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The 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
by tassl 16y ago
The 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.