已知多组数据X,Y,如何用方程y=a+bx^2+cx^(-2)拟合得到a,b,c的值?
1个回答
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f(a,b,c)=Sum_{i=1->N}{y(i)-a-b[x(i)]^2
-
c[x(i)]^(-2)}^2
0=df(a,b,c)/da
=
2Sum_{i=1->N}{y(i)-a-b[x(i)]^2
-
c[x(i)]^(-2)}(-1)
...(1)
0=df(a,b,c)/db
=
2Sum_{i=1->N}{y(i)-a-b[x(i)]^2
-
c[x(i)]^(-2)}{-[x(i)]^2}
...(2)
0=df(a,b,c)/dc
=
2Sum_{i=1->N}{y(i)-a-b[x(i)]^2
-
c[x(i)]^(-2)}{-[x(i)]^(-2)}
...(3)
解由(1),(2),(3)构成的关于a,b,c的三元一次方程组就可以求得a,b,c的值.
-
c[x(i)]^(-2)}^2
0=df(a,b,c)/da
=
2Sum_{i=1->N}{y(i)-a-b[x(i)]^2
-
c[x(i)]^(-2)}(-1)
...(1)
0=df(a,b,c)/db
=
2Sum_{i=1->N}{y(i)-a-b[x(i)]^2
-
c[x(i)]^(-2)}{-[x(i)]^2}
...(2)
0=df(a,b,c)/dc
=
2Sum_{i=1->N}{y(i)-a-b[x(i)]^2
-
c[x(i)]^(-2)}{-[x(i)]^(-2)}
...(3)
解由(1),(2),(3)构成的关于a,b,c的三元一次方程组就可以求得a,b,c的值.
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