Pretty cool function (and its inverse) Watch

Zii
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#41
Report Thread starter 10 years ago
#41
(Original post by pyrolol)
Try something like 0.05. It oscillates between 2 values. (According to wikipedia, this happens when x < e^-e.
Yeah.

Like take the curve y = x^x^x^x^...

We know that for x < e^(1/e), when the x-coordinate is of the form a^(1/a) the y-coordinate must be a.

we already know (e^(1/e), e) is a point on the curve. We know (2^(1/2), 2) is a point on the curve. But what about 4^(1/4) [which is root 2 which is less than e^(1/e) ]? Surely then, at the point x = root 2, there are at least two y-coordinates, 2 and 4.

Take 10^(1/10) . This is less than e^(1/e) [in fact every a^(1/a) will be less than or equal to e^(1/e) ] (10^(1/10), 10) must lie on the curve, but it does not appear to - the greatest y-coordinate is e - - after that, it's infinity.

I'm confused. Does this mean that at every point y< e^(1/e) there are multiple y-coordinates for every x-coordinate? Is that what wikipedia said? There are two y-coordinates for every x?
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Dystopia
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(Original post by Zii)
Yeah.

Like take the curve y = x^x^x^x^...

We know that for x < e^(1/e), when the x-coordinate is of the form a^(1/a) the y-coordinate must be a.

we already know (e^(1/e), e) is a point on the curve. We know (2^(1/2), 2) is a point on the curve. But what about 4^(1/4) [which is root 2 which is less than e^(1/e) ]? Surely then, at the point x = root 2, there are at least two y-coordinates, 2 and 4.

Take 10^(1/10) . This is less than e^(1/e) [in fact every a^(1/a) will be less than or equal to e^(1/e) ] (10^(1/10), 10) must lie on the curve, but it does not appear to - the greatest y-coordinate is e - - after that, it's infinity.

I'm confused. Does this mean that at every point y< e^(1/e) there are multiple y-coordinates for every x-coordinate? Is that what wikipedia said? There are two y-coordinates for every x?
This explains it, doesn't it?

(Original post by Dystopia)
Look at the graph of y = x^{\frac{1}{x}}. For 1 < x < e^(1/e), there are two solutions to x = y^(1/y). If f(x) = x^x^(...), then you've shown that f(x) must be one of the two values, and it is always the smaller value.

This can be shown by considering the sequence x, x^x, x^x^x, ...

Let's say that u_{n} = \underbrace{x^{x^{\cdots ^{x}}}}_{n}

For x>1, it isn't difficult to prove that the sequence is increasing and that if there is some y that satisfies x = y^{\frac{1}{y}}, then u_{n} &lt; y \Rightarrow u_{n+1} &lt; y, so you can prove the result by induction (try it).

For x = e^(1/e), there is only one value of y such that x = y^(1/y).

For x > e^(1/e), there are no values of y, and it diverges.

For x < 1, things are slightly trickier I think (because the sequence isn't increasing).
For example, take the example of x = 2^{\frac{1}{2}} = 4^{\frac{1}{4}}, and define u_{n} as in the quote, with u_{0} = 1, so u_{n+1} = \sqrt{2}^{u_{n}}

Clearly u_0 < 2.
Suppose u_k < 2.
\Rightarrow \sqrt{2}^{u_{k}} &lt; \sqrt{2}^{2}
\Rightarrow u_{k+1} &lt; 2
So u_{n} &lt; 2.

Of course, this argument works exactly the same if you replace 2 by 4, but since 2 is less than 4, the value the sequence converges to will be the smaller one.

Similarly, you can prove that the sequence is increasing.

An increasing function bounded above must tend to a limit, and you have shown that this limit must be either 2 or 4. Since 2<4, the value is 2.

You can do exactly the same thing with all other values of x between 1 and e^(1/e).

To find the value it takes when x = 10^(1/10), draw the graph y = x^(1/x), and the line y = 10^(1/10). This will intersect the graph at two places, and the value will be the x coordinate of the first intersection.
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