EE
102_1_Week-7-Recap

# 102_1_Week-7-Recap - 28 NHAN LEVAN WEEK 7 RECAP 20...

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28 NHAN LEVAN WEEK 7 RECAP 20. Mean (Least) Square Error M(L)SE 20.1. Parseval’s 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 = α i + β j + γ k where (20.2) α := v, i , β := v, j , and γ := 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 PP — connecting the tip P of v and the tip P of v . In other words (20.5) e := PP and E = || PP || = || e || Now, since v x - y -plane, it can be represented by (20.6) v = α i + β j where α and β are real numbers yet to be determined. It follows from the above “set-up” that our problem is simply to determine α and β 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

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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 = α 2 + β 2 + γ 2 (20.12) || v || 2 = α 2 + β 2 (20.13) v, v = ( α i + β j + γ k ) , ( α i + β j + 0 k ) (20.14) = α α + β β (20.15) Therefore (20.11) becomes: E 2 = || v || 2 - 2 v, v + || v || 2 (20.16) = ( α 2 + β 2 + γ 2 ) - 2( α α + β β ) + ( α 2 + β 2 ) (20.17) = ( α - α ) 2 + ( β - β ) 2 + γ 2 (20.18) When does the RHS of (20.18) smallest?
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