Hi everyone,

i need your help with solving the (ii) choice of the quesiton i attached,thanks for your helps .

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- Jan 6th 2013, 07:49 AMsah_matgram-schmidt process help
Hi everyone,

i need your help with solving the (ii) choice of the quesiton i attached,thanks for your helps . - Jan 6th 2013, 07:57 AMsah_matRe: gram-schmidt process help
by the way i confused how to use (u,v) and p(t),q(t),

i really appreciate if you show me the details of obtaining g1 and g2 i can find g3. - Jan 6th 2013, 09:18 AMDevenoRe: gram-schmidt process help
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}.