PRML 第1章 1.1(基本)

---

$$
\begin{array}{ll}
(1.1) & y(x,\mathbf{w}) = w_0 + w_1x + w_2x^2 + \cdots + w_Mx^M = {\displaystyle \sum_{j=0}^M} w_jx^j\\[1em]
(1.2) & E(\mathbf{w}) = \dfrac12 {\displaystyle \sum_{n=1}^N} \{ y(x_n,\mathbf{w}) - t_n \}^2\\[1em]
(1.122) & {\displaystyle \sum_{j=0}^M A_{ij} w_j = T_i}\\[1em]
(1.123) & {\displaystyle A_{ij} = \sum_{n=1}^N (x_n)^{i+j},\qquad T_i = \sum_{n=1}^N (x_n)^i t_n}
\end{array}
$$

---

解答

(1.2)に(1.1)を代入すると，

$$
E(\mathbf{w}) = \dfrac12 \sum_{n=1}^N \left\{ \sum_{j=0}^M w_j(x_n)^j - t_n \right\}^2
$$

となるので，これを$${w_i}$$で微分して0とおくと，

$$
\begin{array}{ll}
0 &= \dfrac{\partial E(\mathbf{w})}{\partial w_i}= {\displaystyle \sum_{n=1}^N \left\{ \sum_{j=0}^M w_j(x_n)^j - t_n \right\} (x_n)^i}\\[1em]
&= {\displaystyle \sum_{n=1}^N \sum_{j=0}^M w_j(x_n)^{i+j} - \sum_{n=1}^N (x_n)^i t_n}\\[1em]
&= {\displaystyle \sum_{j=0}^M \left( \sum_{n=1}^N(x_n)^{i+j} \right) w_j - T_i}\\[1em]
&= {\displaystyle \sum_{j=0}^M A_{ij} w_j - T_i}
\end{array}
$$

ゆえに，誤差関数を最小にする係数$${\{w_i\}}$$は(1.122)の解として与えられる．