→ ntust661:推= =! 12/10 07:06
推 zain523:原來如此啊 ~.~ 12/10 14:15
: 至於 (c) 小題我看題應該是說用 圓形 或 二次方程式 去逼近那個多邊形,
: 去求 least square, 但我就不會了 XD
以下是我從林緯老師的講義打上來的供你參考 ⊙ ⊙
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4 2 2 1/2
1.欲求c0 c1 c2 使 E = [ Σ (c0 + c1xi + c2xi - yi) ] 為最小
i=1
令 A= [ 1 x1 x1^2 ] = [ 1 0 0 ]
[ ] [ ]
[ 1 x2 x2^2 ] [ 1 1 1 ]
[ ] [ ]
[ 1 x3 x3^2 ] [ 1 1 1 ]
[ ] [ ]
[ 1 x4 x4^2 ] [ 1 -1 1 ]
X= [ c0 ]
[ c1 ]
[ c2 ]
b= [ y1 ] = [ -1/2 ]
[ y2 ] [ 0 ]
[ y3 ] [ 1 ]
[ y4 ] [ 0 ]
T T
則欲求 x 使|| Ax-b ||為最小,相當於解A Ax = A b,
T -1 T -1
而因為 A 行獨立,故 x = (A A) A b = [ 4 1 3 ] [ 1 1 1 1 ] [-1/2 ]
[ 1 3 1 ] * [ 0 1 1 -1] * [ 0 ]
[ 3 1 3 ] [ 0 1 1 1 ] [ 1 ]
[ 0 ]
[-4 ]
= 1/8 * [ 2 ]
[ 6 ]
[ 0 ]
又誤差為|| Ax - b || = || [ 1/2 ] ||
[-1/2 ]
[ 0 ]
= (√2)/2
2 2 2
2. 設所求 least square circle 為 (x-a) + (y-b) = r ,
a,b,r 為待求常數,
4 2 2 2 2
則 a,b,r 須滿足 Σ [ r -(xi-a) - (yi-b) ] 為最小
i=1
其中,r^2 - (xi-a)^2 - (yi-b)^2 = 2axi + 2byi + (r^2 - a^2 - b^2) - xi^2 - yi^2
為方便計算,令(r^2 - a^2 - b^2) = c,
令 A= [ 2x1 2y1 1 ] = [ 0 -1 1 ]
[ ] [ ]
[ 2x2 2y2 1 ] [ 2 0 1 ]
[ ] [ ]
[ 2x3 2y3 1 ] [ 2 2 1 ]
[ ] [ ]
[ 2x4 2y2 1 ] [-2 0 1 ]
X= [ a ]
[ b ]
[ c ]
b= [ x1^2 + y1^2 ] = [ 1/4 ]
[ x2^2 + y2^2 ] [ 1 ]
[ x3^2 + y3^2 ] [ 2 ]
[ x4^2 + y4^2 ] [ 1 ]
T T
則欲求 x 使|| Ax-b ||為最小,相當於解A Ax = A b,
T -1 T -1
而因為 A 行獨立,故 x = (A A) A b = [ 12 4 2 ] [ 0 2 2 -2 ] [ 1/4 ]
[ 4 5 1 ] * [-1 0 2 0 ] * [ 1 ]
[ 2 1 4 ] [ 1 1 1 1 ] [ 2 ]
[ 1 ]
[ -1 ]
= 1/80 * [ 46 ]
[ 74 ]
而其誤差為|| Ax-b || = (√10)/20
∵ (√10)/20 < (√2)/2
∴ 圓比較近似
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