五次多项式的相关系数的计算公式,其中公式为tex格式
时间: 2024-03-03 10:52:24 浏览: 26
设五次多项式为 $y = ax^5 + bx^4 + cx^3 + dx^2 + ex + f$,有 $n$ 组数据 $(x_1, y_1), (x_2, y_2), ..., (x_n, y_n)$,则相关系数可以通过以下公式计算:
$$
S_{xx} = \sum_{i=1}^n (x_i - \bar{x})^2 \\
S_{xy} = \sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})
$$
其中 $\bar{x}$ 和 $\bar{y}$ 分别为 $x$ 和 $y$ 的平均值。
然后可以计算出相关系数 $a$、$b$、$c$、$d$、$e$ 和 $f$:
$$
a = \frac{n\sum_{i=1}^n x_i^5 - \sum_{i=1}^n x_i^4 \sum_{i=1}^n x_i}{S_{yy}} - \frac{n\sum_{i=1}^n x_i^4y_i - \sum_{i=1}^n x_i^3y_i \sum_{i=1}^n x_i}{S_{xy}} + \frac{\sum_{i=1}^n x_i^4 \sum_{i=1}^n x_i^2 y_i - \sum_{i=1}^n x_i^3 \sum_{i=1}^n x_i^3 y_i}{S_{xx}} \\
b = \frac{n\sum_{i=1}^n x_i^4y_i - \sum_{i=1}^n x_i^3y_i \sum_{i=1}^n x_i}{S_{yy}} - \frac{n\sum_{i=1}^n x_i^3y_i \sum_{i=1}^n x_i^2 - \sum_{i=1}^n x_i^2y_i \sum_{i=1}^n x_i^3}{S_{xy}} + \frac{\sum_{i=1}^n x_i^3 \sum_{i=1}^n x_i^2 y_i - \sum_{i=1}^n x_i^2 \sum_{i=1}^n x_i^3 y_i}{S_{xx}} \\
c = \frac{n\sum_{i=1}^n x_i^3y_i \sum_{i=1}^n x_i^2 - \sum_{i=1}^n x_i^2y_i \sum_{i=1}^n x_i^3}{S_{yy}} - \frac{n\sum_{i=1}^n x_i^2y_i \sum_{i=1}^n x_i^3 - \sum_{i=1}^n x_iy_i \sum_{i=1}^n x_i^4}{S_{xy}} + \frac{\sum_{i=1}^n x_i^2y_i \sum_{i=1}^n x_i^3 - \sum_{i=1}^n x_iy_i \sum_{i=1}^n x_i^4}{S_{xx}} \\
d = \frac{n\sum_{i=1}^n x_i^2y_i \sum_{i=1}^n x_i^3 - \sum_{i=1}^n x_iy_i \sum_{i=1}^n x_i^4}{S_{yy}} - \frac{n\sum_{i=1}^n x_iy_i \sum_{i=1}^n x_i^4 - \sum_{i=1}^n y_i \sum_{i=1}^n x_i^5}{S_{xy}} + \frac{\sum_{i=1}^n x_iy_i \sum_{i=1}^n x_i^4 - \sum_{i=1}^n y_i \sum_{i=1}^n x_i^5}{S_{xx}} \\
e = \frac{n\sum_{i=1}^n x_iy_i \sum_{i=1}^n x_i^4 - \sum_{i=1}^n y_i \sum_{i=1}^n x_i^5}{S_{yy}} - \frac{n\sum_{i=1}^n y_i \sum_{i=1}^n x_i^5 - \sum_{i=1}^n x_i \sum_{i=1}^n x_i^4y_i}{S_{xy}} + \frac{\sum_{i=1}^n y_i \sum_{i=1}^n x_i^5 - \sum_{i=1}^n x_i \sum_{i=1}^n x_i^4y_i}{S_{xx}} \\
f = \frac{n\sum_{i=1}^n y_i - \sum_{i=1}^n x_iy_i \sum_{i=1}^n x_i^4}{S_{yy}} - \frac{n\sum_{i=1}^n x_iy_i \sum_{i=1}^n x_i^4 - \sum_{i=1}^n x_i^2y_i \sum_{i=1}^n x_i^3}{S_{xy}} + \frac{\sum_{i=1}^n x_iy_i \sum_{i=1}^n x_i^4 - \sum_{i=1}^n x_i^2y_i \sum_{i=1}^n x_i^3}{S_{xx}}
$$
其中 $S_{yy} = \sum_{i=1}^n (y_i - \bar{y})^2$。