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

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The exponential equation e^x=0 has no solutions, not even in the complex world, but e^e^x=1 does have solutions! I was surprised to see how WolframAlpha actually gave the solutions to this seemingly impossible equation and I would like to show you how to solve it! The trick is to write 1 in the complex polar form. Subscribe to ‪@blackpenredpen‬ for more math for fun videos. #math #complexnumbers #blackpenredpen #fun #tutorials
🛍 Euler's Identity e^(iπ)+1=0 t-shirt: amzn.to/427Seae
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30 сен 2024

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Комментарии : 748

@blackpenredpen Год назад

Can 1^x=2? Solution here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-9wJ9YBwHXGI.htmlsi=dx-XGv_Wf3_0VDH2 🛍 Euler's Identity e^(iπ)+1=0 t-shirt: amzn.to/427Seae

@PauloChacal Год назад

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

@doodle1726 Год назад

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

@onesecondbaobab Год назад

I giggled out loud xD

@Hardcore_Remixer Год назад

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 11 месяцев назад

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

@Questiala124 10 месяцев назад

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

@MathFromAlphaToOmega Год назад

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 Год назад

thats so cool!!!

@Isvakk Год назад

Are you sure it's holomorphic?

@MathFromAlphaToOmega Год назад

Yes, it's infinitely differentiable everywhere.

@jakobullmann7586 Год назад

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

@blackpenredpen Год назад

Wow! This is super cool!!

@haashir7312 11 месяцев назад

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.

@acykablyatley Год назад

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 10 месяцев назад

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

@_cran 8 месяцев назад

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 7 месяцев назад

@@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 5 месяцев назад

i like your funny words magic man

@acykablyatley 5 месяцев назад

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

@Sg190th Год назад

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

@mqb3gofjzkko7nzx38 Год назад

Black pen red pen using a blue pen?

@hungry-sandwitch1355 4 месяца назад

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

@hyperbroli6672 4 месяца назад

Calculate the concentration of opium that is in your bloodstream

@SimpdePaint 4 месяца назад

Im waiting for blackpenredpenbluepengreenpenorangepenpinkpen

@timothyrosenvall1496 Год назад

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

@stephanelem822 Год назад

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.

@passager683 Год назад

Man, you can't just compose complex exponential and complex logarithm with that 'real' ease. More rigor, please.

@benmyers4279 Год назад

This! And he didn’t even point out that there’s a complex log in the wolfram alpha solution, which can easily be simplified away. The solution is x=ln(2nπ)+((2m+1)π/2)*i for natural numbers n and integers m

@passager683 Год назад

@@benmyers4279 Indeed mate, what a shame.

@shoutplenty 11 месяцев назад

i’m not convinced the guy is really a mathematician lol

@dougdimmedome5552 Год назад

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

@leoniekrenzer7716 Год назад

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 11 месяцев назад

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

@dethfr491 7 месяцев назад

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

@DanoshTech 8 месяцев назад

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

@alexeynezhdanov2362 6 месяцев назад

Because it is - cool and interesting.

@DanoshTech 6 месяцев назад

@@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

@donwald3436 Год назад

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

@deathmight2uuotba987 11 месяцев назад

Same bro

@CliffSedge-nu5fv 2 месяца назад

Sweet dreams!

@-wx-78- Год назад

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

@zhenhuazhao6100 Год назад

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

@cristrivera Год назад

True😂

@tomctutor Год назад

∞ {🖋🖍} 😎

@isilverboy Год назад

@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 Год назад

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 Год назад

@@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 Год назад

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

@XJWill1 Год назад

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 6 месяцев назад

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!

@camelloy Год назад

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.

@raghavsharma8160 Год назад

CHALLENGE FOR BLACKPENREDPEN If 1+sin^2(A)=3sin(A)cos(A) then prove that A=π\4 and tan(A)=1 or 1/2. Scroll down for hint . . . . . . . . . . . . Hint: Quadratic Equation in tan(A)

@dentonyoung4314 Год назад

Wow. That was an amazing explanation.

@aliariftawfq5354 Год назад

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

@bobh6728 Год назад

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

@epikherolol8189 Год назад

@@bobh6728Bros just Messing around lol

@scoutgaming737 Год назад

But wouldn't you be able to just add an infinite amount of e^2iπc and have an infinite amount of variables?

@ddsqadod2994 Год назад

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.

@robsmith9696 10 месяцев назад

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.

@GMPranav 11 месяцев назад

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

@faresk3168 Год назад

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 Год назад

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

@faresk3168 Год назад

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

@semiconnerd Год назад

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

@1どらごん 11 месяцев назад

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

@rainerzufall42 6 месяцев назад

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?

@MATHMASTERPRO Год назад

hi bprp, I really like watching your videos. By the way, I'm from Indonesia and I'm still in grade 8

@xinpingdonohoe3978 Год назад

What age is grade 8?

@MATHMASTERPRO Год назад

@@xinpingdonohoe3978 14

@rainbye4291 7 месяцев назад

Complex numbers are saviors and sadists at the same time.

@ayssinaattori9313 Год назад

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.

@Ligatmarping Год назад

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!

@NibbaHibba 11 месяцев назад

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) )

@nicolastorres147 11 месяцев назад

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

@michaelz2270 Год назад

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 Год назад

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

@abhijithcpreej Год назад

Usually I have a tough time figuring out my own solutions during some of the "math for fun" videos. But this time, I could tell from the thumbnail. So weird😊

@ScenicFlyer4 Год назад

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 Год назад

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 Год назад

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

@aadityarao8694 Год назад

But the i*2pi*C1 term can be multiplied by 1 as many times as we want which will just end up giving us more constants. Why is it necessary to multiply it by 1 only once?

@xinx9543 Год назад

Because integer C1 already means how many times we multiply other 1 on it. So C1 could be any interger and actually we have infinitely many solutions

@aadityarao8694 Год назад

But then we might as well stop at C1 right? What is the necessity to multiply it by the term with C2? C1 has already dealt with the infinite answers.@@xinx9543

@vivianriver6450 Год назад

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 Год назад

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 Год назад

@@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 Год назад

@@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.

@pikatheminecrafter Год назад

e^x = 0 may not have a solution, but we all know it simplifies to x = ln(0)

@ankat_ 17 дней назад

Holdup... if ln(1) = ln(e^(2iπ)) Then 0 = 2iπ

@lucanina8221 Год назад

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.

@itsviggisbot2149 Год назад

Can anyone integrate -ydx + xdy,.Wolfram alpha says 0

@SuperDeadparrot Год назад

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 10 месяцев назад

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

@rockapedra1130 6 месяцев назад

Ouch ... You damaged something in my brain with this one!

@mikejurney9102 Год назад

Does this also have a solution in the quaternions and octonians? Does i equate to some quaternion number?

@xinpingdonohoe3978 Год назад

i would equate to the quaternion number i. Quaternions are just complex numbers with the j and k axes added to the mix, and octonions follow that pattern again.

@xlorrix-6320 Год назад

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?

@jimschneider799 Год назад

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.

@daniwalmsley611 11 месяцев назад

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 11 месяцев назад

Reminds me of this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-MP3pO7Ao88o.htmlsi=T40ITkfSMril8ZGG

@Sidnv 10 месяцев назад

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 8 месяцев назад

this was taught in schools.

@punpcklbw 11 месяцев назад

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.

@simonekentish7491 Год назад

I wonder if DeMoivre got this giddy when he discovered complex solutions using the complex plane and polar coordinates.

@eric_enter9141 9 месяцев назад

I am just comfused why you multiply by one in that one step. That does nothing for the ecuation as the end result (2*i*pi*c1) allways ends up being 1. I am not slandering. I am in 11th grade. Just asking as this is way beyond what we are going through.

@thetaomegatheta 4 месяца назад

Apologies, I misunderstood what you were referring to, initially. The point is that the relevant number can be represented as 2*Pi*i*c_1*e^(2*Pi*i*c_2).

@woodpecker4044 Год назад

What about x = ln(2*pi*n) + (1/2 + 2k)*pi*i where n,k are whole numbers and n not zero ?

@davidhowe6905 Год назад

This makes checking easier for someone (like me until recently!) who doesn't know logs of complex numbers (though they were needed in the derivation).

@davidhowe6905 Год назад

Just realised it should be ln|2 pi n| + (1/2 + 2k)pi i as the solution allows negative n.

@woodpecker4044 Год назад

@@davidhowe6905 the natural logarithm is also defined for negative numbers because we are in the complex world.

@davidhowe6905 Год назад

Thanks, I saw another video where they said that zero and the negative real number line do not have logs defined; something to do with getting a different value when approaching the negative real axis clockwise and anticlockwise in the complex plane (where the principal value range is taken as -pi to +pi). This is not my speciality, so I have no fixed opinion on this. @@woodpecker4044

@lucanina8221 Год назад

@@woodpecker4044that won't make it the natural log😂.

@lukekim7012 Год назад

You should define ln(conplex number) before u use it

@Lqskxffmwldlxlfmemwmdmf Год назад

Hi sir good day can you teach is differential calculus

@VKHSD Год назад

its cuz ln 0 is just negative infinity but nobody wants to admit it xd

@1dayofmusic748 Год назад

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...

@danencel157 8 месяцев назад

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

@russellchido Год назад

I dont see the point in introducing the term with C_2

@Kv0rpse Год назад

I have a question though. Isnt negative infinity a solution? e^(-inf) = 0 and e^(0) = 1

@jensraab2902 Год назад

-∞ isn't a number, though. If I'm not mistaken you can say that the limit of e^e^x is 0 when x goes to -∞ but that's not a proper solution because +∞/-∞ aren't numbers.

@lammatt 11 месяцев назад

@@jensraab2902infinity is not a number It is more like an idea

@rajaskasar6094 8 месяцев назад

I just hate it when you have to do something completely unexpected to get the result and unfortunately that is most of maths

@alslaboratory570 Год назад

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 Год назад

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.

@NouchG 5 месяцев назад

I accidentally read it 0 has no soul😢

@BurningShipFractal Год назад

Third (Not including bprp himself)

@Tanishk-ot7dx Год назад

Bruh I didnt think about '1' in this way

@johnbutler4631 Год назад

This is really wild. I actually tried this on a TI-84, which js nowhere near as powerful as Wolfram Alpha, and it worked.

@okWishFull Год назад

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

@soulswordobrigadosegostar Год назад

by that logic,doesn't e^x=0 have infinitely many solutions? because you can just say that 0=ln(1) and solve for that also using the same logic

@gmtv_levi8113 Год назад

my fyp really thought i became a genius after watching a few math classes on youtube. They really be recommending me this but i like it tho

@josephyap4981 Год назад

My math homework be like:

@UCXEO5L8xnaMJhtUsuNXhlmQ Год назад

"Just let x be negative infinity" -engineers

@Supdry-nv8we 7 месяцев назад

How about e^e^ln(0)=1 ???? Very easy solution

@faheem4977 7 месяцев назад

Ln(0) is not defined nice try tho

@OptimusPhillip 7 месяцев назад

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).

@orisphera Год назад

My answer is x = (ln (2πn)) + (π/2+πm)i

@mhm6421 Год назад

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

@alexandershapiro28 Год назад

Interesting how it's only e^e^x=1 iff when there's at least one full revolution, while it's impossible when there's inaction. I think this is because of some deep algebraic topology stuff although pi(SO(2,C)) even at no rotation is well defined, maybe someone expert on the matter would give me a light on my fog of thoughts

@burzummmmm Год назад

I am no expert but it is easy to see when c1, where c1 is the amount of rotations, is 0; x = ln(2ipic1)+A = ln0 + A. ln0 is undefined therefore in this case e^x^x is undefined.

@Sidnv 10 месяцев назад

Yes there is a connection with algebraic topology here. You can look at at the exponential function as a map from the complex plane to itself minus 0. The fundamental group of the plane minus 0 is Z (and in fact this space is homotopic to SO(2, R), which is the circle S1). So the fundamental group acts on the preimage of the map above any point, essentially via rotation.

@Gabbyreel Год назад

so then where is the error in taking the natural log of both sides? if we have 2 different valid answers we have a contradiction somewhere don't we? I'm confused!

@SeekingTheLoveThatGodMeans7648 Год назад

Is it an issue of a primary branch? Otherwise natural log wouldn't qualify as being a function.

@tschumacher Год назад

I believe the error is assuming that ln(1)=0 which is not taking into account complex numbers. ln(1) is defined as the exponent we have to raise e to to get 1 and considering the complex numbers all integer multiples of 2ipi work, not just zero. See also en.m.wikipedia.org/wiki/Complex_logarithm

@열럽 26 дней назад

in fact, exp(x) = 0 has no solution, while exp(x) = 1 has infinite solutions. it could be 0, but generally it has 2pi*i*n(for all integers), so exp(exp(x)) =1 has also infinite solutions, and they have 2 degrees of freedom. this video clearly explains above, and when c1=0, there's no solution, because there's no such thing suffices exp(x)=0.

@andrejsasd8904 2 месяца назад

Stupid question: why wouldn't "x= - infinity" work? Shouldn't e^(- infinity)=0? And then e^e^(-infinity)=e^0=1? Or am I missing something here?

@Kambyday 6 месяцев назад

Haha we know e^2iπ=1 e^e^x=e^2iπ Taking the natural log of both sides ((log e)=1) e^x=2iπ Taking the natural log again x=log(2iπ)

@hadhamalnam Год назад

Does it make sense to take the natural log of a complex number though? Wouldnt that have multiple solutions, hence making it not function? For example, I can say that ln(i) is i(pi/2), but it can also be any i(pi/2) +2npi.

@Ironpecker 5 месяцев назад

You don't really need to treat the ln as a function in this case though, it's just an operation you apply, same thing with stuff like arcsin or arcos normally they wouldn't be functions (unless you restrict their codomains) but in an equation it's not important. I'm not 100% sure myself though, but this is the logic that makes the most sense to ms

@misterroboto1 4 месяца назад

Think of the log as the "inverse" of the exponential map. If it makes sense to compute the exponential of a complex number, then it can make just as much sense to ask yourself "what complex number(s) x could I input in the exponential map in order to get a given complex number y as the ouput".

@DoxxTheMathGeek 6 месяцев назад

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

@SeanAlunni Год назад

What is the purpose of "+ 2*i*π*c2"? The result is the same regardless

@raykirushiroyshi2752 6 месяцев назад

Ooooh. At 1st i was wandering why e^x cant be 0, after all for e^n = 1 ,n must be equal to 0 right?NO,NOT IN THE COMPLEX PLANE. The definition of logarithm in the complex plain is Ln(|z|) + i arg(z), ln(1)=z has infinitly many solutions not just 1.

@Ytterbium176 5 месяцев назад

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?

@MATHMASTERPRO Год назад

Can you make a video about the Riemann hypothesis in the next video? I'm very curious about that problem

@rubikaz 6 месяцев назад

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

@hashtag9990 7 месяцев назад

Why does the extra e^(2iπ) = 1 matter? You can have infinitely many solutions for just ln(2iπc), or keep multiplying with 1 over and over and have solution like ln(2iπc1) + 2iπc2 + 2iπc3 and so on......, why stop at only just one more multiplying by 1?

@skinnyladd Год назад

can't you just keep multiplying '1' at 3:05 ? why do you only do it once?

@gitanjalideb454 5 месяцев назад

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 4 месяца назад

'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 4 месяца назад

@@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 4 месяца назад

'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 4 месяца назад

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

@danielduranloosli 8 месяцев назад

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?

@SanctBlack Год назад

This is not good to have imaginary number inside log function. Better to see that i=e^(i*(pi/2+2*pi*c2)). The solution will be better in the form: x=i*(pi/2 +2*pi*c2)+log(2*pi*c1)

@danielmoylan3033 Год назад

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.

@Cagri2001 5 месяцев назад

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

@youngjoelroberts 11 месяцев назад

Yeah, but ln(2i*pi) isn’t a real number… That number is x in e^x=(2i*pi) = e^(i*pi/2)*e^ln(2*pi) So x = i*pi/2 + ln(2*pi) Just worked it out

@Black_Hole_Institute 7 месяцев назад

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.

@drdiegocolombo Год назад

It is not clear why we need to write e^x=i2πc as e^x=i2πc x 1 to solve the equation. From e^x=i2πc it is not possible to immediately solve and therefore x=ln(i2πc)? Thank you