Y^x = x^y

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cronxeh

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An interesting function. If plotting you would probably omit the 0,0 point too.

How do I get the general solution for this? like y = [function of x]? Is it just y=x ?

What about the boundary conditions near origin? How does it look like there??
 

HallsofIvy

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If you mean y being defined implicitely as a function of x, then it certainly is not "y= x". I doubt that it could be solved in terms of elementary function but you might be able to use a version the the Lambert's W function.
 

saltydog

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Hello guys. I don't think it's just y=x. I've attached a rough plot of a curve I think all real solutions (in first quadrant) are on but can't prove it.

Oh yea, I think it would be an interesting exercise to figure out the complex solutions.
 

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cronxeh

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Some particularly interesting similar function was [tex]y=x^x[/tex] When I plotted it from 0 to 1 there was what seemed like an inflection point - although I dont think its really called that but from a graph it does look pretty neat.

Could this be the plot for y^x = x^y from 0 to 1 ?
 
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Icebreaker

It's not just y=x: here's a counterexample! 2^4 = 4^2
 
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Icebreaker said:
It's not just y=x: here's a counterexample! 2^4 = 4^2
there are an infinite number of counterexamples. slatydog graphed them in #3
 

Icebreaker

jdavel said:
there are an infinite number of counterexamples. slatydog graphed them in #3
Yes, I know. I never said it was the only counterexample, now did I?
 

Hurkyl

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I've attached a rough plot of a curve I think all real solutions (in first quadrant) are on but can't prove it.
I think there should be a straightforward proof that there are exactly two (real) solutions for y, for each x, except for when x is... (reaches into my magic hat)... e. I just can't remember what the proof is. :biggrin:

If you happen to recall the proof that 2^y = y^2 has exactly two solutions, that should generalize to this problem.
 

saltydog

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BicycleTree said:
Hello Bicycle. I checked out your reference. Thanks. I found it an interesting exercise to convert:

[tex]x^y=y^x[/tex]

using the substitution y=kx yielding the parametric equations:

[tex]x=k^{\frac{1}{k-1}}[/tex]

[tex]y=k^{\frac{k}{k-1}}[/tex]

However, when I plotted this, I did not obtain the line y=x. Attached is the parametric plot.
 

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Hurkyl

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The reason you don't get the y=x solution is that you assumed k was not 1 when you took the (k-1)-th root of the equation.
 
my simple attempt at proving nothing...

y^x = x^y

xlogy=ylogx

(logy) = (ylogx)/x

10^(ylogx) = y

ylogxlog10 = y

logxlog10 = 1

log(x+10) = 1

10^1 = x+10

10 = x + 10

x = 0

y^0 = 0^y

y = 0

0^0 = 0^0

1 = 1

so, the only solutions I can find for y and x are = 0

(don't flame, i'm a noob and i'm probably wrong)

however...

y^0 = 0^y

0logy = ylog0 <-- that doesn't work...

0 = undefined ?
 

Icebreaker

y=x=0 is the only point on y=x that's not a solution.
 

cronxeh

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z = f(x,y) = x^y - y^x = 0

Can anyone plot this. I got some weird looking graph, maybe someone else will have a better one

Oh man this is exciting. Check out this plot!
 
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Zurtex

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Given some value fixed value of x then a solution of y will be:

[tex]y = - \frac{x \text{ProductLog} \left( -\frac{\log x}{x} \right)}{\log x}[/tex]

According to mathematica. This gives you a graph like so:
 

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cronxeh

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Forget the positive side its too boring! Plot all stuff in QIII !

Same interesting effect but without the positive side can be observed in [tex]x^y=-y^x[/tex]
 
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I think if x^y=y^x and x,y both integer then y=x

If u need the proof please tell
 

Hurkyl

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A proof would be interesting to see, since your claim is false...
 

rachmaninoff

I think if x^y=y^x and x,y both integer then y=x

If u need the proof please tell
[tex]2^4=4^2=16[/tex]

edit: Just noticed Icebreaker said it before me in this same thread, giving him the credit.
 

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