Skip to main content
4 events
when toggle format what by license comment
Dec 7, 2022 at 23:42 comment added user496046 Projecting on normal to $S^d$ gives $(a.x) |x|^2/2 - (a.x) |x|^2 = - (a.x) |x|^2/2$ and by removing that normal component from $X(x)$ we get a $|x|^2 - (a.x) x$ in agreement with stated claims on this webpage for $|x|=1$.
Dec 7, 2022 at 23:17 review Late answers
Dec 8, 2022 at 1:47
S Dec 7, 2022 at 22:57 review First answers
Dec 8, 2022 at 1:47
S Dec 7, 2022 at 22:57 history answered user496046 CC BY-SA 4.0