I can't increase the degree of the Taylor polynomial because G is C2, so the second degree term is the remainder written in matrix notation.
HG(c) is the Hessian matrix of G evaluated at c, where c lies on the segment that goes from (0,0) to (x,y).
Homework Statement
Let G:R2\rightarrowR be a C2 function such that G(tx,ty)=t2G(x, y). Show that:
2G(x,y)=(x,y).HG(0,0).(x,y)t
The Attempt at a Solution
G is C2, so its Taylor expansion is:
G(x,y) = G(0,0) + \nablaG(0,0).(x,y) + \frac{1}{2}(x,y).HG(c).(x,y)t,
where c lies on...
if B(x,y) is an open ball with center (x,y) and radius r, i show that if (a,b) belongs to the ball B, then (a,b) belongs to the set, so a^2 + b^2 < 7
|(a,b)| = |(a,b) - (x,y) + (x,y)| = |(a-x, b-y) + (x,y)| ≤ |(a-x, b-y)| + |(x,y)|
|(a-x, b-y)| < r, so
|(a-x, b-y)| + |(x,y)| < r +...
I think r < min {√7-|(x,y)| ; |(x,y)| - 1} works.
I tried to separate the problem in two sets, because the intersection of two open sets is an open set, and proved that x^2+y^2<7 is open, but the method I used with that proof don't work with the set 1 < x^2 + y^2
mmmm i think in this case the centripetal force is the one applied by the spring, so
(constant)x(4cm) = (m)x(v^2/r)
solve and get v, and then
work of the disk = (1/2)x(m)x(v^2) + (1/2)x(I)x(W^2)
= (1/2)x(m)x(v^2) + (1/2)x(1/2x m xR^2)x(W^2)...