how is e^e^x=1 solvable??
ฝัง
- เผยแพร่เมื่อ 24 ก.ย. 2023
- 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
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Thank you all!
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
Can you solve (e^i)+(i^e) pls
@@ISoldBinLadensViagraOnEbay where the equation?
@@MATHMASTERPRO I am asking for the solution of the answer, this is not an equation, since e and i are numbers
@@ISoldBinLadensViagraOnEbay (e^i)+(i^e)=cos(1)+cos(i*pi/2)+i(sin(1)+sin(i*pi/2))
@@ISoldBinLadensViagraOnEbay
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 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
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.
It's nice seeing the complex world being used more.
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.
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.
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.
I love your excitement! Loved every moment of this!!!
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 .
Wow. That was an amazing explanation.
The greatest thing about complex analysis is slowly overtime making more insane infinite expressions to approximate 1.
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
Love your videos!
There are many great bits on this channel ... but I really loved this one. Still chuckling ... sincere thanks. Cheers
Love it! Thank you.
So glad!
@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!
It's 2am why am I watching this lol.
Same bro
This made me smile! 😊
loved the video!
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.
That was beautiful thank you
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.
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.
Hi Dr. Pen!
Very cool!
Fascinated by the way you hold and switch between markers
I love the excitement at 6:30 😂
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.
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
Great video ! What is the goal of putting at 3:13 the e^i2πC2 ?
Great job great tricks he has
Thanks Sir
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
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...
I understood very little of this as a high school student but it was very enjoyable
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
I desperately dream to be as excited to unravel mathematics as he is.
Really crazy🙈- great👍
Complex analysis is mind-blowing! I wish I had more time to study that area of mathematics.
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
Nobody:
The number 1 - "Now I am become death, the destroyers of worlds".
Nice solution
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😊
Best channel on TH-cam idc
Amazing👍
Great teacher
Brilliant!
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
Please sent the name of book to read wolfram
Where did we get the second “1” which is e^i2πC2? Where did it come from?! As initially it did appear in the exponent!
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 😅
Math is so f***ing satisfying.
this is satisfying answer to the equation
That omg in the end
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
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) )
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.
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?
“True Love, Priceless. For everything else there’s Wolfram Alpha.” ^.^
Why specifically stop at c_2? You could keep multiplying by 1 an infinite amount of times, so wouldnt c_(n>1) be more fitting for conciseness?
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: th-cam.com/video/MP3pO7Ao88o/w-d-xo.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.
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.
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.
Amazing.
Still amazing explanations professor
This is really wild. I actually tried this on a TI-84, which js nowhere near as powerful as Wolfram Alpha, and it worked.
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
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.
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.
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
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".
Its been years since i had any complex analysis, but i remember ln(x) can be a bit didgy sometimes.
Does ln(i) make sense?
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
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
understood like 10% but love your passion and energy🫶
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!
When we have f^-1(f(x)) we say it equals x but this makes sense if f has an inverse i.e. f is an 1-1 function. e^x is not an 1-1 function in the complex plane. So how do we do this?
How can we use ln func, when it's defined from the non negative to the real numbers?
I agree
Why multiply by 1 again and get the second constant c2? What would be the issue with having just c1 constant and not include the c2 constant?
Why do u need to multiply by 1? Would x = ln(i2pi k) not be a solution , or is it simply that I would be missing all other solutions, as there are infinite?
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.
If you do fourier series/fourier transforms often, then youd agree with this video. I don't know what the reason for the 2ipi term is though. Ln(2ipi) seems like a solution.
It is because ln(2i pi) is itself multi-valued. 2ipi can be written as 2pi e^[i(2n + 1/2) pi] for any integer n. This calculation is just taking a fixed value for this logarithm and then adding in all the other solutions.
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
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
Loved it!
this guy never fails to bring out what i like the most out of arithmetics and also what i hate the most
Was the inclusion of c2 at all relevant here? It got included in the equation via *1 and 1 = e^(2pi i c2) shenanigans, but it ended up being factored out the same way, as in the power of e it just turns into *1 again. Surely leaving out the c2 and reducing the solutions to a single dimension (rather than 2 dimensions) of infinite possible answers would have been simpler and just as viable?
Does this also have a solution in the quaternions and octonians? Does i equate to some quaternion number?
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.
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?
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?
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
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).
x=ln(2πN1)+(2N2+1)πi/2, where N1 and N2 are integers
Thats so cursed i love it
Hello sir,
Here I have a question.
What's the value of tan inverse of tan(3x/5) sir?
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.
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.
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
Why not convert the i inside the ln to polar form? To avoid creating another variable?
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).
How about we take Natural log?