數(shù)據(jù)擬合的非線性模型多項(xiàng)式擬合的正交化方法超定方程組最小二乘解溫度數(shù)據(jù)的平滑處理《數(shù)值分析》18數(shù)據(jù)擬合的非線性模型求擬合函數(shù)

f(x,c0,c1,···,cn

)滿足觀測(cè)數(shù)據(jù)2/16例1.已知人口統(tǒng)計(jì)數(shù)據(jù)利用最小二乘法求指數(shù)擬合

y=ceax

x

x1

x2··········

xm

f

y1

y2··········

ym年1991199219931994

19951996數(shù)量11.5811.7211.8511.98

12.1112.24指數(shù)函數(shù)擬合人口統(tǒng)計(jì)數(shù)據(jù)(單位：億)(1)計(jì)算對(duì)數(shù)值yk=

lnNk

(k=1,2,…,6)(2)列出未知數(shù)a、b的超定方程組

a+btk=yk

(k=1，2，……，6)(3)求超定方程組的最小二乘解,得

N=ea+bx,并預(yù)測(cè)

N(2000),N(2008)

N2000=12.7971,N2008=13.9783

設(shè)

N=ea+bt

lnN=a+bt

令

y=lnN,

有

y(t)=a+bt

t199119921993199419951996

lnN2.45

2.46

2.47

2.48

2.492.503/16例2.利用極坐標(biāo)觀察值確定慧星軌道其中，p

為參數(shù)，e為偏心率

令k=1/rr2.702.001.611.201.02486783108126

k

0.37040.50000.62110.83330.9804

0.83781.16941.44861.88502.1991線性方程組(j=1,2,3,4,5)

p=?,e=?4/16r=[2.702.001.611.201.02];fai=[486783108126]*pi/180;n=length(r);b=ones(n,1);G=[1./r'cos(fai')];X=G\b;p=X(1);e=X(2);

p=1.4521,e=0.7027殘差平方和:0.00225/16fa=pi/4:.1:3.5*pi/2;ro=p./(1-e*cos(fa));polar(fai,r,'ro'),holdonpolar(fa,ro)R=r-p./(1-e*cos(fai))R=sum(R.^2)例3.溫度數(shù)據(jù)的擬合問題:中午1:00到第二天中午12:00溫度記錄數(shù)據(jù)如下66,66,65,64,63,63,62,6160,60,59,58,58,58,58,5857,57,57,58,60,64,67,68用最小二乘法求擬合函數(shù)

f(x)=E+Acos(Cx)+Bsin(Dx)E=?,A=？,B=?

,C=?

,D=?6/16定義

設(shè)

j(x)(j=0,1,···,n)是多項(xiàng)式系列,

是一點(diǎn)列,多項(xiàng)式i(x),j(x)在該點(diǎn)列上的值分別為則稱

k(x)(k=0,1,···,n)為點(diǎn)列

上的正交多項(xiàng)式系.(j≠i)如果i=[i(x1),i(x2),······,i(xm)]Tj=[j(x1),j(x2),······,j(xm)]T7/16·····························

p0(x)=1,

pj(x)=xpj-1(x)(j=1,2,····,n)(j=2,3,···,n)8/16給定數(shù)據(jù)表

x

x1

x2··········

xmf(x)y1

y2··········

ym超定方程組:正交多項(xiàng)式系:,······,組成擬合函數(shù)

9/16正規(guī)方程:由正交性得10/16超定方程組最小二乘解X*的幾何意義11/16例向量組平面正規(guī)方程Q—列正交矩陣;R—單位上三角矩陣矩陣的正交三角分解:G=QR12/16矩陣的LDLT分解DLTX=L-1bA=LDLT超定方程組AX=b記LTX=D-1L-1bX=(LT)-1D-1L-1b矩陣G的QR(正交三角)分解算法

G=(

gij)m×nm>n將矩陣按列分塊，記為

G=[g1,g2,······,gn

]

分解算法①q1=g1②q2=g2–r12q1,其中

r12=(g2，q1)/(q1，q1)

···········································n其中······13/16G=[g1

g2

······

gn

]=[q1(r12q1+q2)······(r1nq1+···+rn-1,nqn-1+qn)]G=[q1,q2,······,qn

]RG=

QR

g1=

q1

g2=r12q1+q2··························記Q=[q1,q2,······,qn

]其中14/16多項(xiàng)式擬合用于數(shù)據(jù)平滑處理五點(diǎn)拋物線擬合公式設(shè)擬合函數(shù)為:P(t)=a0+a1t+a2t2

a0=yk–3(yk-2–4yk-1+6yk–4yk+1+y