I never said you said it was a sum. Your argument is that
must go to 1 for x in the interval (0,1) because the sequence is increasing and bounded above by 1. Well, that sum is increasing and bounded above by 1, but does not go to 1. Therefore, your argument can not be valid.
Since this is being done an infinite number of times, it is bounded above by 1 and that for each x, the sequence
is increasing, the limit is 1.
The sequence actually isn't even increasing. The sequence alternates between increasing and decreasing for x in (0,1). It's strictly increasing for
, but is still bounded for
. In fact...more on this at the end of my post.
What I want to say is that for any x in the interval (-1, 0), you aren't even guaranteed that the answer is real! Take
. Raising this to the -.5 power gives you a complex number! Then you get into complex analysis and which branch of the logarithm you have to take to take the square root and such.
That's why I haven't said anything about how f(x) behaves on negative numbers.
BGronin wrote:The point of this long blathering is that the function isn't continuous and so can't be differentiable.
Sure, if you're going to look at negative inputs, f just becomes a complete mess, but if you restrict f to the interval
, I still say it should be continuous. The function is a composition of continuous functions (more specifically, repeated application of x^x), so it would make sense for it to be continuous.
Let's consider for for the moment some arbitrary value a in the interval (0,1). Since
(both terms in the final product are positive) and
(and since the function
is continuous), by the intermediate value theorem, there must be some x in (a,1) such that
. Rearranging, there is a value of x in the interval (a,1) satisfying
is a differentiable function, let's take a derivative. We get
, which is the interval from which we've chosen a, the derivative is negative for all x and
is therefore a decreasing function.
We may now combine these two facts. Let y be the unique real number such that
is a decreasing function, if
. Similarly, if
Keeping the same letters, we know from the first paragraph that y is in (a,1) and therefore, a<y and
. Considering the sequence of power towers of a of increasing height, we see that the sequence starts off increasing. Call the
term in this sequence
. Let's say you reach a term
. From our previous paragraph,
. That is to say, the sequence will decrease. Likewise, it will continue to decrease until it is less than y. At that point, it will increase until it is greater than y. If the sequence converges, it therefore MUST converge to y. (I'm pretty sure the sequence actually alternates between greater than and less than y, but I have to go to class now and don't have time to try to prove it).
The proof that the sequence oscillates every term turns out to be really simple and I have some time now soooo....here ya go!
Consider a value a from the interval (0,1) and y as defined above. Now consider the function
. Since y is a constant for fixed a, the derivative of this function is just
. This derivative is, like in the section above, going to be negative for all x, so g is strictly decreasing over the entire real line. From the way we defined y, we must have that
. Therefore, since g is strictly decreasing, we must have that
. Therefore, since our sequence of
, we know that
, then we must have that