Talk:Homogeneous function

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What the hell is x times gradient of f(x) supposed to mean, dot product?

It means that for a vector function f(x) that is homogenous of degree k, the dot production of a vector x and the gradient of f(x) evaluated at x will equal k * f(x). CodeLabMaster 12:12, 05 August 2007 (UTC)
Yes, as can be seen from the furmula under that one. I've added the dot and changed vector symbols to bold. mazi 18:04, 22 February 2006 (UTC)

[edit] Not all homogeneous functions are differentiable

The article, before I changed it a moment ago, implied that all homogeneous functions are differentiable. Here's a counterexample: f: R^2 --> R, k=1, with f(x,y)= either x (if xy>0) or 0 (otherwise). --Steve 03:23, 6 August 2007 (UTC)

[edit] Is the derivative theorem correct?

According to planetmath [1], the theorem about derivatives is not correct unless we replace "homogeneous" by "positive homogeneous" throughout. Their counterexample is wrong (I just submitted a correction on the site), but could that claim be correct? Does anyone have a reference, or a proof, or a proper counterexample?

Update: The person maintaining that planetmath page responded to my correction by taking away the counterexample but keeping the claim. Again, a reference, proof, or proper counterexample is needed to resolve this. --Steve 15:51, 5 October 2007 (UTC)

The result is correct for functions which are homogeneous of degree k. I've added the elementary proof of this result to the page (and merged "Other properties" with "Euler's theorem" as the proofs are very similar). Is the planetmath contributor worried about α = 0? Clearly the definition of homogeneous of degree k for k < 0 has to be modified so that the condition holds for all \alpha \neq 0 , and I've just changed this too. Mark (talk) 16:31, 11 February 2008 (UTC)

[edit] Notation

Notation such as

\frac{\partial f}{\partial x_1} (\alpha \mathbf{x})

can be confusing: Are we differentiating the expression with respect to the first component of \mathbf{x} or do we mean the partial derivative of f with respect to its first argument evaluated at the point \alpha \mathbf{x} ?

It's therefore better to write

\frac{\partial f}{\partial x_1} (\alpha \mathbf{y}) :

now it's clear that x_1,\ldots, x_n are the arguments of f and we are differentiating with respect to the first argument of f evaluated at the point \alpha \mathbf{y} .

I've cleaned up the notation in my proof of Euler's theorem accordingly. 91.21.25.30 (talk) 21:50, 11 February 2008 (UTC)

I agree. Sorry about my incorrect edit to that effect earlier, thanks for reverting :-) --Steve (talk) 18:00, 11 February 2008 (UTC)