\HHT Method of Lagrange Multipliers h\\b\w
\bMethod of Lagrange Multipliers.b\
\bProblem.b\ For given function f(x) where x=(x\[1 \ ,x\[2 \ ,...,x\[n \ ) - n variables,
find an extrema if x belongs to domain H which is set by one or more
constraints:
g(x) = 0 or
g\[1 \ (x) = 0
g\[2 \ (x) = 0
...
g\[m \ (x) = 0.
\bNote.b\ Summation rule x\[1 \ y\[1 \ + x\[2 \ y\[2 \ + ... = x\[i \ y\[i \ is assumed below.
\bIdea.b\
Let's start from case m=1.
As for three dimesional case,
let's call domain H and contour f(x)=const as "hypersurface".
As in two-dimesional case n=2, suppose that contour of f is "tangent"
to the hypersurface H in point of extrema x=e.
We need to develop more precisely what does this mean "tangent".
In neighbourhood of p, an arbitrary point of hypersurface H,
equation of hypersurface g can be written as:
Gx=0
if to choose the point p as an origin,
denote j-th partilal derivative of g with Gj,
and shortcut inner product G\[j \ x\[j \ with Gx.
The last equation means that H is approximated with S, where
S is vector subspace of IR\]n \ ,
S is orthogonal to G, and dim S = n-1.
Contour's C equation f(x)-const=0 can be written in similar form
Fx=0,
where F = (F\[1 \ , F\[2 \ , ... , F\[n \ ), F\[j \ is partial derivative f by x\[j \ , and
contour C is approximated with subspace T which is orthogonal to F.
Now, we can precisely say that T and S are tangent iff
F and G are parallel, or
if G is not vanished, there exist number L that (*)
F - L*G = 0, or
grad(f-Lg) vanishes in point p.
But, why T and S must be tangent?
At this point, it can be seen that developed language allows to
formulate condition of extrmality without geometrical language
which brought us so far.
Indeed, for p=e, for each x from S, Fx must be zero.
Otherwise, f(x)-f(0) and f(-x) - f(0) have
different sign and p is not an extremum.
This means that F and S are orthogonal which implies (*).
(As F and G are vectors in n-dimensional space which both
are orthogonal to (n-1)-dimensional subspace.)
h\