Cool Math: The Lambert W Function and Infinite Tetration 

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Manderbolt?

It isn’t clickbait. That’s what the function looks like in the complex plane when color-coded by escape

​@november666 it is clickbait because he didn't show that

Hey get used to IFS

Huntress Archer That's a good observation, and since we know e^(inf. tetration of 1/e) and 1/e^(inf. tetration of 1/e) converge, we're ultimately dealing with finite quantities. But suppose you apply this argument to 2^(inf tetration of 2) * (1/2)^(inf tetration of 2)? Although at each finite truncation of the tetration, you get the product of reciprocals, your argument might require that we say that (2^2^2^2...) + -(2^2^2^2^2...) = 0, both terms diverging, unlike (1/e ^ 1/e ^ 1/e ...). Perhaps introducing a limit would strengthen your argument, since that's normally how this sort of thing would be treated in calculus.

ohhh wow you're right, i thought this last statement was totally crazy and hard to proofe and you just explained it with a simple addition theorem

+Jojo160992 Welcome! If you have any video requests or areas of math you find interesting, let me know. With regards to you question, one other commenter asked the same question. Here was my answer and the paper I drew this material from is at the bottom: If we consider the positive numbers, there are regions of (0, inf) to deal with: 1) (0,e^-e) 2) [e^-e, 1) 3) [1,e^(1/e)] 4) (e^(1/e), inf) It is easiest to explain the convergence in interval 3 and divergence in interval 4. In interval 4, the argument is as follows: If infinite tetration on some x in (e^(1/e), inf) converged to some c, then the following equation would hold: x^c = c. This can be rewritten as: x = c^(1/c) However, if you graph x as a function of c, you will find that the maximum of this graph occurs at c = e; x = e^(1/e), which can be shown using calculus on the function x= c^(1/c). This means that for any conceivable c, x cannot be greater than e^(1/e). Next, the convergence in [1,e^(1/e)] is explained by the observation that if you pick some number, x, in this interval, you will produce a monotonically increasing sequence which can be shown to be bounded above by e. That would show the convergence of that interval. In fact, if you iterate e^(1/e), it will converge to e and anything less than e^(1/e) will converge to something less than e. Showing divergence in interval 1 and convergence in interval 2 are a bit more involved and I would have to defer you to the following paper for a more detailed discussion: www.maa.org/sites/default/files/pdf/upload_library/22/Chauvenet/Knoebelchv.pdf Essentially what happens is that in interval 1, there are two subsequences that converge to two different values and in interval 2, the iteration oscillates but eventually converges.

+randomjunkohyeah1 I like this way of solving the problem and I agree it's more direct than involving the Lambert W function. It also makes the identity between Ssqrt and W obvious: exp(W(ln(x))= ssqrt(x). This also means that the lambert W function can be defined in terms of the square super root just by rearranging that identity. I think this is nice because it would probably be easier to explain to someone what a square super root is than the Lambert W.

+mdphdguy1 Thanks man. I'm sure there's something unique about Lambert W that justifies it's existence (haven't read up on it much), but here I don't see a good reason to use it.

+randomjunkohyeah1 Yeah in this application, the super square root seems more natural, but there are cases in differential equations where the W function is more natural. My personal experience with this sort of thing was with the Michaelis-Menten differential equation, which is used to describe how quickly an enzyme converts one chemical to another. If you try to solve the differential equation you need to invert something of the form x*e^x if I remember correctly, so it's more natural than inverting something of the form x^x. There isn't too much written on this function compared to other math topics, which is a good and bad thing; bad because you have to struggle to learn about it, but good because you can discover new things about it on your own.

Did you invent that symbol for the super square root, or copy it from the "Notation, Nonstandard" chapter of _Mathematical Cranks_ by Underwood Dudley?

@@tomkerruish2982 Holy mother of necropost, Batman! …uhhhh, I came up with it myself as a high schooler, lol. Have never heard of this book before. I was able to find a pdf of everything up to the table of contents and _wow_ is it out of date… right in the introduction it blanket dismisses anyone claiming to have proven Fermat’s Last Theorem! Anyway, off to have a minor personal crisis thinking about how different the version of me that wrote this comment is from the version of me that exists today…

Well maybe he saw my video, or we're both drawing from standard examples, or similar sources.

They are all valid and will give the same result. The first one is just easier to deal with.

Yes.

Except that it doesn't converge

lets try i... 0.4382885693+0.3605881546i lets try -1... Its just -1. lets try -sqrt(2)... Surprisingly, -sqrt(2). lets try -2... -0.1271591767-0.01057215205i. actually, every negative number works. lets try 0 (or rather a number infinitesimally close to O (the number needs to be negative though).) Approaches 1. Cool fact! If you plot in something like -0.000000001 (lets call that n), the real part approaches 1 and the imaginary part approaches -pi/(1/x) . Thats really cool. If you wanna know how i calculated this, i just wrote a simple program in casio-BASIC to do the infinite tetration for me. Hope i could help ya out a bit!

clickbait

+Brett Eskrigge Sorry for my delay in responding. It's been a while since I've thought about the reasons for this and I wanted to give a decent response to your question: If we consider the positive numbers, there are regions of (0, inf) to deal with: 1) (0,e^-e) 2) [e^-e, 1) 3) [1,e^(1/e)] 4) (e^(1/e), inf) It is easiest to explain the convergence in interval 3 and divergence in interval 4. In interval 4, the argument is as follows: If infinite tetration on some x in (e^(1/e), inf) converged to some c, then the following equation would hold: x^c = c. This can be rewritten as: x = c^(1/c) However, if you graph x as a function of c, you will find that the maximum of this graph occurs at c = e; x = e^(1/e), which can be shown using calculus on the function x= c^(1/c). This means that for any conceivable c, x cannot be greater than e^(1/e). Next, the convergence in [1,e^(1/e)] is explained by the observation that if you pick some number, x, in this interval, you will produce a monotonically increasing sequence which can be shown to be bounded above by e. That would show the convergence of that interval. In fact, if you iterate e^(1/e), it will converge to e and anything less than e^(1/e) will converge to something less than e. Showing divergence in interval 1 and convergence in interval 2 are a bit more involved and I would have to defer you to the following paper for a more detailed discussion: www.maa.org/sites/default/files/pdf/upload_library/22/Chauvenet/Knoebelchv.pdf Essentially what happens is that in interval 1, there are two subsequences that converge to two different values and in interval 2, the iteration oscillates but eventually converges.

+mdphdguy1 I had a feeling that the limit of e^(1/e) was derived through calculus. I will give that paper a thorough reading, thank you

+sabhrant sachan I don't even know if you could as it's not within the limits. But I too would be interested to see if it's possible

+sabhrant sachan Sounds like an interesting idea and I think I'll steal your idea and make a video on it (unless someone beats me to it). I've only thought about this question a little but to give you some hints on thinking about this sort of thing, first ask how one calculates something like i^i. We can always change bases to rewrite i^i as e^(ln(i)*i) just like we can change bases for real-valued power functions, like changing 2^x to e^(ln(2)*x). Now that we have i^i=e^(ln(i)*i), what should ln(i) be equal to? After all, usually one only talks about taking logarithms of positive numbers. We now remember that ln(x) is asking e to what power gives me x? So, the answer to ln(i) should be the answer to the question: e to what power gives me i? I won't spoil the answer for you but if you have some experience with Euler's formula, you'll be able to figure it out. I think such a computational procedure can make some sense out of calculating complex numbers to complex number powers. Now, getting back to tetration, we could ask for what complex numbers, z, the quantity z^z^z^... converges. I'm not sure if the answer is as simple as that small interval for real numbers and I wouldn't be surprised if the region over which z^z^z^... converges has a fractal structure or some other complicated pattern near the boundary of converging and diverging. To give a simple example, i^i^i^... seems to converge if you try the calculation and if you extend the lambert W to complex values, if i^i^i... converges, it converges to about 0.4383+0.3606i, so we do appear to have at least one example where the infinite tetration of a complex number converges.

Evaluate 0.001^0.00105. I think the result will show you that the sequence didn't converge.

Or a negative base.

Typo.

@@tomkerruish2982 thanks

I am using x**x to represent exponentiation, x^x

It's not necessary to be a mathematician to either know math or to do well, even if it were true that it requires an IQ above 150. Just always work at it, always be reading and learning something. God bless.

@@Math_oma God bless you!

@@Math_oma btw evolution happened to be a big problem of mine for several reasons: - I believed in irreducible complexity. - The person that helped me convert was a creationist and a well spoken one. - life seems impossible to have originated on its own. - I needed absolute proofs of God. - It did, and still does, pose a problem in terms of evil and corruption (sickness, diseas, etc) How can God create a perfect world where disease, evil murder exist since the begin. These are some and I articulated them in a very vague way, but they give you the idea of what I mean. I've been really struggling beacuse of various reasons, but I also have some strong reasons to believe: ultimate proofs that the mind is not derived from the brain, the universe can only be originated from God and it seems very unlikely that abiogenesis/evolution could have happened the way atheists mean it.