Is this a fallacious derivation?

Question

I wish to find the value of the infinite power tower y = x^x^x^x^x.... for x = √2. I check to see if y could be 4, so that we have

y = 4 = x^x^x^x^x... = x^y = (√2)^4 = 4

thus proving that y = 4 is the value yielded when x = √2. Can you show why this is necessarily erroneous?

If you say, "well, if I plug this into my calculator for x^x^x^x... powers", remember, you're not really computing an INFINITE power tower, but a finite one. Who says we can expect to smoothly converge to the value in the infinite case?

donaldgirod, what you've found is the other value for the function y, which is 2. There are lots of functions which are multi-valued. What I'm asking you, then, prove that the infinite tower function is NOT multi-valued. Or show why y = 4 is invalid.

Derek, in fact the function y = x^x^x^x... has real values for 0 < x < e^(1/e), but it does have some really odd properties. I'm just asking if anyone can show that this function is necessarily NOT multi-valued for 1 < x < e^(1/e). And, by the way, it's multi-valued for 0 < x < (1/e)^e.

Zanti3, this is "argument from incredulity", in which you reach a point, and say, "now that sounds just too silly!". But go look up the Banach-Tarski theorem and tell me if that isn't the silliest thing you've ever heard. And yet it's true! Now, let's look at another pair of y values, 1.00698 and 1000. For x = 1.0069317, x^(1.00698) = 1.00698 and x^(1000) = 1000. So, you can see that even for x approaching 1, such values "fit", even though it seems to boggle the mind that an infinite power tower should generate such high values for x near 1. And, again by the way, the reason why the function seems to "blow up" right after x = e^(1/e) can be better understood if you see that the inverse of the infinite power tower is the function y^(1/y). Why this should be so is easy to show: Let y = x^x^x^x... Then y = x^y, and y^(1/y) = x.

Answer 1

OK, I'll contribute my two cents:

For x > y > 1, it is obvious that x^x > y^y, x^x^x > y^y^y, etc. The relationship will hold regardless of how many exponents are in the power tower.

Now let's make the jump to having an infinite number of exponents. We know x^x^x^... converges to a finite number if (1/e)^e ≤ x ≤ e^(1/e). In fact, the largest finite value this power tower can take is e, and that is when x = e^(1/e). By the way, another odd property of this function is that, as soon as x > e^(1/e), the function jumps to infinity. There is no very rapid but continuous climb as in a function like f(x) = 1/x when x approaches zero -- in this power tower, the function gets to e, then the graph simply disappears.

Getting back to the notion that the power tower can have two solutions if x > 1, let's try solving another one: x^x^x^... = 1000. The solution would have to be x = 1000^(1/1000), i.e. a number very close to 1. In fact, change the 1000 to 1,000,000, and it is clear the solution is a number even closer to 1. And that is the logical absurdity when we try to solve x^x^x^... = y for a value y > e: the bigger we make y, the closer to 1 that x gets. It leads to the conclusion that 1^1^1^... approaches infinity as well as 1.

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Well, that Banach Tarski Theorem is just another proof that 1 = 2, right? :)

Yes, coming up with bogus proofs that 1 = 2 is easy when you start playing with infinite sets. For example, we know half of all the integers are even. However, we also know that the set of even integers has the same cardinality as the set of integers, which means both sets are the same size. Hence, 1 = 2.

Now, if you're saying that the power tower x^x^x^... can yield larger and larger results as x approaches 1, well, I suppose that is sort of like the paradox with integers just mentioned. In the case of the integers, the even integers are coming in at only half the rate of the integers so long as the set is finite, but somehow at infinity the set of even integers catches up. So, do the power towers do something similar? Does 1.01^1.01^1.01^... lag behind 1.4^1.4^1.4^... for as long as the power tower has a finite number of exponents, but when the number of exponents is infinite, the first power tower zooms past the second one? I don't know, that still sound too illogical. That's why I don't believe in infinity.

Answer 2

I wouldn't say your derivation is necessarily fallacious, but I would say that it is not completely rigorous, which is why you came up with an erroneous result.

To go from y = 4 = x^x^x^x^x... to x^x^x^x^x... = x^y, you have assumed that the limit of 4 exists in the power tower function. I haven't done much research on the power tower function, but I do know it only converges for e^-e < x < e^(1/e) , or something like that. And 4 might or might not exist in the range of this function, which is the key to your derivation.

Answer 3

what your derivation shows is that if x^x^... = 4 has a solution, then the only possible value for x is x=sqrt(2). But there is no particular reason to think that x^x^... = 4 does have a solution. Note that if you try to solve y=2, the same argument gives

y=2 = x^2; x = sqrt(2)

Thus, if x^x^x.... does have a solution, the only possible solution is 2. In fact, y=2 does have a solution (one has to prove this), and so sqrt(2)^sqrt(2)..... = 2, not 4.

Note that I have not proven anything here, but simply pointed out that your argument is not complete. It is often possible to do calculations which start out

Suppose blah has a solution. Then that solution must be zog.

The remainder of the discussion has to then establish that blah does have a solution.