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

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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 $f:\mathbb {R} ^{n}\to \mathbb {R} ,x\in \mathbb {R} ^{n},I\subset \mathbb {R} ^{n}$. Assume that f is continuous at $I-\{x\}$ (the "minus" sign denotes set difference), and that $\forall v\in \mathbb {R} ^{n}$ the limit $lim_{a\to 0}f(x+av)$ exists (and finite). Then, the multivariate limit $lim_{y\to x}f(y)$ 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 $f(x)={\begin{cases}2y/x^{2}-y^{2}/x^{4}&x>0,0<y<2x^{2}\\0&{\textrm {Otherwise}}\end{cases}}$. Here, not only does the limit $\lim _{a\to 0}f(x+av)$ 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 $f(x)={\begin{cases}1&x>0,y=x^{2}\\0&{\textrm {Otherwise}}\end{cases}}$ 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)