Question

Perform the marquis de Laplace process on the basis {f(x) = x2 - 4x +1, g(x) = 2x + 4, h(x) = x2 + +3} + to create an orthogonal basis for the space of polynomial functions of degree 5 2.

Answer 1

Answer:-

Let, S = c f(u), g(x), h(x)} = 4 x24x+1, 2x+4, x?+3} Now, where g'(x) 9"(x) f'wo f"lu)= 422 f (w 11 omd so on g (1) h() f'(W) h'(x) f"() h"(W) 4²-4x+1 2014 x²3 (4) (2+4) (+3) d2 4x+1)(1+) 23) x?-4x+1. 2014 4273 au-4 3 du 17 [10") = n*"'] 3 20 2 C Expanding with respect to third row) =(-1*5.5 | ******* +0 + 47,5"? 2 roky 2014 = 1:3. (2162x+4) – 3 (x+3)) +1.2 [2(43-4x+1)= (4x16) = 2 (4x2-24?+81-6) + ?(242-42284 +2+16) = 2 ( 24? +81-6) + (-242-84+18) 2:-(3x² +84-6 282-34 +18) = 7.12=34 FO (So, s set it linearly independent (As D to for any u) (By de Laplace furorex)

Now -ron s we will perform Grahn Smidth orthogonalization to find an ortho normal bain ofs on [1] Let,Bs' = – F(), GC4), H (H)} be the stepwird. orthogonal bouis for P2 ( space for polynomial. (2) F(x) = f(1) = x2-44+1 <9(w), FCW))FW) G(x) = g(x) F (N), FOUD) = (21+4) - S(21+4) (x²4x+1) de (4x+1) (4²-44+1) 2.dn =(22+4) - (1²4) =(22+4) + 36 (+244+1) = 115 m2 - +x+387 6 23 6 115

<h(n), Flu)) <h(1), Glu) H(x) =h (n)- F(n) G(4) F(u), Flu) LG(4), G(")) Cu? + 3) (un antidu x²+3 - (024+) Ś Cu² 4+1 52 da Sc73) ( 14 2 4 9 7 + 27 de Śwa? 4,7u +25aou (x²4x+1) - 108 + 'xt 9 36 92 37 153 6 5 967 2837 216 2 = (x+3) + 1934 2827 259 ut 36 37 37 101772 37 (m? 41+) - ( 155 4 2 - atu- ) *2+(-74 +44)* = (1+34 +222016 at 3+ 37 18 501165 101 772

5.241 u 2 +23 nt 0131 upto 9 (correct three decima place) soßs' = f, x£4x+1, x2- 4** +39,5-241u++ 3 + +013 an orthogonal bawn of folynomial of degree <2 is

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