Carry on towering
What is the range of convergence of x^(x^(x^(x^(x... ?
What are the corresponding minimum and maximum values of the expression?
You will find the following very useful:
http://en.wikipedia.org/wiki/E_%28mathematical_constant%29
Go half way down the page to x^(1/x). Have a glance, then follow the tetration link.
This is slightly modified from the original post in view of the info provided.
Only basic differential calculus is required.
What are the corresponding minimum and maximum values of the expression?
You will find the following very useful:
http://en.wikipedia.org/wiki/E_%28mathematical_constant%29
Go half way down the page to x^(1/x). Have a glance, then follow the tetration link.
This is slightly modified from the original post in view of the info provided.
Only basic differential calculus is required.
Labels: mathschallenge
posted by Chris at
4:56 PM
17 Comments:
well, max. shouldn't be the problem ->oo ^^
but why are you sure about the positive value for minimum? Because of the alternating signs? thats where i am still thinking about
Hi t:b:H. I'd just thought of that when having a coffee. Sorted.
It's looking quite tricky. So I've updated the problem with all the info I have. Good luck, I think we're gonna need it.
it reminds me of:
http://xkcd.com/356/
great stuff
Rats! the tetration page contains the details for the upper bound.
Still leaves the lower bound.
Lovely link. LOL. I'm going to go through it now.
Define a(0)=x and a(n+1)=x^a(n). We want to know when this sequence converge. If (and when) it does, a(n+1)=a(n)=A, which means A=x^A.
Take the ln, and have ln(A)=A.ln(x).
To simplify, let's say it this way: we want to find x such as x=k.ln(x). If we draft both graphs, it is easy to see that there is a point in y=ln(x) where the tangent of the graph crosses the origin, and any value of k bigger than this will make the equation impossible. Now it is easy to show that if we write the generic tangent as y=m.x+b and calculate m and b (easy), we find that only when x=e the tangent crosses the origin, and in that case m=1/e. Hence this is the upper bound (I hadn't research for this, excuse me if the answer is already in any source you provided)
Hi Miguel, you can answer any way you wish. I won't insist you have fun my way.
I'm sure that the links provide more than enough clues and that the problem can be answered with straightforward math.
I just love these crazy looking questions.
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I think you guys are doing well, especially as you're not taking advantage of the link(s).
I'll confirm that the upper limit on x is e^(1/e) and then the tower max value is e.
I hope you both (all) are as amazed as I am that the tower simply isn't unbounded for all x > 1 (at least).
t:b:H. I know there's a lower bound because Euler proved it :)
I hope you're not suffering too much with the unreliability of the postings. It's driving me nuts.
....it's also surprising that there is a lower bound.
One interesting result:
if f(x)=x^(x^(x...)), then
x=f(x)^(1/f(x))
Using the result on the problem "xth root of x" posted below, then x is maximum when f(x)=e. Thus, x maximum is e^(1/e)
Hi Miguel. That's exactly the reason that I thought the problem was fair game to post.
I've always had a "thing" about these "recursive" problems. But probably couldn't do them at the time. Still not sure if I can get te lower limit. Been too busy posting in other blogs. Greetz.
Hi guys. OK, I've been doing some research on the lower bound. My eyes misted over. Conclusion, forget doing by maths; Euler's proof is beyond the scope of this web site, which really means "I can't do it" :)
However, the following link contains a graph of the function:
http://wapedia.mobi/en/Euler%27s_infinite_tetration_theorem
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