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by **DeathRowKitty** » Tue Feb 15, 2011 9:17 am

Why isn't the function continuous? I'm not positive it is, but it sure seems to me like it should be....and f(x) is certainly not identically 0 on the interval (0,1)! Play around with the function a bit. Approximate a few values. f(x) is nonzero for all x in that range.

Edit:

BGronin wrote:However, in the interval (0, 1), the value should be 1. If you take any number in (0, 1) and raise it to a power larger than one, then the value becomes smaller. So, raising it to a value between (0, 1) would actually make that number bigger. Doing this a countable amount of times would send the limit to 1.

This is also not true. Consider the sequence $a_n=\sum_{k=1}^n\frac{1}{4}^n$ . Even though each term is larger than the last, the sequence never reaches 1 (the limit of this sequence is $\frac{1}{3}$ ).

by **BGronin** » Tue Feb 15, 2011 12:46 pm

I'm not saying take a sum. I'm saying that if you raise a number in the interval (0, 1), and raise it to the power of some number in the interval (1, $\infty$ ), you will get a number that is smaller than your original number. i.e. $f(x)=x^{r}, r>1$ is concave up in the interval (0, 1) and so is smaller than $g(x)=x$ . However, if you use the same f(x) as from before, but allowing $0 < r \leq 1$ , you get a function that is concave down and so is larger than g(x). Specifically, if you choose x in the interval (0, 1) and raise it to itself, the value will be larger. Since this is being done an infinite number of times, it is bounded above by 1 and that for each x, the sequence $x, x^{x}, x^{x^{x}}, ...$ is increasing, the limit is 1. (This isn't rigorous, but it's still true.) Anyways, so I was wrong for saying that it went to zero, but here it's identically 1 in the interval (0, 1).

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 $x=-\frac{1}{2}$ . 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.

The point of this long blathering is that the function isn't continuous and so can't be differentiable.

by **DeathRowKitty** » Tue Feb 15, 2011 1:52 pm

I never said you said it was a sum. Your argument is that $x^{x^{x...}}$ 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.

BGronin wrote:Since this is being done an infinite number of times, it is bounded above by 1 and that for each x, the sequence $x, x^{x}, x^{x^{x}}, ...$ 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 $x>1$ , but is still bounded for $x \leq e^{\frac{1}{e}}$ . In fact...more on this at the end of my post.

BGronin wrote: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 $x=-\frac{1}{2}$ . 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 $(0,e^{\frac{1}{e}}]$ , 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 $a^a-a=a^a(1-a^{1-a})>0$ (both terms in the final product are positive) and $a^1-1=a-1<0$ (and since the function $a^x-x$ is continuous), by the intermediate value theorem, there must be some x in (a,1) such that $a^x-x=0$ . Rearranging, there is a value of x in the interval (a,1) satisfying $a^x=x$ .

Since $a^x-x$ is a differentiable function, let's take a derivative. We get $a^x \cdot \ln{a}-1$ . Since $\ln{a}<0$ for $0<a<1$ , which is the interval from which we've chosen a, the derivative is negative for all x and $a^x-x$ is therefore a decreasing function.

We may now combine these two facts. Let y be the unique real number such that $a^y-y=0$ . Since $a^y-y$ is a decreasing function, if $z<y$ , $a^z-z>0$ . Similarly, if $z>y$ , $a^z-z<0$ .

Keeping the same letters, we know from the first paragraph that y is in (a,1) and therefore, a<y and $a^a>a$ . Considering the sequence of power towers of a of increasing height, we see that the sequence starts off increasing. Call the $n^{th}$ term in this sequence $a_n$ . Let's say you reach a term $a_m$ such that $a_m>y$ . From our previous paragraph, $a^{a_m}<a_m$ . 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).

Edit:
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 $g(x)=a^x-y$ . Since y is a constant for fixed a, the derivative of this function is just $g'(x)=a^x\cdot\ln{a}$ . 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 $a^y-y=0$ . Therefore, since g is strictly decreasing, we must have that $a^x-y>0$ for all $x<y$ and $a^x-y<0$ for all $x>y$ . Therefore, since our sequence of $a_n$ satisfies $a_{n+1}=a^{a_n}$ , if $a_m>y$ , we know that $a_{m+1}<y$ and if $a_m<y$ , then we must have that $a_{m+1}>y$ .

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