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![[parent]](http://images.planetmath.org:8080/images/uparrow.png) Viewing Correction to 'Lagrange multiplier method, proof of'
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Proof needs regularity condition on g by stevecheng
Correction id: 6941
Filed on: 2005-07-28 22:12:58
Status: Rejected on 2006-03-01 06:46:40
Type: Erratum
Correction text:
Hi,
In summary, it is necessary to assume grad g != 0
for your proof to work. You probably want to clarify this.
The problem step is that r' may be zero,
in that case we cannot conclude grad f is parallel to grad g[*].
Moreover, even if we could, if it happens that grad g = 0
but grad f != 0, one is forced to write
\lambda grad f = grad g instead.
Actually the problem of grad g = 0 and r' = 0
are related --- grad g != 0 is precisely the condition
that guarantees that there exists a curve $r(t)$
such that $r' != 0$ at the stationary point in question.
The formal proof requires the implicit function theorem,
though I realize that you might be reluctant to introduce
these complications in your entry --- that's life, I guess :)
FYI, I discuss some counterexamples involving grad g = 0
in:
http://planetmath.org/?op=getobj&from=objects&id=7276
(Sorry if I seem to be picking on your proof --- actually
in studying my own proof for the Lagrange multiplier method,
I was curious why your proof didn't seem to
need the usual regularity condition...)
[*] Meta correction:
Actually, you say grad f cross grad g is zero.
I think this is confusing, because the cross product
is usually used only in three dimensions, whereas your proof
is for two dimensions. Though I understand what you are saying,
the appropriate operator to use here is det, the determinant.
Yet, det(grad f, grad g) = 0 would look very baffling in this context,
so I suggest simply saying that "grad f and grad g are linearly
dependent", or simply "grad f and grad g are parallel vectors".
Thanks,
// Steve
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No comment from object owner aplant.
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