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Does summation over repeated indices need to be made explicit ?: it's implicit in Einstein summation notation
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==Link does not work==
The site referenced at the end
http://www.exampleproblems.com/
is not functional. It would be better to remove it.

==Statement Section Unclear==
==Statement Section Unclear==
Many of the symbols in the statement section are undefined. Particularly, I have no idea what R, X, and Y are. In fact, the entire line
Many of the symbols in the statement section are undefined. Particularly, I have no idea what R, X, and Y are. In fact, the entire line

Revision as of 08:37, 8 February 2008

The site referenced at the end http://www.exampleproblems.com/ is not functional. It would be better to remove it.

Statement Section Unclear

Many of the symbols in the statement section are undefined. Particularly, I have no idea what R, X, and Y are. In fact, the entire line

is very confusing. Can anyone expand it to make it clearer?--132.239.27.145 19:14, 21 September 2007 (UTC)[reply]

Proof Feedback

I'd appreciate feedback on this proof. Is it too long? Too technical? Not technical enough? A simple proof is very appropriate for this page, I'm just not sure if this is that proof. --Dantheox 02:36, 14 December 2005 (UTC)[reply]

How exactly do we come in Proof in from partial derivative of F with respect to to partial derivatives of F with respect to and ? I don't get it.

It's a standard application of the chain rule -- you can expand out a total derivative with respect to in terms of the derivatives with respect to other quantities. --Dantheox 04:02, 7 March 2006 (UTC)[reply]
And where does the sum come from? In http://en.wikipedia.org/wiki/Chain_rule#Chain_rule_for_several_variables there is a sum, since the f is a sum of u and v. And here we just have F. I'm not very familiar with partial derivatives, just know how to compute simple ones.
The sum comes from the matrix product in the multidimensional case: Have a look at http://en.wikipedia.org/wiki/Chain_rule#The_fundamental_chain_rule. For the first example with in http://en.wikipedia.org/wiki/Chain_rule#Chain_rule_for_several_variables choose E = R, F= R2, G=R; you have to regard as to define a single function R → R2. The derivative of this function is (at each point) a linear mapping from R to R2 represented by a column vector (the two entries being the derivatives of each component funtion resp.). The derivative of is (at each point) a linear map from R2 to R represented by a row vector (the two entries being the partial derivatives in direction of the two coordinates). The matrix product of the two matrices gives exactly the formula in http://en.wikipedia.org/wiki/Chain_rule#Chain_rule_for_several_variables.

Euler-Lagrange Methods & Lagrange multipliers

Is anybody able to give me some hints about the connection between Euler-Lagrange Methods and Lagrange multipliers? At first glance they seem to be closely related. Maybe somebody can clarify this and maybe add a short note to the articles. Cyc 12:37, 22 September 2006 (UTC)[reply]

Does summation over repeated indices need to be made explicit ?

My concern is with this section:

\partial_\mu \left( \frac{\partial \mathcal{L}}{\partial ( \partial_\mu \psi )} \right) - ...

   where ...
   \partial\, is a vector of derivatives:
       \partial_\mu = \left(\frac{1}{c} \frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right). \,

Seems like it could be interpreted as a set of N equations (N being the dimension of the "vector"), rather that the sum over mu (do we always assume summation over repeated indices ? Apologies if this is nit-picking.

Many thanks

—Preceding unsigned comment added by JM516 (talkcontribs) 23:03, 4 October 2007 (UTC)[reply]

In quantum mechanics, the sum is implicit due to Einstein summation notation which is taken as read. — ras52 (talk) 14:41, 5 December 2007 (UTC)[reply]