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Unformatted text preview: 28 NHAN LEVAN WEEK 7 RECAP 20. Mean (Least) Square Error M(L)SE 20.1. Parsevals Relation. 1 T T 2- T 2 | f ( t ) | 2 dt = n =- | F n | 2 . 20.2. A Simple Problem: Estimating the Length of a Tree. A tall tree is represented by a vector v in R 3 . Moreover, we know that v admits the orthonormal representation (20.1) v = i + j + k where (20.2) := v, i , := v, j , and := v, k and i, j, k are orthonormal vectors on the x , y , z , axes respectively. Our problem is to measure the length (or the norm) || v || of v . However, due to the wind and since one gets pain in the neck for looking up for a long time, we cannot get the true value of || v || even after 102 measurements! Because of the above reasons, we turn to the problem of finding a vector v in the x- y-plane, i.e., R 2 (hence no need to look up, hence no pain in the neck) so that the norm of the error vector (20.3) e := v- v is smallest. It is clear that the length of e , denoted by (20.4) E := || e || = || v- v || is simply the length of the vector P P connecting the tip P of v and the tip P of v . In other words (20.5) e := P P and E = || P P || = || e || Now, since v x- y-plane, it can be represented by (20.6) v = i + j where and are real numbers yet to be determined. It follows from the above set-up that our problem is simply to determine and so that the length E of the error vector e is smallest. Another way of saying this is: to minimize the quantity (20.7) E := || e || = || v- v || over all v i.e., all vectors in the x- y-plane . However, we know that the norm || w || of a vector w is defined as (20.8) || w || := w, w || w || 2 = w, w Therefore, instead of minimizing E we can also minimize (20.9) E 2 = || v- v || 2 = v- v, v- v 29 An advantage of this is the fact that we can open up the RHS to obtain E 2 = v, v- v, v- v, v + v, v (20.10) = || v || 2- 2 v, v + || v || 2 (20.11) keeping in mind that since we are dealing with real scalars: v, v = v, v But we know that || v || 2 = 2 + 2 + 2 (20.12) || v || 2 = 2 + 2 (20.13) v, v = ( i + j + k ) , ( i + j + 0 k ) (20.14) = + (20.15) Therefore (20.11) becomes: E 2 = || v || 2- 2 v, v + || v || 2 (20.16) = ( 2 + 2 + 2 )- 2( + ) + ( 2 + 2 ) (20.17) =...
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This note was uploaded on 04/01/2008 for the course EE 102 taught by Professor Levan during the Spring '08 term at UCLA.
- Spring '08