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## xtatic

Posted by Chris on July 3, 2010 – 3:29 pm

When it converges, what is the maximum possible value of x^(x^(x^(x^(…)))) and what is x at that condition?

That’s an infinite power tower of x’s.

This post is under “MathsChallenge” and has 15 respond so far.

### 15 Responds so far- Add one»

1. 1. Random Guy Said：

If the value of x is 2, then I would say the value of the equation would be… infinity. But if it were 1, I’d say the value of the equation would be 1. So 1 and 1.

2. 2. Chris Said：

Hi Random Guy. I’ve modified the question. You are right in that x = 2 gives an infinte value and x = 1 gives 1.

There is a definite maximum value of x such that 2 > x > 1 that gives a finite value for the power tower, and if you make x any larger, the power tower becomes infinitely big.

3. 3. Chris Said：

You will almost certainly find it useful to consider the separate problem of finding the maximum value of y^(1/y) when answering the posted problem.

4. 4. slavy Said：

Maybe it is e^((ln2)/2)

5. 5. The Winner Said：

1.0000…1 the smallest possible # thats greater than one.
that is to say if you add 0.9999… to that it will = 2

6. 6. Chris Said：

The Winner. 1.000…1 in the way you suggest is undefined. However, 1.000…1 = 1 exactly, where … implies an infinite number of 0s. Also 0.999… = 1 exactly. Both are due to the real numbers being Archimedean => the infinities and the infinitesimals are not included in the real numbers: something that I have been aware of for less than 6 months.

In this problem x > 1. My initial instinct was than x > 1 would lead to x^(x^(x… -> ∞. That’s partly why I like this problem. There is also a least value for x with the series being convergent, but the determination of that is beyond my capabilities.

Slavy. Sorry, but good guess.

7. 7. BearSprite Said：

I learned these lessons about ten years ago in high school math, so forgive me my advantage – and my disadvantage at not being able to recall the more eloquent descriptions and proofs of my teachers. Are we not dealing with limits here? That is:

While X > 1, the answer should be infinite.
While X = 1, the answer is 1.
While 1 > X > 0, the answer approaches 1 but never reaches it (this is called a limit). We know this because the square root of 1/2, or (1/2)^(1/2) is larger than 1/2, and a few quick calculations show that if this process continues it quickly approaches 1 but never quite reaches it (the Archimedean number Chris talks of). However, because it never reaches 1, it is the LIMIT approaching 1.
While X = 0, the answer is 0.
Having done a few quick calculations I realized that while 0 > X > -1 we leave the real number system and have to enter the IMAGINARY number system. I gave up trying to figure out why the numbers react the way they do, but I’m pretty sure it’s uniform throughout this range that the real coordinate hops from above 1 to below 1 getting ever closer without ever reaching 1 (giving 2 LIMITS at once). On the imaginary coordinate I’m pretty sure it simply becomes negative infinity.
While X = -1 the answer is -1
While X < -1 and is an integer and is even, it's numerical quality gets larger and larger, but it is constantly flipped from being the denominator of a fraction with 1 in the numerator to a larger number with the numerical quality in the numerator and 1 in the denominator. In short, these numbers approach the limit of both 0 and infinity.
While X < -1 and is an integer and is odd, it's numerical quality gets bigger as an even number, with the same negative exponent fraction reversal quality, only because the initial negative number cannot cancel itself out due to a persistent odd number of multiplications, the number is negative. It approaches the limit of 0 and negative infinity.
I will have to leave numbers where X < -1 and not integers by saying that they are imaginary numbers that make my head hurt. I'm not even going to try and derive a pattern based in these mixed fraction negative exponents. Such numbers should never exist in order to save some of us the headache.

8. 8. Chris Said：

Hi BearSprite. I have only seen this problem dicussed with real numbers. The minimum value of x is approx 0.066 and the maximum is approx 1.44. Between these 2 values the power tower has a definite finite value. Outside those values, the value is undefined i.e. it is infinite.

I doubt that (m)any visitors to this site will be able to analytically verify the lower value (I certainly can’t). But anyone with a basic knowledge of calculus (and taking advantage of the clue I gave in post 3) should be able to crack it.

9. 9. BearSprite Said：

I’d love to see the proof on this one. The laws of exponents hold (written below as my proofs for my preceding entry), I don’t see how any finite digit numbers come out of this save 1, 0 and -1. If you could tell me the real values that come out of these numbers I’d love to know, because even if I use a calculator I get infinite values from 1 – 1.44 (having tried such numbers as 1.43 and 1.0000000001).

The laws to start us off are:

x^(a/b) = the b root of x^a
x^-a = 1/(x^a)
therefore x^(-a/b) = the b root of 1/(x^a)

so (-1/2)^(-1/2) = the square root of (-2/1), or the square root of -2.
The square root of negative numbers always turns into imaginary numbers, as any even numbered root, since no numbers multiplied by themselves an even number of times can become negative.
Imaginary numbers are presented in the format X + Yi, where X is the real value, Y is the imaginary value and i is the square root of -1 (so Y multiplied by the root of -1 is the imaginary component).
Interestingly, if the denominator of the negative value of X is odd, it doesn’t turn into an imaginary number (I missed this above).

Really, though, you got me stumped.

10. 10. Wizard of Oz Said：

I’m just as much in the dark as everyone else.
I think that the maximum value of y^(1/y) (following the hint in post # 3) is about 1.4446676… and this occurs when y = e (= 2.71828…).
Putting this value of y into the power tower expression leads to a limit which seems close to e again. (Here I used Excel to generate the iterations – very much a brute force approach).
This result just comes from following the hint in post # 3. I have no idea how to prove it (if indeed it is the correct one).
This problem seems vaguely familiar. Have we had this one or a similar one before?

11. 11. Chris Said：

Wiz, you did good enough for the proverbial cigar. I have posted this one before. But because of Vago’s comment on the “Picking Numbers” problem, I thought I’d give him a treat.

Here’s my “solution”: NB I’m only working with real numbers and I’m blissfully ignoring that e^(2πi) = 1 and it’s implications.

Let y = x^(x^(x^(… = x^y => x = y^(1/y) = e^(ln(y)/y) and this is defined for all real y > 0.

Differentiating => dx/dy = e^(ln(y)/y) (-ln(y)/y² + 1/y²)
At a critical point, dx/dy = 0 => ln(y) = 1 => y = e
It turns out that this corresponds to the maximum value of x
i.e. x = y^(1/y) ≤ e^(1/e) ≈ 1.44466786 and y = e
I have nothing to say about the lower limit.

So the max value of y = x^(x^(x… = e when x = e^(1/e).
At this point dx/dy = 0 => dy/dx = ∞ => if x increases y → ∞

For a graph see http://en.wikipedia.org/wiki/Tetration
(3rd diagram at the time of publishing).

For x^(1/x) see http://en.wikipedia.org/wiki/E_%28mathematical_constant%29
The graph is about half way down the page.

12. 12. slavy Said：

OK I give up! Could you give me a link where I can find the computations for the lower bound? What I can show is that for any x\in(0,1) the even subsequence is monotonically decreasing, while the odd subsequence is monotonically increasing. They are both bounded from the proper side (by 0 and 1 respectively) and hence convergent. The previous computations for y^(1/y) show that for each x there exists a unique y, such that x^y=y (in this case y is the common bound for both the subsequences). However, there may be cases, when the two bounds differ and then the initial sequence does not converge. This is equivalent to showing that x^(x^y)=y has more than one solution in (x,1) for the bad choices of x (the number of solutions is always odd number, so they should be at least 3). This is very ugly to analyze and I don’t see where 0.066 comes from…

13. 13. Chris Said：

Hi slavy. 0.066… = e^(-e) and was determined by Euler. I haven’t tried (this time) to find out how he got that, but I recall, from my previous efforts, that quite advanced maths was involved. It took quite a lot of determined Googling to find that out and I haven’t got the time to do that now. It is largely because of the lack of time that I have been reposting problems from the old ToM site.

As far as I’m aware, when x is outside the range (e^(-e),e^(1/e)), the series diverges to infinity.

The tower can be extended to complex x. See the Lambert W function on the tetration page.

14. 14. slavy Said：

Thanks for the remarks Just one comment – there is no way in (0,0.066) the sequence to go to infinity, since every element of the sequence is positive and thus x to a positive power is always less than 1 (hence, the sequence is bounded by 1). In this case, the two convergent subsequences should simply have different bounds…

15. 15. Chris Said：

Hi slavy. My bad, you’re right, the sequence does oscillate between two values in range (0, 0.066…). I hadn’t noticed the comment under the second diagram on the titration link and I hadn’t looked at the graph closely enough.

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