QUOTE(VengenZ @ Jan 15 2012, 11:32 PM)
Its the same thing.take k(x) for example.
h(x) = x or we say the function h(x), graphed as y = x
hence h(k(x)) = k(x)
Given k(x) = x + 2
we know hk(x) = (x + 2)
formalising an idea,
for an unknown function z(x),
if k(x) is a known function.
and zk(x) can be expressed as y = a + b(k(x)) + c or a function of k(x), for a,c, are known constants/parameters.
then z(x) is simply y = a + bx + c
which is how TS, used the "completing the Square method, to obtain the equation as a factor of (k(x))
and work out h(x). An alternative method from inversing the identity.
Jan 15 2012, 11:49 PM

Quote
0.0211sec
0.87
6 queries
GZIP Disabled