Insight on a particular class of curves. (1 Viewer)

qwert73 · Dec 30, 2015

So are you going to staterank MX1 and MX2?

Sien · Dec 30, 2015

qwert73 said:
So are you going to staterank MX1 and MX2?

Yes he is

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qwert73 · Dec 30, 2015

Paradoxica do you go to a selective school?

Paradoxica · Dec 30, 2015

qwert73 said:
Paradoxica do you go to a selective school?

Whether or not I go to a selective school has nothing to do with my natural curiosity.

glittergal96 · Dec 30, 2015

Nothing all that special is going on here, you will get this kind of behaviour with a lot of functions.
I will just give a brief partial answer illustrating where some of the properties come from.

Noting that f(x)=x^x has a minimum at x=1/e, we have x^x+y^y >= 2f(1/e) = 1.3844...

So for k_1 smaller than this quantity, our first curve will vanish, and for k_1 slightly larger than this quantity we expect our curve to be close to (1/e,1/e). In fact, as we are looking at level sets near a local minimum, these curves will look like ellipses about the minimum. (You can find the aspect ratio of these ellipses in terms of the second order local behaviour of x^x+y^y. I haven't done this but it might even be circular as you are suggesting.)

A similar phenomenon is going on with g(x)=x^(-x), it is just a local maximum at 1/e instead with 2g(1/e)=2.8893...

With curves 2 and 4, we still have a stationary point at (1/e,1/e), but this time it is a saddle point. When you take the preimage of saddle points you will get things that look like hyperbolas (again you can look at second order behaviour to find exactly what these hyperbola are). This explains the two branches meeting tangentially in the limit.

Analogous things happen with your second quadruple of functions.

Paradoxica · Dec 30, 2015

glittergal96 said:
Nothing all that special is going on here, you will get this kind of behaviour with a lot of functions.
I will just give a brief partial answer illustrating where some of the properties come from.

Noting that f(x)=x^x has a minimum at x=1/e, we have x^x+y^y >= 2f(1/e) = 1.3844...

So for k_1 smaller than this quantity, our first curve will vanish, and for k_1 slightly larger than this quantity we expect our curve to be close to (1/e,1/e). In fact, as we are looking at level sets near a local minimum, these curves will look like ellipses about the minimum. (You can find the aspect ratio of these ellipses in terms of the second order local behaviour of x^x+y^y. I haven't done this but it might even be circular as you are suggesting.)

A similar phenomenon is going on with g(x)=x^(-x), it is just a local maximum at 1/e instead with 2g(1/e)=2.8893...

With curves 2 and 4, we still have a stationary point at (1/e,1/e), but this time it is a saddle point. When you take the preimage of saddle points you will get things that look like hyperbolas (again you can look at second order behaviour to find exactly what these hyperbola are). This explains the two branches meeting tangentially in the limit.

Analogous things happen with your second quadruple of functions.

At the exact point where the hyperbola branches degenerate, what function best describes the smooth intersecting curves?

glittergal96 · Dec 30, 2015

Paradoxica said:
At the exact point where the hyperbola branches degenerate, what function best describes the smooth intersecting curves?

Well if I wanted to describe/study that union of two curves I would literally just use x^y+y^x-1.3844... . The branches should be symmetric, so finding a way of representing a single one of them but not the other probably wouldn't be particularly useful.

As for what this looks like globally, I am not sure (have you tried using something like mathematica?).

I don't really see a reason that the answer will necessarily be nice (where "nice" means can be parametrised using elementary functions or is at least very simple geometrically).

Paradoxica · Jan 3, 2016

How to find closed form for the saddle point of this curve?

$z = x^2 y+xy+xy^2+x^2+y^2+x^3+y$

$(x,y,z) \approx (-0.4165564424007429267408885089,-0.648702704353748793567371902,-0.1442832165549094621412856336)$

InteGrand · Jan 3, 2016

Paradoxica said:
How to find closed form for the saddle point of this curve?

$z = x^2 y+xy+xy^2+x^2+y^2+x^3+y$

$(x,y,z) \approx (-0.4165564424007429267408885089,-0.648702704353748793567371902,-0.1442832165549094621412856336)$

$Given: $z=f(x,y) = z = x^2 y+xy+xy^2+x^2+y^2+x^3+y$.$

$Compute partial derivatives:$

$\frac{\partial f}{\partial x}(x,y)=2xy + y + y^2 + 2x + 3x^2$

$\frac{\partial f}{\partial y}(x,y)= x^2 + x + 2xy +2y + 1 .$

$Set both partials to 0 to obtain simultaneous equations:$

$\begin{cases}2xy+y+y^2 + 2x + 3x^2 = 0 \quad (1)\\ x^2 + x + 2xy +2y + 1 = 0. \quad (2) \end{cases}$

$From (2),$

$\begin{align*} y\cdot (2x+2) &= - \left(x^2 + x + 1 \right) \quad (\text{observe that }x=-1 \text{ is not a solution so we can divide through by }2x+2 = 2(x+1)) \\ y&= -\frac{\left(x^2 + x + 1 \right)}{2(x+1)} \quad (3).\end{align*}$

$Sub. in (1):$

$\begin{align*}2x\left( -\frac{\left(x^2 + x + 1 \right)}{2(x+1)}\right) -\frac{\left(x^2 + x + 1 \right)}{2(x+1)} +\left(\frac{\left(x^2 + x + 1 \right)}{2(x+1)} \right)^2 + 2x + 3x^2 &= 0, \quad x\neq -1. \end{align*}$

$Now, use algebraic manipulation (multiply by $(x+1)^2$) to obtain$

$\begin{align*} -x (x+1) \left(x^2 + x + 1\right) - \frac{x+1}{2} \left(x^2 + x + 1\right) +\frac{ \left(x^2 + x + 1\right)^2}{4}+2x(x+1)^2 + 3x^2 (x+1)^2 &= 0\end{align*}.$

$Using tedious expansion or WolframAlpha, this is equivalent to $\frac{9}{4}x^4 +6x^3 + \frac{19}{4}x^2 + \frac{1}{2}x-\frac{1}{4}=0 \Longleftrightarrow 9x^4 + 24x^3 + 19x^2 + 2x -1 = 0$. This is a quartic equation, so we can obtain closed-form expressions for the roots using the tedious quartic formula. (Using WolframAlpha, we can see that there are only two real roots.) Then remember to check that the one you wanted is indeed a saddle point by substituting it into the Hessian matrix after finding $y$ from Equation (3): $H_f (x_0,y_0)= \begin{bmatrix} f_{xx}(x_0,y_0) &f_{xy}(x_0,y_0) \\ f_{yx}(x_0,y_0)&f_{yy} (x_0,y_0) \end{bmatrix}$. Then evaluate the determinant $\det \left( H_f (x_0,y_0) \right)$ and hope that it's negative, which implies a saddle point (best just to use decimal approximations rather than exact closed-form expressions for this).$

$In summary, the $x$-value you are after turns out to be the one and only negative root of the quartic equation $9x^4 + 24x^3 + 19x^2 + 2x -1 = 0$.$

Paradoxica · Jan 10, 2016

$x^xy^y=0.4791417088$

The curve vanishes near this value. Wondering what it's closed form is.

https://www.desmos.com/calculator/bjkcdx9gwc

Do any of these enclosed areas have closed form numerical values? Only for the sections of the curve that converge, obviously.

InteGrand · Jan 23, 2016

Paradoxica said:
$x^xy^y=0.4791417088$

The curve vanishes near this value. Wondering what it's closed form is.

https://www.desmos.com/calculator/bjkcdx9gwc

Do any of these enclosed areas have closed form numerical values? Only for the sections of the curve that converge, obviously.

$\noindent For the curve in your post, let us consider the curve $x^x y^y = c$, where $c>0$ is a constant and $x,y>0$. First of all, note that the minimum value of the function $\phi (t):= t\ln t$ is $\phi _{\text{min.}}=-\mathrm{e}^{-1}$ (occurring at and only at $t=\mathrm{e}^{-1}$). This is very easy to prove using HSC calculus. For reference/convenience, here is a link to the graph of $y=t\ln t$:$

https://graphsketch.com/?eqn1_color..._lines=1&line_width=4&image_w=850&image_h=525 .

$\noindent Now consider $x^x y^y = c$ for $x,y>0$. We have real positive solutions iff there exist $x,y>0$ such that$

$\begin{align*}x^x y^y &= c \\ \Longleftrightarrow \ln \left(x^x y^y \right)&= \ln c \\ \Longleftrightarrow \ln \left(x^x \right) + \ln \left(y^y \right)&= \ln c \\ \Longleftrightarrow x\ln x + y\ln y &= \ln c. \end{align*}$

$\noindent Now since $t\ln t \in \left[- \mathrm{e}^{-1},\infty\right)$ for $t>0$ (we can refer to the graph above to see this), we see that if $\ln c> 2\cdot \left(- \mathrm{e}^{-1}\right)$, there will exist solutions, and if $\ln c< 2\cdot \left(- \mathrm{e}^{-1}\right)$, there will be no solutions. If $\ln c=2\cdot \left(- \mathrm{e}^{-1}\right)$, then there will be the single solution $(x,y) = \left(\mathrm{e}^{-1},\mathrm{e}^{-1} \right)$ (since $t\ln t$ attains its minimum at and only at $t=\mathrm{e}^{-1}$). So the critical value of $c$ you are after (where the curve will disappear once $c$ goes below) is the $c$ such that $\ln c = -2\mathrm{e}^{-1}$. Solving for $c$, we see that this is $c_{\text{critical}}=\mathrm{e}^{-2\mathrm{e}^{-1}}\approx 0.479141...$.$

$\noindent I haven't looked at your other curves, but maybe you can approach them in a similar manner to this one.$

Insight on a particular class of curves. (1 Viewer)

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