solving the tetration equation x^x^x=2 by using Newton's Method

Here's an interesting tetration equation x^x^x=2 that looks solvable with the Lambert W function but is not.

#calculus #blackpenredpen #math

[OonHan]

But what is the exact value? Time for the *LAMBERT W FUNCTION!!!*
...




?

[shobhitsinha]

The approximation reminds me of an old joke.
An engineer and a mathematician are racing towards an attractive girl who is 100m away.
In their first step, they travel 50m i.e. half the initial distance.
The constraint is that the distance they can travel in successive time-steps should be half of the previous time step.
 The mathematician gives up, realizing it as an infinite geometric series sum.
 The engineer continues to move for he is confident that he will soon get close enough for all practical purposes.

[clubstepdj]

2 is one less than π
-Blackpenredpen, 2018

[semiawesomatic]

Guys, he definitely memorized all of those approximations. He was just looking at the camera because he wanted to make eye contact with his audience.

[neilgerace]

I think it's fascinating that a simple looking function has such a long expression for the derivative.

[Bodyknock]

I seem to recall you did a video that √2 ^ √2 ^ √2 ^ √2 ... = 2. If so then it would make sense that √2 ^ √2 ^ √2 is likewise going to be fairly close to 2 so x will be near √2.

[wolframalpha]

Damn! When did I start developing a craze for math...I just cannot miss a day without watching your videos :)

[light]

1:24 most glorious moment in the history of blackredpen

[fluffymassacre]

Aproximations what are we engineers????

[mikety]

I agree with where another poster's suggestion below was going. Since tetration is to be considered in some sense a natural extension of the arithmetical operations, then so its inverse operations should be considered as such, too. In particular, for the nth power tower x^^n (or ^n x), we should also have that its inverse, the nth tower root, say twrt(n, x), should be a valid special function alongside of it. Moreover, the square tower root can effectively replace the Lambert W function altogether, since they are exactly equivalent: if

x^x = a

then

x = e^W(ln(a))

i.e.

twrt(2, a) = e^W(ln(a))

and thus

W(x) = ln(twrt(2, e^x)).

So W(x) proves unnecessary/redundant once twrt(n, x) - nth tower root - is accepted, and twrt(n, x) gives you more solving power than old W. The interesting question is if there are any more interesting, non-trivial equations that can be solved with higher-order tower roots like twrt(3, a) than just x^x^x = a and suitably increasing extensions.

[GreenMeansGOF]

For all the people who want an exact form, here you go. However, we must ask: "At what point does defining these productlog type functions become arbitrary?"


Let f: R to R, f(x) = x*e^(x*e^x). Define V(x) to be the inverse of f(x).


x^(x^x) = 2
ln(x)*x^x = ln(2)
ln(x)*e^(ln(x)*e^ln(x)) = ln(2)

ln(x) = V(ln(2))
x = e^V(ln(2)) = 1.47668...

[blue_blue-]

I remember my teacher saying that there are a lot of problems which you can‘t e.g. integrate exactly. You have to apply numeric approximations. But thanks for this video.
Bprp (almost) integrates everything.

[XAE-ycrr]

I am not sure if someone else in the comments has the solution using Lambert's W-function, but here is mine:

Given: x^x^x=2.

Raise both sides to the power of 1/x:

Apply the natural log to both sides and execute log properties:

Multiply both sides by x: (x^2)*ln(x)=ln(2)

Replace x inside the log by (x^2)^(1/2), then execute its properties:

Multiply by 2 and use log property:

Replace x^2 with e^(ln(x^2)): ln(x^2)*e^(ln(x^2))=ln(4)

Use W-function:

Use "e" as a base on both sides and solve for x:

Thus: (+/-)sqrt(2)

[gloystar]

Excellent video that demonstrates how useful Newton's method is for approximating solutions very accurately.

[kingbeauregard]

Many years ago I came up with a numerical method for finding extrema: take the values of a function at x, x-dx, and x+dx, and figure out where the extremum of the corresponding parabola is. I never found any actual use for it, but it was pretty cool.

[awertyuiop]

What if you take the limit, as n goes to infinity, of X(n)?
Is it even possible to get a limit at all?

[filip-kochan]

“That is two - so just one less than pi” xd xd

[bobdole]

You could just cheat and and say x = super-cubed-root of 2 and leave that as an exact answer!

[grzegorzkosim]

Question fo BPRP: how long 1/2in garden hose one can fit on a 30 by 30in table, if the hose is rolled into spiral (single layer)?

[crazyjee]

can x^(x^x) be actually solved using the lambert function??
Or lambert function doesnt apply to this>??
PLease tell me..otherwise I may not be able to sleep at night