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Lecturenotes1-4April2011

# Lecturenotes1-4April2011 - Examples 1 Let p(x = x2 2x 3 q(x...

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Examples: 1. Let p ( x ) = x 2 + 2 x + 3 , q ( x ) = 3 x + 2 . With the inner product < p, q > = a 0 b 0 + a 1 b 1 + a 2 b 2 for p = a 0 + a 1 x + a 2 x 2 , q = b 0 + 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 || ≥ 0 with equality if and only if v=0 (b) || kv || = | k ||| v || (c) d(u,v)=d(v,u) (d) d ( u, v ) 0 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 0 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 Cauchy-Schwarz Inequality If u and v are vectors in a real inner product space V, then | < u, v > | ≤ || u |||| v || Definition: Two vectors u and v in an inner product space are called orthogonal if < u, v > = 0 .

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