- f(x,mx) = m x^2/ ( x^2 + 4 m^2 x^2) = m/(1+4m^2).
- The level set for -1/4 is found by setting f(x,y) = -1/4 and
solving for the (x,y) values that satisfy that equation. We
find out that the level set is the line
y = -x/2.
The level set of 0 is the union of the lines x=0 and y=0.

The level set of 1/4 is the line y=x/2.

The level set of 1/4 is the union of the lines y=1/4 x and y=-x.

- Evaluating the gradient along y= +/- x/2, we discover that the gradient is the zero vector on those lines. Thus these graph of f is flat along those lines, having a horizontal tangent plane at each point along the lines (with the possible exception of the origin). Equivalently, f is at a maximum or minimum all along those lines.
- Along any line y=mx, the function has the constant value m/(1+4m^2). In particular, the limit as we approach the origin along the line y=mx is m/(1+4m^2).
- The function cannot be continuous at (0,0) since we get different
limiting values depending on how we approach the origin.
The function is constant along radial lines through the origin.
the graph has a big ridge along y=x/2 at height1/4;
it has a big valley (or channel) along y=-x/2 at height -1/4.
The function increases as we sweep out angles between -Pi/8 and Pi/8, then decreases from P/8 to Pi-Pi/8. This pattern continues from the angle Pi-Pi/8 until we return to -Pi/8.

It seems that it is easiest to visualize the function on circles. For example, if we restrict the function to the circle parametrized by (cos(t), sin(t), the function becomes

cos(t)sin(t) ------------------------ (cos(t))^2 + 4(sin(t))^2

I can graph this on my graphing calulator to see the two peaks and two valleys of f restricted to this circle.[A 2D graph such as described above, or a 3D sketch is also required.]

- The definition of continuity says that we need to
define g(0,0) so that it matches the limits that we get
as we approach the origin IN ANY WAY. But, heck, we can't even
approach the origin along STRAIGHT LINES without getting different
limits!
For example, suppose we chose to define g(0,0)=0. While it is true that we have equality between g(0,0) and the limit of g as we approach the origin along a coordinate axis, it is NOT true that the limit as we approach (0,0) along any OTHER line will give us g(0,0).

Last modified: Fri Nov 22 10:06:06 1996