Consider the sequence {x, x^x, x^x^x,...} for which values of x the series is converging???

]]>Wouldn't the cut-off point be y=e?

It makes sense that the cut-off point is e, because that's when the gradient of x starts becoming negative. But maybe I've missed something.

Edit: I've done some research in Excel and you're right.

The 10th root of 10 is implied to be the solution of x for y=10, but using that value makes y converge to 1.371288574.

]]>Well, not entirely, but it is right if you've roundeded it. But the

realquestion is... what is 1.414 more commonly expressed as?

Mathsy, yes, I did round it and I do I know what it is expressed as - just wanted to leave something in the puzzle for someone else!

]]>I think I can declare that there is no solution for the above equation. The highest value of y for

[align=center][/align]

is y=e or 2.7182818284 approximately and the highest value of x for finite y is x=1.444667861 approximately. I am sure it can be proved that

[align=center][/align]

has no solution.]]>

But if you think about it, for x and y > 1, x < y implies that x^x^x^... is always less than y^y^y^... , so this implies that 10 < 2, which is nonsense. I'm still thinking about this, but it should mean that there is a cutoff value for y above which there is no x solution.

So my final puzzle related to this is: find the largest value of y such that

[align=center]

[/align]has a solution. Unfortunately, I think you'll need to know calculus to be able to figure this out, but the answer makes me wonder if there's something really deep going on here that I don't understand. I find this really interesting!

]]>Let

Therefore,

***someone continue from where I have left...***

]]>Oh, and the solution for 10 is actually lower. Get your head around that one!

]]>[align=center]

[/align]]]>[align=center]

[/align]That's x raised to the power of x raised to the power of x, going on forever equals 2. Solve for x.

]]>