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
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.
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...
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
@@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
@@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
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
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.
@@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).
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
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!
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.
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)
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)).
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.
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.
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.
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?
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.
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.
@@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.
@@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.
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) )
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.
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.
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.
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
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).
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
I have a different answer: e^e^x = 1 e^e^x = e^i2πn, n is an integer They have the same base so we can assume their exponents are equal e^x = i2πn, n is an integer x = ln(i2πn), n is an integer
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
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.
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.
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
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=-∞.
'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.
@@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.
'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.
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 🤔
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.
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.
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.
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...
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.
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.
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.
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!
With the help of DeMoivre Who's theorem we love ya There's cube roots all over the plane Yes they're complex, but do not perplex, A new kind of numbers we gain
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
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?
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.
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.
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
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😊
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).
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
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
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
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!
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?
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?
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
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?
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.
Can 1^x=2?
Solution here: ruclips.net/video/9wJ9YBwHXGI/видео.htmlsi=dx-XGv_Wf3_0VDH2
🛍 Euler's Identity e^(iπ)+1=0 t-shirt: amzn.to/427Seae
Can you solve (e^i)+(i^e) pls
@@ISoldßinLadensViagraOnEbayఔ where the equation?
@@MATHMASTERPRO I am asking for the solution of the answer, this is not an equation, since e and i are numbers
@@ISoldßinLadensViagraOnEbayఔ (e^i)+(i^e)=cos(1)+cos(i*pi/2)+i(sin(1)+sin(i*pi/2))
@@ISoldßinLadensViagraOnEbayఔ
cos(1)+cos(πe/2)+i sin(1)+i sin(πe/2)
It won't really contract into any nice form.
Number “1” will never be the same after your explanation
Real true man now i can't get it out of my head arrrrghh😱😱🤯
I giggled out loud xD
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 😂
Ahh I gave this comment the 1000th like…the transition from 999 to 1k seemed surreal
Instead of saying “I am number one” say “ I am number e^(2i(Pi)n)”.
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
thats so cool!!!
Are you sure it's holomorphic?
Yes, it's infinitely differentiable everywhere.
@@Isvakk It’s a composition of holomorphic functions, this holomorphic itself.
Wow! This is super cool!!
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.
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...
lol, you definitely would have never guessed this answer, be humble.
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
@@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
i like your funny words magic man
@@seja098 i did not say that i guessed that answer, and my comment clearly says it was easy to see /after/ watching the video.
I love how passionate he is he makes math seem cool and interesting
Because it is - cool and interesting.
@@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
It's nice seeing the complex world being used more.
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
The greatest thing about complex analysis is slowly overtime making more insane infinite expressions to approximate 1.
@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.
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.
@@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).
Log of a complex number is many-valued function, therefore it is preferable not to use it when it is possible
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
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!
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.
Wow. That was an amazing explanation.
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)
Wow okay so this is the reason why the simple approach misses solutions!
That's why there is term called "principal logarithm" of complex numbers .
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)).
I simply brute-forced my way to e^(e^(0.5*pi*i+1.8378770664093455)) lmao
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.
I love your excitement! Loved every moment of this!!!
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.
Take integral both side to get (c1,c2) value
Thank you
The c’s can be any non-negative integer. How does a integrating find a value?
@@bobh6728Bros just Messing around lol
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.
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.
0:50 *a very familiar whistle*
😂 whistle baby
@@blackholesun4942 here we go
😙😙😙😙😙
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
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.
ln doesn't become log
instead ln becomes Ln and log(a)b becomes Log(a)b - capital letters
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?
Black pen red pen using a blue pen?
I know how impossible this sounds, but black pen red pen is using a blue marker
Calculate the concentration of opium that is in your bloodstream
Im waiting for blackpenredpenbluepengreenpenorangepenpinkpen
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.
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.
@@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.
@@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.
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!
There are many great bits on this channel ... but I really loved this one. Still chuckling ... sincere thanks. Cheers
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) )
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.
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.
I understood very little of this as a high school student but it was very enjoyable
Fascinated by the way you hold and switch between markers
Love it! Thank you.
So glad!
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.
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
And you got the wrong answer.
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).
It's 2am why am I watching this lol.
Same bro
Sweet dreams!
Complex analysis is mind-blowing! I wish I had more time to study that area of mathematics.
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
The solution without using complex logarithms is ln(2πn)+(2m-1)πi/2 with n, m integer numbers
I have a different answer:
e^e^x = 1
e^e^x = e^i2πn, n is an integer
They have the same base so we can assume their exponents are equal
e^x = i2πn, n is an integer
x = ln(i2πn), n is an integer
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
Reminds me of this video: ruclips.net/video/MP3pO7Ao88o/видео.htmlsi=T40ITkfSMril8ZGG
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.
this was taught in schools.
this is satisfying answer to the equation
Nobody:
The number 1 - "Now I am become death, the destroyers of worlds".
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.
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
“True Love, Priceless. For everything else there’s Wolfram Alpha.” ^.^
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=-∞.
'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.
@@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.
'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.
@@thetaomegatheta Okay, I am sorry about my context. Thank you.
But I won't stop working (∞ &0) 🗿
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 🤔
Love your videos!
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.
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.
x=ln(2πN1)+(2N2+1)πi/2, where N1 and N2 are integers
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.
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...
Nice solution
For the final answer i don't like to be inside a log
Great video ! What is the goal of putting at 3:13 the e^i2πC2 ?
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.
Lim(n->∞): n.sin(π-e/n) = n.sin(π) = n -> ∞
if I am reading your input correctly!
It's e right?
@@livinglogically8180 yes
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.
@@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.
ln(1) = e^x. The only way to have this is with limits, as x approaches -infinity
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.
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!
ln(2iπ) is actually equal to ln(2π) + iπ/2, not undefined.
I wonder if DeMoivre got this giddy when he discovered complex solutions using the complex plane and polar coordinates.
With the help of DeMoivre
Who's theorem we love ya
There's cube roots all over the plane
Yes they're complex,
but do not perplex,
A new kind of numbers we gain
In complex world, things are… complex? 😉
Complex numbers are marvelous. Cf. Gauss numbers.
Hi Dr. Pen!
Very cool!
Man, you can't just compose complex exponential and complex logarithm with that 'real' ease. More rigor, please.
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
@@benmyers4279 Indeed mate, what a shame.
i’m not convinced the guy is really a mathematician lol
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?
understood like 10% but love your passion and energy🫶
This made me smile! 😊
That was beautiful thank you
I love the excitement at 6:30 😂
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?
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.
Best channel on RUclips idc
This is really wild. I actually tried this on a TI-84, which js nowhere near as powerful as Wolfram Alpha, and it worked.
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.
I am not sure if anyone has commented on it already. There are literally boxes and boxes of "blackpenandredpen" under the table. 😂😂😂
True😂
∞ {🖋🖍} 😎
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
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 😅
they should add this to the math Olympiad
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😊
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?
almost there, note that c1 cannot be negative inside your logarithm
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
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
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).
Loved it!
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
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
'e^eˣ = 1
⇒eˣ = In(1)/In(e)'
No such implication if x is complex.
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
My answer is
x = (ln (2πn)) + (π/2+πm)i
was i the only one that thought he was going to use the W Lambert function at 3:27 hahaha
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
The expression 'e^x' is undefined in this context.
Also, functions do not have asymptotes. Curves do.
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!
Or for that matter, why not a solution with 3 constants or 4 constants or…?
@@jessewolf7649 exactly, 1 could be exploded into a product of exp^2piC_n
I didn't get this either. Maybe it was in the solution wolfram alpha gave?
Math is so f***ing satisfying.
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?
Once you see what 1 actually is, you can't unsee it.
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?
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
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
ln(2nπi)
n includes all integers
How about we take Natural log?
Omg hes evolved to black pen red pen AND blue pen, this is insane
Limx->-infinity (e^x) =0. => e^e^x=1
Has anser -infinity
Incorrect.
e^x = 0 has no solutions.
'e^(-inf)' is an undefined expression in this context.
Also, google 'continuous function'.
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?
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.