There are a lot of videos like this on the internet and they all make the same mistake: Since the complex exponential function is not injectiv, the law ln(pow(x,y)=y*ln(x) does not apply for the side branches. Therefore this "solution" ist wrong.
The video glosses simplistically over a large number of mathematical subtleties, but at the end it arrives at the correct solution. In fact, it does not arrive at one solution but at an infinite number of solutions depending on the two parameters m and k. The first difficulty is that a complex exponential with the base different from e is not only not injective, but it's not even a single-valued function. In fact x^y is defined as e^(y*Log(x)) where Log(x) denotes the complex logarithm and it is a multi-valued function as for each complex number x Log(x) = ln(|x|) + i (phase(x) + 2k*pi), that is an infinite number of values as k varies in Z. Applying to our case, the definition of exponential function with base different from e, we have 1^x = e^(x*Log(1)). Since Log(1) = ln(1) + i (0 + 2k*pi) gives an infinite number of values it follows that 1^x = e^(x*Log(1)) = e^(i 2k*pi*x). As a consequence, the equation 1^x = -2 can be rewritten as e^(i 2k*pi*x) = -2, which shows that it is not a single equation but an infinite family of equations parameterized by k. Note that the so called principal value, corresponding to k=0, gives an unfeasible equation as e^0 = 1, never equal to -2, so k must be a non-zero relative nuumber. Let's solve one equation of the family by fixing the value of k, say k=1: e^(i 2*pi*x) = -2. The non-injectivity of the complex exponential is accounted for by using the complex logarithm, which as we saw is multi-valued: Log( e^(i 2*pi*x) = Log(-2) which becomes i*2*pi*x = ln(2) + i (pi + 2m*pi) and hence x = [(2m+1)*pi - i ln(2)]/(2*pi). Hence every equation of the family has an infinite number of solutions parameterized by m varying in Z. Considering all the equations in the family, all the solutions are given by x = [(2m+1)*pi - i ln(2)]/(2*pi*k) for every pair or relative numbers (k,m), with k different from zero.
@@italnsd No, the only "correct" solution is that no solution exists to this problem. BY DEFINITION, if a is a positive real number, and x is complex, then a^x = e^( x ln(a) ). Thus 1^x = (by definition) e^( x ln(1) ) = e^( x (0) ) = e^0 = 1 for ALL complex x. Thus 1^x = -2 has no solutions for any x, real or complex.
k belong to Z means -2, -1, 0, 1, 2, ... but in your example you use N for k range, not Z. I know this will not change anything, just a small fix to understanding the integers and natural numbers
The problem is that calculators are not programmed to return all the infinite values of the complex logarithm function, but they return only the principal logarithm, with phase in (-pi, pi] which in this case corresponds to the non-existing real solution.
@@italnsd you are just confidently incorrect. There are several mistakes in his "proof", but the simplest one is the assumption that (b^x)^y=b^(x*y), which doesn't necessarily hold for complex numbers. There are several other mistakes that make this "proof" invalid as well, but this is the simplest one to explain
That is because calculators are programmed to return only the principal value of the complex logarithm, which in this case does not lead to a solution. To verify a solution x, you should enter its value in the expression exp(i 2k*pi*x)
This snippet of Matlab code can be used to verify the solutions: i=sqrt(-1); %Choose a value of k as non-zero positive or negative integer k=50 %Choose a value of m as positive or negative integer m=-32 %Compute the solution for chosen k and m z=(log(2)+j*pi*(2*m+1))/(j*2*k*pi) %insert in the equation and verify the result is -2 result=exp(i*2*k*pi*z)
in my anti-math brain you have to get -0.5 as x. -1/2 more literally:) if you multiple 1 on -2 it's same as divide on -1/2 so -0.5 is way to get -2... but i don't sure because i'm really math moron.... no... it's something not right... by some reason when i was kid,,, well not kid just not so adult, only alcohol give me a chance do something with math more complicated than 2x2 lol... accidently i have a bear in the fredger, so may be later:)
Guys, there is solution: Let's rewrite the equation: 1^z = -2, z = x+i*y 1^(x + i*y) = -2 In exponential form: e^((x + i*y)*ln(e)) = (e^x)*(cos(y) + i*sin(y)) = (e^x)*cos(y) + i*(e^x)*sin(y) = -2 + i*0 Two complex numbers are equal to each other, if their Real and Imaginary Parts are equal to each other. We will have the system: (e^x)*cos(y) = -2 (e^x)*sin(y) = 0 e^x = -2 / cos(y) y = pi*k, k ∊ Z We have two systems, if cos(y) = 1, y = 2*pi*k and if cos y = -1. y = pi + 2*pi*k e^x = -2 / 1 y = 2*pi*k (1) or e^x = -2 / (-1) y = pi + 2*pi*k (2) System (1) doesn't have solutions because e^x can't be negative number System (2) has a solution: x = ln(2) y = pi + 2*pi*k and z = ln(2) + i*(pi+ 2*pi*k), k ∊ Z Let's check the solution: 1^(x+i*y) = 1^(ln(2) + i*(pi + 2*pi*k)) = e^((ln(2) + i*(pi + 2*pi*k))*ln(e) = e^((ln(e))*ln(2)+i*(ln(e))*(pi + 2*pi*k)) = e^(ln(2))*(cos(pi + 2*pi*k) + i*sin(pi + 2*pi*k)) = -2 This solution is correct. Hello from Switzerland (Russian Education, BMSTU)
@methatis3013 , The main problem is that the logarithm in complex numbers is multivaluable function. Hence, for one of the cases this equation will not have solutions. But for another cases we can have solutions because of 1 = e^(i*2*pi×k), k ∈ Z. Insert e^(i*2*pi*k) instead of 1 in the equation and you can also solve it.
@@sub_stance-c5d logarithms are not an inverse function on complex numbers, that's the problem you're having here. Another issues in the proof is that (b^x)^y is not necessarily equal to b^(x*y) Another consequence of that fact is that ln(a*b) ≠ ln(a) + ln(b) "Multi-valued function" is not a thing. There is a thing called a relation, but logarithms aren't that. You may argue there are multiple branches of a logarithm, but each branch is just a function with a different codomain. Precisely because of that, some log rules we take for granted in reals don't actually hold in complex numbers
Dude, you gotta stop doing this. You keep posting this same damn problem over and over, and they're all wrong (links at bottom). I and others have explained this to you multiple times, and yet you keep doing it. *Stop it - you're misleading people!* You've posted what's essentially the same problem NINE TIMES in the past year, making your same mistake each time. You've also been corrected in the comments of most of these videos (including by me). You've also LIED by twice claiming (and a third time implying) that these were IMO Problems (International Math Olympiad), when in fact they are not. (I check the only time you came close to citing the IMO problem, your December 28, 2023 where you claimed it was an IMO 2019 problem. Nope - not according to the IMO database. But of course, it's independently obvious that these problems are not IMO Problems, because these problems have NO SOLUTION.) And your multiple reposting of essentially the same problem (and same mistake) for over a year now is weird. It makes me wonder, are you actually _trying_ to mislead people? You also once referred to these basic problems as an "Unsolved Problem" (which would be weird to put on an IMO test, dontcha think?). That suggests the kind of ego grandisoity often found among math cranks, so maybe that's what's going on with you. I don't know. What I DO know that you need to stop doing this, whatever your motivations, because you're misleading students! 1^x = -2 ... Aug 13, 2024 ... "Everything is possible" ... watch?v=27RVC2djbIQ 1^x = 3 ... Aug 10, 2024 ... "Math is not for beginners" ... watch?v=6diuf1GmRpM 1^x = 5 ... Jul 22, 2024 ... "Everything is possible in math" ... watch?v=Df8NPsiQCqI 1^x = 3 ... Mar 1, 2024 ... "Unsolved Problem in Mathematics | Solution" ... watch?v=jCkGtYHncBw 1^x = 3 ... Dec 28, 2023 ... "One of The Hardest International Maths Olympiad Problem" ... watch?v=Taf9Cxs7VH4 1^x = 2 ... Dec 20, 2023 ... "International Maths Olympiad Problem | Algebra" ... watch?v=G3shcIcsJFE 1^x = 2 ... Sep 20, 2023 ... "A Tricky Algebra Problem for Geniuses | Can You Solve?" ... watch?v=GIc6fzbEcck 1^x = 3 ... Sep 2, 2023 ... "A Tricky Algebra Problem for Geniuses | Olympiad Maths" ... watch?v=a_4hLwcHrxo 1^x = 2 ... Aug 4, 2023 ... "A Tricky Algebra Problem for Geniuses | Can You Solve This?" ... watch?v=ioDCrRzawQs
