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?
x=[27.27032.52932.69934.49537.97538.31743.21946.551]y=[3.07413.08673.08703.10613.1432...
x=[27.270 32.529 32.699 34.495 37.975 38.317 43.219 46.551] y=[3.0741 3.0867 3.0870 3.1061 3.1432 3.1449 3.2098 3.2627] x实为一组频率值,数量级为10的14次方。y 为介电常数。
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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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