Replace problem1.ipynb

homework1/problem1/problem1.ipynb

@@ -88,17 +88,78 @@
    ]
    ]
   },
   },
   {
   {
-   "cell_type": "code",
-   "execution_count": null,
+   "cell_type": "markdown",
    "metadata": {},
    "metadata": {},
-   "outputs": [],
-   "source": []
+   "source": [
+    "$L(w)=\\sum_{i}^{m}{y_{i}^{}log\\tilde{y}}+(1-y_{i})log(1-\\tilde{y}),\\tilde{y}=\\frac{1}{1-e_{}^{-(wx+b)}}$"
+   ]
   },
   },
   {
   {
-   "cell_type": "code",
-   "execution_count": null,
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    " "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\frac{\\partial L}{\\partial w}=\\sum_{i}^{n}{(A_{i}^{}-y_{i}^{})*x_{i}^{}},A=\\frac{1}{1-e^{-(wx+b)}}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    ">用向量表示即 $\\frac{\\partial L}{\\partial w}=X*(A-Y)^{T}$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$\\frac{\\partial^{2}L}{\\partial^{2}w}=\\sum_{i}^{n}\\frac{x_{ij}x_{ik}e^{-(wX_{i}+b)}}{A^{2}}=\\sum_{i}^{n}x_{ij}x_{ik}A_{i}(1-A_{i})$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "$H=\\left(\n",
+    "  \\begin{array}{ccc}\n",
+    "    x_{11}& ……&…… & x_{1n} \\\\\n",
+    "    ……  &&&…… \\\\\n",
+    "    ……  &  &&……   \\\\\n",
+    "x_{n1} &……&…… &x_{nn}\n",
+    "  \\end{array}\n",
+    "\\right)*\\left(\n",
+    "  \\begin{array}{ccc}\n",
+    "    A_{1}(1-A_{1})& …… & …… & 0 \\\\\n",
+    "    0 &A_{2}(1-A_{2})&&  \\\\\n",
+    "    ……  &  &……&  \\\\\n",
+    "0 &……&…… &A_{2}(1-A_{2})\\\\\n",
+    "  \\end{array}\n",
+    "\\right)*\\left(\n",
+    "  \\begin{array}{ccc}\n",
+    "    x_{11}& ……&…… & x_{n1} \\\\\n",
+    "    ……  &&&…… \\\\\n",
+    "    ……  &&&……   \\\\\n",
+    "x_{1n} &……&…… &x_{nn}\n",
+    "  \\end{array}\n",
+    "\\right)=X^{T}*V*X$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    ">由于V中对角线上的值＞0，因此矩阵H>=0,故逻辑回归函数的损失函数为凸函数"
+   ]
+  },
+  {
+   "cell_type": "markdown",
    "metadata": {},
    "metadata": {},
-   "outputs": [],
    "source": []
    "source": []
   },
   },
   {
   {
@@ -125,7 +186,7 @@
    "name": "python",
    "name": "python",
    "nbconvert_exporter": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
    "pygments_lexer": "ipython3",
-   "version": "3.7.1"
+   "version": "3.7.3"
   }
   }
  },
  },
  "nbformat": 4,
  "nbformat": 4,