f(x) = x^x
as x --> 0, f(x) --> 1 (from the lower side)
as x --> 1, f(x) --> 1 (from the lower side)
Therefore, there must be some minimum value of f(x) in between x=0 and x=1
[Editor's Note: I think that value occurs somewhere within a thousandth of x = 0.368]
The curve of f(x) = x^x must be at least somewhat bowl-shaped in between x=0 and x=1
[Editor's Note: I graphed it in Excel, below, y-axis is x^x, x-axis is just x]
So imagine drawing a horizontal line at f(x) = 1 and sweeping it down. You would pass through literally an infinite number of f(x) values that could be produced by 2 values of x.OK, so now that I've proven there are lots more of those numbers, a few questions remain: what are they? How do we find them? What else do they have in common? What else makes them interesting?
I'll do more work on it later; I'm hoping I'll be back at my desk at work tomorrow (which means sleep for now). Good luck, fellow nerds!
PS - If this turns out to be a thing, I reserve naming rights. Don't worry, I'll pick something totally awesome for it.
7 comments:
Here's my take on how to find the numbers, but I'm not sure if it works.
At the local minimum of f(x), there is no solution to x^x=y^y, y =/= x. On either side of the local minimum is a part of a curve {different equation than x^x over the range (min[f(x)], 1] or (0, min[f(x)]) } that could probably be ascertained based on the points it describes (best fit? least squares? fourier analysis?). Equating the two formulae for the partial curves might provide a solution.
-A
I thought about that, but when you're dealing with numbers and increments so small, an approximation makes it impossible to tell when you've got a pair without actually raising both numbers to their own power, and then trying it for all the numbers near it (trial and error, essentially). It might hit the big ones like 1/4 and 1/2, which is the one I've found, but if you tried to find the x and y for f(x) = 0.9, it becomes a lot trickier.
You wind up with 0.9^(1/x) = x, solve for both values of x...hey, does anyone have a TI-89 handy? Is there a quick way to run these computations? I just tried a quick Excel model, but it gets tripped up on #DIV/0! errors. I'll work on it some more later.
I feel like that's a good approach, though, once we know enough about the properties of these pairs to be able to intelligently say what looks right or figure out what a number close to the right one must be an approximation for.
*digs out god-box*
Yeah I have an '89 handy, but I'd say use Matlab or some similar program where you can put a full computer's power against the problem instead of a Motorola 6800. But that's just me.
Proof requires that the function be continuous which you can show by proving it's differentiable since differentiability => continuity. Once that is shown, you're in good shape.
f(x) = x^x
log f(x) = x log x
f'/f = log x + 1
f'(x) = x^x(log x + 1)
Since the derivative exits, then it should be continuous.
What's interesting is that it seems that you always get irrational values for f(x). If the function is truly continuous, how do you get rational values of f(x) in (0,1)?
Just one thing for now (maybe more later if I think more and haven't totally lost my ability to do math): The minimum value of f(x) is 1/e, which is, as you noted, more or less equal to .368. (My TI-83 gets me as far as 1/e=.3678794412...)
f(x) = x^x
log f(x) = x log x
f'/f = log x + 1
Here is where you lost me, it looks like you did remember chain rule on the left, but its in a confusing notation. It looks like you didn't take the derivative of the right side of the equation.
f'(x) = x^x(log x + 1)
I'm going to look like a huge arrogant fool if I'm wrong, but here's my attempt at the derivative:
I'm going to call f(x) "y" for simplicity
ln(y)=x ln(x)
don't forget chain rule
1/y*dy/dx=
product rule
1*ln(x) + x(1/x)
so (1/y)(dy/dx)=ln(x)+1
sub in original function for y
(1/(x^x))(dy/dx)=ln(x)+1
so dy/dx=(X^X)(ln(x)+1)
f'(x)=x^x(ln (x))+x^x
I may be a little out of my league here, but I'm pretty sure the +1 needs to be outside of the ln.
So how else can we tinker with this?
f'(x)=ln(x^x^x)+x^x
This is a pretty cool looking function, but I don't see anything here.
Unless I typed it into my calculator wrong, Rhyno's derivative can't yield a negative value on (0,1] which wouldn't make sense considering that the function is clearly decreasing on (0,1/e)
-Pifry's "lil'est" brother, Sheba
PS: I don't yet have the tools in my mathematical toolbox necessary to integrate this function. If someone could post that, it might be helpful to the analysis.
this is why I shouldn't do math at 12:30 am. Sorry, misread your notation (ln x + 1) does not equal ln (x+1)
The prophecy has come true, I am a huge arrogant fool!
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