Hi everyone,
i need your help with solving the (ii) choice of the quesiton i attached,thanks for your helps .
ok, since we can pick any basis vector to start with, i'll choose g_{1} = b_{1} = 1 (the constant polynomial, not the number) that is: b_{1}(t) = 1, for all t.
now to find g_{2}, we have the find the projection of b_{2} in the direction of 1. we have to use the given inner product for this, but the general formula is:
proj_{g1}(b_{2}) = [(g_{1},b_{2})/(g_{1},g_{1})]g_{1}.
note that g_{1}(0) = g_{1}(1) = g_{1}(2) = 1 (it's a constant function). thus (g_{1},g_{1}) = (1)(1) + (1)(1) + (1)(1) = 3.
the real action comes in when we compute (g_{1},b_{2}). b_{2}(t) = t, so:
b_{2}(0) = 0, b_{2}(1) = 1, b_{2}(2) = 2.
thus (g_{1},b_{2}) = g_{1}(0)b_{2}(0) + g_{1}(1)b_{2}(1) + g_{1}(2)b_{2}(2) = (1)(0) + (1)(1) + (1)(2) = 3.
hence proj_{g1}(b_{2}) = (3/3)(1) = 1 (the constant polynomial, not the number).
to find g_{2}, we simply subtract the projection of b_{2} in the direction of g_{1} from b_{2}:
g_{2}(t) = b_{2}(t) - proj_{g1}(b_{2})(t) = t - 1.
let's verify that g_{2} is orthogonal to g_{1}:
(g_{1},g_{2}) = (1,t-1) = g_{1}(0)g_{2}(0) + g_{1}(1)g_{2}(1) + g_{1}(2)g_{2}(2) = (1)(-1) + (1)(0) + (1)(1) = 0. see?
finding g_{3} will be the same, except now you have to find 2 projections, and subtract them both from b_{3}(t) = t^{2}.