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Unformatted text preview: Examples: 1. Let p ( x ) = x 2 + 2 x + 3 , q ( x ) = 3 x + 2 . With the inner product < p,q > = a b + a 1 b 1 + a 2 b 2 for p = a + a 1 x + a 2 x 2 ,q = b + b 1 x + b 2 x 2 . Find length of p,q and distance between p and q. Solution:  p  = √ < p,p > = √ 1 + 4 + 9 = √ 14 ,  q  = √ 9 + 4 = √ 13 d ( p,q ) =  p q  = √ < p q,p q > = p < x 2 x + 1 ,x 2 x + 1 > = √ 3 2. Let f ( x ) = x 2 and g ( x ) = x be continuous functions on [1,2]. Consider the inner product < f,g > = R 2 1 f ( x ) g ( x ) dx . Find length of f,g and distance between f and g. Solution:  f  = √ < f,f > = q R 2 1 x 4 dx = p 31 / 5 ,  g  = √ < g,g > = q R 2 1 x 2 dx = p 7 / 3 d ( f,g ) =  f g  = p < x 2 x,x 2 x > = p 31 / 30 . Theorem 6.1.1: if u and v are vectors in a real inner product space V, and if k is a scalar, then: (a)  v  ≥ with equality if and only if v=0 (b)  kv  =  k  v  (c) d(u,v)=d(v,u) (d) d ( u,v ) ≥ with equality if and only if u=v. Section 6.2 Angle and orthogonality in Inner Spaces Recall that for vectors in R n we defined angle between two vectors u and v as cos ( θ ) = u · v  u  v  hence we said u is orthogonal to v if and only if u · v = 0 . We will generalize this formula for inner product spaces. That is for a given inner product <,> we have cos ( θ ) = < u,v >  u  v  where ≤ θ ≤ π and u is orthogonal to v if and only if < u,v > = 0 . This formula is valid because of Theorem 6.2.1 CauchySchwarz InequalityTheorem 6....
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This document was uploaded on 04/26/2011.
 Spring '11

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