bio | website | sma.epfl.ch/~moric |
---|---|---|
location | ||
age | ||
visits | member for | 4 years, 7 months |
seen | Feb 4 '13 at 13:05 | |
stats | profile views | 408 |
Jul 2 |
awarded | Curious |
Apr 23 |
awarded | Yearling |
Feb 3 |
asked | another diameter-perimeter-area inequality |
Feb 2 |
accepted | a diameter-perimeter-area inequality for convex figures |
Feb 2 |
comment |
a diameter-perimeter-area inequality for convex figures
oh i see, you are right. so there is no non-trivial inequality of this type, which explains why i couldn't find it in the literature :) |
Feb 1 |
comment |
a diameter-perimeter-area inequality for convex figures
I had checked before both references (by Scott and Awyong and by Burago and Zalgaller), but I couldn't find what I need. |
Feb 1 |
asked | a diameter-perimeter-area inequality for convex figures |
Nov 9 |
comment |
two polynomial equations
in fact, there is a "big" chance, maybe up to some small technical assumptions on $f$, for this reason: the partial differential equation we get for $f$ by eliminating $t$ is precisely the equation that characterizes ruled surfaces $z=f(x,y)$, see mathoverflow.net/questions/77865/… however, in the old book by salmon a shorter deduction of the characterizing equation is given (that uses no diff geo), but i find it incomplete. the statement above, if it has a short proof (that doesn't use the fact we are actually proving!), would clarify it. |
Nov 9 |
comment |
two polynomial equations
maybe that part was equally convoluted, but i need the following: fix a point $(x_0,y_0)$; that point will give us some $t$, say $t=15$. now we are looking at the set of all other points $(x,y)$ that would give us also $t=15$, and i would like to claim that that set must contain a line through $(x_0,y_0)$. |
Nov 9 |
comment |
two polynomial equations
@fedja: why is $f(x,y)=x^2-y^2$ a counterexample? I want the set of $(x,y)$ for which $t$ is equal to a constant $c$ to be either empty or contain a line. in this case, for $c=1$ we get the whole plane, for $c\neq1$ it's empty. |
Nov 9 |
asked | two polynomial equations |
Apr 23 |
awarded | Yearling |
Apr 2 |
accepted | dimension of a real affine variety |
Apr 2 |
revised |
intersections of real algebraic sets (a bezout-type question)
added 269 characters in body |
Jan 1 |
awarded | Popular Question |
Oct 13 |
accepted | partial differential equation for ruled surfaces |
Oct 12 |
comment |
partial differential equation for ruled surfaces
@Robert(concerning your last comment): that's exactly the kind of answer i was looking for, except that i don't understand what is the "II-trace-free part of III" and "discriminant" (is it the same as here en.wikipedia.org/wiki/Discriminant?). also, if you want you can put your comment as an answer. thx. |
Oct 12 |
comment |
partial differential equation for ruled surfaces
@michael: ok, sorry, i see you're right. in fact, i'd be satisfied with the form $z=f(x,y)$, i.e. if the surface is described parametrically as $(x,y,f(x,y))$. then i'm looking for a partial equation in $f$. |
Oct 12 |
comment |
partial differential equation for ruled surfaces
here faculty.fairfield.edu/jmac/rs/halftw.htm they mention an equation $x^2z_{xx}+2xyz_{xy}+y^2z_{yy}=0$, but it's not the right one, it misses e.g. $z=xy$ (probably this equation gives only a subclass of the ruled surfaces). |
Oct 12 |
comment |
partial differential equation for ruled surfaces
yes, i'm looking for a partial equation that characterizes the ruled surfaces, but preferably for surfaces given by $f(x,y,z)=0$ and not in the parametric form. so if we're given a surface $f(x,y,z)=0$ we just plug in $f$ in the partial equation and we know that it's ruled iff we get 0. |