real^real^real^... = imaginary? Wow, by pure coincident, Mathedidasko also did a video on this. Check it out: • Imaginary Unit As An I... Patreon: / blackpenredpen
It can get crazier btw. I go through an infinite 'power tower' of functions in ruclips.net/video/tFcAl40nJeM/видео.html. When you plug blackpenredpen's answer into the general solution you get i (so I don't understand the comments saying he's wrong).
@@jamesexplainsmath - I love the parenthetical part of this comment. Far too often, people assume "I don't understand" == "they are obviously wrong". As Gamefreak129127 said, the symptom that something is wrong is that the limit function f(x) of the sequence of functions f[i+1](x) = x^f[i](x), f[0](x) = x does not converge for e^pi/2. f[0](e^(pi/2)) = e^(pi/2) =~ 4.81, f[1](e^(pi/2)) = e^(pi e^(pi/2)/2) =~ 1912.717, f[2](e^(pi/2)) =~ 6.81 x 10^1304, etc. Another clue that x=e^(pi/2) is not the correct answer is the fact that it is positive; for all real a, b, such that a > 0 and b > 0, the principal value of a^b is a positive real number. Therefore, for any positive real x, f(x) as defined by the sequence above is either a positive real number, or does not exist. The fact that blackpenredpen derived this nonsensical solution means that his solution methodology was flawed. Gamefreak129127 posits that the problem is that i is not in the range of x^x^x^x^... I suspect this is the case, but I'd like to see a proof.
@@jimschneider799 Proofs of several related theorems were provided by Euler (1783) and Eisenstein (1844). The infinite tetration converges if and only if e^(-e)
Sadly #disagree. The sequence x^x^x^... diverges for x=e^pi/2. The mistake was the substitution of x^x^x^...=i into the equation itself. By doing so we assume that our original claim that x^x^x...=i holds. But this premise is wrong. The whole calculation is therefore only implicative not equivalent. IF there was a solution to x^x^x^...=i then it would be e^pi/2 but as there is no solution, this is false. Or in other words, x=e^pi/2 is not in the domain of the infinite power tower function f(x)=x^x^x^... As we only take real numbers as x, this also doesn't change when assuming a complex number as a result
Gamefreak129127 Convergence has nothing to do with, since infinite tetration for complex numbers is not defined as the limit of a sequence (nor should it be).
@@christianvukadin7747 if you say y=x^x^x^..., then you can say y=x^y And therefore, raising both sides to the 1/y power, you get y^(1/y)=x. This is the inverse of y=x^(1/x). This is why personally, I believe that defining the infinite power tower as being the inverse of x^(1/x) is the best way to go.
I don't believe you can do that because the function x^x^x^... only converges for a very small range. Specifically, [1/(e^e) , e^(1/e)], corresponding to y values of [1/e , e]. If you did the same trick to find x^x^x^...=4 you would get x=sqrt(2), which is the same value you get if you did the trick for x^x^x^...=2. However, it's only valid for sqrt(2)^sqrt(2)^...=2 because both sqrt(2) and 2 fall in the domain and range for convergence, while 4 is greater than e (approximately 2.718). The graph x^x^x^... can be represented in terms of the Lambert W Function, but that is also equal to the graph for x=y^(1/y) for the specifically domain listed above, as this isn't a function because it would have 2 outputs for certain values. Including x=sqrt(2), where y=2,4. Regardless, neither e^(π/2) nor i fall within the domain or range respectively, thus making this false. Saying this would be valid is the same as saying that 1 + 2 + 4 + 8 + ... = -1 because the geometric series 1 + x + x^2 + x^3 + ... = 1/(1-x). This is untrue because this only diverges for (-1 < x < 1). tldr; This video seems to be do steps that are mathematically invalid
Flamingpaper I agree with what you said in the "real world". In fact, just like in the video, I wanted to ask you guys if you guys agree this makes sense in the complex world or not. I actually don't have the answer for it yet.
As far as I’m concerned, only in the set of real numbers is infinite tetration defined via convergence. In the complex numbers, the definition is independent of convergence, and it uses the Lambert W function, similar to what was done in the video. So you do not have much of a point.
Also, in advanced mathematics, the result 1 + 2 + 4 + ••• = -1 is made rigorous even though the series is divergent. So that is not much of an objection eitrr
Zoltán Kürti I think it’s actually smart ass people who say the series isn’t equal to number. Convergent series aren’t equal to number either, by your argument. Want to know why? Because convergent series aren’t convergent because the infinite sum is well-defined. If you want to tell me that the summation of infinitely many elements is equal to 1, such as in the case of 1/2 + 1/4 + 1/8 + •••, then I will say “bullshit”, because that isn’t addition. Addition is defined as an operator in the extensions of the Peano axioms, and last time I checked, none of the axioms allowed for infinities or limits of any type. You want to say that a series equals to the limit of the sequence of partial sums? Sure. That’s an assigned value. But trying to pretend that this somehow is equal to some summation of some sort is stupid. That is what rigor says. We can say that infinite summation is undefined - because the axioms of summation say it is - but then we can still assign some value to the undefined expression for the sake of convenience. And if we want to do it via convergence, then cool. But pretending this assigned value is an equality in the same sense that 2 + 1 = 3 is an equality is nonsense. If you can accept the informal argument I presented here, then we have an agreement. But if you want to pretend that my reasoning is false when concerning convergent series yet true when talking about divergent series, then no, we cannot have an agreement, because such inconsistency is due to pure lack of rigor, misuse of terminology, and it goes again basic principles of mathematical logic. This is part of the reason why at some point in time mathematicians pushed fir standard analysis to get replaced by hyperreal numbers analysis (non-standard). I think the fact that someone can say 1 + 2 + 3 is not -1/12 under anything but we can still assign it this value, yet still say 1 + 1/2 + 1/4 + ••• is exactly equal to 2 in the actual sense of equality, that is what I consider smartass, and also dishonest. M This also ignores the fact that the words “equality” and “assigning a value” really mean the same thing. Equality is a special case of equivalence relation. Equivalence relation are things we assign and define within sets. The definition of equality already varies from going to the natural numbers to the rational numbers. Yet no one ever says “0.999.... = 1 is not an equality in the same sense that 1 + 1 = 2 is an equality”. You bet they aren’t similar types of equality, nor should we pretend they are the same type of equality. But if people are going to pretend they are the same type of inequality, then it’s actually hypocritical to say 1 + 2 + 3 + ••• = -1/12 isn’t an equality in the same sense (it’s an equality in divergent summation theory). I rest my case.
Zoltán Kürti Well then, I suppose we do have an agreement. I simply abhor the mischaracterizations of infinite series that have become so popular as to be impossible to undo.
The conclusion is incorrect. x^x^x^...=y implies that x^y=y, but x^y=y doesn't imply that x^x^x^...=y. Therefore, finding a solution to x^y=y only shows that it is a potential solution to the original equation. In this case, the potential solution e^π/2 is extraneous, and the original equation has no solutions. This is clearly shown by noting that, when you iterate a(n+1)=(e^π/2)^a(n), with a(1)=e^π/2 (a sequence that is always equal to e^π/2↑↑n), the result diverges to infinity as n increases. It doesn't converge to i or even become nonreal at any step. Using the method in this video without checking whether the potential result actually converges to the right value can prove that 2=4. If x^x^x^...=2, then x^2=2, and we get the solution x=√2, which turns out to be correct. If x^x^x^...=4, we get x^4=4, and so we also get √2 as a solution. In this case though, it is clearly not correct because √2^√2^√2^... can't be equal to both 2 and 4. #Disagree
Interestingly, you can analitycally continue the infinite power tower to most of the complex plane using the identity z^(z^(z^...)))=W(-ln(z)/-ln(z), where W is Lambert's function. However, being multivalued, I'm nor entirely sure which branch you should use.
I'm also in the #disagree camp but I think it's pretty easy to fix. After all, e^{i\pi/2} is pretty much equal to e^{i\pi/2+n2\pi}, right? If you get e^{-3\pi/2} instead of e^{\pi/2} then the convergence radius is no longer an issue, is it?
I promised to myself I would go to bed early today. It's 3 a.m. in my timezone, and I find out real numbers raised to a real number can be an imaginary number. I guess I'm done
One thing you should mention is that i^i has multiple values, as cos and sin both have 2pi as their periods. The X can be e^(pi/2 + 2k×pi) where k is an integer.
It’s implied that we only use the principal branch, so there really isn’t a need to mention since he isn’t trying to explain how to complex exponentiayion.
A common mistake, when solving equations, is to assume that a solution exists. The argument in the video is: if x is a solution to x^x^x^... = i, then e^pi is a solution to x^x^x^... = i however (e^pi)^(e^pi)^(e^pi)^... diverges, thus the equation had no solutions to begin with (since a true statement cannot imply a false statement).
Leander Tilsted Kristensen Actually, with a non-truth-preserving set of inference schema, you can in fact imply a false statement from a true statement. Also, to be strictly technical, it was never stated that the solution is in fact e^π.
@@angelmendez-rivera351 you are correct, didn't want to do too many details in a RUclips comment. The statement was rather: If x is a solution to x^x^x^...=i Then x is also a solution to x^i= i However cocluding the opposite, that a solution to the latter is a solution to the first is a big logical no-no. In order to prove, that there are no solutions, one should prove, that all solutions to x^i= i are not solutions to x^x^x^...=i .
@@frimi8593 No matter how many times you replace i with (e^π/2)^i, you always have i in both sides of equation. You can push i infinitely far, but you don't get rid of it!
This is similar to the approach Numberphile took to 1 + 2 + 3... = -1/12 (which is correct, but the way they showed it wasn't correct) The series of x^x^x^x... (or x^^infinity) is divergent for any x greater than e, since it approaches infinity, and e^(pi/2) is approximately 5. e^-(pi/2), another "solution", doesn't work either, as for any x from 0 to but not including 1/e, the series converges to 0, and e^-(pi/2) is less than 1/e. #disagreeToMethod
#Agree. The reality is, people keep presenting the straw man argument concerning convergence, but infinite tetration is not defined by the limit of a sequence, not in the complex numbers - because tetration on the complex numbers is a continuation of that of the real numbers, necessarily so as a^^b is only well-defined for a nonzero and b an integer with b > -2. Nor should it actually be defined by such a limit, because doing so represents to many problems, as argued using mathematical logic. (1) Mathematical theories of limits to infinity and about infinity in general in relation to arithmetic and algebra are not effectively axiomatized formal theories. In fact, they cannot even be called axiomatized formal theories, to start with. (2) Standard analysis is a theory about the vector spaces of sequences and functions over the field of real numbers. The limit operator is a linear functional, and the derivative is a linear operator. However, professors and even sometimes mathematicians themselves conflate these with the algebraic-arithmetic operations of multiplication, addition, exponentiation, etc. The best example of this comes from the topic of infinite series. In standard analysis, we would say that 1/2 + 1/4 + 1/8 + ••• = 1, yet this is strictly incorrect, because this equality as proven in standard analysis actually has nothing to do with addition or multiplication, which are group operators defined by the proper extensions of the Peano axioms to the field of real numbers, or the extended real numbers, whatever be the case. This equality is instead a statement about sequences and linear functionals. There exists some linear functional Σ which operates on A = (1/2, 1/4, 1/8, •••) such that Σ[A] = 1, and Σ is defined as (lim ~n -> \infinity\~)[S(n)], where S(n) is the sequence of partial sums of A with respect to the cut-off ñ(n/N) = 1[n, m), where 1 is an indicator function. This composition of linear operators on the sequence is most certainly not a statement about arithmetic. As such, translating this as 1/2 + 1/4 + ••• = 1 is horrendous abuse of notation, one which is accepted even in the most rigorous discussions and works. Calling it a summation is far from true. In the same vein, the claim 1 + 2 + 3 + ••• = -1/12 is not a statement about arithmetic either, and criticisms of this claim come from mathematicians or mathematics students pretending that it is indeed a statement about arithmetic. Furthermore, the arguments that are made in the criticism are non-rigorous and they fail to account for (A) the fact that the status of convergence or not can only be determined after a cut-off over which the partial sums of a sequence one wants to calculate is specified, which is conveniently ignored by everyone in the discussion. (B) a theory about limits to infinity in standard analysis is not meaningful without a rigorous treatment of asymptotic expansions, especially if we work with the non-extended real line, which is usually the case, and which should be the case, since no axioms or semantics are constructed and added to the prior theories to specify otherwise - which is in itself an aberration in mathematical logic. (C) the relationship between the linear functionals over these vector spaces and the group operations of the sequence-generating fields. (3) Now we see that the problem here is the lack of constancy, and the fact that people want to pretend the all mathematical claims should be centered on standard analysis and that it is the default paradigm on which we talk is abominable. By this same inference scheme, one should never accept the existence of transfinite numbers in any situation EVER unless one explicitly, in exact words, says “Let’s talk transfinite theory”, and similarly, one should never ever state that complex numbers exist either unless a very unique context is specified in a verbatim formula. It is nonsense to ask for this when mathematicians themselves never obey such a rule in their formalities in other domains of discussion. There is no default theory. There is nothing wrong with assigning values with divergent sequences, as we already do it with convergent sequences, even when it is uncalled for. The statement 1 + 2 + 3 + 4 + ••• = -1/12 should never be taken to be a statement of arithmetic as long as we work with the Peano axioms in the extended real line. Neither should 1/2 + 1/4 + ••• = 1. These are statements about the linear functionals of certain sequences and nothing more. Now, if we want to work with an extended real line or extended complex plane, it must be specified which extension is desired, such as the Riemann sphere or the transreal set, as well as an algebraic structure which includes such extended sets, with a proper specified extension of the Peano axioms for arithmetic. This is the only way in which expressions such as a + b + c + ••• ad infinity and x^x^x••• have meaning without constituting an abuse of notation. Since we assume this the case for this video, we then have it that we should define infinite tetration in terms of arithmetic, algebraic operations, or via functions and formulas, without recur to vector spaces and linear functionals. This is precisely what BPRP introduced in this video, and what he showed is what would be the heuristic consequence of such treatments of tetration. In advanced mathematical fields of study, infinite tetration is well-defined for complex numbers, and it is not via the limit of a sequence, unlike the case with infinite series. Rather, it is defined via the Lambert-W function. By this treatment, what BPRP did is correct. Sure, he never stated the rules here or stated that he was using such unintuitive definition. However, as I said earlier, mathematicians often never specify their axioms or give context in other domains of discourse, and will still discuss results that by Peano axioms or standard analysis is strictly incorrect, and this is done even when there is no background context to be known, almost as if there were telepathy. So, the conclusion is, either BPRP is wrong with what he is doing, but the mathematicians are wrong with how they treat the issue of convergence as if it were the be-all of mathematics... or we can accept what mathematicians do for a living and their little informalities, but then for consistency, we must then be able to accept what BPRP did here - notice that he never even claimed to be a channel of rigorous mathematical demonstrations, and this channel is not meant to substitute a mathematical paper for obvious reasons, so y’all are silly for thinking he is supposed to be rigorous in the first place too. This is why I must disagree with #Disagree camp. Their notions are simply misguided if we argue using mathematical logic and pragmatism.
I mostly agree, but why is 1/2 + 1/4 + 1/8 + ... = 1 an abuse of notation? Is it because the Peano axiom's do not define summation for an infinite number of summands?
@@xCorvus7x It is abuse of notation for two reasons: 1. because Peano axioms indeed do not define summation for infinite summands. Now, this by itself shouldn't be a problem, but... 2. the reason we say that series is equal to 1 is not because we extended the definition of summation for infinite summands. IN fact, there is not much summation involved in that proof. Rather, the result is derived from obtaining the limit of sequence of partial sums. And in calculus, it's fine to do that, but that is a statement about sequences, not really about arithmetic summation. And the fact that some try to disguise it as a result of summation is why I consider it to be misleading and bad notation. Because it's not really a summation, it's just an operation we perform on a sequence to get a number, and the operation uses a limit. I don't know if I made myself clear enough on this, but hopefully that answers your question.
All this video shows is that If that infinite power sequence converges then it converges to i. However “if” is the key word - that sequence actually diverges. To see that it diverges you just have to notice that e > 2 and pi/2 > 1. Therefore every member of the power sequence is greater than the previous member by greater amounts so the series diverges toward positive infinity. This is similar to the when people say that “1+2+3+... v= -1/2”. The number -1/2 is what the sequence would converge to if it converged, but it doesn’t converge to anything, it diverges to infinity. P.S. The range of convergence of this sequence is (e^-e, e^(1/e)), or about 0.07 to 1.44. E is obviously more than 1.44 so it diverges.
Based on the lambert W function definition, I'd say that [e^(π/2)]^[e^(π/2)]^[e^(π/2)]... is actually -i. i would be the extraneous solution that you would get from assuming that if y=x↑↑∞ then y=x^y kind of like how ∛3^∛3^∛3... is not actually 3.
(real)^(real) = non-real? - Of course! (-1)^½ = i "Of course, they (real) don't have to be the same..." - But they can be: (-½)^(-½) = (-2)^½ = i√2 (real)^(real)^(real)^...[∞'ly] = non-real? - In light of the first answer - seemingly yes. But, ... Ah, another commenter has answered this simply: (-1)^½^1^1^1^... = i Now having watched, you ask: "x = e^(½π) Agree or disagree?" Well, you've bypassed an essential thing - establishing that this particular "power tower" (PT) converges. If it doesn't, your result is invalid. In particular, e^(½π) > 2; and an infinite PT built on 2 (or any real no. > 2) definitely diverges. So I have to disagree. Very cute, nevertheless! Fred
I disagree because if e^(pi/2)^....=i then their absolute value must be equal for instance and for the right side it's 1 and left side infinity. But this counter examples only says that it is impossible but it's not saying where the mistake is....well that's a hard question I'm not able to answer 1)I suppose e^(pi/2) is not allowed in x-convergence values but how to calculate that interval? 2) if e^(pi/2) was allowed then I'm not sure if we can power both sides to the (1/i) and keep having an equality for example ample if -1=-1 and you power both sides to (1/2) the left side could be i and the right either i or -i more far if u start with a wrong assumption 1=-1 raising to power 2 u could end with non true results 1=1 but it looks true at all but anyhow it's an awesome topic what u showed us thanks
@@federicopagano6590 The mistake is this. We have an expression, in this case, an infinite "power tower," set = to some definite value, C. In order to evaluate it, we substituted C into the power tower, That step is valid only if the power tower *has* an actual value; that is, if the finite tower of n terms, converges as n→∞. In this case, that criterion is not met. Ergo, you can't make that substitution. Fred
ffggddss thank you for the answer very clear and I agree with that but how are u sure that this value C is not allowed before calculating this in the first place?
@@federicopagano6590 It's really a _reductio ad absurdam_ argument. You suppose for the moment, that the substitution *is* valid. You solve for x. You find that, for that value of x, the power tower diverges. You then conclude that the assumption of validity of the substitution is false. Fred
After using the fact that the infinite tetration is equal to -W(-ln(x))/ln(x), if you plug in the value e^(pi/2), you actually get i as presented in the video. And according to Wikipedia, the only restriction on the equation is x=1. I understand why W(-pi/2) = i * pi/2, but why is the only restriction for the infinite tetration equation x = 1 and there isn't more restrictions?
It is not okay to actually do. The value of x^x^x^... when x=e^pi/2 doesn't converge to i. It increases without bound, and it always stays real. Simply finding all the solutions to x^y=y is not a valid way to find the solutions to x^x^x^...=y. You must also check for extraneous solutions.
Joseph Noonan I see your point and it makes the most sense, though Wikipedia still says the equality with the Lambert W function and infinite tetration still holds for all x not equal to 1. Either there was a typo in the page or it has a point, but again Wikipedia isnt that reliable of a source lol
I also have seen a college level math research paper saying it is possible for numbers outside of the domain of the infinite tetration to be equal to a complex number or any complex number in some cases, I can post a link to a paper here if you want.
e^(pi/2) is about 4.8. surely a positive real number larger than 1 would keep increasing in magnitude when you raise them to themselves as powers and it would diverge off to infinity,and definitely not decrease the magnitude down to |i| = 1?
x=√2 also holds, giving x^x^x...=2 In fact, any x^(1/x) x^(x^x^x...) = x^x^x... x^y=y, x=√2, y=2 x=3^(1/3), y=3 i^(1/i) is just one of the many solutions e^pi/2=i^(1/i)
Well, it's an absurd because we can demonstrate that x^x^x^... converges only if 1/(e^e) < x < e^(1/e) but e^(pi/2) is bigger than e^(1/e), so the equation has no solution.
Your second example (infinite power tower) is wrong. You did not consider convergence. The hyperpower function (also known as infinite power tower and infinite tetration) x^x^x^... = k only converges for e^(-e) < x < e^(1/e). The implication here is that you can only apply the "trick" of taking x = k^(1/k) when k < e. So x^x^x^... = 2 implies x = sqrt(2) but x^x^x^... = 3 does *not* imply x = cuberoot(3) since $k = 3 > e$ here. Similarly x^x^x^... = i does not imply x = e^(pi/2).
If you can find an example for any case of a finite number of indices, the infinite version is true as well as you can just follow it up by raising the finite case to the 1st power infinitely many times.
Solution holds theoretically but it is a bit silly to say i^i is a real number(even by Euler’s formula ) because that is an undefined expression, what is an ith power , the limit of e^pi/2 tetrated n times as n goes to infinity is just infinity so essentially we are saying that i or the square root of -1 is infinity
Another way to see why the second equation fails for x=e^(π/2) is define recursively the sequence: x₀=e^(π/2), xₙ₊₁=(e^(π/2))^xₙ. If it converges, since ℝ is complete it would converges to a real number, not complex. :^)
Personally, I disagree. It is just because by defining a sequence a_i with a_0 = e^(pi/2), a_i = a_{i-1}^a_0. It is easy to prove that this sequence is divergence. Note that this issue is similar to a trick that sum of all integer = -1/12. You cannot substitute a value into a series causing its divergency.
that "trick" about -1/12 is not a trick, its only because those bozos at numberphile made it sound mysterious and a joke. Its a well known fact in complex analysis and by methods of analytic continuation that the riemann zeta function evaluated at -1 which is the same as saying all the natural numbers added together does equal -1/12. But only in the complex plane not the real plane. That is where the confusion is you have i believe.
You are assuming that divergent expressions have no value by virtue of being divergent. This a misnomer. Divergence is a claim about sequences and the limit operator, not a statement about algebraic operations of the real numbers with themselves, which is what iterated exponentiation is.
I know another way to do the real power tower. Take the tower as x^y^t^t^t^t...... let x:-1. Let y= 1/2. And finally, t=1. I mean, no one said it coudlnt be anything in the powers. Just put a single 1 anywhere to reset the value and get a finite tower
Using your first example, substitute into the second: (-1)^(1/2)^real^real... replace all real with 1. Since 1^1^1... =1, therefore the second eqtn = i
It is true that it diverges in the real number plane, however in the complex plane it converges, which is how it equals to 'i'. its a similar case to the summation of (1/n^-1) to infinity. Obviously that would be a divergent series if we limited our scope to the reals only. But in the complex plane it does converge to a value of -1/12 or -1/12 + 0i to be exact, the infamous number no one understand XD
Can't you take X to be for example (-1/2)? then x^x = (-1/2)^(-1/2) = 1/sqrt(-2) = imaginary then if you keep raising it to the -1/2 power, you just keep on taking square roots, so it stays an imaginary number right?
In the second example you raised e^(pi/2) to itself, but in the first example you did not rais the real number to itself, you raised it to a different real number 1/2.
We can use e^(i*pi/2) = i to construct this infinite power tower. Sure, i has to be at the zenith of the power tower, but since it can be constructed infinitely high, so well...
#agree as u have taken a infinite tetration function,the domain of it is (1/e,e).as here x is e^π/2 the x value is out of domain and definitely we will have a complex answer.
I believe it's been pointed out that this doesn't work due to e^π/2 being outside the convergence range for infinite tetration, but there is another not entirely trivial solution that does work: Instead of assuming x^x^x^x^... = i, we only want to know if we're using an infinite power tower with entirely real numbers, so they need not all be the same, so let's instead assume: y^x^x^x^x^... = i Now, we can look at the equation we already know works: (-1)^1/2 = i, and we can set the values of some of these variables: y = -1 x^x^x^x^... = 1/2 Now, we still need to solve that last one, but we can use the same method as in the video for that: x^(x^x^x^...) = 1/2 x^1/2 = 1/2 (x^1/2)^2 = (1/2)^2 x = 1/4 Unlike in the video, 1/4 IS within the range e^−e ≤ 1/4 ≤ e^1/e, so we have a valid, convergent, non-trivial solution which is: (-1)^(1/4)^(1/4)^(1/4)^... = i
The real question is how to do it when the reals are all the same. The first one is easy, the other one idk: (-1/2)^(-1/2)= 1/((-1/2)^(1/2))= 1/((-1)^(1/2)*(1/2)^(1/2))= 1/(i*sqrt(1/2))= 1/(i*sqrt(2)/2)= -2i/sqrt(2)= -sqrt(2)i (definitely non-real)
One of the solutions is -1 - i, if the power tower has has height % 4 = 2. Or -1 + i for height % 4 = 0. For uneven heights, it gets weird. So I think there’s no fixed solution possible. Clearly the given solution is wrong, but that’s already explained by others in comments.
-1^1=e^pi i? I’m only in precalc (junior) but I love math and I’ve always kind of felt like it’s my calling. That being said I’m not the first person you should ask for this, but is this right? Or is it no longer imaginary because it’s transcendental?!?
The limit of a sequence need not have the properties of any element in the sequence itself. Take irrational numbers as an example. They are limits of sequences where every member is strictly a rational number. Also, infinite tetration is not defined by limits of sequences in the generalization to complex numbers.
@@easymathematik I understand the answers and can prove them and everything -- but it's still weird. And -1/12 is a very similar case. For any finite amount amount of iterations, it explodes. But for an infinite number of iterations, it is seemingly paradoxically defined.
Limax Chauhan There is no mathematical fallacy being committed here. You are allowed to evaluate a root operation on both sides of the equation at any given point so long as it is well-defined.
@@angelmendez-rivera351 for complex numbers,these rules are not allowed! Check on wikipedia, Google >> mathematical fallacy (Wikipedia) >> power and root >> complex exponent
Angel Mendez-Rivera yes it is allowed you are right. The Wikipedia page is not saying that you can't raise to an imaginary power, it's saying that you can't compare two exponential function in the set of complex numbers as the exponential function is not bijective in the complex numbers. So yes, since that has nothing to do with that, then it is allowed. And I also have studied this too and trust me it's allowed.
The problem with the power tower method is that you need to assume that there is actually a solution. If there isn't a solution, you'll get an answer but it'll be invalid. For example, let's say we want to find when the infinite tower = 2. X^X^X^X... = 2 By power tower trick X^2 = 2 Square root both sides X = SQRT(2) So this implies that sqrt(2)^sqrt(2)^sqrt(2)... is = 2 In fact, this is a true solution. But now let's try 4 instead... x^x^x^x... = 4 By infinite power tower trick x^4 = 4 4th root both sides x = sqrt(2) Which implies that sqrt(2)^sqrt(2)^sqrt(2)... = 4. But from the first equation, sqrt(2)^sqrt(2)^sqrt(2)... = 2. 2=4?????? Clearly there's a paradox here, and the cause of this is a lack of an existing solution for an infinite tower of "X"s equaling 4. If there is in fact no solution for an infinite tower of real "X"s equaling i, then the math breaks down, since you might as well start your math problem off saying that 1 = 0. Any conclusions drawn off of an invalid statement are probably invalid. For example: 1 = 0 Add one to both sides 2 = 1 The pope and I are two men 2=1, so The pope and I are one man I am the pope. 😎
Thank you! Always very interesting! I think if you can make your camera so it doesn't change the background light automatically that would help a lot. The video is too dark.
This is flawed unfortunately... if we have sequence x_0 = e^pi/2 with x_n = (e^pi/2)^x_(n-1) then this sequence diverges. It diverges if you start with x_0 = any real number in fact. It converges to i as stated in the video if and only if you start with x_0 = i. Otherwise it diverges
Okay but for the second one we didn't even made sense out of the equation we didnt' see if it was convergent, what i think you did was : if it converges toward i then, it is e^whatever, but it might not even converge idk tho im still a student
This is only true if the sequence converges. Edit: well, it looks like just about everyone is pointing that out... You could probably extend the function, but then you're no longer dealing with an infinite sequence outside its radius of convergence
I MAY be wrong, but I don’t think this is correct. You have to show that your limit is a complex number before you can manipulate it the way you do in this video. It seems to be sort of like the trick where the sum of all positive integers is -1/8...it just doesn’t work. Edit: I like your videos, but this result is incorrect. Using this line of reasoning, we also have (x^x)^i = i. Then we have x^(ix) = i => x^x = i^(-i) = e^(pi/2). But you've already said that x = e^(pi/2). Therefore, (e^pi/2)^(e^pi/2) = e^pi/2. Take ln of both sides and cancel the factor of pi/2 and we show that e^(pi/2) = 1. This is obviously a contradiction...so the answer can't be correct.
This sounds wrong, since looking at the series a_n=(x^x^... n times) - every element is real (using your "solution"), so it sounds unreasonable that the limit would be imaginary. Still - infinity is weird, so this isn't a "proof" of wrong-doing. Instead, I want to point out that the function f(z)=z^(1/i) is **not** injective, and so saying "(x^i)^(1/i)=y therefore x=y" is patently false. It would be the same as saying "(-1)^2=1^2 therefore -1=1". In other words - you have proven **some** solutions to the equation x^i=i satisfy the relation x^x^x^...=i, but that doesn't necessarily mean all of them do, nor does your specific chosen one.
I think the answer is wrong because the function of x^x^x^x^x^... stop being useful from somewhere (the values does not converge), and I'm quite sure that somewhere is less then e^1.57
I'm pretty sure it can't be correct, but I'm having hard time proving it : ( I'm not good at math... e^(π/2) is a positive real number greater than 1. making a power operation (with exponent, again, positive greater than 1) gives a bigger positive real number. repeating the operation will always increase the result... I don't see how it can converge to anything. And if it does, then why it doesnt for any positive real number? why 2^2^2^2^... doesnt converge also? What the product (or also the sum) of "n^n" converge to, for n integer greater than 1? Could it be that the "trick" is in applying substitutions only for principal values? Or, does the power operation with an imaginary exponent (you use "i" and "1/i" in this case) follow the same rules as with a real exponent? Again, I'm too noob to spot the error, but I have the feeling there must be one... @blackpenredpen PLEASE DUDE, enlighten us! Ps: I dont even know how to "tag" you (T__T)
x^x^x^... -> Infinity which is simply a NaN! Try computing it with any supercomputer, it will still produce a NaN even if the supercomputer was the size of the entire cosmos raised to the value of mole! You can not calculate any infinite series, but you can calculate their limits, their domains, and their ranges!
The (-1)^(1/2)^1^1^1^... is certainly a nice and simple solution, but I can suggest another one: The solution to the power tower equation x^x^x^... = 1/2 is x = 1/4, so the equation (-1)^(1/4)^(1/4)^(1/4)^... must be i.
e^(pi/2) is positive so there is no way a positive infinite power tower can give a complex number, also solving the x^x^x^... = i by doing x^i=i doesn't work because the infinite power tower diverges at some point and there would be no solution to this equation. Indeed a power tower of e^(pi/2) diverges. As an example of an infinite power tower diverging. Trying to solve x^x^x^... = 4 using the substitution method x^4 = 4 gives you sqrt(2) as a result; however, there is in fact no solution for this equation, and an infinite power tower of sqrt(2) actually converges to 2
2^(1/2)^(1/3)^(1/4)^(1/5)^..... = 1.5 ¿? It´s strange because when is (1/pair) is 1.6...... but when is not pair is 1.5.... its like is converging from de rigth and left at same time. I put in Wolfram (1/n)^(1/n+2) to see what happen to the last numbers of the series, looks OK but when i put (1/n)^(1/(n+1))^(1/(n+2)) looks like e^(-x) Sin(x) Looks like es equal to 1.6
Is e^(pi/2) actually a real number? I'm not a mathematician but I thought e and pi were different categories of number outside of the set of real numbers (unless raising e to pi makes it real?)
All this proves is that, if x ^ ... = i for some x, then x is necessarily of the form e ^ (pi/2 - 2 pi n) for integer n. This is not a sufficient proof that (e ^ (pi/2)) ^ ... = i, or even that (e ^ (pi/2 + 2 pi n)) ^ ... = i for any n. Indeed, it seems all such x are outside the known domain of convergence e^(-e)
If you consider -1 as the first base then 1/2 as the first exponent then have all other exponents in the sequence be 1, you will indeed get an infinite sequence of real values that have a non-real value. However, the result of e^(pi/2) repeatedly exponentiated will simply grow to infinity. #disagree. The error, I believe was in the assumption that there exists a single real number that when repeatedly exponentiated will converge to i. I don't think that's the case: a value < -1 will diverge. A value in the range (-1, 0) will converge to 1 (This needs a proof, of course), values in the range [0, 1) will converge to 0, 1 will converge to 1, and values greater than 1 will diverge.
i want to ask something , does i (the imaginary number) in euler's formula indicate the quadrant position of the trig ? since i and -i located in the vertical axis of complex plane , then 1 and -1 in the horizontal axis
I am on the #disagree side, too. With the same method, if I want to demonstrate than the tetration x^x^x^x^x^... is equal to a negative number, I can say : "try x^x^x^x^... = -5". So x^(-5)=-5 => 1/x = 5thRoot(-5) => x = -1/(5thRoot 5). Because we are not inside the interval of convergence of the infinite tetration (seems around [0, 1,4446...], but why ???) But why the bad commentaries ? Steve only wants us to think !
CORRECTIONS: 3:50 (e^(pi/2))^(e^(pi/2))^(e^(pi/2))^... is not equal to i. We then have to plug in i into the derivative. (pi/2)*(e^(pi/2))^i=i*pi/2, |i*pi/2|=pi/2>1 Since the absolute value is bigger than 1, it is unstable and does not converge. #disagree
I disagree because for it to converge, x has to be larger than 1/e^e and smaller than e-th root of e so for x = e^pi/2; x^x^x^x^x^... does not converge But I think there is a number that does it (but not all numbers are the same there) let a = -1 and b = x^x^x^x^x^... I'm looking for a^b = i so b = 1/2 x^x^x^x^x^... = 1/2 x^1/2 = 1/2 x = 1/4 so (-1)^(1/2)^(1/2)^(1/2)^(1/2)^(1/2)^... = i
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
Im telling you, I think I’ve seen it all and then u go and show me some straight crazy shit.
Yaboylemon lovely comment!!!
@@blackpenredpen well, please do something interesting like this again, i like it!
It can get crazier btw. I go through an infinite 'power tower' of functions in ruclips.net/video/tFcAl40nJeM/видео.html. When you plug blackpenredpen's answer into the general solution you get i (so I don't understand the comments saying he's wrong).
@@jamesexplainsmath - I love the parenthetical part of this comment. Far too often, people assume "I don't understand" == "they are obviously wrong". As Gamefreak129127 said, the symptom that something is wrong is that the limit function f(x) of the sequence of functions f[i+1](x) = x^f[i](x), f[0](x) = x does not converge for e^pi/2. f[0](e^(pi/2)) = e^(pi/2) =~ 4.81, f[1](e^(pi/2)) = e^(pi e^(pi/2)/2) =~ 1912.717, f[2](e^(pi/2)) =~ 6.81 x 10^1304, etc. Another clue that x=e^(pi/2) is not the correct answer is the fact that it is positive; for all real a, b, such that a > 0 and b > 0, the principal value of a^b is a positive real number. Therefore, for any positive real x, f(x) as defined by the sequence above is either a positive real number, or does not exist.
The fact that blackpenredpen derived this nonsensical solution means that his solution methodology was flawed. Gamefreak129127 posits that the problem is that i is not in the range of x^x^x^x^... I suspect this is the case, but I'd like to see a proof.
@@jimschneider799 Proofs of several related theorems were provided by Euler (1783) and Eisenstein (1844). The infinite tetration converges if and only if e^(-e)
Sadly #disagree. The sequence x^x^x^... diverges for x=e^pi/2. The mistake was the substitution of x^x^x^...=i into the equation itself. By doing so we assume that our original claim that x^x^x...=i holds. But this premise is wrong. The whole calculation is therefore only implicative not equivalent. IF there was a solution to x^x^x^...=i then it would be e^pi/2 but as there is no solution, this is false. Or in other words, x=e^pi/2 is not in the domain of the infinite power tower function f(x)=x^x^x^... As we only take real numbers as x, this also doesn't change when assuming a complex number as a result
Gamefreak129127 Convergence has nothing to do with, since infinite tetration for complex numbers is not defined as the limit of a sequence (nor should it be).
@@frimi8593 ...except that there's still that last i at the end. Where did you hide it?
So x to the x to the x to the x... is not equal to the Limit of the sequence where first I got x^x, then x^x^x and so on?
But e^pi/2 is a real number
@@christianvukadin7747 if you say y=x^x^x^..., then you can say y=x^y
And therefore, raising both sides to the 1/y power, you get y^(1/y)=x.
This is the inverse of y=x^(1/x). This is why personally, I believe that defining the infinite power tower as being the inverse of x^(1/x) is the best way to go.
1:43 you would like to solve equation, but I want to say
(-1)^(1/2)^1^1^1^... is simpler at all, right?
R8
Reighr?
@@rafciopranks3570 R-eight ~> Right
right?
That was my first thought too.
That was excellent and I feel stupid that I didn't get the real^real immediately!!
lol thanks, now I’m not embarrassed for feeling like the only one who didn’t find the real^real immediately
It happens lol
I also didn't get it hahaha unu
Same, I neglected to consider fractional exponents.
I was instantly looking for a general solution for some reason
I don't believe you can do that because the function x^x^x^... only converges for a very small range. Specifically, [1/(e^e) , e^(1/e)], corresponding to y values of [1/e , e].
If you did the same trick to find x^x^x^...=4 you would get x=sqrt(2), which is the same value you get if you did the trick for x^x^x^...=2. However, it's only valid for sqrt(2)^sqrt(2)^...=2 because both sqrt(2) and 2 fall in the domain and range for convergence, while 4 is greater than e (approximately 2.718). The graph x^x^x^... can be represented in terms of the Lambert W Function, but that is also equal to the graph for x=y^(1/y) for the specifically domain listed above, as this isn't a function because it would have 2 outputs for certain values. Including x=sqrt(2), where y=2,4.
Regardless, neither e^(π/2) nor i fall within the domain or range respectively, thus making this false.
Saying this would be valid is the same as saying that 1 + 2 + 4 + 8 + ... = -1 because the geometric series 1 + x + x^2 + x^3 + ... = 1/(1-x). This is untrue because this only diverges for (-1 < x < 1).
tldr;
This video seems to be do steps that are mathematically invalid
Flamingpaper
I agree with what you said in the "real world". In fact, just like in the video, I wanted to ask you guys if you guys agree this makes sense in the complex world or not. I actually don't have the answer for it yet.
As far as I’m concerned, only in the set of real numbers is infinite tetration defined via convergence. In the complex numbers, the definition is independent of convergence, and it uses the Lambert W function, similar to what was done in the video. So you do not have much of a point.
Also, in advanced mathematics, the result 1 + 2 + 4 + ••• =
-1 is made rigorous even though the series is divergent. So that is not much of an objection eitrr
Zoltán Kürti I think it’s actually smart ass people who say the series isn’t equal to number. Convergent series aren’t equal to number either, by your argument. Want to know why? Because convergent series aren’t convergent because the infinite sum is well-defined. If you want to tell me that the summation of infinitely many elements is equal to 1, such as in the case of 1/2 + 1/4 + 1/8 + •••, then I will say “bullshit”, because that isn’t addition. Addition is defined as an operator in the extensions of the Peano axioms, and last time I checked, none of the axioms allowed for infinities or limits of any type. You want to say that a series equals to the limit of the sequence of partial sums? Sure. That’s an assigned value. But trying to pretend that this somehow is equal to some summation of some sort is stupid. That is what rigor says. We can say that infinite summation is undefined - because the axioms of summation say it is - but then we can still assign some value to the undefined expression for the sake of convenience. And if we want to do it via convergence, then cool. But pretending this assigned value is an equality in the same sense that 2 + 1 = 3 is an equality is nonsense.
If you can accept the informal argument I presented here, then we have an agreement. But if you want to pretend that my reasoning is false when concerning convergent series yet true when talking about divergent series, then no, we cannot have an agreement, because such inconsistency is due to pure lack of rigor, misuse of terminology, and it goes again basic principles of mathematical logic. This is part of the reason why at some point in time mathematicians pushed fir standard analysis to get replaced by hyperreal numbers analysis (non-standard). I think the fact that someone can say 1 + 2 + 3 is not -1/12 under anything but we can still assign it this value, yet still say 1 + 1/2 + 1/4 + ••• is exactly equal to 2 in the actual sense of equality, that is what I consider smartass, and also dishonest. M
This also ignores the fact that the words “equality” and “assigning a value” really mean the same thing. Equality is a special case of equivalence relation. Equivalence relation are things we assign and define within sets. The definition of equality already varies from going to the natural numbers to the rational numbers. Yet no one ever says “0.999.... = 1 is not an equality in the same sense that 1 + 1 = 2 is an equality”. You bet they aren’t similar types of equality, nor should we pretend they are the same type of equality. But if people are going to pretend they are the same type of inequality, then it’s actually hypocritical to say 1 + 2 + 3 + ••• = -1/12 isn’t an equality in the same sense (it’s an equality in divergent summation theory). I rest my case.
Zoltán Kürti Well then, I suppose we do have an agreement. I simply abhor the mischaracterizations of infinite series that have become so popular as to be impossible to undo.
The conclusion is incorrect. x^x^x^...=y implies that x^y=y, but x^y=y doesn't imply that x^x^x^...=y. Therefore, finding a solution to x^y=y only shows that it is a potential solution to the original equation. In this case, the potential solution e^π/2 is extraneous, and the original equation has no solutions. This is clearly shown by noting that, when you iterate a(n+1)=(e^π/2)^a(n), with a(1)=e^π/2 (a sequence that is always equal to e^π/2↑↑n), the result diverges to infinity as n increases. It doesn't converge to i or even become nonreal at any step.
Using the method in this video without checking whether the potential result actually converges to the right value can prove that 2=4. If x^x^x^...=2, then x^2=2, and we get the solution x=√2, which turns out to be correct. If x^x^x^...=4, we get x^4=4, and so we also get √2 as a solution. In this case though, it is clearly not correct because √2^√2^√2^... can't be equal to both 2 and 4.
#Disagree
Thanks for your well-reasoned response.
Interestingly, you can analitycally continue the infinite power tower to most of the complex plane using the identity z^(z^(z^...)))=W(-ln(z)/-ln(z), where W is Lambert's function. However, being multivalued, I'm nor entirely sure which branch you should use.
@@modestorosado1338 That is easy. The correct one, of course. And with "correct" I mean the one that gives you the desired answer.
Wow, really nice reasoning! 👍
Infinity is a flawed concept
I'm also in the #disagree camp but I think it's pretty easy to fix. After all, e^{i\pi/2} is pretty much equal to e^{i\pi/2+n2\pi}, right? If you get e^{-3\pi/2} instead of e^{\pi/2} then the convergence radius is no longer an issue, is it?
I promised to myself I would go to bed early today.
It's 3 a.m. in my timezone, and I find out real numbers raised to a real number can be an imaginary number.
I guess I'm done
Here I am doing the same thing you did 2 years later
@@jhonnyrock Here I am doing the same thing 2 years later
Isn't this outside the radius of convergence?
10 likes + Heart thanks
One thing you should mention is that i^i has multiple values, as cos and sin both have 2pi as their periods. The X can be e^(pi/2 + 2k×pi) where k is an integer.
Young Lin
Yes I should have done that.
It’s implied that we only use the principal branch, so there really isn’t a need to mention since he isn’t trying to explain how to complex exponentiayion.
A common mistake, when solving equations, is to assume that a solution exists.
The argument in the video is:
if x is a solution to x^x^x^... = i,
then e^pi is a solution to x^x^x^... = i
however (e^pi)^(e^pi)^(e^pi)^... diverges, thus the equation had no solutions to begin with (since a true statement cannot imply a false statement).
Leander Tilsted Kristensen Actually, with a non-truth-preserving set of inference schema, you can in fact imply a false statement from a true statement. Also, to be strictly technical, it was never stated that the solution is in fact e^π.
@@angelmendez-rivera351 you are correct, didn't want to do too many details in a RUclips comment.
The statement was rather:
If x is a solution to x^x^x^...=i
Then x is also a solution to x^i= i
However cocluding the opposite, that a solution to the latter is a solution to the first is a big logical no-no.
In order to prove, that there are no solutions, one should prove, that all solutions to x^i= i are not solutions to x^x^x^...=i .
@@frimi8593 No matter how many times you replace i with (e^π/2)^i, you always have i in both sides of equation. You can push i infinitely far, but you don't get rid of it!
#disagree on the result because a convergence test is needed, but definitely #agree on the fun!
Dani Borrajo Gutiérrez : )
nice tetration, also holy crap you've been uploading a lot recently!! awesome!!
Goku17yen
Thank you Goku! I am on super saiyan mode now too!
just love the new intro man
I'm glad you corrected your "right?" to "isn't it?" I nearly had a heartattack.
This is similar to the approach Numberphile took to 1 + 2 + 3... = -1/12 (which is correct, but the way they showed it wasn't correct)
The series of x^x^x^x... (or x^^infinity) is divergent for any x greater than e, since it approaches infinity, and e^(pi/2) is approximately 5. e^-(pi/2), another "solution", doesn't work either, as for any x from 0 to but not including 1/e, the series converges to 0, and e^-(pi/2) is less than 1/e. #disagreeToMethod
#Agree. The reality is, people keep presenting the straw man argument concerning convergence, but infinite tetration is not defined by the limit of a sequence, not in the complex numbers - because tetration on the complex numbers is a continuation of that of the real numbers, necessarily so as a^^b is only well-defined for a nonzero and b an integer with b > -2. Nor should it actually be defined by such a limit, because doing so represents to many problems, as argued using mathematical logic. (1) Mathematical theories of limits to infinity and about infinity in general in relation to arithmetic and algebra are not effectively axiomatized formal theories. In fact, they cannot even be called axiomatized formal theories, to start with. (2) Standard analysis is a theory about the vector spaces of sequences and functions over the field of real numbers. The limit operator is a linear functional, and the derivative is a linear operator. However, professors and even sometimes mathematicians themselves conflate these with the algebraic-arithmetic operations of multiplication, addition, exponentiation, etc. The best example of this comes from the topic of infinite series. In standard analysis, we would say that 1/2 + 1/4 + 1/8 + ••• = 1, yet this is strictly incorrect, because this equality as proven in standard analysis actually has nothing to do with addition or multiplication, which are group operators defined by the proper extensions of the Peano axioms to the field of real numbers, or the extended real numbers, whatever be the case. This equality is instead a statement about sequences and linear functionals. There exists some linear functional Σ which operates on A = (1/2, 1/4, 1/8, •••) such that Σ[A] = 1, and Σ is defined as (lim ~n -> \infinity\~)[S(n)], where S(n) is the sequence of partial sums of A with respect to the cut-off ñ(n/N) = 1[n, m), where 1 is an indicator function. This composition of linear operators on the sequence is most certainly not a statement about arithmetic. As such, translating this as 1/2 + 1/4 + ••• = 1 is horrendous abuse of notation, one which is accepted even in the most rigorous discussions and works. Calling it a summation is far from true. In the same vein, the claim 1 + 2 + 3 + ••• = -1/12 is not a statement about arithmetic either, and criticisms of this claim come from mathematicians or mathematics students pretending that it is indeed a statement about arithmetic. Furthermore, the arguments that are made in the criticism are non-rigorous and they fail to account for (A) the fact that the status of convergence or not can only be determined after a cut-off over which the partial sums of a sequence one wants to calculate is specified, which is conveniently ignored by everyone in the discussion. (B) a theory about limits to infinity in standard analysis is not meaningful without a rigorous treatment of asymptotic expansions, especially if we work with the non-extended real line, which is usually the case, and which should be the case, since no axioms or semantics are constructed and added to the prior theories to specify otherwise - which is in itself an aberration in mathematical logic. (C) the relationship between the linear functionals over these vector spaces and the group operations of the sequence-generating fields.
(3) Now we see that the problem here is the lack of constancy, and the fact that people want to pretend the all mathematical claims should be centered on standard analysis and that it is the default paradigm on which we talk is abominable. By this same inference scheme, one should never accept the existence of transfinite numbers in any situation EVER unless one explicitly, in exact words, says “Let’s talk transfinite theory”, and similarly, one should never ever state that complex numbers exist either unless a very unique context is specified in a verbatim formula. It is nonsense to ask for this when mathematicians themselves never obey such a rule in their formalities in other domains of discussion. There is no default theory. There is nothing wrong with assigning values with divergent sequences, as we already do it with convergent sequences, even when it is uncalled for.
The statement 1 + 2 + 3 + 4 + ••• = -1/12 should never be taken to be a statement of arithmetic as long as we work with the Peano axioms in the extended real line. Neither should 1/2 + 1/4 + ••• = 1. These are statements about the linear functionals of certain sequences and nothing more. Now, if we want to work with an extended real line or extended complex plane, it must be specified which extension is desired, such as the Riemann sphere or the transreal set, as well as an algebraic structure which includes such extended sets, with a proper specified extension of the Peano axioms for arithmetic. This is the only way in which expressions such as a + b + c + ••• ad infinity and x^x^x••• have meaning without constituting an abuse of notation. Since we assume this the case for this video, we then have it that we should define infinite tetration in terms of arithmetic, algebraic operations, or via functions and formulas, without recur to vector spaces and linear functionals. This is precisely what BPRP introduced in this video, and what he showed is what would be the heuristic consequence of such treatments of tetration. In advanced mathematical fields of study, infinite tetration is well-defined for complex numbers, and it is not via the limit of a sequence, unlike the case with infinite series. Rather, it is defined via the Lambert-W function. By this treatment, what BPRP did is correct. Sure, he never stated the rules here or stated that he was using such unintuitive definition. However, as I said earlier, mathematicians often never specify their axioms or give context in other domains of discourse, and will still discuss results that by Peano axioms or standard analysis is strictly incorrect, and this is done even when there is no background context to be known, almost as if there were telepathy. So, the conclusion is, either BPRP is wrong with what he is doing, but the mathematicians are wrong with how they treat the issue of convergence as if it were the be-all of mathematics... or we can accept what mathematicians do for a living and their little informalities, but then for consistency, we must then be able to accept what BPRP did here - notice that he never even claimed to be a channel of rigorous mathematical demonstrations, and this channel is not meant to substitute a mathematical paper for obvious reasons, so y’all are silly for thinking he is supposed to be rigorous in the first place too.
This is why I must disagree with #Disagree camp. Their notions are simply misguided if we argue using mathematical logic and pragmatism.
Man how much did it take to write that?
@@teraflonik idk
@@teraflonik limit x --> 0 y --> 0 x^y was how long it took me. A.k.a 0^0
I mostly agree, but why is 1/2 + 1/4 + 1/8 + ... = 1 an abuse of notation?
Is it because the Peano axiom's do not define summation for an infinite number of summands?
@@xCorvus7x It is abuse of notation for two reasons: 1. because Peano axioms indeed do not define summation for infinite summands. Now, this by itself shouldn't be a problem, but...
2. the reason we say that series is equal to 1 is not because we extended the definition of summation for infinite summands. IN fact, there is not much summation involved in that proof. Rather, the result is derived from obtaining the limit of sequence of partial sums. And in calculus, it's fine to do that, but that is a statement about sequences, not really about arithmetic summation. And the fact that some try to disguise it as a result of summation is why I consider it to be misleading and bad notation. Because it's not really a summation, it's just an operation we perform on a sequence to get a number, and the operation uses a limit. I don't know if I made myself clear enough on this, but hopefully that answers your question.
All this video shows is that If that infinite power sequence converges then it converges to i. However “if” is the key word - that sequence actually diverges. To see that it diverges you just have to notice that e > 2 and pi/2 > 1. Therefore every member of the power sequence is greater than the previous member by greater amounts so the series diverges toward positive infinity.
This is similar to the when people say that “1+2+3+... v= -1/2”. The number -1/2 is what the sequence would converge to if it converged, but it doesn’t converge to anything, it diverges to infinity.
P.S. The range of convergence of this sequence is (e^-e, e^(1/e)), or about 0.07 to 1.44. E is obviously more than 1.44 so it diverges.
Based on the lambert W function definition, I'd say that [e^(π/2)]^[e^(π/2)]^[e^(π/2)]... is actually -i. i would be the extraneous solution that you would get from assuming that if y=x↑↑∞ then y=x^y kind of like how ∛3^∛3^∛3... is not actually 3.
I think the last one only makes sense if it is the analytical continuation of the x^x^x^x.... function
(real)^(real) = non-real? - Of course! (-1)^½ = i
"Of course, they (real) don't have to be the same..." - But they can be: (-½)^(-½) = (-2)^½ = i√2
(real)^(real)^(real)^...[∞'ly] = non-real? - In light of the first answer - seemingly yes. But, ...
Ah, another commenter has answered this simply: (-1)^½^1^1^1^... = i
Now having watched, you ask:
"x = e^(½π) Agree or disagree?"
Well, you've bypassed an essential thing - establishing that this particular "power tower" (PT) converges. If it doesn't, your result is invalid.
In particular, e^(½π) > 2; and an infinite PT built on 2 (or any real no. > 2) definitely diverges.
So I have to disagree.
Very cute, nevertheless!
Fred
I disagree because if e^(pi/2)^....=i then their absolute value must be equal for instance and for the right side it's 1 and left side infinity. But this counter examples only says that it is impossible but it's not saying where the mistake is....well that's a hard question I'm not able to answer 1)I suppose e^(pi/2) is not allowed in x-convergence values but how to calculate that interval? 2) if e^(pi/2) was allowed then I'm not sure if we can power both sides to the (1/i) and keep having an equality for example ample if -1=-1 and you power both sides to (1/2) the left side could be i and the right either i or -i more far if u start with a wrong assumption 1=-1 raising to power 2 u could end with non true results 1=1 but it looks true at all
but anyhow it's an awesome topic what u showed us thanks
Fred just served you, my dude
@@federicopagano6590 The mistake is this.
We have an expression, in this case, an infinite "power tower," set = to some definite value, C.
In order to evaluate it, we substituted C into the power tower,
That step is valid only if the power tower *has* an actual value; that is, if the finite tower of n terms, converges as n→∞.
In this case, that criterion is not met.
Ergo, you can't make that substitution.
Fred
ffggddss thank you for the answer very clear and I agree with that but how are u sure that this value C is not allowed before calculating this in the first place?
@@federicopagano6590 It's really a _reductio ad absurdam_ argument.
You suppose for the moment, that the substitution *is* valid.
You solve for x.
You find that, for that value of x, the power tower diverges.
You then conclude that the assumption of validity of the substitution is false.
Fred
for the x^x^x^....= i equation, is that actually ok to do in math? I ask because I know that the function f(x) = x^x^x^.... has a domain of e^(-e)
After using the fact that the infinite tetration is equal to -W(-ln(x))/ln(x), if you plug in the value e^(pi/2), you actually get i as presented in the video. And according to Wikipedia, the only restriction on the equation is x=1. I understand why W(-pi/2) = i * pi/2, but why is the only restriction for the infinite tetration equation x = 1 and there isn't more restrictions?
It is not okay to actually do. The value of x^x^x^... when x=e^pi/2 doesn't converge to i. It increases without bound, and it always stays real. Simply finding all the solutions to x^y=y is not a valid way to find the solutions to x^x^x^...=y. You must also check for extraneous solutions.
@@plasmaballin I think I'm gonna have to agree with you
Joseph Noonan I see your point and it makes the most sense, though Wikipedia still says the equality with the Lambert W function and infinite tetration still holds for all x not equal to 1. Either there was a typo in the page or it has a point, but again Wikipedia isnt that reliable of a source lol
I also have seen a college level math research paper saying it is possible for numbers outside of the domain of the infinite tetration to be equal to a complex number or any complex number in some cases, I can post a link to a paper here if you want.
e^(pi/2) is about 4.8. surely a positive real number larger than 1 would keep increasing in magnitude when you raise them to themselves as powers and it would diverge off to infinity,and definitely not decrease the magnitude down to |i| = 1?
i don't want to be on the bottom!
no because that means negative i (edit: typo)
This is crazy..🐒🐒 by the way the Black pen red pen theme in the starting is nice❤️
Any negative real number to an irrational will result in a non real
You're wrong. Any negative real number raises to 1/2n where n a real number will result in a non real number.
DO462 n has to be an integer. If you say n is a real number then I set n =1/2 making it -1^1
-1^1/(2*pi) stills non real
Some counter examples, if you choose n to be the reciprocals of any even integer will result in a real
x=√2 also holds, giving x^x^x...=2
In fact, any x^(1/x)
x^(x^x^x...) = x^x^x...
x^y=y,
x=√2, y=2
x=3^(1/3), y=3
i^(1/i) is just one of the many solutions
e^pi/2=i^(1/i)
Where's "Blackpenredpen yayyyyyy !" ?
Yayyyyy (child noise)
Well, it's an absurd because we can demonstrate that x^x^x^... converges only if 1/(e^e) < x < e^(1/e) but e^(pi/2) is bigger than e^(1/e), so the equation has no solution.
Truly, you make magic with imaginary numbers.
Your second example (infinite power tower) is wrong. You did not consider convergence. The hyperpower function (also known as infinite power tower and infinite tetration) x^x^x^... = k only converges for e^(-e) < x < e^(1/e). The implication here is that you can only apply the "trick" of taking x = k^(1/k) when k < e. So x^x^x^... = 2 implies x = sqrt(2) but x^x^x^... = 3 does *not* imply x = cuberoot(3) since $k = 3 > e$ here. Similarly x^x^x^... = i does not imply x = e^(pi/2).
A requirement for the result to be imaginary is for the base to be negative, so x
If you can find an example for any case of a finite number of indices, the infinite version is true as well as you can just follow it up by raising the finite case to the 1st power infinitely many times.
Is there a video about convergence of the power tower I'd really be interested in that.
Yea this one should help ruclips.net/video/_xxPckmh9Fc/видео.html
@@blackpenredpen Thank you! It was an interesting video, but it is about the limit, not about the convergence. ;-)
Damn it Steve! I'm switching from physics to a math major because of you. Please put a 'proceed with caution ' notice on your videos. Thank you
Avi Sharma
: ))))
You made my day!
@@blackpenredpen you made me take a good decision sir :). Keep making great videos.
Avi Sharma thanks
Don't do it!!
a troll answer for the second part would be to take the first answer and just raise it to the power tower of 1, thus still getting i.
Solution holds theoretically but it is a bit silly to say i^i is a real number(even by Euler’s formula ) because that is an undefined expression, what is an ith power , the limit of e^pi/2 tetrated n times as n goes to infinity is just infinity so essentially we are saying that i or the square root of -1 is infinity
Another way to see why the second equation fails for x=e^(π/2) is define recursively the sequence: x₀=e^(π/2), xₙ₊₁=(e^(π/2))^xₙ.
If it converges, since ℝ is complete it would converges to a real number, not complex. :^)
The imaginary part is by definition (-1)^0.5 which is real to the real, speaking of which, I like to move it move it.
You can only put equation into itself if it has roots, otherwise you will get incorrect
Is the intro a CGI animation? And did you use a deep-learning based sound generator to make the matching sounds?
It's maybe an answer that works under some contexts, but we all know that the sequence of partial exponentiations is divergent.
You keep on blowing my mind in each and every video...I cant be asked...
Elegant simplicity as usual ...
Personally, I disagree.
It is just because by defining a sequence a_i with a_0 = e^(pi/2), a_i = a_{i-1}^a_0.
It is easy to prove that this sequence is divergence.
Note that this issue is similar to a trick that sum of all integer = -1/12. You cannot substitute a value into a series causing its divergency.
that "trick" about -1/12 is not a trick, its only because those bozos at numberphile made it sound mysterious and a joke. Its a well known fact in complex analysis and by methods of analytic continuation that the riemann zeta function evaluated at -1 which is the same as saying all the natural numbers added together does equal -1/12. But only in the complex plane not the real plane. That is where the confusion is you have i believe.
You are assuming that divergent expressions have no value by virtue of being divergent. This a misnomer. Divergence is a claim about sequences and the limit operator, not a statement about algebraic operations of the real numbers with themselves, which is what iterated exponentiation is.
@@angelmendez-rivera351 you legit replied to every comment I found lol
I know another way to do the real power tower. Take the tower as x^y^t^t^t^t...... let x:-1. Let y= 1/2. And finally, t=1. I mean, no one said it coudlnt be anything in the powers. Just put a single 1 anywhere to reset the value and get a finite tower
Using your first example, substitute into the second:
(-1)^(1/2)^real^real... replace all real with 1.
Since 1^1^1... =1, therefore the second eqtn = i
#disagree e^(pi/2) > 1, it doesn't converge when raised to itself infinitely many times.
It is true that it diverges in the real number plane, however in the complex plane it converges, which is how it equals to 'i'. its a similar case to the summation of (1/n^-1) to infinity. Obviously that would be a divergent series if we limited our scope to the reals only. But in the complex plane it does converge to a value of -1/12 or -1/12 + 0i to be exact, the infamous number no one understand XD
Therefore #agree
Well it doesn't converge for any real value because e^(π/2) > e^(1/e), but that doesn't necessarily hold for complex values :)
@Cameron Maloney yea lol, i typed it that way cause that is actually the reimann zeta function evaluated at -1.
Ah ok. I don't know anything about complex analysis yet.
Can't you take X to be for example (-1/2)?
then x^x = (-1/2)^(-1/2) = 1/sqrt(-2) = imaginary
then if you keep raising it to the -1/2 power, you just keep on taking square roots, so it stays an imaginary number right?
Could be
That's a good idea, but if you try it in a calculator it keeps bouncing between a few different values and doesn't approach anything.
@@ShanePenick When done in a calculator, e^pi/2 raised to itself repeatedly doesn't converge either.
You're too generous on new intros last time... good job! Beautiful intro! I love 3D animations!
In the second example you raised e^(pi/2) to itself, but in the first example you did not rais the real number to itself, you raised it to a different real number 1/2.
These kind of videos are the best!!!
We can use e^(i*pi/2) = i to construct this infinite power tower. Sure, i has to be at the zenith of the power tower, but since it can be constructed infinitely high, so well...
#agree as u have taken a infinite tetration function,the domain of it is (1/e,e).as here x is e^π/2 the x value is out of domain and definitely we will have a complex answer.
I believe it's been pointed out that this doesn't work due to e^π/2 being outside the convergence range for infinite tetration, but there is another not entirely trivial solution that does work:
Instead of assuming x^x^x^x^... = i, we only want to know if we're using an infinite power tower with entirely real numbers, so they need not all be the same, so let's instead assume:
y^x^x^x^x^... = i
Now, we can look at the equation we already know works: (-1)^1/2 = i, and we can set the values of some of these variables:
y = -1
x^x^x^x^... = 1/2
Now, we still need to solve that last one, but we can use the same method as in the video for that:
x^(x^x^x^...) = 1/2
x^1/2 = 1/2
(x^1/2)^2 = (1/2)^2
x = 1/4
Unlike in the video, 1/4 IS within the range e^−e ≤ 1/4 ≤ e^1/e, so we have a valid, convergent, non-trivial solution which is:
(-1)^(1/4)^(1/4)^(1/4)^... = i
I think that there’s something incorrect with the e^(pi/2) answer because it’s approximately 4.81 and should diverge, but I’m not sure why yet.
The real question is how to do it when the reals are all the same. The first one is easy, the other one idk:
(-1/2)^(-1/2)=
1/((-1/2)^(1/2))=
1/((-1)^(1/2)*(1/2)^(1/2))=
1/(i*sqrt(1/2))=
1/(i*sqrt(2)/2)=
-2i/sqrt(2)=
-sqrt(2)i (definitely non-real)
One of the solutions is -1 - i, if the power tower has has height % 4 = 2. Or -1 + i for height % 4 = 0. For uneven heights, it gets weird. So I think there’s no fixed solution possible. Clearly the given solution is wrong, but that’s already explained by others in comments.
-1^1=e^pi i? I’m only in precalc (junior) but I love math and I’ve always kind of felt like it’s my calling. That being said I’m not the first person you should ask for this, but is this right? Or is it no longer imaginary because it’s transcendental?!?
Can somebody explain how a number that approaches infinity becomes finite all of a sudden
-1/12. We live in a weird world.
Infinity is a flawed concept. Gives you all kind of contratictions
HYEOL They are not contradictions. Intuition does not define truth.
The limit of a sequence need not have the properties of any element in the sequence itself. Take irrational numbers as an example. They are limits of sequences where every member is strictly a rational number.
Also, infinite tetration is not defined by limits of sequences in the generalization to complex numbers.
@@easymathematik I understand the answers and can prove them and everything -- but it's still weird.
And -1/12 is a very similar case. For any finite amount amount of iterations, it explodes. But for an infinite number of iterations, it is seemingly paradoxically defined.
2:20
there is a solution
of x^i = i
x = e^(π/2 + 2 * π * n) where n is the set of all integers
checked with wolframalpha
#Disagree
Search ''mathematical fallacy'' on wikipedia! After 2:30 you can't take that power on both sides...
Limax Chauhan There is no mathematical fallacy being committed here. You are allowed to evaluate a root operation on both sides of the equation at any given point so long as it is well-defined.
@@angelmendez-rivera351 for complex numbers,these rules are not allowed! Check on wikipedia, Google >> mathematical fallacy (Wikipedia) >> power and root >> complex exponent
Limax Chauhan I know the rules of exponentiation of complex numbers. They are allowed.
Angel Mendez-Rivera yes it is allowed you are right. The Wikipedia page is not saying that you can't raise to an imaginary power, it's saying that you can't compare two exponential function in the set of complex numbers as the exponential function is not bijective in the complex numbers. So yes, since that has nothing to do with that, then it is allowed. And I also have studied this too and trust me it's allowed.
The problem with the power tower method is that you need to assume that there is actually a solution. If there isn't a solution, you'll get an answer but it'll be invalid.
For example, let's say we want to find when the infinite tower = 2.
X^X^X^X... = 2
By power tower trick
X^2 = 2
Square root both sides
X = SQRT(2)
So this implies that sqrt(2)^sqrt(2)^sqrt(2)... is = 2
In fact, this is a true solution. But now let's try 4 instead...
x^x^x^x... = 4
By infinite power tower trick
x^4 = 4
4th root both sides
x = sqrt(2)
Which implies that sqrt(2)^sqrt(2)^sqrt(2)... = 4.
But from the first equation,
sqrt(2)^sqrt(2)^sqrt(2)... = 2.
2=4??????
Clearly there's a paradox here, and the cause of this is a lack of an existing solution for an infinite tower of "X"s equaling 4.
If there is in fact no solution for an infinite tower of real "X"s equaling i, then the math breaks down, since you might as well start your math problem off saying that 1 = 0. Any conclusions drawn off of an invalid statement are probably invalid. For example:
1 = 0
Add one to both sides
2 = 1
The pope and I are two men
2=1, so
The pope and I are one man
I am the pope.
😎
This makes absolutely no sense to me but I have no choice but to believe it
Thank you! Always very interesting! I think if you can make your camera so it doesn't change the background light automatically that would help a lot. The video is too dark.
This is flawed unfortunately... if we have sequence x_0 = e^pi/2 with x_n = (e^pi/2)^x_(n-1) then this sequence diverges. It diverges if you start with x_0 = any real number in fact. It converges to i as stated in the video if and only if you start with x_0 = i. Otherwise it diverges
e^\pi ,as well as the e^{- \frac{\pi}{2}}, is a transcendental number, is that relevant to the second problem?
Okay but for the second one we didn't even made sense out of the equation we didnt' see if it was convergent, what i think you did was : if it converges toward i then, it is e^whatever, but it might not even converge
idk tho im still a student
Dont see anyone saying how good your intro is
The last one is divergent, and thats why you get the paradoxical positive ^ positive ^....=imaginary, which shouldnt be possible to happen
The logarithm is not well defined in C until you pick a branch. The expression i^i generally has no meaning
He did pick a branch. He also said that he would be using the principle value.
2:44 if both sides are multiplied by an additional i^2 you get i^i? doesn't that just mean neither option is correct
This is only true if the sequence converges.
Edit: well, it looks like just about everyone is pointing that out...
You could probably extend the function, but then you're no longer dealing with an infinite sequence outside its radius of convergence
Coolest intro man!
Bhargav Modak
Thanks
I MAY be wrong, but I don’t think this is correct. You have to show that your limit is a complex number before you can manipulate it the way you do in this video. It seems to be sort of like the trick where the sum of all positive integers is -1/8...it just doesn’t work.
Edit: I like your videos, but this result is incorrect. Using this line of reasoning, we also have (x^x)^i = i. Then we have x^(ix) = i => x^x = i^(-i) = e^(pi/2). But you've already said that x = e^(pi/2). Therefore, (e^pi/2)^(e^pi/2) = e^pi/2. Take ln of both sides and cancel the factor of pi/2 and we show that e^(pi/2) = 1. This is obviously a contradiction...so the answer can't be correct.
This sounds wrong, since looking at the series a_n=(x^x^... n times) - every element is real (using your "solution"), so it sounds unreasonable that the limit would be imaginary. Still - infinity is weird, so this isn't a "proof" of wrong-doing.
Instead, I want to point out that the function f(z)=z^(1/i) is **not** injective, and so saying "(x^i)^(1/i)=y therefore x=y" is patently false. It would be the same as saying "(-1)^2=1^2 therefore -1=1".
In other words - you have proven **some** solutions to the equation x^i=i satisfy the relation x^x^x^...=i, but that doesn't necessarily mean all of them do, nor does your specific chosen one.
Creo que la asunción que haces en el minuto 2:10 es dudosa o incorrecta.
I think the answer is wrong because the function of x^x^x^x^x^... stop being useful from somewhere (the values does not converge), and I'm quite sure that somewhere is less then e^1.57
I'm pretty sure it can't be correct, but I'm having hard time proving it : ( I'm not good at math...
e^(π/2) is a positive real number greater than 1. making a power operation (with exponent, again, positive greater than 1) gives a bigger positive real number. repeating the operation will always increase the result... I don't see how it can converge to anything. And if it does, then why it doesnt for any positive real number? why 2^2^2^2^... doesnt converge also? What the product (or also the sum) of "n^n" converge to, for n integer greater than 1? Could it be that the "trick" is in applying substitutions only for principal values? Or, does the power operation with an imaginary exponent (you use "i" and "1/i" in this case) follow the same rules as with a real exponent? Again, I'm too noob to spot the error, but I have the feeling there must be one... @blackpenredpen PLEASE DUDE, enlighten us!
Ps: I dont even know how to "tag" you (T__T)
x^x^x^... -> Infinity which is simply a NaN! Try computing it with any supercomputer, it will still produce a NaN even if the supercomputer was the size of the entire cosmos raised to the value of mole! You can not calculate any infinite series, but you can calculate their limits, their domains, and their ranges!
Isn't (a^b)^c only valid for complex a and b when c is an integer? So when you write (e^-ipi/2)^i = e^ipi that's not true.
analytic continuation strikes again.
The (-1)^(1/2)^1^1^1^... is certainly a nice and simple solution, but I can suggest another one:
The solution to the power tower equation x^x^x^... = 1/2 is x = 1/4, so the equation (-1)^(1/4)^(1/4)^(1/4)^... must be i.
e^(pi/2) is positive so there is no way a positive infinite power tower can give a complex number, also solving the x^x^x^... = i by doing x^i=i doesn't work because the infinite power tower diverges at some point and there would be no solution to this equation. Indeed a power tower of e^(pi/2) diverges. As an example of an infinite power tower diverging. Trying to solve x^x^x^... = 4 using the substitution method x^4 = 4 gives you sqrt(2) as a result; however, there is in fact no solution for this equation, and an infinite power tower of sqrt(2) actually converges to 2
2^(1/2)^(1/3)^(1/4)^(1/5)^..... = 1.5 ¿?
It´s strange because when is (1/pair) is 1.6...... but when is not pair is 1.5.... its like is converging from de rigth and left at same time.
I put in Wolfram (1/n)^(1/n+2) to see what happen to the last numbers of the series, looks OK but when i put (1/n)^(1/(n+1))^(1/(n+2)) looks like e^(-x) Sin(x)
Looks like es equal to 1.6
alright i call shenanigans on this 😂
Is e^(pi/2) actually a real number? I'm not a mathematician but I thought e and pi were different categories of number outside of the set of real numbers (unless raising e to pi makes it real?)
e and pi are real numbers.
All this proves is that, if x ^ ... = i for some x, then x is necessarily of the form e ^ (pi/2 - 2 pi n) for integer n. This is not a sufficient proof that (e ^ (pi/2)) ^ ... = i, or even that (e ^ (pi/2 + 2 pi n)) ^ ... = i for any n. Indeed, it seems all such x are outside the known domain of convergence e^(-e)
Wow what is with this awesome new intro!?
If you consider -1 as the first base then 1/2 as the first exponent then have all other exponents in the sequence be 1, you will indeed get an infinite sequence of real values that have a non-real value. However, the result of e^(pi/2) repeatedly exponentiated will simply grow to infinity. #disagree. The error, I believe was in the assumption that there exists a single real number that when repeatedly exponentiated will converge to i. I don't think that's the case: a value < -1 will diverge. A value in the range (-1, 0) will converge to 1 (This needs a proof, of course), values in the range [0, 1) will converge to 0, 1 will converge to 1, and values greater than 1 will diverge.
I miss the piano music in the background tbh
i want to ask something , does i (the imaginary number) in euler's formula indicate the quadrant position of the trig ? since i and -i located in the vertical axis of complex plane , then 1 and -1 in the horizontal axis
Do we get a follow up video where you explain why we can't do the second one?
I am on the #disagree side, too. With the same method, if I want to demonstrate than the tetration x^x^x^x^x^... is equal to a negative number, I can say : "try x^x^x^x^... = -5".
So x^(-5)=-5 => 1/x = 5thRoot(-5) => x = -1/(5thRoot 5).
Because we are not inside the interval of convergence of the infinite tetration (seems around [0, 1,4446...], but why ???)
But why the bad commentaries ? Steve only wants us to think !
CORRECTIONS:
3:50
(e^(pi/2))^(e^(pi/2))^(e^(pi/2))^... is not equal to i. We then have to plug in i into the derivative.
(pi/2)*(e^(pi/2))^i=i*pi/2, |i*pi/2|=pi/2>1
Since the absolute value is bigger than 1, it is unstable and does not converge. #disagree
I disagree because for it to converge, x has to be larger than 1/e^e and smaller than e-th root of e
so for x = e^pi/2; x^x^x^x^x^... does not converge
But I think there is a number that does it (but not all numbers are the same there)
let a = -1 and b = x^x^x^x^x^...
I'm looking for a^b = i
so b = 1/2
x^x^x^x^x^... = 1/2
x^1/2 = 1/2
x = 1/4
so (-1)^(1/2)^(1/2)^(1/2)^(1/2)^(1/2)^... = i
x^x^x^x^x^... only converges for x
Ramanujan Summation: 1+2+3+... = -1/12
if x=e^(π/2), x^x^x^...^x = i
So, should this be called Ramanujan power tower?
What does e^(pi/2)^(e^(pi/2)^..... converge to? if it is infinity then don't you just have (e^(pi/2))^infinity?
Steve, para la ecuación
y = x tan(x), existe un análogo a la Lambert W function?
?
@@blackpenredpen Es posible definir una función T tal que T(xtanx)=x, análogo a W(xe^x)=x?