3 6 1 3 1 3 2 3 As w 3 6 3 we have v 3 3 6 1 3 1 3 2 3 1 6 1 6 2 6 w 4 u 4 u 4

# 3 6 1 3 1 3 2 3 as w 3 6 3 we have v 3 3 6 1 3 1 3 2

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3 6 = 1 3 1 3 - 2 3 0 . As || w 3 || = 6 3 we have v 3 = 3 6 1 3 1 3 - 2 3 0 = 1 6 1 6 - 2 6 0 . w 4 = u 4 - u 4 , v 1 v 1 - u 4 , v 2 v 2 - u 4 , v 3 v 3 = 1 0 0 0 - 1 2 1 2 1 2 1 2 1 2 - 1 2 3 1 2 3 1 2 3 1 2 3 - 3 2 3 - 1 6 1 6 1 6 - 2 6 0 = 1 0 0 0 - 1 4 1 4 1 4 1 4 - 1 12 1 12 1 12 - 3 12 - 1 6 1 6 - 2 6 0 = 1 2 - 1 2 0 0 . 1
As || w 4 || = 1 2 we have v 4 = 2 1 2 - 1 2 0 0 = 2 2 - 2 2 0 0 . 3. Let u 1 = 2 + 3 x 2 , u 2 = 5 x 2 - 1 and u 3 = 10 - 7 x + 2 x 2 . Now v 1 = 1 || u 1 || u 1 = 1 13 (2 + 3 x 2 ). w 2 = u 2 - u 2 , v 1 v 1 = 5 x 2 - 1 - (2 + 3 x 2 ) = 2 x 2 - 3 . As || w 2 || = 13 we have v 2 = 1 13 (2 x 2 - 3). w 3 = u 3 - u 3 , v 1 v 1 - u 3 , v 2 v 2 = 10 - 7 x + 2 x 2 - 2(2 + 3 x 2 ) + 2(2 x 2 - 3) = - 7 x. As || w 3 || = 7 we have v 3 = - x . 4. We start with the basis of P 2 given by { u 1 = 1 , u 2 = x, u 3 = x 2 } and use the Gram-Schmidt process to construct an orthonormal basis { v 1 , v 2 , v 3 } . Set v 1 = u 1 || u 1 || . As || 1 || 2 = 1 - 1 1 dx = [ x ] 1 - 1 = 2 , we have v 1 = 1 2 . Set w 2 = u 2 - u 2 , v 1 v 1 . As u 2 , v 1 = 1 - 1 x 2 dx = x 2 2 2 1 - 1 = 0 , we have w 2 = u 2 = x. Now we set v 2 = w 2 || w 2 || . As || w 2 || 2 = 1 - 1 x 2 dx = x 3 3 1 - 1 = 2 3 we have v 2 = 3 2 x.

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