这是一份southampton南安普敦大学MATH6168W1-01作业代写的成功案例

At this point, we can gain some insight into the role of the bias parameter $w_{0}$. If we make the bias parameter explicit, then the error becomes
Setting the derivative with respect to $w{0}$ equal to zero, and solving for $w_{0}$, we obtain
$$
w_{0}=\bar{t}-\sum_{j=1}^{M-1} w_{j} \overline{\phi_{j}}
$$
where we have defined
$$
\bar{t}=\frac{1}{N} \sum_{n=1}^{N} t_{n}, \quad \overline{\phi_{j}}=\frac{1}{N} \sum_{n=1}^{N} \phi_{j}\left(\mathbf{x}_{n}\right)
$$

MATH6168W1-01 COURSE NOTES ：

Setting the derivative with respect to $\mu_{1}$ to zero and rearranging, we obtain
$$
\boldsymbol{\mu}{1}=\frac{1}{N{1}} \sum_{n=1}^{N} t_{n} \mathbf{x}{n} $$ which is simply the mean of all the input vectors $\mathbf{x}{n}$ assigned to class $\mathcal{C}{1}$. By a similar argument, the corresponding result for $\mu{2}$ is given by
$$
\boldsymbol{\mu}{2}=\frac{1}{N{2}} \sum_{n=1}^{N}\left(1-t_{n}\right) \mathbf{x}_{n}
$$