For a general function F(x,y) with a stationary point at (x_0,y_0) we have the Taylor series expansion
F(x_0+xi,y_0+eta) = F(x_0,y_0) + 1/(2!)(F_{x x} xi^2+F_{yy}eta^2+2F_{xy}xi eta ) +ldots
For the function
f(x)=x y\ e^{-x^2-y^2}
we have
(del f)/(del x) = ye^{-x^2-y^2}+x y (-2x)e^{-x^2-y^2}
qquad = y(1-2x^2)e^{-x^2-y^2}
(del f)/(del y) = xe^{-x^2-y^2}+x y (-2y)e^{-x^2-y^2}
qquad = x(1-2y^2)e^{-x^2-y^2}
It is easy to see that the both the first derivatives vanish at the following ponrs
- (0,0)
- (0,pm 1/sqrt2)
- (pm 1/sqrt2, 0)
- (pm 1/sqrt2, pm 1/sqrt2)
To examine the nature of these stationary points, we need to look at the behavior of the second derivatives there.
Now
(del^2 f)/(del x^2) = y(-4x)e^{-x^2-y^2}+y(1-2x^2) (-2x)e^{-x^2-y^2}
qquad = x y(4x^2-6)e^{-x^2-y^2}
and similarly
(del^2 f)/(del y^2) = xy(4y^2-6)e^{-x^2-y^2}
and
(del^2 f)/(del xdel y) = (1-2y^2)e^{-x^2-y^2}+x(1-2y^2) (-2x)e^{-x^2-y^2}
qquad = (1-2x^2-2y^2+4x^2y^2)e^{-x^2-y^2}
qquad = (1-2x^2)(1-2y^2)e^{-x^2-y^2}
So for (0,0) we have (del^2 f)/(del x^2) = (del^2 f)/(del y^2) = 0 and (del^2 f)/(del x del y)=1 - hence
f(0+xi,0+eta) = f(0,0) + xi eta = xi eta
If you approach (0,0) along the line x=y, this this becomes
f(0+xi,0+xi ) = xi^2
and so (0,0) is obviously a minimum if you approach from this direction. On the other hand, if you approach along the line x=-y we have
f(0+xi,0-xi ) = -xi^2
and so (0,0) is a maximum along this direction,
Thus (0,0) is a saddle point.
For (1/sqrt2,1/sqrt2) it is easily seen that
(del^2 f)/(del x^2) = (del^2 f)/(del y^2) = -2e^{-1/2}<0 and (del^2 f)/(del x del y)=0
which means that
f(1/sqrt 2 + xi,1/sqrt 2+eta) = f(1/sqrt 2,1/sqrt 2) -e^{-1/2(xi^2+eta^2)}
So, the function decreases whichever way you move away from (1/sqrt 2,1/sqrt 2) and this is a local maximum. It is easily seen that the same goes for (-1/sqrt2,-1/sqrt2) (this should have been obvious, since the function stays the same under (x,y) to (-x,-y)!
Again, for both (1/sqrt2,-1/sqrt2) and (-1/sqrt2,1/sqrt2) we have
(del^2 f)/(del x^2) = (del^2 f)/(del y^2) = 2e^{-1/2}>0 and (del^2 f)/(del x del y)=0
So, both these points are local minima.
The four points (0,pm 1/sqrt2) and (pm 1/sqrt2, 0) are more problematic - since all second order derivatives vanish at these points. We have to now look at higher order derivatives. Fortunately, we do not really need to work very hard for this - the very next derivative yields
(del^3 f)/(del x^3) = -2y(3-12x^2+4x^4)e^{-x^2-y^2}
which is non-zero for both (0,pm 1/sqrt2) and (pm 1/sqrt2, 0). Now, this means that, for example
f(0+xi,1/sqrt 2) = f(0,1/sqrt 2) +1/3((del^3 f)/(del x^3) )_{(0,1/sqrt2)}xi^3+...
which shows that this will increase from f(0,1/sqrt 2) in one direction, and decrease from it in the other. Thus (0,1/sqrt2) is a **point of inflection. The same argument works for the other three points.
