Solving an Exponential Tower Equation

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[kobethebeefinmathworld]

6:30 Rigorously, the infinite height would have no real solution (suppose we set our domain and codomain be subsets of the real numbers) as the range of the infinite height is [1/e, e]. In other words, if we substitute x = 5^{1/5} back into the infinite height and take the limit, we will not be able to get back 5 as the answer.

[alextang]

The 2nd method is awesome! I like it! 🤗🤗🤗🤗🤗🤗

[geoffreyparfitt]

If you're looking for rigor, you're in the wrong place... Humour beats rigor any day. Definitely the best mathematics channel!!

[MichaelRothwell]

The 2nd method is very appealing, but unfortunately is wrong as it stands.

The good news is that it can easily be "fixed"!

The problem is that the "^5" at the end of the power towers x^x^x^...^x^5 makes a difference.

It is true (and easy to see) that if x=5^(1/5) or ⁵√5, then:
x⁵=5
x^x⁵=x⁵=5
x^x^x⁵=x⁵=5 and so on.

But for the same value of x the sequence: x, x^x, x^x^x etc. takes completely different values, always ≤ e (see proof later).

Therefore, the sequence x^x^x^...^x has a limit ≤ e, different from the limit of the sequence x^x^x^...^x^5 which is 5.

When we write the infinite power tower x^x^x^... we mean the limit of the sequence x^x^x^...^x. However, there is no standard notation for the limit of the sequence x^x^x^...^x^5.

But, if we were to invent a new notation such as x^x^x^...[^5] for this limit, the problem would be fixed: the proof would be far from rigorous, but it would no longer be wrong!

To shed some light on this, we need to understand a bit about the infinite power tower x^x^x^..., namely why it is true that if it converges, then its value is ≤ e.

Here is a proof.

First of all, we need to clarify what we mean by the value of the infinite power tower x^x^x^.... It is the limit lim[n→∞]uₙ of the recursively defined sequence uₙ, given by u₁=x, uₙ₊₁=x^uₙ.

Now suppose that, for a given value of x>0, the power tower x^x^x^... converges, to y say.
Informally, we have x^x^x^...=y and x^(x^x^...)=y, so xʸ=y.
More rigorously, (the 2nd step holds as the function x^t is a continuous function of t for a given x>0).

As xʸ>0 for x>0 and any real y, we have y=xʸ>0.
Thus xʸ=y ⇒ x=y^(1/y)

We define the function f by f(x)=x^(1/x) for x>0. So f(x)>0 for x>0.

Then ln f(x)=1/x ln x, so
f'(x)/f(x)=1/x×1/x-1/x² ln x=1/x²(1-ln x)
f'(x)=f(x)/x²(1-ln x)

So for 0<x<e (where ln x<1) f'(x)<0, f'(e)=0 and for x>e (where ln x>1) f'(x)>0.
Hence f has a (unique) global maximum f(e)=e^(1/e).

Now we shall prove that the infinite power tower x^x^x^... converges for 1<x ≤ e^(1/e), with value at most e, but diverges for x > e^(1/e).

Together with the fact that for 0<x<1, x^x^x^...^x<1, and for x=1, x^x^x^...^x=1, this will allow us to conclude that (for x>0) if the infinite power tower x^x^x^... converges, then its value is at most e, and, in particular, cannot be 5.

At the end, we shall explain how the 2nd method, incorrect as it stands, led to the correct answer, and how it can easily be fixed.

Step 1, for 1<x≤ e^(1/e):
1.1 uₙ is strictly increasing (this applies for any x>1). We prove by induction that uₙ₊₁>uₙ. For the base case n=1, u₂=xˣ>x¹=x=u₁ (as, for x>1, xᵗ is a strictly increasing function of t). For the inductive step, uₙ₊₁>uₙ⇒x^uₙ₊₁>x^uₙ⇒uₙ₊₂>uₙ₊₁ (for the same reason). So the result is true by induction.

1.2 uₙ ≤ e. Again, we prove this by induction. For the base case, u₁=x≤ e^(1/e)≤e¹=e, as 1/e<1 (for the same reason used in 1.1)
For the inductive step as uₙ≤e by hypothesis (as for a>0, xᵃ is an increasing function of x for x>0, and for the same reason used above). So the result is true by induction.

1.3 From steps 1.1 and 1.2 we deduce by the monotone convergence theorem that uₙ converges and its limit y satisfies y ≤ e.

1.4 If x=e^(1/e), then y=x^x^x^... converges (by 1.3), and f(y)=e^(1/e). But f(e)=e^(1/e) and as f has a unique global maximum at y=e, y=e is the only value of y such that f(y)=e^(1/e). So y=e.

1.5 For 1<x<e^(1/e), y=x^x^x^... satisfies f(y) = x, which we saw earlier has two solutions for y, one with 1<y<e and the other with y>e. As we have shown in 1.3 that y=x^x^x^...≤e, it is the solution less than e (i.e. the smaller solution) that gives the actual value of y for such values of x.

Step 2, for x>e^(1/e): We show that uₙ → ∞, so that y=x^x^x^... = ∞.
We saw uₙ is strictly increasing for any x>1. So either the sequence uₙ is limited above and converges to a finite value, or it is not limited above and tends to infinity. If uₙ were to converge to a finite value y, then we would have f(y)=x. But we have also seen that f(y) has a global maximum value of f(e)=e^(1/e), so that x ≤ e^(1/e), which contradicts x > e^(1/e). Hence the sequence uₙ does not converge to a finite value, so it tends to infinity.

Step 3, For x>0, If x^x^x^... converges, then its value is less than or equal to e.
If x>e^(1/e), then by Step 2, x^x^x^... diverges, so x must satisfy 0<x<e^(1/e).
If 0<x<1, then any positive real power of x is less than 1. So every uₙ<1, and its limit y (if this exists) satisfies y≤ 1, hence y≤e.
If x=1, x^x^x^... converges to 1≤e.
If 1<x<e^(1/e), then by 1.3, x^x^x^... converges to a value ≤e.
So if x^x^x^... converges, then its value is less than or equal to e.

This is what we set out to prove, but we can now also understand the curious fact that the 2nd method, although incorrect, led to the right answer.

From 1.5, applied to the case x=5^(1/5), we have x^x^x^...=y, where f(y)=x. There are two values of y satisfying f(y)=x, i.e. y^(1/y)=5^(1/5), one less than e and the other greater than e. The obvious value is y=5. However, as 5>e, it is the other value, with 1<y<e, that is in fact the value of x^x^x^....

As we saw from the 2nd method, if we add a "^5" to the end of each term x^x^x^...^x to get x^x^x^...^x^5, this flips the limit to the larger value 5.

A nice way to look at this is to note that the limit of the recursively defined sequence uₙ depends on the starting term. We can think of the standard sequence as starting with u₀=1, which converges to the smaller value of y satisfying f(y)=x, but if we change the starting term to u₀=5 (for this value of x), the limit changes to the larger value of y which is 5.

So, if we use notation such as x^x^x^...[^5] for an infinite tower with the initial term specified as 5, the 2nd method becomes correct (as the equation still holds), but, of course, lacks rigour.

[riccardofroz]

a) x^x^x^5=5
expand both sides as "x to the power of" to get
x^x^x^x^5=x^5
if we set y=x^5 then
x^x^x^y=y
which is equal to a) when y=5 so
5=x^5
x=5^(1/5)

[pwmiles]

If you take a free variable u you can have a function
f(u) = x^u
The equation becomes
f(f(f(u)) = 5
Put u = 5. f(5)=5 is a solution and then
x^5 = 5, x = 5^(1/5)

You have to check that the iterated f(u) doesn't have a 3-cycle (5, v, w, 5, v, w...). Thus
There can't be solutions with x<=1 so x>1 and f(u) is increasing.
Then if f(5)>5, f(f(5)) > f(5) i.e. w > v.> 5. By the same reasoning 5 > w, a contradiction.
Similarly if f(5) < 5 then 5 < w < v < 5, also a contradiction.

[natanytzhaki]

As for the second method, just to clarify that the x^x^x^x... = 5 has no solution, the equation in the video does have solution because of the 5 on the top of the tower.
In general x^x^x^x... = y has solution just for y in between e and 1/e

[dihey]

Based on the second method the solution for every x^x^n=n equation where n is greater than 0, the solution would be n^(1/n)? Very cool!

[Roq-stone]

You should’ve done the second method first. I love when the viewers lose hair.
As a matter of fact, there should be three methods and then do the third method second and create mayhem!

[BorisRaifler]

If we will solve the similar tower equation for 4 (X^X^X^... ^X^4= 4) you will find that X=4^(1/4) = 2^(1/2) => X^X^X^...^X^2 = 2.
Moreover, for each Y > e exists Z < e, that X^X^X^...^X^Y = Y that gives X=Y^(1/Y) that is equal to Z^(1/Z)
Now, try to solve X^X^X^... = 2 and you will be able to solve it (X= 2^(1/2) even without given 2 as a last power, because lim(X^X^X^...) exists if it is less than e.
But lim (X^X^X^...) = 4 does not exist because 4 > e, but trying to make the substitution will "work", causing the paradox of ambiguity.

[briannataylor]

@2:48, there were only three ys. Where is that extra y coming from?

[tedgiann]

nice content thank you.

what softwre do you use for the blackboard??

[hmedina]

I enjoy your videos and have subscribed to your channel. Keep the videos coming!

[eeetube]

The result of 2nd method should be checked in general. When you replace 5 with x^x^x^5 you assume that equasion x^x^x^5 = 5 have a solution.

[misterdubity]

I first tried: what would solution be if this was an infinite tower? Answer: 5^(1/5). Plugged that in to the finite version, and it worked. For any height exponential tower, it works.

[loqutisborg]

Anything over 10 and I gotta take my socks off.

[ardiris]

I bet having multiple distinct methods like this has a name.
(:

[manuelgonzales]

Buena observación para la solución por el segundo método, me gusto.

[marcocappiello]

Unfortunately the second method doesnt work that well because when you have an infinite tetration you cant have real solutions for x^x^x^x^...>e

[oenrn]

If you have x^x^x^...^x^n = n, regardless of the number of x, the answer is always x = nth root of n.

Proof (working backwards is easier):
x = n^(1/n)
x^n = n
x^ *x^ n* = x^ *n* = n
x^ *x^ x^ n* = x^ *n* = n
x^ *x^ x^ x^ n* = x^ *n* = n
x^ *x^ x^ x^ x^ n* = x^ *n* = n
...