neural_network/numpy_nn/softmax_derivative.ipynb

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  {
   "cell_type": "markdown",
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   "source": [
    "对softmax求导要比对分类的交叉熵损失函数求导要难得多。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e45006fb",
   "metadata": {},
   "source": [
    "$$S_{i,j}=\\frac{e^{z_{i,j}}}{\\sum^L_{l=1}e^{z_{i,l}}} \\rightarrow \\frac{\\partial S_{i,j}}{\\partial z_{i,k}}=\\frac{\\partial\\frac{e^z_{i,j}}{\\sum^L_{l=1}e^{z_{i,l}}}}{\\partial z_{i,k}}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6c3ddae7",
   "metadata": {},
   "source": [
    "$S_{i,j}$:第$i$个数据的softamx的第$j$个输出。  \n",
    "$z:$是一个向量，上一层的输出。  \n",
    "$z_{i,j}:$第$i$个数据，$z$向量中第$j$个输入。  \n",
    "$L$:输入的数量。  \n",
    "$z_{i,k}$:第$i$个数据，$z$向量中的第$k$个输入。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7796014f",
   "metadata": {},
   "source": [
    "回顾一下，对分数进行求导如下。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "81ecc7e6",
   "metadata": {},
   "source": [
    "$$f(x) = \\frac{g(x)}{h(x)}\\rightarrow f'(x) = \\frac{g'(x)\\cdot h(x)-g(x)\\cdot h'(x)}{[h(x)]^2}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "24cc1e33",
   "metadata": {},
   "source": [
    "那么我们对softmax进行求导。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "935665a2",
   "metadata": {},
   "source": [
    "$$\\frac{\\partial S_{i,j}}{\\partial z_{i,k}}=\\frac{\\partial\\frac{e^z_{i,j}}{\\sum^L_{l=1}e^{z_{i,l}}}}{\\partial z_{i,k}} =\\frac{ \\frac{\\partial}{\\partial z_{i,k}}e^{z_{i,j}}\\cdot\\sum^L_{l=1}e^{z_{i,l}}-e^{z_{i,j}}\\cdot\\frac{\\partial}{\\partial z_{i,k}}\\sum^L_{l=1}e^{z_{i,l}}}{[\\sum^L_{l=1}e^{z_{i,l}}]^2}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7baa2962",
   "metadata": {},
   "source": [
    "我们先看右侧的公式:$\\frac{\\partial}{\\partial z_{i,k}}\\sum^L_{l=1}e^{z_{i,l}}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "97811b46",
   "metadata": {},
   "source": [
    "对指数函数$e^n$求导。\n",
    "$$\\frac{d}{dn}e^n = e^n\\cdot\\frac{d}{dn}n=e^n\\cdot 1=e^n$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "01238382",
   "metadata": {},
   "source": [
    "$$\\frac{\\partial}{\\partial z_{i,k}}\\sum^L_{l=1}e^{z_{i,l}}=\\frac{\\partial}{\\partial z_{i,k}}e^{z_{i,1}} + \\frac{\\partial}{\\partial z_{i,k}}e^{z_{i,2}} + \\cdots \\frac{\\partial}{\\partial z_{i,k}}e^{z_{i,k}} + \\cdots + \\frac{\\partial}{\\partial z_{i,k}}e^{z_{i,L-1}} + \\frac{\\partial}{\\partial z_{i,k}}e^{z_{i,L}}\\\\\n",
    "=0 + 0 + \\cdots + e^{z_{i,k}} + \\cdots + 0 + 0 = e^{z_{i,k}}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a0adce34",
   "metadata": {},
   "source": [
    "对加和符号求导，除了相关项，其他看出常数。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "cd2ebf37",
   "metadata": {},
   "source": [
    "对$\\frac{\\partial}{\\partial z_{i,k}}e^{z_{i,j}}$进行求导。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "492aec4d",
   "metadata": {},
   "source": [
    "$$\\frac{\\partial}{\\partial z_{i,k}}e^{z_{i,j}} = \\frac{e^{z_{i,j}}\\cdot \\sum^L_{l=1}e^{z_{i,l}} - e^{z_{i,j}}\\cdot e^{z_{i,k}}}{[\\sum^L_{l=1}e^{z_{i,l}}]^2}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3dd002e2",
   "metadata": {},
   "source": [
    "如果j=k。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f0472238",
   "metadata": {},
   "source": [
    "$$\\frac{e^{z_{i,j}}\\cdot \\sum^L_{l=1}e^{z_{i,l}} - e^{z_{i,j}}\\cdot e^{z_{i,k}}}{[\\sum^L_{l=1}e^{z_{i,l}}]^2} = \\frac{e^{z_{i,j}}}{\\sum^L_{l=1}}\\cdot\\frac{\\sum^L_{l=1} - e^{z_{i,k}}}{\\sum^L_{l=1}e^{z_{i,l}}}=\\\\\n",
    "\\frac{e^{z_{i,j}}}{\\sum^{L}_{l=1}e^{z_{i,l}}}\\cdot (\\frac{\\sum^L_{l=1}e^{z_{i,l}}}{\\sum^L_{l=1}e^{z_{i,l}}}-\\frac{e^{z_{i,k}}}{\\sum^L_{l=1}e^{z_{i,l}}})=S_{i,j}\\cdot(1-S_{i,k})$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b6adb4c7",
   "metadata": {},
   "source": [
    "如果j$\\neq k$。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "31985b0a",
   "metadata": {},
   "source": [
    "$$\\frac{0\\cdot \\sum^L_{l=1}e^{z_{i,l}} - e^{z_{i,j}}\\cdot e^{z_{i,k}}}{[\\sum^L_{l=1}e^{z_{i,l}}]^2} = \\frac{-e^{z_{i,j}}}{\\sum^L_{l=1}e^{z_{i,l}}}\\cdot \\frac{e^{z_{i,k}}}{\\sum^L_{l=1}e^{z_{i,l}} } = -S_{i,j}\\cdot S_{i,k}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "795a4717",
   "metadata": {},
   "source": [
    "综上。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "409d8390",
   "metadata": {},
   "source": [
    "$$\\frac{\\partial S_{i,j}}{\\partial z_{i,k}}=\\begin{cases}\n",
    "S_{i,j}(1-S_{i,k})& \\text{j=k}\\\\\n",
    "-S_{i,j}\\cdot S_{i,k}& \\text{j $\\neq$ k}\n",
    "\\end{cases}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "b9dfea11",
   "metadata": {},
   "outputs": [],
   "source": []
  }
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