Talk:Lagrange multiplier

Notation change

Hi,

I just changed looking for an extremum of g to looking for an extremum of h although I'm not absolutely sure. But I think it is the right term.

Thanks for catching that; that occurrence of g seems to have been missed when the notation was changed in December.--Steuard 20:51, Jan 28, 2005 (UTC)

Strictly looking for an extremum of $h({\vec {x}},{\vec {\lambda }})$ also implies the original ${\vec {g}}({\vec {x}})={\vec {0}}$ via ${\frac {\partial {\vec {h}}}{\partial {\vec {\lambda }}}}={\vec {0}}$. 84.160.236.56 19:29, 6 Feb 2005 (UTC)

Reformulating Lagrangian as Hamiltonian

Citation from the article: "One may reformulate the Lagrangian as a Hamiltonian, in which case the solutions are local minima for the Hamiltonian. This is done in optimal control theory, in the form of Pontryagin's minimum principle." This seems a very important statement, and the article should include detailed explanations and an example of such transform "Lagrangian to Hamiltonian". Links here redirect to general theory of Hamiltonian dynamics and do not explain how this reformulation can be done

Puzzling assertion

The section Modern formulation via differentiable manifolds contains the following sentence:

"In what follows, it is not necessary that $M$ be a Euclidean space, or even a Riemannian manifold."

But it is not stated what is necessary for $M$ to be.

I hope someone knowledgeable about this matter can fix this, by stating some reasonable condition(s) that $M$ must satisfy.

Surely it must satisfy *some* condition(s) for these operations to make sense.

possible typo

The section about "Modern formulation via differentiable manifolds" uses $\Lambda$. but the link in that section for Exterior Algebra uses $\bigwedge$. Not a mathematician so figured I'd write about it and get a second opinion. Thanks for working on it

mistake in example 4

The derivative of p log p is wrong. ~2026-13955-08 (talk) 11:19, 4 March 2026 (UTC)

"Rather naturally"

In the section Summary and rationale: "The relationship between the gradient of the function and gradients of the constraints rather naturally leads to a reformulation of the original problem, known as the Lagrangian function or Lagrangian." My bolding and italicizing emphasis of "rather naturally".

Nothing appears "rather natural" to anyone wishing to learn about the Lagrange multiplier and therefore visiting this Wiki-entry to do so. The aforementioned relationship between various gradients leads "rather naturally" to the Lagrangian only for people who don't need to visit this page.

I am not one of those latter people (I am dumb as a bag of rocks; my mind is still in its "rather natural" untutored state). I suggest that the article be rewritten so that it isn't aimed only at people who don't need to read the article. Oikosmonaut (talk) 10:45, 15 May 2026 (UTC)