For 0 < x < 1, the expression x^x^x^x^x.... oscillates between two values B < A, where A = x^x^x^x^... with an even number of terms and B = x^x^x^x^x... with an odd number of terms. For example, 0^0 = 1, while 0^0^0 = 0^(1) = 0. Show that for small x approaching 0, the infinite expression x^x^x^x.... approaches 0.
2007-08-22 21:25:34 · 1 answers · asked by Scythian1950 7 in Science & Mathematics ➔ Mathematics
Zanti3, this "power tower" function has some pretty unusual properties, doesn't it? But please dig deeper.
2007-08-23 03:34:23 · update #1
I was the one who had to dig deeper. After 100,000,000 reiterations, I could see that there was a definite split in the convergence where x < (1/e)^e. I just couldn't find the "upper" convergence near x = 0, and I thought I had a proof that there wouldn't be any. So, you get best answer.
2007-08-23 03:54:38 · update #2
From what I remember about this "power tower", the interval of convergence is between (1/e)^e and e^(1/e), or from about .0659 to about 1.4446. For 0 < x < (1/e)^e, the power tower acts weird, apparently converging to two different numbers simultaneously. Since it is not converging to a single value, that means it is NOT convergent.
2007-08-23 03:07:31 · answer #1 · answered by Anonymous · 2⤊ 0⤋