (even though I suspect it might be past my mathematical understanding).

  • @TheMemesofDestruction
    @TheMemesofDestruction ปีที่แล้ว

    “True Love, Priceless. For everything else there’s Wolfram Alpha.” ^.^