1.Gadients and Hessians
a.
由第一项，我们可以得到
$$ f_1(x)=\frac{1}{2}x^TAx+b^Tx $$因为$\nabla_x(x^TAx)=(A+A^T)x$，因此我们可以得到
$$ \nabla_x(\frac{1}{2}x^TAx)=\frac{1}{2}(A+A^T)x $$因为A是对称的，$A^T=A$，所以
$$ \frac{1}{2}(A + A^T)x=\frac{1}{2}(A+A)x=Ax $$由第二项，我们可以得到
$$ f_2(x)=b^Tx=\sum_{i}b_ix_i $$直接对x求偏导：
$$ \nabla_x(b^Tx)=b $$所以**$\nabla f(x)=Ax+b$**
b.
令$z=h(x)$，则$f(X)=g(z)$ $$ \frac{\partial f}{\partial x_i} = g'(h(x)) \cdot \frac{\partial h(x)}{\partial x_i} $$ 所以 $$ \nabla f(x)
\begin{pmatrix} \frac{\partial f}{\partial x_1} \ \frac{\partial f}{\partial x_2} \ \vdots \ \frac{\partial f}{\partial x_n} \end{pmatrix}
\begin{pmatrix} g’(h(x)) \frac{\partial h}{\partial x_1} \ g’(h(x)) \frac{\partial h}{\partial x_2} \ \vdots \ g’(h(x)) \frac{\partial h}{\partial x_n} \end{pmatrix}= g’(h(x)) \begin{pmatrix} \frac{\partial h}{\partial x_1} \ \vdots \ \frac{\partial h}{\partial x_n} \end{pmatrix}
g’(h(x)) \nabla h(x) $$
c.
由a得
$$ (\nabla f(x))_i = \sum_{j=1}^{n} a_{ij} x_j + b_i $$$$ (\nabla^2 f(x))_{ij}
\frac{\partial}{\partial x_j} \left( \frac{\partial f}{\partial x_i} \right)
\frac{\partial}{\partial x_j} \left( \sum_{k=1}^{n} a_{ik} x_k + b_i \right) $$
由
$$ \frac{\partial}{\partial x_j} (a_{ik} x_k) = a_{ik} \frac{\partial x_k}{\partial x_j} = a_{ik} \delta_{kj} $$所以
$$ (\nabla^2f(x))_{ij}=\sum_{k=1}^{n}a_{ik}\delta_{kj}=a_{ij} $$因此
$$ \nabla^2f(x)=A $$