You can take the natural logarithm (ln) of both sides: ln(e^(e^x)) = ln(1) Using the property that ln(e^a) = a: e^x = 0 Now, you have e^x = 0, which has no real solutions because you cannot raise a positive number (e) to any power and get 0.
Notice that 1^x = 2 and 1 = 2 ^(1/x) are, actually, two diffferent equations with different domain of x. You solved the second equation and not the first one. Edit: that is EXACTLY why wolfram can solve the second one.
To simplify: {1^x-2=y, y=0} ≠ {1-2^(1/x)=y, y=0} Similar example: x^2-1=0 and x-sqrt(1)=0 will give different graph. [Search on google] x^2-1=0 x^2=1 x=sqrt(1) x-sqrt(1)=0 ■
1) x real -> No solution. OK 2) Let's suppose x complex -> x = a + ib (a, b Real numbers !) -> 1^x = 1^(a + 1^ib) = 1^a . 1^ib = 1 . 1^ib = 1^ib Applying ln on complex numbers: ln(z) = ln|z| + i arg(z), we get : ln (1^ib) = ln|1| + ib = 0 + ib = ib Hence ln(1^x) = ib = ln(2). Pure imaginary = pure real -> Impossible. x is not a complex number Solution: NO SOLUTION. What about that ?
@@vaarmendel1657 close, but the issue is you've used the wrong definition of arg(z). arg(z) requires base e, not base 1. So ln(1^ib)=ln(e^(2πni×ib)) =ln(e^2πkb)=2πkb. Then 2πkb=ln(2) has solutions for b.
@@zenedhyr7612bro forgot the plus or minus. Ofc x^2-1 is different from x-1. Its not just about degree. The solutions arent even the same like equivalent equations. Thats just blatant nonsense
@KiwiTV It doesn't defeat itself. It just happens to be inconvenient for humans. Mathematics has never been intuitive, though. Human brains didn't evolve to be able to easily deal with complex numbers. They evolved so that we could do 3rd grade elementary school arithmetic. Everything else is just us making ourselves more miserable against our own evolution for the sake of additional benefits.
@KiwiTV Complex Numbers allow us to solve real word phenomenons, like apparent/reactive power in electrical systems, pretty elegant. It doesn't defeat itself, it only offers multiple perspectives of a problem
If you do it the way you did it, I get: x = ln(2)/(ln(0)) and ln0 is ln(0) + i*n Maybe if there is a different way it's possible but not with how you did it in the video.
This is very similar to the equation sqrt(x)=-1. If you put that into Wolfram, it will tell you that it has no solutions. You can try to argue that well, actually, one of the square roots of 1 is -1, but the thing is that's not what sqrt(_) actually is. The same is true under complex exponentiation: the principal branch is used by definition and as such, 1^x=1 no matter which x you plug in. As others have pointed out, this does not contradict that 2^(1/x)=1 does have solutions in C (even when we are taking the principal brach). So no, Wolfram's right here.
I think wolframalpha is right here. 1^x=2 has no solution but 1=2^(1/x) does. Grinding this down to fundamentals you see that (a^b)^c is not equal to a^(bc) for complex numbers, exactly because the change of branch of log you expertly portrayed in the video. Another instance that messes with this is my all time favourite: a = e^(log a) = e^((2\pi i / 2\pi i ) log a ) = (e ^ (2\pi i) ) ^ (log a / 2\pi i) = 1 ^ (whatever) = 1.
Actually Wolfram-Alpha is correct. Too understand why we will need some function-theory/complex analysis (for example: Complex Analysis, Elias M. Stein S. 97-100). At first we will need a definition of z^w with w,z in C. For any z in C\(-∞,0] we can define a function z^: C --> C by z^w:=exp(log(z)•w) where log is the principal branch of the logarithm (that means that log(1)=0). Of course you can choose another branch but in this case the definition does not match with the exponetialfunction with a real basis. Using this definition we get: 1^x =exp(log (1)•x)=exp(0•x)=1 which states that the equation 1^x = 2 got no solution. Now we take a look at the Question: "Can we finde a x in C such that 2^(1/x)=1?" Using the definition we get 2^(1/x)=exp(log (2)•(1/x)) which is equal to 1 whenever log (2)/x=2πi•k, for any k in Z. This gives the solutions you are getting too. After clearing this we should talk about the "contradiction" at 06:23. What you are writing there is correct but its not a contradiction to the above: 1^x=2 => 1=2^(1/x) means "every solution of the first expression is also a solution of the second Expression" (which is correct cause the left expression got no solutions). The other direction 1^x=2 x²=1 is correct but x²=1 => x=√1 is wrong (the solutions -1 "gets lost"). And this gets even "worst" when complex numbers are involved...
The main error is conflation of the notion of a function with the notion of a relation, by using what are sometimes referred to in complex analysis books as "multi-valued" functions. There is no such thing. A function, by definition, is decidedly NOT "multi-valued". This leads to an aberration involving a failure to understand how equal signs work in a coherent presentation of mathematics. There are two fundamental ways they can be used coherently, and all other coherent uses are definable from these. (a) The main fundamental semantic use of an equal sign is to write a mathematical statement that is interpreted as true in the context of some given model of some theory because it pertains to the facts in that context, AND that statement that is interpreted as true in that context BECAUSE the one and only thing that is described to the left of that equal sign is exactly the same thing in that model as the one and only thing described to the right of that one and the same equal sign. This is written to convey clearly to the reader some accurate information about the context provided by the given model. Mistakes can be forgiven, but persistent incoherent abuses of notation should be excised from mathematical vocabularies. (b) The main fundamental syntactic use of an equal sign is to write a mathematical statement that is to be tested for truth in the context of some given model or class of models of some theory because it pertains to the facts in that context, AND that statement that is an hypothetically testable assertion in that specific context BECAUSE the descriptions to the left and right of that particular equal sign are interpretable in the context of the model or class of models to be considered, and the question of whether some model or class of models satisfies that particular statement is a coherent question in that context. Such a syntactic use of an equal sign is written to convey clearly to the reader some problem (i.e. it is a mathematical query) aimed to elicit some accurate information (a clearly formulated and completely explained solution of the problem and answer to the question) about the context provided by the given model or class of models. Mistakes can be forgiven, but persistent incoherent abuses of notation should be excised from mathematical vocabularies. Sensationalism should be rooted out and excised, just like incoherent abuses of notation should be. Otherwise, we our logic system will prove absurdities like 0=1 in the real number system. A logical system that allows such nonsense is not useful, because from such things, the notion of "provable equation" and "equation" are indistinguishable. By a variant of Occam's Razor, we should not invent a terminology that pretends to distinguish things that are indistinguishable.
@@nicktravisano7152 There's a distinction between log applied as a function and the relation called the inverse image of a point in a space via a function. Both are relations between elements of sets but a function has the property that if x=y then fx=fy . The relation named "inverse image" has not in general such property. What blackpen writes on the board is formally incorrect. You cannot use the equal sign if applying something which is not a function on both sides of the equation, such writing down "log" but in fact meaning inverse image of the exponential function in the complex plane. The apparently revolutionary results you find in the complex plane are not so much revolutionary but abuses of the inverse relation treated as a function when it should not be. Still fun though.
You know what. This came in my recommended, and man let me tell you, my algebra 2 teacher must be doing a great job because I don't know how I willingly clicked on this and was genuinely interested lol
Okay if you solve for x in 1=2^(1/x) using wolfram alpha, you indeed get x=-i*ln(2)/(2pi*n), but as I have learned in math competitions, you should -always- usually plug it in to the original equation. And plugging x=-i*ln(2)/(2pi) as 1^x in wolfram alpha, you get 1. So x=-i*ln(2)/(2pi*n) are all extraneous solutions, which is why they are not solutions to the original equation 2^x=1.
@extraterrestrial46 the original equation has a domain of all real x, and a certain range. The equation 1 = 2^(1/x) has a more restricted domain, and a different range. This changes the solutions, producing new solutions that do not work for eqn 1.
Regarding this video, the two answers from wolfram alpha are consistent. b^z when b is real and positive and z is complex, is being taken in both equations as Exp[z Log[b]], where Log is a branch in which Log[b] is real. Hence, you get different answers in Wolfram depending on whether you ask Solve[1^x == 2, x] or Solve[2 \[Pi] I x m + 2 \[Pi] I n == Log[2], x]. Given the fact that you have the possibility to reach big audiences, you could make a bigger effort to present these topics in a more precise fashion, and use this topic to explore its nuances, and you are intentionally choosing not to do this.
@@secavara This is dishonest criticism, as the purpose of these videos are not to present the topic rigorously as understood by mathematicians. The purpose of these video are to showcase what happens when one is not careful, and to present topics heuristically, which is necessary before a student can begin to approach a complex topic with rigor. I have my disagreement with BPRP regarding how some subjects should be presented, but implicitly accusing him of being dishonest via your tone and wording when the criticism is not even applicable is itself dishonest and hypocritical.
It is a rule that 1 to any power is 1. I assume this is true for all numbers but whatever tricks he is doing, you should always get a div by zero error.
another way to solve this is 1=i^4n, n being an integer different than 0, you then have ln(i^4n)=4nln(i)=4nln(e^pi/2*i)=4n*pi/2*i, so ln(2)/ln(1)=ln(2)/4n*pi/2*i=ln(2)/2n*pi*i=-i*ln(2)/2n*pi, same result with a slightly different method
All good until "ln(2)/ln(1)". ln(1) = 0, so you are juat dividing by 0. Which is the same mistake in the video (just that in the video it's packaged differently). Wolfram alpha is correct, the equation has no solutions (including complex ones). The only way you get solutions is by doing something illegal (like dividing by 0).
@@gabrielfoos9393 ln(1) is not 0 for other branches of the logarithm in C, so it can be divided by for those, it's only the one that aligns with the real-valued logarithm (the principal logarithm) that has ln(1) = 0, it's usually the most convenient to use but *not the only one*
In general, we can find the solution of x 1 = P(n) where P(n) is a polynomial in R or C the solution is: -i(2 *Pi* k + Log ||(P(n)||)/2*Pi with ||(P(n)|| the modulus of P(n) and k a relative integer. it is the magic of complex numbers that allows this in particular the possibility of writing: 1= exp(2*Pi*k)
Wolfram Alpha does not assume x in R. Wolfram Alpha assumes the principal branch of exponentiation. In the principal branch, 1^x = 1 for all complex x. In order to get any other value, you have to assume the non-principal branch of Log(1).
base 1 is undefined in any Field, what he's doing is messing with the fact that log function on C is not a function by definition (one value of z leads to infinitely many values of log(z)), we have to use the principal value of log, the Log function, instead, which locks the n value to 0, and is bijective.
@@PicaroPariah Not really. The expression in Spanish is: "sacar agua de las piedras". I preferred to use "juice" instead of "water". A failed poet, as you can see.
Question: Do you need the negative sign in the numerator? If n runs through all positive and negative integers, it's the same "solution set" with or without the negative sign ... ... isn't it ?? PS - Your gentle emotional delivery here (moments of disappointment, exasperation, near-defeat) is a refreshing new contribution to the art of math videos!
It's a similar issue as when you rewrite the following diffEQ solution: y = e^(-x + C) as: y = C*e^(-x) The C came from a constant of integration that was completely arbitrary, and when we rewrite the final solution, the C is still an arbitrary constant. However, it is NOT the same arbitrary constant as you originally generated. Some professors insist you either assign subscripts or a tilde on top of one of them to tell them apart, or use a different letter entirely. If it were my choice, I'd accept a reuse of C, as long as write "Reassign C", or something else that means the same thing, to indicate that it isn't the same letter. Likewise, with this solution of: -i*ln(2)/(2*pi*n) The value of n is an arbitrary integer, so you could just as easily assign a different letter like m, and write it as: i*ln(2)/(2*pi*m) Heehee....pi m. Dr Peyam, this one is for you.
I would guess 2 possibilities: 1. No. 1^x cannot equal 2. 2. If 1^x can equal 2, then 1^x can equal anything, because there is nothing special about 2.
The 1st one is correct :) 1^x will ALWAYS equal 1, even if x is a complex number. The video is just tricking you, it’s like those old 1=2 videos when the guy slily hides the fact that he is breaking the math axioms somewhere
@@speedyx3493 Both possibilities are wrong, the 3rd correct possibility that 1^z can have a countably infinite amount of solutions (not indeterminant like 0/0), but only when z is not rational.
@@glitchy9613 I feel you're misunderstanding his point. He's saying 1^x can equal anything given a sufficient value of x. And by branches, it's correct. 1^x=y, take the right branch and you can get x=ln(y)/2πni. Maybe take different branches of the ln(y) and you'll get even more solutions. The exception will be 0. Branches don't matter, taking a logarithm of 0 will cause some sort of issues. Maybe on a Riemann sphere you can argue it, but even then not necessarily.
@@glitchy9613 I reread my thing, and I can only assume I wasn't referring to Speedy Gonzales here. I think I was referring to BPRP, just from where I stand. Sure, for each x it is true that 1^x can equal 1, but for complex x, 1^x may be something else too.
@@ejb7969 Oh, wow! Thanks for noticing the reference. NOBODY has EVER noticed it before you! I extracted these notes from the 3rd movement of the Concerto for Solo Piano (Op.39, No.10). In addition to Quasi-Faust, there are 2 occurrences of a triple-sharp in this movement, after it modulates to the parallel major, F-sharp major. Again, thanks for noticing!! 👍😀
@@ejb7969 Yeah, for the longest time, I also only knew about just the triple-sharp of Quasi-Faust. That’s the example that’s widely used in mentioning Alkan’s use of triple-sharps. I just happened to hear about the ones in the Concerto. They’re very easy to overlook in the torrent of notes in the score!
I think this is probably my favorite problem you've done. It just shows how completely messed up things can get when you get to the complex plane, to the point where 1 raised to a power does not equal 1. Complex analysis is just bizarre. But in the end, this problem is so dang easy. You can get to things like -1^x = 10 x = i ln 10/pi How crazy is that?
All depends of the branch of logarithm chosen because k^x=exp(x*"log"(k)) where you need specify the 2*pi magnitude interval of the imaginary part of function "log" how is defined; if it contains 0 there are no solutions. For example, the case of principal Log doesn't work because Log(1)=0; thus, for powers of principal branch 1^x is always 1
Thank you. This is one of the cases my Math Prof. warned me about. Does the imaginary unit really enable such relations or are the exponential / logarithmic laws more limited in the complex world than one might first time think?
This answers a question I've had for a long time: does some math treat e^2 and e^(2+tau*i) differently? I learned something today. Thank you so much :D
I came here to see how many people undestand this, because boi i really weak in math. It took me hours to understand this, and open bunch of books about ln, log, and how they works. By the way i'm in 8th grade, and your vids helped me to understand lots more of how to calculate. So i thanked you for that 💙
The funniest thing is that the video is just wrong :). So if you really claim you understood it, then you didn't understand it enough :P. Hint: think about what it means to take log base 1.
A complex number raised to a complex power has multiple values. If we treat 1^x as such a multi-value expression, and the question is to find x so that 2 is one of values of 1^x, then x=-i*ln2/(2*pi) as stated in the video is a solution. Actually there are multiple solutions: x=-i*ln2/(2*pi*k) with k being any non-0 integer.
There should be more solutions. You can also split 2=2*exp(2*pi*m*i) and use the additive rule of the logarithm. The general solution is x=-iln(2)/2/pi/n+m/n,n is not 0,m,n are integers.
In general, if 1^x = z, with x,z ∈ *C* and z≠0, then x = k + [(ln z)/(2πm)] i for some k,m ∈ *N* with m≠0. In this case, the values of 1^x are the principal values of z^(-n/m) for all integers n. So the principal value of 1^x is always z^0 = 1, but the desired value is always in there. Specifically, 1^x = z on the branch n = -m.
There are many things wrong with this video, but they serve as good indicators of why we need the concept of branches in complex analysis. Let us first see that some things are certainly not right: Suppose the reasoning by BPRP works, and that one can indeed write ln(1) = ln(e^(2pi*i*n)) = 2*pi*i*n, for any integer n. Then we run into the problem that ln(1) = ln(1^2) = 2ln(1). This immediately implies that ln(1) = 0, because that is the only solution to x = 2x. So either BPRP is wrong, or ln(1^2) = 2ln(1) is wrong. However, this same property of logarithms is used by BPRP himself when he writes ln(e^(2pi*n*i)) = (2pi*n*i)*ln(e), thus in any case BPRP's argument is not self-consistent. This property of logarithms should be familiar and we would certainly want this to be true. Let us now get to the heart of the problem. BPRP did not consider that in the case of inverting the complex exponential, you may not use all properties that we are used to when dealing with logarithms of real numbers. To 'invert' the complex exponential, you need to choose a specific branch, precisely to deal with problems like the one that we see in the video, namely that 1= e^2pi*i = e^4pi* i = e^6pi*i = e^(2pi*n*i) for any integer n. What does choosing a branch mean? Well from these equalities we see that there is no single inverse value to the complex expontential for the value 1: We need to choose one of the values 2*pi*n*i to get a step closer to defining something like an 'inverse' function to the complex exponential. Making such choices in order to always know which value to pick is (crudely speaking) what mathematicians call choosing a branch. The natural logarithm for complex numbers is an example where such a choice of branch has been made: The natural logarithm is defined for complex numbers by choosing the principal value branch, which restricts to the interval (-π, π]. This means that even though 1 = e^2pi*n*i for any integer n, when we use ln(e^(iθ)), we choose the value inside (-π, π] (even if θ is outside this interval!). In the case of ln(1) = ln(e^(2pi*n*i)) for any integer n, the natural logarithm then simply gives 0. This last point stresses that ln(e^(2pi*n*i)) = 0, thus the argument in the video does not work. Then one might be tempted to defend BPRP's argument by saying that he implicitly chose a different branch for the natural logarithm, precisely by asserting that ln(1) = 2pi*n*i for some integer n other than 0. However, even then one encounters the problem we discussed above: ln(1) = ln(1^2) = 2ln(1). This does not mean that one cannot take different branches for logarithms. Instead it means that when we do take a different branch, we cannot expect precisely the same rules to hold for our logarithm. In particular the rule for logarithms that log_a(b^c) = c*log_a(b) does not hold anymore if we choose a branch corresponding to ln(1) = 2pi*n*i for n ≠ 0. Of course this is a cherished property of logarithms, and motivates even more why mathematicians prefer to choose the principal value branch: It is the only branch in which this property holds. Thus in conclusion, BPRP's algebraic gymnastics to solve the equation 1^x = 2 is not correct, and upon further inspection Wolfram Alpha turns out to be exactly right: There is no solution to this equation. However there are some solutions to the equation 1 = 2^(1/x), which is NOT an equivalent equation. But my comment is long enough as is. If anyone is interested, I can elaborate more on this later.
Pretty interesting comment. I would like to see why 1 = 2^(1/x) its not an equivalent equation to 1^x=2. I'm guessing that it is because we also have to choose a branch of the function f(z)=z^(1/x) to apply on both sides?
Thanks for bringing this up! In an engineering course I'm taking, the instructors were very up front about telling us to give our answers in terms of this interval. Now I'm starting to see why. Time to do some more research/learning.
It's much easier than that... the video is wrong from the first minute where they take log base 1 of both sides. You can't do that, that's equivalent to dividing by 0.
1^x = 2, So x = Log（base1）2 = Log(2)/Log(1) The answer will be a complex number a + b i, where a is any rational number, and b = -Log(2)/(2 n Pi), n be an integer.
You should make a video on x^x=x^2 Obviously 1 and 2 are the only real solutions but I did this with some code I wrote, it would be interesting to see how the actual math works. Also what about complex solutions
@@blackpenredpen Hey thanks for the feedback! I appreciate it so much. I edited the title and description I'm trying the whole math youtuber thing, I like it so far (inspired by you and others)
1*1^(x - 1) = 2 | divide by 1 1^(x - 1) = 2 | but 2 also equal to 1^x x - 1 = x -1 = 0 | and we can do it for any other integer not just -1, they are all equal to 0 1^x = 2 is not the same as 1 = 2 ^ (1/x). Just like 1 = -1 is not the same as 1² = (-1)². Raisng equation to the power might change it's answer.
Table at 6:03 it should be Ln, not Log I guess? Btw, like this you can get any number by replacing ln(2) with ln(k) where is the number (or it's power) that you want.
Yes, if x = (-ln(2)*i)/(2nπ), where n is any non-zero integer. But only for non-principal values of 1. The principle value of 1 is e^(0*i), or e^0. And raising e^0 to any power will yield e^0, which is 1. So entering 1^x into a calculator will yield a result of 1, even if you use this x value. But if you use this value of x when entering 2^(1/x) into a (sufficiently sophisticated) calculator, you will, in fact, get 1. EDIT: Cool, I didn't miss any details. I actually solved this slightly differently from how you did. I actually jumped straight into polar form, restating the equation as (e^((0+2nπ)*i))^x = 2 right off the bat.
I came here because I was curious of how you would have solved the equation, to discover that behind it there was a story of a change of perspective. Great video, I had fun, even though I can't do all that math :)
log functions are not defined for base 1 . in fact its in the definition that base of log is positive and can't be equal to 1 . so how can you take log with base 1 on both sides 🤔
Don't you acknowledge it in the first place? Suppose look up how "i" was invented? looking up for complex value, then one may have to go through complex solution just as doing things with √(-1)
For increased effect you should have asked if 1^x can equal e. Then following the same reasoning you would have concluded that 1^(-i/2pi) = e, which is quite beautiful, and correlates directly with euler's formula: [1^(-i/2pi) = e]^(2pi*i) -> 1 = e^(2pi*i) I know the math is ultimately not correct, but it's just for good fun.
It's not "ultimately" correct, but it's not particularly wrong either. The principal value of 1^x is 1. Always. But if x is not an integer, you will be getting either finitely many or infinitely many more values. 1^(ln(2)/2πni) has principal value 1, but can take on other values such as 2. 1^(1/2) has principal value 1, but can also take on -1.
The definition of a^b for a, b complex numbers is a^b=e^(b*ln(a)). So 1^i=e^(i*ln(1))=1. Now, if there is a complex number z=x+iy, with x, y real numbers, such that 1^z=2, we have 2=1^(x+iy)=(1^x)*((1^i)^y)=1*1^y, which means that 1^y=2, with y a real number. This is impossible.
In what environment are you encountering toxicity? Personally, I just didn't really find maths fun until I got to higher maths like Mathematical Reasoning, Graph Theory and Group Theory. It really pays off when you finally see your work in one subject area began helping you understand other areas (ie, some understanding of computer datastructure helps a lot with Graph Theory, and Vector/Matrix stuff is all in Group Theory). That just never happened for me when I did high-school stuff. I also don't usually talk with other people about maths and instead rather figure stuff out from scratch myself with help from textbooks or forums, so that might be why I never felt any peer-pressure or competition
@@iCarus_A like I remember doing a ton of stupid and easy mistakes on a maths paper, people thought oh I must be a genius or something and then I overheard people saying that they thought I was smart and it really hurt to be honest everyone is just so competitive idk why and i feel my passion for maths getting dimmer-its not the maths thats hard its just overcoming the mental barrier when doing maths, for some reason when I do maths I just assume I can't do it since thats what some of my teachers said
@@zeroitedono2547 Something I learned is that the more passion you hold for a specific hobby/profession/pursuit, the more likely it is for you to get stuck in a mentality of perfectionism. At some point, you just become so paralyzed with the fear of failure that even taking a single step feels difficult. The first thing to know when recovering from something like this is to realize that you're not perfect and the only way to get better is to keep doing it. I'm usually fairly good at maths and tend to lead math-related projects in small groups, but I also make tons of small mistakes regularly. A mistake fixed is not a mistake at all, and it's far more important to get the overarching goal right -- try to get a friend or someone else to talk your stuff over without getting too involved and see if you can explain your thought progress clearly to someone who's not even deeply familiar with your field.
I would like to note that this problem has some additional weird properties. For example, if you attempt to solve this via a limit, you find that if you try to solve for 2^x=1, you end up with this being true as x approaches 1/infinity, essentially as the infinitieth root. Or just use the identity where raising something to the power of 0 makes it into 1 However on the other hand, you can prove that for all real exponents of x, 1^x is 1. Imaginary numbers are useful, yes, but they also kinda messy and can pretty commonly give non practical solutions(like in the real world, there is no real operation to do i rotations or whatever) Long story short, solutions like this never really see use except when i is representative of some specific axis. Like the phase of an AC circuit. And even then, the most you really see of i is e^i(angle) or a basic complex number. Or at least, in the classes ive been in so far. College junior rn
@@The_Commandblock I didn't use the 0 power property since the video didn't really go over them and I wanted to use a way other people on the comments didn't use. I will add that in though
Can e^e^x=1?
Sol here th-cam.com/video/ckc9F0VjZ3k/w-d-xo.html
Hello😊
You can take the natural logarithm (ln) of both sides:
ln(e^(e^x)) = ln(1)
Using the property that ln(e^a) = a:
e^x = 0
Now, you have e^x = 0, which has no real solutions because you cannot raise a positive number (e) to any power and get 0.
e=0 x=1 easy
0^0^1=1
@@dragondg6412bro studied from kfc toilet 💀
@@dragondg6412bro graduated from the 15 year old marker board in social studies
Imaginary numbers are the math equivalent of going into the shadow dimension to get through obstacles
Fourier transforms are more like that
This
I like that analogy
@@albertoaltozano8354 Fourier transforms are within the shadow dimension of imaginary numbers, to be fair
It is straight up that)
When blue pen gets in involved, you know it's serious...
Good one xd
Really serious is the purple pen.
@@WerewolfLord you don't want to see the purple pen
@@GianniCampanale purple is for thanos
@@godson200 this was blackpenredpen infinity war. Next is blackpenredpen endgame...
“360, but we are adults so we use 2pi” I felt that
@Memes shorts 1 like for you from me
Nobody like James Fleming’s comment. It needs to stay this way
What about tau?
360 likes
420 likes
Notice that 1^x = 2 and 1 = 2 ^(1/x) are, actually, two diffferent equations with different domain of x. You solved the second equation and not the first one. Edit: that is EXACTLY why wolfram can solve the second one.
To simplify:
{1^x-2=y, y=0} ≠ {1-2^(1/x)=y, y=0}
Similar example:
x^2-1=0 and x-sqrt(1)=0 will give different graph. [Search on google]
x^2-1=0
x^2=1
x=sqrt(1)
x-sqrt(1)=0 ■
1) x real -> No solution. OK
2) Let's suppose x complex -> x = a + ib (a, b Real numbers !)
-> 1^x = 1^(a + 1^ib) = 1^a . 1^ib
= 1 . 1^ib = 1^ib
Applying ln on complex numbers: ln(z) = ln|z| + i arg(z), we get :
ln (1^ib) = ln|1| + ib = 0 + ib = ib
Hence ln(1^x) = ib = ln(2). Pure imaginary = pure real -> Impossible. x is not a complex number
Solution: NO SOLUTION.
What about that ?
@@vaarmendel1657 close, but the issue is you've used the wrong definition of arg(z). arg(z) requires base e, not base 1. So ln(1^ib)=ln(e^(2πni×ib))
=ln(e^2πkb)=2πkb. Then 2πkb=ln(2) has solutions for b.
@@zenedhyr7612bro forgot the plus or minus. Ofc x^2-1 is different from x-1. Its not just about degree. The solutions arent even the same like equivalent equations. Thats just blatant nonsense
"Can 1^x = 2?"
- No.
*video ends*
x=log base 1 of 2
@@abhaysharmafitness log base 1 of 2 = indefinite, so no
@@abhaysharmafitness no such thing as log base 1
@@Pirater666l *indefinite in real numbers
@@korayacar1444 yes there is, in the complex world.
“We need two things. The first thing is the distance. The next thing is to erase the equal sign better. The third thing is the angle.”
Lol
@@blackpenredpen I have question for you
4^x+6^x=9^x
FIND THE VALUE OF X
@@lordmomstealer haha dude this is video by mind ur decisions
@@lordmomstealer very basic u = 2^x and v = 3^x substitution
@@lordmomstealer this is from mind your decision
He would do it in minutes
My man starting to look like an ancient philosopher who lives on a mountain, I dig it
Like a sage or a monk or something
He keeps on a cave meditating about the marvels of math...
Nah
He's just a muslim
With a sacred pokeball
Thus spoke BlackPenRedPen
Mathematicians whenever they wanna look complicated : Let's talk about complex numbers
Physicists whenever they wanna look complicated: let's talk about Quantum physics.
Chemists whenever they wanna look complicated : let's talk about chemistry!
To be fair, modern chemistry based on quantum physics
And QM involve hilbert space so everybody talks about complex number 😂
Chem is not hard
Chemistry is applied quantum mechanics, quantum mechanics is applied mathematics. And as always, math is king.
Complex numbers? Why not quaternions, octonions, sedenions and the Clifford algebras?
I like how this whole time he was holding a poke ball and half of us were too busy having our brains crushed to realise
It's actually his mic (in case you didn't notice)
WTF 😮
Isn't that usual though? He has it in every video
I clicked the video because of the Pokeball lmao
I actually noticed immediately and soon figured that it must be either his mic (somehow) or a random thing he holds as a gag for all his videos.
That felt like a mathematical crime.
Nah. It's perfectly legal, as long as you DON'T consider the wrong logarithm branch.
More like exploiting loopholes
@@bagochips1208 it's not a loophole though. The problem lies in the argument function.
@KiwiTV It doesn't defeat itself. It just happens to be inconvenient for humans. Mathematics has never been intuitive, though. Human brains didn't evolve to be able to easily deal with complex numbers. They evolved so that we could do 3rd grade elementary school arithmetic. Everything else is just us making ourselves more miserable against our own evolution for the sake of additional benefits.
@KiwiTV Complex Numbers allow us to solve real word phenomenons, like apparent/reactive power in electrical systems, pretty elegant. It doesn't defeat itself, it only offers multiple perspectives of a problem
Can 0^x=2?
Love you from India 🥰🥰
Nah
No. Ln 0 is undefined, and r=0 on the complex plane, so you are always stuck dividing by zero.
If you do it the way you did it, I get:
x = ln(2)/(ln(0))
and ln0 is ln(0) + i*n
Maybe if there is a different way it's possible but not with how you did it in the video.
I came back to your channel. It is so funny the topics related.
"I don't like to be on the bottom, I like to be on the top."
Now that's what everyone wants to be... Underrated comment...
"i don't like to be on the bottom, i like to be on the top."*
*mm nice*
That phrase leads to two different endings
We got the same surname😂
This is very similar to the equation sqrt(x)=-1. If you put that into Wolfram, it will tell you that it has no solutions. You can try to argue that well, actually, one of the square roots of 1 is -1, but the thing is that's not what sqrt(_) actually is. The same is true under complex exponentiation: the principal branch is used by definition and as such, 1^x=1 no matter which x you plug in.
As others have pointed out, this does not contradict that 2^(1/x)=1 does have solutions in C (even when we are taking the principal brach). So no, Wolfram's right here.
I think wolframalpha is right here. 1^x=2 has no solution but 1=2^(1/x) does. Grinding this down to fundamentals you see that (a^b)^c is not equal to a^(bc) for complex numbers, exactly because the change of branch of log you expertly portrayed in the video.
Another instance that messes with this is my all time favourite:
a = e^(log a) = e^((2\pi i / 2\pi i ) log a ) = (e ^ (2\pi i) ) ^ (log a / 2\pi i) = 1 ^ (whatever) = 1.
You are evolving into one of these chinese big beard philosophers lol
@Sierox same xD
@@zaheersuhabuth2677 Pakistani
@Sierox me too
@@ADeeSHUPA 🇵🇰
th-cam.com/video/vUcNDYlBtoc/w-d-xo.html
Actually Wolfram-Alpha is correct. Too understand why we will need some function-theory/complex analysis (for example: Complex Analysis, Elias M. Stein S. 97-100). At first we will need a definition of z^w with w,z in C. For any z in C\(-∞,0] we can define a function
z^: C --> C by z^w:=exp(log(z)•w) where log is the principal branch of the logarithm (that means that log(1)=0). Of course you can choose another branch but in this case the definition does not match with the exponetialfunction with a real basis.
Using this definition we get:
1^x =exp(log (1)•x)=exp(0•x)=1 which states that the equation
1^x = 2 got no solution.
Now we take a look at the Question: "Can we finde a x in C such that 2^(1/x)=1?"
Using the definition we get
2^(1/x)=exp(log (2)•(1/x))
which is equal to 1 whenever log (2)/x=2πi•k, for any k in Z. This gives the solutions you are getting too.
After clearing this we should talk about the "contradiction" at 06:23. What you are writing there is correct but its not a contradiction to the above:
1^x=2 => 1=2^(1/x) means "every solution of the first expression is also a solution of the second Expression" (which is correct cause the left expression got no solutions).
The other direction 1^x=2 x²=1 is correct but
x²=1 => x=√1 is wrong (the solutions -1 "gets lost").
And this gets even "worst" when complex numbers are involved...
The main error is conflation of the notion of a function with the notion of a relation, by using what are sometimes referred to in complex analysis books as "multi-valued" functions. There is no such thing. A function, by definition, is decidedly NOT "multi-valued". This leads to an aberration involving a failure to understand how equal signs work in a coherent presentation of mathematics. There are two fundamental ways they can be used coherently, and all other coherent uses are definable from these.
(a) The main fundamental semantic use of an equal sign is to write a mathematical statement that is interpreted as true in the context of some given model of some theory because it pertains to the facts in that context, AND that statement that is interpreted as true in that context BECAUSE the one and only thing that is described to the left of that equal sign is exactly the same thing in that model as the one and only thing described to the right of that one and the same equal sign. This is written to convey clearly to the reader some accurate information about the context provided by the given model. Mistakes can be forgiven, but persistent incoherent abuses of notation should be excised from mathematical vocabularies.
(b) The main fundamental syntactic use of an equal sign is to write a mathematical statement that is to be tested for truth in the context of some given model or class of models of some theory because it pertains to the facts in that context, AND that statement that is an hypothetically testable assertion in that specific context BECAUSE the descriptions to the left and right of that particular equal sign are interpretable in the context of the model or class of models to be considered, and the question of whether some model or class of models satisfies that particular statement is a coherent question in that context. Such a syntactic use of an equal sign is written to convey clearly to the reader some problem (i.e. it is a mathematical query) aimed to elicit some accurate information (a clearly formulated and completely explained solution of the problem and answer to the question) about the context provided by the given model or class of models. Mistakes can be forgiven, but persistent incoherent abuses of notation should be excised from mathematical vocabularies.
Sensationalism should be rooted out and excised, just like incoherent abuses of notation should be. Otherwise, we our logic system will prove absurdities like 0=1 in the real number system. A logical system that allows such nonsense is not useful, because from such things, the notion of "provable equation" and "equation" are indistinguishable. By a variant of Occam's Razor, we should not invent a terminology that pretends to distinguish things that are indistinguishable.
bruh what
@@nicktravisano7152 There's a distinction between log applied as a function and the relation called the inverse image of a point in a space via a function. Both are relations between elements of sets but a function has the property that if x=y then fx=fy . The relation named "inverse image" has not in general such property. What blackpen writes on the board is formally incorrect. You cannot use the equal sign if applying something which is not a function on both sides of the equation, such writing down "log" but in fact meaning inverse image of the exponential function in the complex plane. The apparently revolutionary results you find in the complex plane are not so much revolutionary but abuses of the inverse relation treated as a function when it should not be. Still fun though.
tf am I reading
@@HDitzzDH University students having a discussion
You know what. This came in my recommended, and man let me tell you, my algebra 2 teacher must be doing a great job because I don't know how I willingly clicked on this and was genuinely interested lol
ikr lmao
He went from:
guys I have pen and I do math
to:
筆子曰：「無實數解既找虛數解」。
"no real number solution then go find complex number solution"
@@yehe297doing gods work
Okay if you solve for x in 1=2^(1/x) using wolfram alpha, you indeed get x=-i*ln(2)/(2pi*n), but as I have learned in math competitions, you should -always- usually plug it in to the original equation. And plugging x=-i*ln(2)/(2pi) as 1^x in wolfram alpha, you get 1. So x=-i*ln(2)/(2pi*n) are all extraneous solutions, which is why they are not solutions to the original equation 2^x=1.
How exactly extraneous, wdym by that
@extraterrestrial46 the original equation has a domain of all real x, and a certain range. The equation 1 = 2^(1/x) has a more restricted domain, and a different range. This changes the solutions, producing new solutions that do not work for eqn 1.
The math police have issued a warrant for your arrest.
2:59 I was gonna say 360 degrees like the child that I am. I can't wait to be an adult and say 2 pi!
Radians 4 adults 😎
2 7.18808272898
funny factorial joke haha (or rather abuse gamma function for a factorial joke joke)
Or you can take it a step further and say tau!
Whaddaya mean 360°? 2pi? I only know 400 GRAD
@@josefmuller86 Every thing, that don't give you a round number, if it's a right angle, is stupid. ^^
“Yes, but not all the time”
YOU’VE BROKEN MATH
After finally having learnt complex numbers, it feels so good being able to understand these types of videos! Keep up the great work.
Answer from google: Logarithm is not defined for base 1.
2^x=-1 vs. (-1)^x=2 but in ONE minute
th-cam.com/video/pBnS7K-uB-g/w-d-xo.html
Regarding this video, the two answers from wolfram alpha are consistent. b^z when b is real and positive and z is complex, is being taken in both equations as Exp[z Log[b]], where Log is a branch in which Log[b] is real. Hence, you get different answers in Wolfram depending on whether you ask Solve[1^x == 2, x] or Solve[2 \[Pi] I x m + 2 \[Pi] I n == Log[2], x]. Given the fact that you have the possibility to reach big audiences, you could make a bigger effort to present these topics in a more precise fashion, and use this topic to explore its nuances, and you are intentionally choosing not to do this.
@@secavara This is dishonest criticism, as the purpose of these videos are not to present the topic rigorously as understood by mathematicians. The purpose of these video are to showcase what happens when one is not careful, and to present topics heuristically, which is necessary before a student can begin to approach a complex topic with rigor. I have my disagreement with BPRP regarding how some subjects should be presented, but implicitly accusing him of being dishonest via your tone and wording when the criticism is not even applicable is itself dishonest and hypocritical.
i love how you hold a pokeball, because i love pokemon
th-cam.com/video/vUcNDYlBtoc/w-d-xo.html
@@secavara No! Bprp’s just having fun for the viewer’s entertainment!!
"ok so we'll just take a logarithm and set the base to 1"
this triggers me
@Tessellating Tiger Lol then you can divide. He had to hide it lol.
Log base 1 is so cursed
log base one isn't defined in mathematics
@@prashant2650 in complex numbers too?
It is a rule that 1 to any power is 1. I assume this is true for all numbers but whatever tricks he is doing, you should always get a div by zero error.
As a random 13 year old, my mind imploded and exploded at the same time
th-cam.com/video/vUcNDYlBtoc/w-d-xo.html
So it cancels out and nothing happen to your mind.
@@jondo7680 nah, it just disintegrated in place
@@vibaj16 hey, I'm just someone trying to make a 13 years old feel dumb. Don't come with such high level stuff to me.
@Hans von Zettour Nobody knows
another way to solve this is 1=i^4n, n being an integer different than 0, you then have ln(i^4n)=4nln(i)=4nln(e^pi/2*i)=4n*pi/2*i, so ln(2)/ln(1)=ln(2)/4n*pi/2*i=ln(2)/2n*pi*i=-i*ln(2)/2n*pi, same result with a slightly different method
All good until "ln(2)/ln(1)". ln(1) = 0, so you are juat dividing by 0. Which is the same mistake in the video (just that in the video it's packaged differently). Wolfram alpha is correct, the equation has no solutions (including complex ones). The only way you get solutions is by doing something illegal (like dividing by 0).
Yeah you are right but breaking the law is fun
@@gabrielfoos9393 ln(1) is not 0 for other branches of the logarithm in C, so it can be divided by for those, it's only the one that aligns with the real-valued logarithm (the principal logarithm) that has ln(1) = 0, it's usually the most convenient to use but *not the only one*
Oh okay well I’m pretty new to complex logarithms, thanks for clarifying !
In general, we can find the solution of
x
1 = P(n) where P(n) is a polynomial in R or C
the solution is:
-i(2 *Pi* k + Log ||(P(n)||)/2*Pi
with ||(P(n)|| the modulus of P(n) and k a relative integer.
it is the magic of complex numbers that allows this in particular the possibility of writing:
1= exp(2*Pi*k)
bprp: 1^x=2, x=?
Wolframalpha: *By assuming x€R,...*
Wolfram Alpha does not assume x in R. Wolfram Alpha assumes the principal branch of exponentiation. In the principal branch, 1^x = 1 for all complex x. In order to get any other value, you have to assume the non-principal branch of Log(1).
@@angelmendez-rivera351 This man complexes
what is this principle and non principle branch?
@@abdulkabeer7313 Principle just means restricting our polar form to have an argument between 0 and 2pi.
@@tupoiu isn’t it from -pi to pi?
Wait a minute,
Log base 1 is undefined
Anyways,
The pokemon in his hand is more important
So is division by 0, but he managed to get past it
@@Kokurorokuko L'hopital would be proud.
base 1 is undefined in R , he's working in C
base 1 is undefined in any Field, what he's doing is messing with the fact that log function on C is not a function by definition (one value of z leads to infinitely many values of log(z)), we have to use the principal value of log, the Log function, instead, which locks the n value to 0, and is bijective.
Literally doing that, if z1 = z2 then log(z1) = log(z2) in complex field, which is false (log in lowercase is not bijective!!!).
You have just extracted juice from a stone. It´s beautiful.
is this an idiom in your country?
@@PicaroPariah Not really. The expression in Spanish is: "sacar agua de las piedras".
I preferred to use "juice" instead of "water". A failed poet, as you can see.
@@edgardojaviercanu4740 in portuguese is "tirar leite de pedra" which means extract milk from stones lol
@@yoonsooham3261 f
@@porfiriodev in Bulgarian it's water too
Question: Do you need the negative sign in the numerator? If n runs through all positive and negative integers, it's the same "solution set" with or without the negative sign ...
... isn't it ??
PS - Your gentle emotional delivery here (moments of disappointment, exasperation, near-defeat) is a refreshing new contribution to the art of math videos!
It's a similar issue as when you rewrite the following diffEQ solution:
y = e^(-x + C)
as:
y = C*e^(-x)
The C came from a constant of integration that was completely arbitrary, and when we rewrite the final solution, the C is still an arbitrary constant. However, it is NOT the same arbitrary constant as you originally generated. Some professors insist you either assign subscripts or a tilde on top of one of them to tell them apart, or use a different letter entirely. If it were my choice, I'd accept a reuse of C, as long as write "Reassign C", or something else that means the same thing, to indicate that it isn't the same letter.
Likewise, with this solution of:
-i*ln(2)/(2*pi*n)
The value of n is an arbitrary integer, so you could just as easily assign a different letter like m, and write it as:
i*ln(2)/(2*pi*m)
Heehee....pi m. Dr Peyam, this one is for you.
I would guess 2 possibilities:
1. No. 1^x cannot equal 2.
2. If 1^x can equal 2, then 1^x can equal anything, because there is nothing special about 2.
The 1st one is correct :) 1^x will ALWAYS equal 1, even if x is a complex number. The video is just tricking you, it’s like those old 1=2 videos when the guy slily hides the fact that he is breaking the math axioms somewhere
@@speedyx3493 Both possibilities are wrong, the 3rd correct possibility that 1^z can have a countably infinite amount of solutions (not indeterminant like 0/0), but only when z is not rational.
@@glitchy9613 I feel you're misunderstanding his point. He's saying 1^x can equal anything given a sufficient value of x. And by branches, it's correct.
1^x=y, take the right branch and you can get x=ln(y)/2πni. Maybe take different branches of the ln(y) and you'll get even more solutions.
The exception will be 0. Branches don't matter, taking a logarithm of 0 will cause some sort of issues. Maybe on a Riemann sphere you can argue it, but even then not necessarily.
@@xinpingdonohoe3978 he literally says "1^x will ALWAYS equal 1" I doubt that was his point
@@glitchy9613 I reread my thing, and I can only assume I wasn't referring to Speedy Gonzales here. I think I was referring to BPRP, just from where I stand.
Sure, for each x it is true that 1^x can equal 1, but for complex x, 1^x may be something else too.
Teacher : 1 to the power anything is 1
BPRP: Hold my M A R K E R
The most EPIC beard in all of the TH-cam Mathematics community! Also, LOL at that look-of-disappointment at 2:31! 😆
LOL thanks!
Love your username! The double-sharps in your logo caught my eye. Is that where the triple-sharp is in the Quasi-Faust?
@@ejb7969 Oh, wow! Thanks for noticing the reference. NOBODY has EVER noticed it before you! I extracted these notes from the 3rd movement of the Concerto for Solo Piano (Op.39, No.10). In addition to Quasi-Faust, there are 2 occurrences of a triple-sharp in this movement, after it modulates to the parallel major, F-sharp major. Again, thanks for noticing!! 👍😀
I didn't know that about the Concerto movement, and I've been over that score many times! (As a listener, not a player.)
@@ejb7969 Yeah, for the longest time, I also only knew about just the triple-sharp of Quasi-Faust. That’s the example that’s widely used in mentioning Alkan’s use of triple-sharps. I just happened to hear about the ones in the Concerto. They’re very easy to overlook in the torrent of notes in the score!
*blackandredpen: writes log1(2) to the board*
Me: *wait, that's illegal!*
I think this is probably my favorite problem you've done. It just shows how completely messed up things can get when you get to the complex plane, to the point where 1 raised to a power does not equal 1.
Complex analysis is just bizarre.
But in the end, this problem is so dang easy. You can get to things like
-1^x = 10
x = i ln 10/pi
How crazy is that?
Both beautiful and horrifying😆😆😆
4:15 "I don't like to be on the bottom I like to be on the top"
😂😂😂 omg. I love the fact that you're searching for interesting equation to solve. That's amazing 👏🏻 keep going . This is your folower from Morocco ❤️
hello fellow moroccan
The first thing I noticed is he don't have glasses.
All depends of the branch of logarithm chosen because k^x=exp(x*"log"(k)) where you need specify the 2*pi magnitude interval of the imaginary part of function "log" how is defined; if it contains 0 there are no solutions. For example, the case of principal Log doesn't work because Log(1)=0; thus, for powers of principal branch 1^x is always 1
You can change ln(2) also to ln(2e^(i(0+2pi*m)))
This way you can get more answers
X=(ln(2)+2pi*m*i)/(2pi*n*i)
I think this might be why Wolfram won't give an answer. Too many possibilities.
if a guy holding a pokeball tries to prove something, he is trying to prove nothing
Thank you. This is one of the cases my Math Prof. warned me about. Does the imaginary unit really enable such relations or are the exponential / logarithmic laws more limited in the complex world than one might first time think?
the exponential function in the complex plane is not biejctive, and ln(z) is not the inverse of e^z
This is hard to accept since it's zero division of exponents
This answers a question I've had for a long time: does some math treat e^2 and e^(2+tau*i) differently? I learned something today. Thank you so much :D
e^2 = e^(2 + 2·π·i), but this does not imply 2 = 2 + 2·π·i. z |-> e^z is not injective if z is a complex number.
I came here to see how many people undestand this, because boi i really weak in math. It took me hours to understand this, and open bunch of books about ln, log, and how they works. By the way i'm in 8th grade, and your vids helped me to understand lots more of how to calculate. So i thanked you for that 💙
I hate to do this but......r/imverysmart
@@eekumbokum6770 Not at all lmao. Lots of people are motivated to make positive changes in their lives...
The funniest thing is that the video is just wrong :). So if you really claim you understood it, then you didn't understand it enough :P. Hint: think about what it means to take log base 1.
@@Lightn0x you don't have to take log base 1 though. Just express 1=e^(2πni) and take the natural log.
I love this! I'm taking an EE class right now which revolves around complex numbers, and your videos are super helpful/inspiring.
E
Moving math into higher or adjacent dimensions is always a very neat thing to see. Things you think that are impossible become commonplace
no, jerry, you’re forgetting quaternions. there _is_ a solution for sin(e^(7x^2))=53x-25,000
-elementary schools in the year 900000
A complex number raised to a complex power has multiple values. If we treat 1^x as such a multi-value expression, and the question is to find x so that 2 is one of values of 1^x, then x=-i*ln2/(2*pi) as stated in the video is a solution. Actually there are multiple solutions: x=-i*ln2/(2*pi*k) with k being any non-0 integer.
One of the first videos I was ahead of. Guess studying is paying off.
This proves it, complex numbers were invented by mathematicians who were on some extremely good weed.
Great! Always something surprising.
There should be more solutions. You can also split 2=2*exp(2*pi*m*i) and use the additive rule of the logarithm. The general solution is x=-iln(2)/2/pi/n+m/n,n is not 0,m,n are integers.
Why tf is he holding a pokeball is he trying to get sued by nintendo
it's his mic
Hello there is a mistake from the beginning when u wrote X=log(2) because log(1)=0 so basically X is multiplied by zero
wait, that's illegal!
Dude your flicking motion to flip between the markers is so smooth.
Thank God I still have my 12th grade knowledge to understand this randomly recommended video.
All I could see was the beard.
all I could hear was "stop looking at the beard! concentrate! concentrate!"
And his mic
This is also a clickbait 😂
Cancelling log and exponent so casually gives me anxiety about messing up with branches....
Good.
In general, if 1^x = z, with x,z ∈ *C* and z≠0, then x = k + [(ln z)/(2πm)] i for some k,m ∈ *N* with m≠0. In this case, the values of 1^x are the principal values of z^(-n/m) for all integers n. So the principal value of 1^x is always z^0 = 1, but the desired value is always in there. Specifically, 1^x = z on the branch n = -m.
“But you know me I don’t like to be on the bottom I like to be on the top” bruh
4:16
There are many things wrong with this video, but they serve as good indicators of why we need the concept of branches in complex analysis.
Let us first see that some things are certainly not right:
Suppose the reasoning by BPRP works, and that one can indeed write ln(1) = ln(e^(2pi*i*n)) = 2*pi*i*n, for any integer n. Then we run into the problem that ln(1) = ln(1^2) = 2ln(1). This immediately implies that ln(1) = 0, because that is the only solution to x = 2x. So either BPRP is wrong, or ln(1^2) = 2ln(1) is wrong. However, this same property of logarithms is used by BPRP himself when he writes ln(e^(2pi*n*i)) = (2pi*n*i)*ln(e), thus in any case BPRP's argument is not self-consistent. This property of logarithms should be familiar and we would certainly want this to be true. Let us now get to the heart of the problem.
BPRP did not consider that in the case of inverting the complex exponential, you may not use all properties that we are used to when dealing with logarithms of real numbers. To 'invert' the complex exponential, you need to choose a specific branch, precisely to deal with problems like the one that we see in the video, namely that 1= e^2pi*i = e^4pi* i = e^6pi*i = e^(2pi*n*i) for any integer n. What does choosing a branch mean?
Well from these equalities we see that there is no single inverse value to the complex expontential for the value 1: We need to choose one of the values 2*pi*n*i to get a step closer to defining something like an 'inverse' function to the complex exponential. Making such choices in order to always know which value to pick is (crudely speaking) what mathematicians call choosing a branch.
The natural logarithm for complex numbers is an example where such a choice of branch has been made: The natural logarithm is defined for complex numbers by choosing the principal value branch, which restricts to the interval (-π, π]. This means that even though 1 = e^2pi*n*i for any integer n, when we use ln(e^(iθ)), we choose the value inside (-π, π] (even if θ is outside this interval!). In the case of ln(1) = ln(e^(2pi*n*i)) for any integer n, the natural logarithm then simply gives 0. This last point stresses that ln(e^(2pi*n*i)) = 0, thus the argument in the video does not work.
Then one might be tempted to defend BPRP's argument by saying that he implicitly chose a different branch for the natural logarithm, precisely by asserting that ln(1) = 2pi*n*i for some integer n other than 0. However, even then one encounters the problem we discussed above: ln(1) = ln(1^2) = 2ln(1).
This does not mean that one cannot take different branches for logarithms. Instead it means that when we do take a different branch, we cannot expect precisely the same rules to hold for our logarithm. In particular the rule for logarithms that log_a(b^c) = c*log_a(b) does not hold anymore if we choose a branch corresponding to ln(1) = 2pi*n*i for n ≠ 0. Of course this is a cherished property of logarithms, and motivates even more why mathematicians prefer to choose the principal value branch: It is the only branch in which this property holds.
Thus in conclusion, BPRP's algebraic gymnastics to solve the equation 1^x = 2 is not correct, and upon further inspection Wolfram Alpha turns out to be exactly right: There is no solution to this equation. However there are some solutions to the equation 1 = 2^(1/x), which is NOT an equivalent equation. But my comment is long enough as is. If anyone is interested, I can elaborate more on this later.
Pretty interesting comment. I would like to see why 1 = 2^(1/x) its not an equivalent equation to 1^x=2. I'm guessing that it is because we also have to choose a branch of the function f(z)=z^(1/x) to apply on both sides?
interested
Thanks for bringing this up! In an engineering course I'm taking, the instructors were very up front about telling us to give our answers in terms of this interval. Now I'm starting to see why. Time to do some more research/learning.
Does this video do a disservice to the field?
It's much easier than that... the video is wrong from the first minute where they take log base 1 of both sides. You can't do that, that's equivalent to dividing by 0.
How about "x tetration i = 2" :0
x^^i is not a well-defined operation.
How does that even work
4:15.....why would he assume that WE know "HE" likes to be on the top rather at the bottom?
He considers all his viewers to be his rabid fans and expects us to have done the necessary enquiries so we are prepared if we ever meet him.
i love how he explain in a way that i seem to understand, but then i realized i have no clue on what he's doing
1^x = 2, So x = Log（base1）2 = Log(2)/Log(1)
The answer will be a complex number a + b i, where a is any rational number, and b = -Log(2)/(2 n Pi), n be an integer.
3:02 I’m pretty sure I heard “we are adults now, so we use 2 pi” xD
You should make a video on x^x=x^2
Obviously 1 and 2 are the only real solutions but I did this with some code I wrote, it would be interesting to see how the actual math works. Also what about complex solutions
Love your quick explanation! You would have to rule out 0 since 0^0 is not defined. Also 0^0 isn’t 0^2.
@@blackpenredpen Hey thanks for the feedback! I appreciate it so much. I edited the title and description
I'm trying the whole math youtuber thing, I like it so far (inspired by you and others)
@@blackpenredpen i think it is defined for algebra purposes, but even with that it's not a solution as 0^0 =/= 0^2 => 1 =/= 0
Depend on which branch of 1^x you take
1*1^(x - 1) = 2 | divide by 1
1^(x - 1) = 2 | but 2 also equal to 1^x
x - 1 = x
-1 = 0 | and we can do it for any other integer not just -1, they are all equal to 0
1^x = 2 is not the same as 1 = 2 ^ (1/x). Just like 1 = -1 is not the same as 1² = (-1)². Raisng equation to the power might change it's answer.
Would be lovely to use tau instead of pi here
I love tau
Table at 6:03 it should be Ln, not Log I guess? Btw, like this you can get any number by replacing ln(2) with ln(k) where is the number (or it's power) that you want.
actually "log" could have different meanings in each category we dealing with. Therefore, denote as "Ln" might be more clear👍
Math itself must've felt violated after the problem was solved
Wh-what are you doing, redpenblackpen-chan?
@@viktorramstrom3744 what are doing step brpr
*I don't know if 1^x will be 2 or not,but I surely like your Pokemon mic*
Yes, if x = (-ln(2)*i)/(2nπ), where n is any non-zero integer. But only for non-principal values of 1. The principle value of 1 is e^(0*i), or e^0. And raising e^0 to any power will yield e^0, which is 1. So entering 1^x into a calculator will yield a result of 1, even if you use this x value. But if you use this value of x when entering 2^(1/x) into a (sufficiently sophisticated) calculator, you will, in fact, get 1.
EDIT: Cool, I didn't miss any details. I actually solved this slightly differently from how you did. I actually jumped straight into polar form, restating the equation as (e^((0+2nπ)*i))^x = 2 right off the bat.
I came here because I was curious of how you would have solved the equation, to discover that behind it there was a story of a change of perspective. Great video, I had fun, even though I can't do all that math :)
I have no idea why I'm watching this on a Saturday evening, but here I am
The pokemon in the pokeball in his hands probably learnt more maths than me
log functions are not defined for base 1 . in fact its in the definition that base of log is positive and can't be equal to 1 . so how can you take log with base 1 on both sides 🤔
Don't you acknowledge it in the first place? Suppose look up how "i" was invented?
looking up for complex value, then one may have to go through complex solution just as doing things with √(-1)
For increased effect you should have asked if 1^x can equal e. Then following the same reasoning you would have concluded that 1^(-i/2pi) = e, which is quite beautiful, and correlates directly with euler's formula: [1^(-i/2pi) = e]^(2pi*i) -> 1 = e^(2pi*i)
I know the math is ultimately not correct, but it's just for good fun.
It's not "ultimately" correct, but it's not particularly wrong either. The principal value of 1^x is 1. Always. But if x is not an integer, you will be getting either finitely many or infinitely many more values. 1^(ln(2)/2πni) has principal value 1, but can take on other values such as 2. 1^(1/2) has principal value 1, but can also take on -1.
1ˣ = 2
⇒x = In(2)/In(1)
∵ aˣ = b ⇒ x = In(b)/In(a)
⇒x = In(2)/0
[any no. divided by 0 is undefined]
∴The solution of x is undefined
When Wolfram alpha breaks you know your fkd.
Starting the day with these kinda sums as a jee aspirant feels refreshing.
4:37
My brain: starts melting
The definition of a^b for a, b complex numbers is a^b=e^(b*ln(a)). So 1^i=e^(i*ln(1))=1.
Now, if there is a complex number z=x+iy, with x, y real numbers, such that 1^z=2, we have 2=1^(x+iy)=(1^x)*((1^i)^y)=1*1^y, which means that 1^y=2, with y a real number. This is impossible.
this man can bend reality
Now I'm gonna dream about this, shit.
(It's 2am here)
But here is about 11:30 pm in India
@@rishabhagrawal80 now 12:07 lol we are everywhere!!
@@pravargupta6285 12:09
Its 12:24 am in india date is 8/1/2021
It's 8:11 PM in Czech republic. Still 07\-i^2\(sqrt(4 084 441))
any tips on how to increase self esteem in maths? i was always interested in maths but people are just so toxic and competitive man
In what environment are you encountering toxicity? Personally, I just didn't really find maths fun until I got to higher maths like Mathematical Reasoning, Graph Theory and Group Theory. It really pays off when you finally see your work in one subject area began helping you understand other areas (ie, some understanding of computer datastructure helps a lot with Graph Theory, and Vector/Matrix stuff is all in Group Theory). That just never happened for me when I did high-school stuff. I also don't usually talk with other people about maths and instead rather figure stuff out from scratch myself with help from textbooks or forums, so that might be why I never felt any peer-pressure or competition
@@iCarus_A like I remember doing a ton of stupid and easy mistakes on a maths paper, people thought oh I must be a genius or something and then I overheard people saying that they thought I was smart and it really hurt to be honest everyone is just so competitive idk why and i feel my passion for maths getting dimmer-its not the maths thats hard its just overcoming the mental barrier when doing maths, for some reason when I do maths I just assume I can't do it since thats what some of my teachers said
@@zeroitedono2547 Something I learned is that the more passion you hold for a specific hobby/profession/pursuit, the more likely it is for you to get stuck in a mentality of perfectionism. At some point, you just become so paralyzed with the fear of failure that even taking a single step feels difficult. The first thing to know when recovering from something like this is to realize that you're not perfect and the only way to get better is to keep doing it. I'm usually fairly good at maths and tend to lead math-related projects in small groups, but I also make tons of small mistakes regularly. A mistake fixed is not a mistake at all, and it's far more important to get the overarching goal right -- try to get a friend or someone else to talk your stuff over without getting too involved and see if you can explain your thought progress clearly to someone who's not even deeply familiar with your field.
@@iCarus_A thank you
@@zeroitedono2547 Happy to help even in the slightest :D
"They called me mad for trying to divide by zero! But I'll show them! I'll show them all! ahahahaha!"
this guy can definitely explain to us how gojou satoru's power really worked mathematically
I would like to note that this problem has some additional weird properties. For example, if you attempt to solve this via a limit, you find that if you try to solve for 2^x=1, you end up with this being true as x approaches 1/infinity, essentially as the infinitieth root. Or just use the identity where raising something to the power of 0 makes it into 1
However on the other hand, you can prove that for all real exponents of x, 1^x is 1.
Imaginary numbers are useful, yes, but they also kinda messy and can pretty commonly give non practical solutions(like in the real world, there is no real operation to do i rotations or whatever)
Long story short, solutions like this never really see use except when i is representative of some specific axis. Like the phase of an AC circuit. And even then, the most you really see of i is e^i(angle) or a basic complex number. Or at least, in the classes ive been in so far. College junior rn
You dont need a limit. You just need a zero. Anything to the zero is equal to one
@@The_Commandblock I didn't use the 0 power property since the video didn't really go over them and I wanted to use a way other people on the comments didn't use. I will add that in though
bro chill my girl’s on this app
Is base 1 for logs exist?
I have the same questions...
But seems that exist in complex
No. Also, 1^x = 2 has no solutions.
When watching these videos I feel like I am fcking up whatever little maths I know
@@pendragon7600 then how did he do it using the laws of math?
100 LIMITS, i need that !!!
i got a eaxam in 4 days, i believe in you
What about 100 limits without L'H
With epsilon-delta!
@@harshvardhanpandey8057 jee aspirant ???
@@sakshitandel8572 he means l'hospital rule. It's like a cheat code for limits
Argument of a complex no. are only defined for θ Ε [0,π]U[-π,0]
Since the bases are different...
So the powers should be same..
Therefore 1^x = 2^x
Put x = 0
=> 1^0=1^0
=> 1=1
The title should be "10 ways to (not) write zero"