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It's not a paradox at all. It's simply not mathematical. The first question, which is completely ignored: Is x^x^x^... well defined? And does he mean x^(x^(x^
by albertzeyer 16y ago
It's not a paradox at all. It's simply not mathematical. The first question, which is completely ignored:
Is x^x^x^... well defined?
And does he mean x^(x^(x^...)) or ((x^x)^x)^x...
Because of his second statement, I guess he means x^(x^(x^...)). So he is looking at the limit of the sequence (x, x^x, x^(x^x), ...). So you have the function f
x |-> limit of (x, x^x, x^(x^x), ...)
This function is not defined for all x (e.g. it diverges for x>=2) and (thus) its target set is clearly not R.
You can see that it does converge for x=sqrt(2), because 0 < sqrt(2)^a <= a for all a in 0..2) and thus it is correct that f(sqrt(2)) = 2 (by using his statements).
You can also see that 4 is not contained in the target set, because a^4 > 4 for all a>sqrt(2) and because f is monotone for x>0 and because f(sqrt(2))=2.
Thus, the set {x | f(x)=4} is empty.
I.e., his assumption that there exists such an x results in the wrong conclusion.
- ced 16y agoYou clearly know more than I do about these things, but: 0 < sqrt(2)^a <= a for all a in 0..2) is not true. It fails for a=1. If you can make it right, could you explain the reasoning that gets you there? I never know how to tackle sequences like x, x^x, x^(x^x),...
- albertzeyer 16y agoWhoops, yes you are right. Fixed, more complete version of this part: 1 < sqrt(2)^a <= 2 for all a in [1,2]. I.e. the sequence (with x=sqrt(2)) is lower-bounded by 1 and upper-bounded by 2. Also, because sqrt(2)^a >= a for all a>sqrt(2), the sequence is also monotone. By that, it follows that it converges to some value y. Now, given that it converges, you know that sqrt(2)^y = y and also y in [1,2]. And then you have y=2.
- tassl 16y agoThe function diverges for x>2, converges to 1 if x=1 and converges to 0 if x<1. For the described problem, the incorrect assumption is that you can find a x st b = x^(x^(x^(x... Where b>=4. Edit: as RiderOfGiraffes pointed, the series does converge for x=sqrt(2)
- RiderOfGiraffes 16y agoIt's fairly easy to prove that the sequence x, x^x, x^(x^x), x^(x^(x^x)), ... converges for x=sqrt(2). Further, for x=sqrt(2) the sequence does converge to 2, so you're not entirely correct.
- tmachinecharmer 16y agoYes. It's simply not mathematical. Suppose, x^(x^(x...) = y At first the author says --> y = 2 and when he says "now consider the equation..." he equivalently says --> 4 = y. Then obviously we can say 4 = 2. Means, you are assuming 4=2 and then concluding 4=2. Please point out if I am wrong. :)
- deleted 16y ago[deleted]
- Lozzer 16y agoYou are wrong. When x is 1, y is 1. When x is sqrt(2), y is 2.