Exercise 6.3.23

Answers

1.

Direct calculate that

A^{∗} A = (\begin{matrix} \sum_{i = 1}^{m} t_{i}^{2} & \sum_{i = 1}^{m} t_{i} \\ \sum_{i = 1}^{m} t_{i} & m \end{matrix})

and get the result by

A^{∗} A (\begin{matrix} c \\ d \end{matrix}) = A^{∗} y .

For the second method, we may calculate that

E = \sum_{i = 1}^{m} {(y_{i} - c t_{i} - d)}^{2}

and $\frac{∂E}{∂c} = 0$ and $\frac{∂E}{∂d} = 0$ give the normal equations.

2.

We want to claim that

\bar{y} = c \bar{t} + d

, whrer

\bar{t}

\bar{y}

are defined in the question and

c

d

are a solution of the normal equations. But this is an instant result by dividing the second equation by

m

jephianlin

2011-06-27 00:00