# Is there anyway to see that $h(x,y) = \frac{20}{3+x^2+2y^2}$ represents a graph looking like a mountain?

The function $$h(x,y) = \frac{20}{3+x^2+2y^2}$$ represents the following graph:

Is there anyway to see from the equation, without plotting it, that its graph will look as shown above? Moreover, is there a simple way of coming up with another equation that represents two "mountains", of given heights above the $$xy$$-plane, next to each other (or even more complicated "mountain ranges" with given properties)?

• Is this a function in one or two variables? Jul 29, 2019 at 8:00
• @Dr.SonnhardGraubner It should be two variables, of course. I fixed my mistake now. Jul 29, 2019 at 8:04
• To have two mountains, try $$g(x,y) = h(x,y) + h(x-5, y)$$ Jul 29, 2019 at 8:07
• @K.Claesson Hint: Look at the denominator $d(x,y) = 3 + x^2 + 2y^2$. Where does the minimum value occur and what are the contours of $d(x, y)$ shaped like? Jul 29, 2019 at 8:07
• @ZeroXLR By inspecting the denominator I see that the minimum value of $h$ is 0 and occurs as $x \vee y \rightarrow \pm \infty$. Furthermore, the maximum value of $h$ is $20/3$ (which is the height of the mountain) and occurs at $(x,y) = (0,0)$. Jul 29, 2019 at 8:15

The function $$x^2+y^2+1$$ is known to describe a vertical paraboloid, i.e. the result of rotating the parabola $$z=x^2+1$$ around it axis. Hence it has a single minimum at $$(0,0,1)$$ and goes to infinity in every direction.

You can dilate the coordinates along the axis $$x$$ and $$y$$ to break the rotational symmetry, and every section becomes an ellipse instead of a circle: $$z=ax^2+by^2+1$$. You can also add a mixed term so that the ellipse axis takes any direction: $$ax^2+bxy+cy^2+1$$, making sure that $$b^2<4ac$$ to keep ellipses.

Now the inverse of this function,

$$\frac{z_m}{ax^2+bxy+cy^2+1}$$ is a "mountain-like" function, which has a single maximum at $$(0,0,z_m)$$ and goes down to zero in every direction. You can adjust the height via the parameter $$z_m$$, the steepness via $$a$$ and $$c$$, and the orientation via $$b$$.

You can translate a mountain to another location by shifting the coordinates, giving

$$z=\frac{z_m}{a(x-x_m)^2+b(x-x_m)(y-y_m)+c(y-y_m)^2+1}.$$

Now to obtain a more complex landscape, you can combine several mountains by

• taking the sum of such functions with different parameters,

• taking the maximum.

The sum will result in a blending effect. The maximum keeps the original surfaces and shows them "intersecting". More generally, you can tune the blending by a formula such as

$$\sqrt[\alpha]{z_0^\alpha+z_1^\alpha}$$ where $$\alpha$$ is a free parameter.

For $$r>0$$, the equation $$x^2+2y^2=r$$ is an ellipse centered at the origin. By increasing $$r$$, the ellipse becomes larger and larger. Moreover the function $$r\to \frac{20}{3+r}$$ is decreasing with respect to $$r$$. So the given function $$f$$ attains its maximum value $$\frac{20}{3}$$ at the origin where $$r=0$$ (peak of the mountain). Then, as $$r$$ goes to infinity, the height of the level set (or isoline) of the function, the ellipse $$x^2+2y^2=r$$, decreases to zero.

In order to have two peaks at height $$h>0$$, we may try $$f(x,y)=\frac{h}{1+(x^2+y^2)((x-1)^2+(y-1)^2)}.$$ Now the maximum value $$h$$ is attained at $$(0,0)$$ and $$(1,1)$$.

Here are some considerations that can lead you to conclude several aspects of this mountain shape:

• $$x^2 + 2y^2$$ is almost rotationally symmetric around the origin, except that one unit in $$y$$ direction has the same effect as $$\sqrt2$$ units in $$x$$ direction. From this you can read that the lines of equal height are “squashed circles”, or ellipses.
• For large $$x,y$$ the denominator becomes arbitrarily large, which combined with the fixed numerator makes the surface converge towards zero further out.
• On the other hand, the thing has a peak where the denominator becomes minimal, namely at $$x=y=0$$. The value there is $$\frac{20}3$$ so it's still finite, no funky sign changes passing through infinity.

If you want multiple mountains, just add some such terms. You probably want to move them by using $$x-x_0$$ in stread of $$x$$, and $$y-y_0$$ instead of $$y$$. Perhaps you also want to rotate them. So you would go for one of these forms:

$$\frac{a}{b+c(x-x_0)^2+d(y-y_0)^2+e(x-x_0)(y-y_0)}\qquad \frac{a}{b+cx+dy+ex^2+fxy+gy^2}$$

Each of these has $$7$$ free parameters you can use as knobs to tweak. The left version makes it easier to tell that the center is at $$(x_0,y_0)$$ while the right version with $$\dots,e,f,g$$ has a simpler structure in the denominator. In both versions, you can scale all numbers by a constant factor to achieve the same shape, so geometrically there are only $$6$$ degrees of freedom in picking individual mountains according to this pattern.

If you "stretch" the graph in the $$y$$ direction, that is, replace $$y$$ with $$\frac{1}{\sqrt{2}}y$$, then the function becomes $$g(x,y) = \frac{1}{3+x^2+y^2}.$$

So basically, the graph of the original function is the graph of $$g$$, but stretched along the $$y$$ axis by a factor of $$\frac{1}{\sqrt{2}}$$.

The graph of $$g$$ is simpler, since $$g(x,y)=\frac{1}{3+\|(x,y)\|^2}$$. This (3D) graph is just the (2D) graph of $$\frac{1}{3+x^2}$$, rotated around the $$y$$ axis.

$$h(x,y) = \frac{20}{3+x^2+2y^2}$$ Obviously $$h$$ is maximum when $$x=0$$ and $$y=0$$. So it appears as a peak.

$$h\to 0$$ when $$x\to\infty$$ and/or $$y\to\infty$$. So it appears as plains.

More generally : $$h(x,y) = \frac{H}{1+A(x-a)^2+B(y-b)^2}$$ $$H>0,A>0,B>0,a,b$$ are constants.

The coordinates of the peak are $$(a,b)$$ and its high is $$H$$.

$$A$$ and $$B$$ define the size of the ellipse as shape of the "mountain".

Several "mountains" are obtained for example with : $$h(x,y) = \frac{H_{1}}{1+A_1(x-a_1)^2+B_1(y-b_1)^2}+ \frac{H_{2}}{1+A_2(x-a_2)^2+B_2(y-b_2)^2}+\text{etc.}$$

With some imagination, yes, there is a quick and dirty way.

Know how the function $$y=\frac{1}{x^2}$$ looks like and know how the function of an ellipse in the $$x,y-$$plane and then try to analyze the intersections of the $$3D$$ surface $$z=\frac{20}{3+x^2+2y^2}$$ with the $$y,z-$$ and $$x,z-$$planes.

(1) Assume $$x=0$$, then we are in the $$y,z-$$plane and we have $$z = \frac{20}{3+2y^2}$$.

We can just draw $$\color{red}{z=\frac{1}{y^2}}$$ instead of $$z = \frac{20}{3+2y^2}$$ to get a rough idea about the intersection of the surface with the $$y,z-$$plane:

(2) Assume $$y=0$$, then we are in the $$x,z-$$plane and we have $$z = \frac{20}{3+x^2}$$.

We can just draw $$\color{blue}{z=\frac{1}{x^2}}$$ instead of $$z = \frac{20}{3+x^2}$$ to get a rough idea about the intersection of the surface with the $$x,z-$$plane:

(3) Now at the point $$(x,y)=(0,0)$$ we have $$z(0,0) = \frac{20}{3}$$. So, the point $$(x,y,z)=(0,0,20/3)$$ is on the surface:

(4) We see that the surface doesn't touch the $$x,y-$$plane. That's why we could try $$z=1$$ instead of $$z=0$$ to see the result of the intersection with the plane $$z=1$$ (which has points parallel to the $$x,y-$$plane). Setting $$z=1$$ and plugging in the function yields $$\color{green}{x^2+2y^2=17}$$, which is an equation of an ellipse and we are done

Some hints regarding graphs of $$z= f(x,y)$$ Monge forms: In $$z=\frac {1}{r^2}= \frac {1}{x^2+y^2}$$ it goes infinite at pole $$r=0$$ of a single peak.

To remedy this i.e., to have a flat plateau add some positive real number in denominator.. like

$$z=\frac {1}{4+r^2}= \frac {1}{4+ x^2+y^2}$$

or Gauss probability /Bell curve rotated peak where the finite polynomial series has even terms included in exponential function to force center to become flat.

$$z= e^ {- (x^2+y^2) / (2 \sigma^2)}$$

To get two peaks work around two foci. For example Cassinian Ovals of const. product $$b^4$$, height $$z$$ can replace its single parameter/constant $$b^4$$.

$$b^4 = (x+a)^2 * (x-a)^2 \rightarrow z^4 = (x+a)^2 * (x-a)^2$$

We can raise the power of polynomial to get infinitely many peaks, valleys and cols. Sine function for example produces infinite periodic mountain range.. $$z = \sin x + \sin y$$