Can You Solve x^x^x^... = 2? Infinite Exponent Tower Trick

What is the value of this infinite exponent tower?

See my follow-up video about a potential paradox 4 = 2:

I was asked this problem in a high school math competition, and you had to solve it as fast as possible. Hence I made a short 1 minute video showing this trick. For those wanting a proof suitable for a graduate level real analysis course (which is the only way to rigorously approach this problem), here are references.

Award winning paper Knoebel, R. Arthur. "Exponentials reiterated." The American Mathematical Monthly 88.4 (1981): 235-252.

Direct link to pdf:

Wikipedia tetration

False proof 2 equals 4

Math StackExchange links

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[MindYourDecisions]

See my follow-up video about a potential paradox 4 = 2:

I was asked this problem in a high school math competition, and you had to solve it as fast as possible. Hence I made a short 1 minute video showing this trick. For those wanting a proof suitable for a graduate level real analysis course (which is the only way to rigorously approach this problem), here are references.

Award winning paper Knoebel, R. Arthur. "Exponentials reiterated." The American Mathematical Monthly 88.4 (1981): 235-252.

Direct link to pdf:

Wikipedia tetration

False proof 2 equals 4

Math StackExchange links

[gratefuldude]

Took me less than one second.

Q. Can you solve

A. No.

[guilemaigre]

i feel cheated... I really was expecting more.

[paulchapman]

Here's the clincher.

Suppose instead we solve for x in x^x^x^... = 4?

By your method: x^x^x^... = x^(x^x^x^...) = x^4 = 4. So x = sqrt(2).

By your original result, when x = sqrt(2), x^x^x^... = 2. Therefore 4 = 2. QED.

[timothyswan]

You proved that if it were a unique real, it is root 2. You haven't proven it exists, though.

[JiveDadson]

This only proves that sqrt(2) is a solution IF the exponential series converges. Still to do: Prove that it does converge for sqrt 2, and that sqrt 2 is the only solution.

[bowtangey]

Replace the 2 with 4, and you can "show" that repeatedly exponentiating root 2 yields 4 as well. This emphatically shows that convergence must be considered for a complete solution.

[zhenrad]

Here is the same solution, explained in more detail:

x^x^x^... = 2
log_x(x^x^x^...) = log_x(2)
x^x^x^... = log_x(2)
2 = log_x(2)
x^2 = 2
therefore, x = sqrt(2)

[OkenAndMosh]

For those of you who are interested in the function f(x)=x^x^x...etc. It is only convergent in the values: e^(-e)<x<e^(1/e), so x must be between about 0.066 and 1.44. This is why root2 is a possible answer, being about 1.41. The range of the function is e^-1<f(x)<e. So 2 is within this range of about 0.37 and 2.71.

[Europa_Forever]

wtf, these videos are too short, you must explain your results sir.

[joaquito]

I put this into a Python program, and after 1000 x^x^x^x^... It gives me so, it's true xd

[SmarticlesTheGoof]

For people who are trying to do this with a calculator:

The convention for power towers is to evaluate them top to bottom. Therefore, (2^3)^4 = 8^4 is VERY different from 2^3^4 = 2^(3^4) = 2^81. If it doesn't seem to be converging, that's usually the reason. You're entering it in so that it evaluates from bottom to top rather than top to bottom.

[danjbundrick]

What the heck? Did you have a plane to catch or something? This doesn't help at all!

[praveenkumars]

Disappointed. You didn't ask "Did you figure this out?" 😂😂😂

[wixom]

Maybe this will help a few people "see" the answer is correct, and that it does converge to 2, albeit slowly:
start with x^x = sqrt2^sqrt2 = 1.6325 (i'll just round to 4 digits)
now x^x^x = sqrt2^sqrt2^sqrt2 = sqrt2^(1.6325) = 1.7608
now x^x^x^x = sqrt2^sqrt2^sqrt2^sqrt2 = sqrt2^(1.7608) = 1.8409
now x^x^x^x^x = sqrt2^(1.8409) = 1.8927
and so on. For every line you add, u simply use the answer from the previous line as your exponent on sqrt2. That is, if you continue the pattern, each additional line is sqrt2^(answer from previous line). A spreadsheet could run this out a few thousand lines pretty quickly, and what you will see is that it does indeed approach 2, asymptotically.

[graboy]

Can also be done by taking log_x of both sides, then using change of base.

[Dartnix]

I checked the answer on a calculator adding another exponent to the leaning tower and that's actually right! *MIND BLOWN* I was expecting anything greater than 1 to give infinity as a result but after 25 stacked exponents, the value 1.99993 and the slower and slower rate of ascent makes it clear that the upper bound is 2. Great video, I learned something new.

[flexico]

Here I though it was going to be some massively deep fundamental truth and yet it was so short. XD

[thedocotrL]

I just experimented with Matlab, including making plots of convergence or lack thereof. It appears that x = nth root of n over the range (approximate) from 0.37 to 2.71. I have no idea why. I simply set up the program and tried a range of numbers from 1 to 100. Since 1 and 2 "worked'" I started trying values between 1 and 2 and then just outside this range. I was going as high as 1950 levels of powers to make sure convergence wasn't just being slow. Below 0.37 the results showed oscillations. Above 2.71 the final value converged to a value less than n. This was already noted in the comments for n = 4, which converges to 2.

Edit: I just saw this:
Down below Attila Losonczi states that convergence is only between 1/e and e, that would be 0.3678... and 2.71828... Those are very close to the values I found empirically. (I had a typo of 2.17 that I corrected to 2.71). If Attila Losonczi or someone else could point all of us to a proof, that would be great.

[RickyRoro]

This is the same fallacy as is used with the "-1/12 infinite series sum."
You cannot arbitrarily manipulate an infinite series. If you want to correctly use algebra with infinite series, then you need to use Calculus.