Wikipedia:Reference desk/Archives/Mathematics/2011 November 15

= November 15 =

LaTeX lists
I want to make a list with two columns in it. I know how to use \begin{enumerate} \item question 1 \item question 2 \end{enumerate} But, like I said, I'd like questions 1 and 2 to be on the same line, then 3 and 4 on the next, etc. — Fly by Night  (  talk  )  15:46, 15 November 2011 (UTC)


 * It sounds like you either want a table or a two-column list. If you want a table, I'd just use that instead of a list. If you want a list that spans two columns I'd use multicolumn like

\begin{multicols}{2} \begin{enumerate} \item a   \item b    \item c \end{enumerate} \end{multicols}
 * You will need to include the multicolumn option at the top of your document. -- k a i n a w &trade; 15:56, 15 November 2011 (UTC)
 * How do I do that? — Fly by Night  (  talk  )  16:04, 15 November 2011 (UTC)

Showing differentiability of a multivariate function when the partial derivatives are not continuous.
To prove the differentiability of a multivariate function, it is a sufficient condition that the partial derivatives are continuous. This is not a necessary condition however, a function can be differentiable at a point even if the partial derivatives are not continuous. How does one show the differentiability of a function in this case? Consider the following function:

$$f(x,y) = \begin{cases} (x^2+y^2)\sin(x^2+y^2)^\frac{-1}{2}, & \mbox{if } (x,y) \ne (0,0) \\ 0, & \mbox{if } (x,y) = (0,0) \end{cases}$$

This function is differentiable at (0,0) even though the partial derivatives are not continuous there. How does one show this?Widener (talk) 15:49, 15 November 2011 (UTC)


 * What makes you think that your function is differentiable at (0,0)? It is not locally linear, it behaves locally like a cone. 98.248.42.252 (talk) 16:53, 15 November 2011 (UTC)
 * Perhaps the -1/2 is meant to go inside the sine, so in polar coordinates function is r2 sin (1/r).--RDBury (talk) 21:24, 15 November 2011 (UTC)
 * That's probably it. Widener, when the theorems you know don't help, prove differentiability from the definition. See Differentiability. 98.248.42.252 (talk) 05:09, 16 November 2011 (UTC)

Weierstrass function
If I take the Weierstrass function and consider only the interval [-1,1] (or for those who anally insist on generality, [a,b] where it is not at its global maximum or minimum), can either endpoint be considered a local max or min? The professor says that the endpoints of any continuous (but not necessarily differentiable) function will be a local minimum or maximum unless it is a constant function, but in the Weierstrass function somewhere in the middle isn't it always possible to find a value for the function in any arbitrary given interval that is smaller and larger than the value at the endpoint, due to its fractal nature? Thanks. 122.72.0.41 (talk) 23:54, 15 November 2011 (UTC)
 * Either your professor is wrong or you misunderstood him, and you don't need to go as far as the Weierstrass function. Take $$f(x)=\left\{\begin{array}{cc}x^2\sin1/x&x\neq0\\0&x=0\end{array}\right.$$ on [0,1]. At 0 it is neither a local maximum nor a minimum. This function is even differentiable. -- Meni Rosenfeld (talk) 05:37, 16 November 2011 (UTC)


 * See Maxima and minima, last paragraph. HTH. --CiaPan (talk) 14:29, 16 November 2011 (UTC) PS. Meni, it could be written simpler with {cases} instead of {array} as described here and there.