Is y=x^y an increasing function in (0, e^(-e)]?

Y

yma16

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When x=e^(-e), y=1/e. lny=ylnx. y'=yy/x(1-ylnx).

How can I show y'>0? Thanks.
 
y>0, =>y'<>0. If I can show that y<1/e at a pt in (0,e^(-e)), then it is increasing.

or when x->0, y<1/e.
 
yma16 said:
When x=e^(-e), y=1/e. lny=ylnx. y'=yy/x(1-ylnx).

How can I show y'>0? Thanks.
Click to expand...

My first response to the question was that y = x^y is not a function. That is, it is not written as a function. Once I realized that you mean the function y = f(x) defined implicitly by y = x^y, the question remains: have you proved that it is a function? That is, is there only one value of y for a given x?

Your subsequent posts seem to be about the derivative at just one point in the interval. I don't see how that proves anything about all points in the interval. Perhaps you need to say more about what you are thinking.

But note that, since lny = ylnx, we have y' = y^2/x(1-ylnx) = y^2/x(1-lny) . We know that y^2 > 0 and x > 0; is 1-lny > 0 throughout the interval?
 
if you change it to y-x^y=0, then it

Dr.Peterson said:
My first response to the question was that y = x^y is not a function. That is, it is not written as a function. Once I realized that you mean the function y = f(x) defined implicitly by y = x^y, the question remains: have you proved that it is a function? That is, is there only one value of y for a given x?

Your subsequent posts seem to be about the derivative at just one point in the interval. I don't see how that proves anything about all points in the interval. Perhaps you need to say more about what you are thinking.

But note that, since lny = ylnx, we have y' = y^2/x(1-ylnx) = y^2/x(1-lny) . We know that y^2 > 0 and x > 0; is 1-lny > 0 throughout the interval?
Click to expand...

has form f(x,y)=0 as a function. To show it is increasing, I need to show y'>0. In this case, y<>0. Therefore, y'<>0. So either y'>0 or y'<0. Since at one point y'>0, it implies y'>0 on the interval. I do need to show y' is continuous on the interval. This is to show 1-lny<>0,

1=lny, y=e, e=x^e <=> x=e^(1/e)>e^(-e). This show the denominator of y' is not 0 over the interval because the 0 is outside the interval. Sorry about the jump.
 
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yma16 said:
if you change it to y-x^y=0, then it has form f(x,y)=0 as a function.
Click to expand...

No, you're missing some important points. (I don't have time now, so I'm not looking at your handling of the derivative, only at whether it is a function.) That's unfortunate, because the point of the video is to warn you to be careful of subtleties!

When we say an equation represents a function, we don't just mean that the equation can be written using a function (of two variable); that goes without saying. We mean that y is a function of x. That is especially true when we then say that it is an increasing function, because that can't be said of a multi-valued "function". There must be only one value of y for any given value of x.

That is not true for y = x^y! Go to Desmos.com and type in "y = x^y", and look at the graph. (Not all graphing programs can do this!) One thing you will notice is that (sqrt(2),2) and (sqrt(2),4) are both on the graph. Does that sound familiar from the video?

This graph includes the graph of y = x^x^x^..., which the video is about, but goes beyond it. When they "solve" for both y=2 and y=4 and get x=2 both times, what they are really doing is solving your equation, not the one they claim to be solving, and that is the source of the error.
 
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Slight correction: \(\displaystyle y = x^y\) is indeed not a function. (It doesn't satisfy the vertical line test.) But the points (2,2) and (2, 4) are not solutions!

-Dan
 
topsquark said:
Slight correction: \(\displaystyle y = x^y\) is indeed not a function. (It doesn't satisfy the vertical line test.) But the points (2,2) and (2, 4) are not solutions!

-Dan
Click to expand...

Yes, I was in a hurry, and didn't take the time to proofread. I've fixed that detail.

There's still a lot more to say about both this equation and the video, and their relationship!
 
Thank you for pointing out it is not a function

It seems that the double values of y occurs when x>e^(-e). Can you shows an example that when x is in (0,e^(-e)), there are two ys?

The following is the general form of double y values; y=2, 4 is one case of them.

Let x=(1+1/z)^z(z/(1+z))^z=(1+1/z)^(z+1)(z/(1+z))^(z+1)

y=x^y has y=(1+1/z)^z or y=(1+1/z)^(z+1) to fit the equation. When z=1, y=2 or 4.
 
yma16 said:
It seems that the double values of y occurs when x>e^(-e). Can you shows an example that when x is in (0,e^(-e)), there are two ys?

The following is the general form of double y values; y=2, 4 is one case of them.

Let x=(1+1/z)^z(z/(1+z))^z=(1+1/z)^(z+1)(z/(1+z))^(z+1)

y=x^y has y=(1+1/z)^z or y=(1+1/z)^(z+1) to fit the equation. When z=1, y=2 or 4.
Click to expand...

Where did this come from? Did you find some source that gives you this "general form", or did you derive it? I'd like to see this.

I think an important starting point for exploring this equation (and probably also the infinite exponential x^x^x^...) is to focus not on y as a (non-)function of x, but on its inverse, which IS a function:

y = x^y ==> ln(y) = y ln(x) ==> ln(x) = ln(y)/y ==> x = e^(ln(y)/y) ==> x = y^(1/y)

The graph suggested making (of which a restricted part was shown in the video) is the inverse of this one, and exploring its maximum, minimum, and end behavior will show you where the numbers he gave come from (such as e^(-e)). It will also answer your question above.

By the way, have you looked at the paper recommended in the video? I found it here, but have only glanced at it: https://www.maa.org/sites/default/files/pdf/upload_library/22/Chauvenet/Knoebelchv.pdf .
 
The general form is under the video as a comment and I

Dr.Peterson said:
Where did this come from? Did you find some source that gives you this "general form", or did you derive it? I'd like to see this.

I think an important starting point for exploring this equation (and probably also the infinite exponential x^x^x^...) is to focus not on y as a (non-)function of x, but on its inverse, which IS a function:

y = x^y ==> ln(y) = y ln(x) ==> ln(x) = ln(y)/y ==> x = e^(ln(y)/y) ==> x = y^(1/y)

The graph suggested making (of which a restricted part was shown in the video) is the inverse of this one, and exploring its maximum, minimum, and end behavior will show you where the numbers he gave come from (such as e^(-e)). It will also answer your question above.

By the way, have you looked at the paper recommended in the video? I found it here, but have only glanced at it: https://www.maa.org/sites/default/files/pdf/upload_library/22/Chauvenet/Knoebelchv.pdf .
Click to expand...

replaced the n with z. I understand the video. The only thing I do not understand is where e^(-e) come from. When x=1, y=1 and it has only one solution. I suspect that when x<1, y=x^y is a function.
 
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yma16 said:
replaced the n with z. I understand the video. The only thing I do not understand is where e^(-e) come from. When x=1, y=1 and it has only one solution. I suspect that when x<1, y=x^y is a function.
Click to expand...

So, you have been able to prove that claim for yourself, since no proof is given? I'm not saying it's wrong, just that math is not a matter of believing whatever someone posts in a comment, but of proof. The author does appear to know what he's talking about. On the other hand, maybe there is a better way to generalize. I haven't looked into that yet. (I'm mostly just trying to encourage you to go deeper!)

I believe you are right about y being a function of x for 0<x<=1. You can see that by examining the graph (particularly its end behavior).

I think the restriction of x>e^-e is not a matter of y having a single value, but of convergence, which is a separate issue. I'll have to think more about proving convergence (or read the paper); but just playing with it in a spreadsheet, I see that it converges very, very slowly at or near the boundaries e^-e and e^(1/e). You may find it interesting to do the same; you'll see in what way it fails to converge outside the interval of convergence.
 
yma16 said:
The following is the general form of double y values; y=2, 4 is one case of them.

Let x=(1+1/z)^z(z/(1+z))^z=(1+1/z)^(z+1)(z/(1+z))^(z+1)

y=x^y has y=(1+1/z)^z or y=(1+1/z)^(z+1) to fit the equation. When z=1, y=2 or 4.
Click to expand...

I went back and tried to check this, and I don't even know what it means. If I'm reading it correctly, x = (1+1/z)^z (z/(1+z))^z simplifies to x = ((z+1)/z)^z (z/(1+z))^z = ((z+1)/z)^z ((z+1)/z)^-z = 1! Maybe you meant to say that, for any z, [(1+1/z)^z]^[(z/(1+z))^z]=[(1+1/z)^(z+1)]^[(z/(1+z))^(z+1)], as an example of m^(1/m) = n^(1/n). But you'll have to show that this is true.

But the Knoebel paper has some good material.

On page 3 there is a similar "parametrization of solutions to x^(1/x) = y^(1/y)", given by x = s^(1/(s - 1)) and y = s^(s/(s - 1)), saying "To check that these expressions really satisfy x^y = y^x is a good exercise in the rules of exponentiation." (I haven't tried doing so myself yet.)

On page 6 is the graph from the video, and a proof of the domain of convergence of x^x^x^..., which shows the source of e^(-e) that you asked about. It is not simple!
 
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