Re: [理工] [線代]-中興-96資科所
※ 引述《gn00618777 (123)》之銘言：
: 解答獨缺96年= =，想跟解過的大大對一下線代部分
: 7 [3]
: projection= [4]
: [4]
: 8 [1 -2] [1/√2 1/√2] [3 0] [1/√2 -1/√2]
: [-2 1] = [-1/√2 1/√2] [0 -1] [1/√2 1/√2 ]
: 9 [2 0 0]
: [0 0 0]
: [0 1 0]
B={x^2,x,1}
2
d x 2
---- = 2x = 0x + 2x + 0
dx
d x 2
---- = 1 = 0x + 0x + 1
dx
d 1 2
--- = 0 = 0x + 0x + 0
dx
[0 0 0]
=> [D] = [2 0 0]
B [0 1 0]
: 10 不會~"~ 大大教一下～
1
Let V=C([-1,1]) with inner product <f,g>=∫ f(t)g(t)dt
-1
and W be the subspace P_2(R).
let B={1,x,x^2}, use Gram-Schmidt process to get an orthonormal
basis B'{1/√2, (√3/2)x, (√10/4)(3x^2-1)}, (方便打字 記為 u1 u2 u3)
x
then the least squares linear approximation to f(x) = e
is < f(x),u1>u1 + <f(x),u2>u2 +<f(x),u3>u3 (這邊我就沒仔細算了)
: 11 y=-69/50 X+ 42/10 題目有點看不清楚(4,4)還是(4,-4)?
Let A = [1 1] and y = [ 1 ]
[2 1] [2.4]
[3 1] [3.6]
[4 1] [ 4 ]
t t -1 1
hence A A = [30 10] thus (A A) = ---- [ 4 -10]
[10 4] 20 [-10 30]
1
Therefore x = ----[ 4 -10][1 2 3 4][ 1 ] = [1.02]
0 20 [-10 30][1 1 1 1][2.4] [0.2 ]
[3.6]
[ 4 ]
it follows that the line y=1.02x + 0.2 is the least square line.
剛剛算的不知道有沒有算錯
有錯請指正 謝謝
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 123.195.0.199
推
03/10 23:41, , 1F
03/10 23:41, 1F
→
03/10 23:43, , 2F
03/10 23:43, 2F
→
03/10 23:52, , 3F
03/10 23:52, 3F
討論串 (同標題文章)