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

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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. -- kainaw™ 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:
$f(x,y)={\begin{cases}(x^{2}+y^{2})\sin(x^{2}+y^{2})^{\frac {-1}{2}},&{\mbox{if }}(x,y)\neq (0,0)\\0,&{\mbox{if }}(x,y)=(0,0)\end{cases}}$
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#Differentiability_in_higher_dimensions. 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 $f(x)=\left\{{\begin{array}{cc}x^{2}\sin 1/x&x\neq 0\\0&x=0\end{array}}\right.$ 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#Analytical definition, 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.