Testwiki:Reference desk/Archives/Mathematics/2016 January 15

{| width = "100%"

|- ! colspan="3" align="center" | Mathematics desk |- ! width="20%" align="left" | < January 14 ! width="25%" align="center"|<< Dec | January | Feb >> ! width="20%" align="right" |Current desk > |}

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

January 15

Alzebra

Solve it (a+b)^4 — Preceding unsigned comment added by 49.126.0.41 (talk) 00:39, 15 January 2016 (UTC)

There's nothing to solve, but you can expand it to a^4 + 4(a^3)b + 6(a^2)(b^2) + 4a(b^3) + b^4 using the Binomial theorem. --ℕ ℱ 02:25, 15 January 2016 (UTC)

From the title, I assumed that this was a strip(p)ed-down problem→86.139.120.76 (talk) 12:36, 18 January 2016 (UTC)

Multivariate Limit

Let $f:{ℝ}^n\to ℝ,x\in {ℝ}^n,I\subset {ℝ}^n$. Assume that f is continuous at $I-\{x\}$ (the "minus" sign denotes set difference), and that $\mathrm{\forall }v\in {ℝ}^n$ the limit $li{m}_{a\to 0}f(x+av)$ exists (and finite). Then, the multivariate limit $li{m}_{y\to x}f(y)$ exists? 213.8.204.40 (talk) 08:48, 15 January 2016 (UTC)

I believe the classical counterexample is

$f(x,y)=\frac{xy}{x^2+y^2}.$

around the origin .Dmcq (talk) 10:41, 15 January 2016 (UTC)

Another counterexample is $f(x)=\{\begin{array}{cc}2y/x^2-y^2/x^4& x>0,0<y<2x^2\\ 0& \text{Otherwise}\end{array}$. Here, not only does the limit $\underset{a\to 0}{\lim }f(x+av)$ exist, it's the same (0) for all v. But the function is still discontinuous at the origin.

(I've started with the known example $f(x)=\{\begin{array}{cc}1& x>0,y=x^2\\ 0& \text{Otherwise}\end{array}$ and modified it to be continuous everywhere except the origin. I hope I didn't mess it up.) -- Meni Rosenfeld (talk) 11:24, 15 January 2016 (UTC)