A CHALLENGE

PROVE THAT EACH CLOSED SIMPLE PLANE CURVE HAS AT LEAST 4 POINTS WHERE THE CURVATURE VANISHES

Erm, any closed simple plane curve... like a circle for example? You must be missing a hypothesis somewhere.

Circle?

>>1
Do u mean like where ddy/ddx is 0?

>>2,3
CLOSED SIMPLE PLANE CURVE: BIJECTIVE MAP c: [a,b] -> R^2 SUCH THAT c[a] = c[b]. BIJECTIVITY IMPLIES IT NEVER INSECTS AND dc(s)/ds IS NEVER ZERO, WHERE WE TAKE dc PARAMERTICISED WITH ARC LENGTH s (HENCE ds IS THE ARC LENGTH DIFFERENTIAL)

THE CURVATURE OF A CLOSED SIMPLE PLANE CURVE c PARAMERTICISED TO UNIT VELOCITY IS |c''(t)| (I.E. IT'S ACCELERATION)

PROVE THAT ALL SUCH CURVES HAVE /AT LEAST/ 4 POINTS WHERE THE CURVATURE IS ZERO (IN ANY PARAMETRISATION)

HINT: THE ELLIPSE CENTERED AT THE ORIGIN HAS EXACTLY 4 SUCH POINTS WHERE IT INTERSECTS THE COORDINATE AXES

>>5
I know what a simple plane curve is. I simply gave a counterexample to your assertion (the unit circle). Oh, and you need not type in all caps.

AH FUCK. I MEAN WHERE THE DERIVATIVE OF THE CURVATURE VANISHES :)

CCCAAAAAAAALLLLLLLLLLLLLLLCCCCCCCCCCCUUUUUUUUUUUULLLLLLLLLUUUUUUUUUUUUSSSSSSSSSSSSSSSSSSSS

WHY IS EVERYONE TYPING IN CAPS

>>10
BECAUSE YOU ARE :)

ARE YOU READY TO UNLEASH THE FURY???

>>10
BECAUSE CAPS LOCK IS CRUISE CONTROL FOR COOL

>>1
REALLY TOO MUCH TO LIST HERE ;)

UNLEASH THE FUUUURY!

COUNTEREXAMPLE!!!111!!!:  e^x

>>5
why did this make me hard?