- #1
- 872
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If we have a function f(x,y) for which we wish to find the stationary points, then there is one set of rules for classifying those points.
Let there be an extremum of f(x,y) at (x,y) = (a,b), and D = fxx(a,b)*fyy(a,b) - {fxy(a,b)}², then
1) D > 0 and fxx > 0 => min
2) D > 0 and fxx < 0 => max
3) D < 0 => saddle point
4) D = 0 - indeterminate - needs deeper analysis.
Now I can never remember this lot of rules so I made up my own little set of rules - as follows.
In 2-D maths, fxx > 0 gives a min, and fxx < 0 gives a max, so I extended this to 3-D maths, as below.**
My rules
1) if fxx and fyy > 0 then a min
2) if fxx and fyy < 0 then a max
3) if fxx and fyy of opposite sign then a saddle
Now, this has always worked well for me; I find it easy to transfer the 2-D rules to a 3-D situation, and it saves me having to work out fxy.
The only thing it doesn't cover is; if fxx and fyy are the same sign (which by my rules should be either a max or a min) is it possible for D to be < 0, which would mean the TP was a saddle?
So, my questions are:
1) can anyone find an error in my rules?
2) does anyone know of a situation where fxx and fyy were both of the same sign, but the TP was a saddle?
**: here, fxx should really be d²y/dx²
Let there be an extremum of f(x,y) at (x,y) = (a,b), and D = fxx(a,b)*fyy(a,b) - {fxy(a,b)}², then
1) D > 0 and fxx > 0 => min
2) D > 0 and fxx < 0 => max
3) D < 0 => saddle point
4) D = 0 - indeterminate - needs deeper analysis.
Now I can never remember this lot of rules so I made up my own little set of rules - as follows.
In 2-D maths, fxx > 0 gives a min, and fxx < 0 gives a max, so I extended this to 3-D maths, as below.**
My rules
1) if fxx and fyy > 0 then a min
2) if fxx and fyy < 0 then a max
3) if fxx and fyy of opposite sign then a saddle
Now, this has always worked well for me; I find it easy to transfer the 2-D rules to a 3-D situation, and it saves me having to work out fxy.
The only thing it doesn't cover is; if fxx and fyy are the same sign (which by my rules should be either a max or a min) is it possible for D to be < 0, which would mean the TP was a saddle?
So, my questions are:
1) can anyone find an error in my rules?
2) does anyone know of a situation where fxx and fyy were both of the same sign, but the TP was a saddle?
**: here, fxx should really be d²y/dx²