I Solved A Nice Exponential Equation (NOT REALLY!)
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- Опубліковано 7 січ 2024
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This video is a good example of why the exponent rules of real numbers cannot be blindly applied to complex numbers. In particular, the real identity (a^b)^c = a^(bc) does NOT hold in general in the complex world. (It does work for integer exponents, but not for fractional exponents like is done in this video.) With more care, you wouldn't end up with nonsense results claiming that an expression with modulus clearly equal to 1 equals a value with modulus e.
I agree!
@@SyberMath I feel like you posted this video legitimately and after people pointed out that it's wrong, you've decided to play it off as just a joke.
The distribution rule holds when you explicitly state the paramedic function n, as the video did. That's not an issue.
no you're wrong because i is a very broken number in the first place. you can't deal with such expression with that number i that is equal to its reciprocal, to its opposite, and i is also a number that proves that any number equals any number. + plus it doesn't make sense in geometry so yeah heck that
@@cegexen8191 are you tripping?
I am a bit disappointed. My math hero falls for the same ”trap“! 1 has a modulus of 1, e one of e. Exponentiating complex numbers works by multiplying the argument with the exponent and exponentiating the modulus. You can only reach all n complex n-th roots of 1 by exponentiating 1. They have all a modulus of 1. Otherwise 1 to anything is always 1. I only say: De Moivre. Too sad!
Truly it was quite sad that he fell for the trap
What's the trap?
@@SyberMath The trap is that (a^b)^c = a^(bc) is not generally valid in complex arithmetic.
Oh, I see. Thanks!
@@SyberMath Yes. And you can see the trouble if you reconsider your step at 7:50 where you are checking your candidate solution x=1 - (1 / 2π)i. At that point you chose to represent 1 as e^2π, then misapplying the Laws of Indices outside of where it is valid, you erroneously find that 1^x=e. But what if, when checking the solution with that same value of x, you instead chose to represent 1 as e^0 (a more natural choice as it lies within what is most commonly chosen as the principal branch of the complex logarithm function)? Then, by applying the very same steps as before, you would find 1^x=0. You certainly are not arguing that e=0, so that clearly shows a problem in your methodology.
Makes me think about that one limit definition of e = lim (1+1/n)^n as n->infty and how that's the indeterminate form 1^infty which equals e (unfortunate that 1^x in the video is always 1, even in the complex world if using the principal root)
8:39 "1 ^ 💩💩"
I've never heard of the word "complexify" before today! Thank you, SyberMath, for making it a part of my mathematical vocabulary!! :) :)
Well it can also be used outside of math and is actually less common in math. It means to make something complex so if someone is making things more difficult than they should be you could say that person is complexifying things
My pleasure!
When you “complexify” 1, you get 1, or maybe an alternate expression for 1, but it’s 1. If “complexify” means to make it complex, then it is a redundant operation. Remember, every real number is a complex number. R is a subset of C.
Because it's probably not a real word
Nice problem.
Thinking about the last bit. I think it's right to ignore complex roots because you have already split real and imaginary parts.
If a and b are real numbers, a^(bi)=e^(bi*ln(a))=cos(b*ln(a))+i*sin(b*ln(a)), which has magnitude 1. It follow that for all complex z=u+Iv, 1^z=1^u*1^(iv) which has magnitude 1 and can therefore cannot be e.
You are only considering the priniciple value of the complex logarithm, and also left out the parameter for periodicity in the complex expression of 1. The principle value does have a modulus of 1 but there are other values with a higher value. The video is correct.
@@WhosBean Okay, I will reduce the problem of solving 1^x=e in C to finding a solution when 1 is in the branch, B, of ln containing
[-i*pi,i*pi). In this branch, ln(1) = ln(|1|) +i*arg(1)=ln(1)=0, and if x=a+bi, then for this branch 1^x=1^(a+bi)=1^a*1^(bi)=1^(bi)=e^(bi*ln(1))=e^0=1 (=e^(2*pi*k*i) where on our branch, the integer k is zero). So for 1 on the branch B of ln, 1^x=1 for all x in C. For any other representation of 1 as e^(2*pi*k*i), 1^x=1, since e is 2*pi*i periodic. Therefore 1^x = e has no solutions in C.
In situations where you have to limit the domain or range of a function to where it is one-to-one in order to find an inverse function, you have to decide how to restrict the function’s domain (or range). The definition (branch) of the inverse function depends on that decision. In using an inverse function when there is more than one possibility, you have to specify, either explicitly or by convention, which inverse function you are using. Switching from one possible inverse to another is perilous and most likely leads to errors. You don’t have to leave the real numbers to see that problems can develop. For example, there are two possible branches for the inverse of f(x)=x^2: for a >=0 (1) sqrt(a^2)=a, or (2) msqrt(a^2)=-a. There is a useful identity for each of these possible inverses: sqrt(x^2)=|x| and msqrt(x^2)=-|x|. In the “where’s the error” problem, 1=sqrt(1^2)=sqrt((-1)^2)=~1, one error is that the second application of sqrt is actually an application of the other inverse.
In the real numbers, the branches of sqrt, arcsin, arccos, and arctan are established by convention.
In the video, he goes from e^a = e^b to a=b. This is incorrect as f(x)=e^x is not a one-to-one function. If a and b are in different branches of ln, then getting to a=b will require applying one branch of ln on one side and another branch on the other side. Mixing inverses, in other words. It is true that e^2=e^(2+2*pi*i), but we certainly cannot conclude that 2=2+2*pi*i.
@@tcmxiyw yes what you're saying is right but it doesn't disprove the video because he never swapped from one branch to the other, he used the parametric form which includes all branches at once.
In the principle branch the modulus in 1, that's not debated. If you set n=k=0 in the results then you'll get that too. But there are OTHER branches which do have solutions. All are equally valid. Which branch is principle is a matter of convention, it isn't inheritely "better" in any sense to the other branches.
@@WhosBean Concluding from e^a = e^b that a=b is an error without proving that a and b are on the same branch. All you can conclude from e^a =e^b is that a and b differ by a multiple of 2*pi*i. Moving numbers out of the exponent is taking the Ln of both sides. You have to use the same branch of Ln on each side. He makes this error at around 4:55. This can be cleaned up. Also, for the equation he develops at about 4:55, it would be helpful to ask, “Given values of n and k, for what values of x do the numbers on each side of the equation have arguments that differ by less than 2*pi?” For these values of x we can apply the same branch of Ln to each side.
I made an error in my first response to you: 1^z IS multivalued: if z=a+bi, where a and b are real then the multiple branches of 1^z are given by
1^z=1^(a+bi)=e^((a+bi)*ln(1))=e^((a+bi)*(ln(|1|)+2pi*k*i))=e^(-2*pi*k*b+2*pi*a*k*i).
I erred in looking at the periodic representations of 1 rather than multiple values of the Ln.
@@tcmxiyw I'm not saying that a and b are equal of they're from different branches. I'm saying the opposite. When you take logarithm of e^a you have to return it in parametric form because log is a periodic function. Same with b. Their PARAMETRIC FORMS are what's equal to each other. When you set BOTH parameters equal to the same constant, then that is comparing values along the same branch.
1^x=e
x•Log(1)=Log(e)
Log(1) = 2πki
Log(e)=1+2πni
x=(1/2πi+n)/k, where k,n are the arbitrary integer numbers, k≠0
The equation has the infinite numbers of roots
As mentioned by other comments, the proof is not valid.
When you back substitute the general solution in the equation the most simplified value of LHS is e^(1/n) This can only be equal to e only if n=1.
So the general solution reduces to k +i/2PI ?
Or since the principal value of n is not allows can we say no solutions?
e is defined as
lim ( h --> 0) ( 1 + h) ^ ( 1/h)
Hereby only feasible value of x is Infinity
infinity is not a value
it works in wolfram alpha if you input xth root of e = 1
it did not work for me
@@SyberMath Surd[e,x]=1 ?
yeah but a solid 1 that we have here raise to any power including 0 is always 1 by definition.
WolframAlpha not having the solution really messed me up, did the same method as you for the same answer, but WolframAlpha kept saying my complex number was equal to 1, i didn't think to check it by hand haha
Great question keep them coming
WolframAlpha had it right.
You may wish to look at the comments here by @thomaslangbein297 & @TedHopp (and their replies) which explain where @SyberMath when wrong with this one.
If a and b are real numbers, then 1^(a+bi)=(1^a)(1^(bi)) = e^(bi*ln(1)+2*i*pi*k) = e^(2*i*pi*k) = cos(2*pi*k)+i*sin(2*pi*k) = 1 for every integer k. In other words, for all z in C, 1^z = 1. That is why Wolfram Alpha reports that 1^z = e has no solutions.
if you type e^(1/x) = 1, which is the same equation in a different form, you will get the answer
@@tcmxiyw If you write 1^(-i/(2 pi)) in wolframalpha, then several values will be output, one of which is e
Took the ln on both sides. Very easy. Interestingly with the use of complex numbers, 1^x has a true meaning because it can be anything (except zero)
No, 1^x power cannot “be anything”. Suppose z=a+bi, where a and b are real numbers.
If a and b are real numbers, then 1^(a+bi)=(1^a)(1^(bi)) = e^(bi*ln(1)+2*i*pi*k) = e^(2*i*pi*k) = cos(2*pi*k)+i*sin(2*pi*k) = 1 for every integer k. In other words, for all z in C, 1^z = 1. That is why Wolfram Alpha reports that 1^z = e has no solutions. What I’ve written is somewhat redundant because I’m using ln as a real valued function. This is okay, because for complex z, the ln function is defined in terms of the ln function on the positive real line. By definition for z in C, ln(z) = ln(|z|) + i*arg(z), where you specify which argument you are using by specifying which branch of ln you are using.
What did you get when you took the ln on both sides?
@@QuasiRandomViewer . A lot of answers. Because the ln is multi-valued in complex numbers. Remember that ln(1) is not only zero. So you are not dividing by zero. It boils down to the same answer as in the video. Some raise to the power "e", others take the ln. It is also important to note two auxiliary variables, being "k" and "n". It should be noted that, when referring to the video, most folks don't forget the "k", but DO forget the "n". In other words, the notation "x=e^lnx" is technically incomplete. Sybermath has taken that perfectly into account
@@mathisnotforthefaintofheartWhat are you smoking?
@@j.r.8176Sober as always:) How about yourself?
great!!!
Cool - I got the complex part of the answer, but not the real part.
You got partial credit for that: i points! 😁
@@SyberMath good to know!
1/(ln(1)) with is undefined
Yep, simple ... no solution ... you can find any number of pseudo-solutions, they must all be rejected.
The trick is finding where the bogus step lies in this "proof" that e=1. (Ans: Misapplication of law of indices to domains in which they don't apply.
The problem is that it was presented as a legitimate proof.
4:18 is wrong u can’t put the x inside, u used Moivre which only work for x in integer
Mhm.
Multiply each site with its complex conjugate creates an equation with only real numbers:
x=a+ib
1^(a+ib) * 1^(a-bi) = e * e
1^2ae^2
a is a real number and 1 raised by any real number is 1.
Are 1^(a+ib) and 1^(a-bi) conjugates?
@@SyberMath
Yes.
In general
With a, b, c real
c^(a+bi) =c^a * e^(ln(c) * b *i)
ln(c) * b is the angle. To get the conjugate replace b by - b.
Checking the calendar is it really April Fools' Day today? Fun watching it anyway...
I wanted to think that it was a joke -- that it was going to end with "OK, clearly this answer is incorrect, so where was the faulty step in the logic?"
But alas, that was not to be.
Could someone explain why (a^b)^c = a^(bc) is not generally valid in the complex world?
With usual definitions : a = e, b = 2i pi, c = 1/2 then (a^b)^c = 1^(1/2) = 1 but a^(bc) = e^(i pi) = -1
@@yannld9524 Didn't understand that at all because you used a = 1 in one instance and a = e in another. Can you try again?
@@trojanleo123 a = e, b = 2i pi, c = 1/2
So a^b = e^(2i pi) = 1 and (a^b)^c = 1^(1/2) = 1.
On the other hand bc = i pi so a^(bc) = e^(i pi) = -1.
@@yannld9524 Thank you. I get it now.
I have just invented Perpetuum mobile
Log(1)e
log with the base 1 is always undefined
You are amazing!
Wow, thank you! 😍
@SyberMath Thanks for sharing. I don't mean to pester but I hope you can respond to.mynquestuon on the sum pf squares of integers videos..how tomsolve without using cubes formula..since wouldn't younagree I don't see anyone thinking of thst of they didn't know it beforehand?
lan abisi
ln(1) is only 0, not 2kπi . Function can have only 1 answer for 1 argument, it can be that ln(1)=0 and also 2kπi.
e IS NOT Euler's number.
Whose number is it?
'e' is indeed Euler's number. Perhaps you are confusing 'e' with Euler's constant which is a different mathematical constant denoted as gamma (γ).
Please talk half the speed
sorry
This is wrong, obviously 1^z = exp(z ln(1)) = exp(0) = 1 for any z. You should delete
Clearly, you don't know the power of the complex world.
@@Bruh-bk6yo sorry to tell you that complex world is not that powerful
@@yannld9524 it is. Do you know the definition of the complex logarithm?
Ln(z)=ln(|z|)+i(phi+2πk)
@@yannld9524Have you met my friend Quaternion? He is complex and powerful. If you ever seen anything moving graphics around on your phone/computer, it is complex power behind the scenes.
@@Bruh-bk6yo Yes i know what is "a" (not "the") complex logarithm , but that doesn't change my point. By the very definition of the exponentiation if a > 0 and z complex then a^z = exp(z ln(a)). With a = 1 that gives 1^z = 1.
I'm really annoyed by these videos where a guy who doesn't master the subject he talks about (here math) explains something that everyone knows is false, but in a convincing enough way that a "non expert" can believe him.
first
Complex numbers are fake invented math because (1) the definition of a complex number contradicts to the laws of formal logic, because this definition is the union of two contradictory concepts: the concept of a real number and the concept of a non-real (imaginary) number-an image. The concepts of a real number and a non-real (imaginary) number are in logical relation of contradiction: the essential feature of one concept completely negates the essential feature of another concept. These concepts have no common feature (i.e. these concepts have nothing in common with each other), therefore one cannot compare these concepts with each other. Consequently, the concepts of a real number and a non-real (imaginary) number cannot be united and contained in the definition of a complex number. The concept of a complex number is a gross formal-logical error; (2) the real part of a complex number is the result of a measurement. But the non-real (imaginary) part of a complex number is not the result of a measurement. The non-real (imaginary) part is a meaningless symbol, because the mathematical (quantitative) operation of multiplication of a real number by a meaningless symbol is a meaningless operation. This means that the theory of complex number is not a correct method of calculation. Consequently, mathematical (quantitative) operations on meaningless symbols are a gross formal-logical error; (3) a complex number cannot be represented (interpreted) in the Cartesian geometric coordinate system, because the Cartesian coordinate system is a system of two identical scales (rulers). The standard geometric representation (interpretation) of a complex number leads to the logical contradictions if the scales (rulers) are not identical. This means that the scale of non-real (imaginary) numbers cannot exist in the Cartesian geometric coordinate system.
A person who thinks using words, I see. In math, 'Real' and 'Imaginary' are simply labels - they don't have their conventional meaning. All numbers are imaginary, in your sense of the word. Pls post me a '3' by physical mail if you believe that not to be the case.
@@PaulMurrayCanberra does the concepts "countable" and "uncountable" help you understand?
You’re not making sense. What do you mean essential features negates another concept? Have you seen galios or group theory
You’re not making sense. What do you mean essential features negates another concept? Have you seen galios or group theory
You’re not making sense. What do you mean essential features negates another concept? Have you seen galios or group theory
Hi Cybermath, can you solve this goniometric inequality?
√3sinx≥cosx-1
Thank you so much if you will ❤❤