bauer/Solutions-Coursera-Machine-Learning-Course
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

Merge pull request #55 from frostblooded/patch-1

Browse Source 
Mistaken indexes fix
This commit is contained in:
Gerges Dib
2020-06-06 22:59:01 -07:00
committed by GitHub
parent 29ebe14cd9 461a27df81
commit eaf6822ae8
1 changed files with 2 additions and 2 deletions
Download Patch File Download Diff File Expand all files Collapse all files

4
Exercise8/exercise8.ipynb
View File

@@ -683,7 +683,7 @@
"\n", "\n",
"$$ \\frac{\\partial J}{\\partial x_k^{(i)}} = \\sum_{j:r(i,j)=1} \\left( \\left(\\theta^{(j)}\\right)^T x^{(i)} - y^{(i,j)} \\right) \\theta_k^{(j)} $$\n", "$$ \\frac{\\partial J}{\\partial x_k^{(i)}} = \\sum_{j:r(i,j)=1} \\left( \\left(\\theta^{(j)}\\right)^T x^{(i)} - y^{(i,j)} \\right) \\theta_k^{(j)} $$\n",
"\n", "\n",
"$$ \\frac{\\partial J}{\\partial \\theta_k^{(j)}} = \\sum_{i:r(i,j)=1} \\left( \\left(\\theta^{(j)}\\right)^T x^{(i)}- y^{(i,j)} \\right) x_k^{(j)} $$\n", "$$ \\frac{\\partial J}{\\partial \\theta_k^{(j)}} = \\sum_{i:r(i,j)=1} \\left( \\left(\\theta^{(j)}\\right)^T x^{(i)}- y^{(i,j)} \\right) x_k^{(i)} $$\n",
"\n", "\n",
"Note that the function returns the gradient for both sets of variables by unrolling them into a single vector. After you have completed the code to compute the gradients, the next cell run a gradient check\n", "Note that the function returns the gradient for both sets of variables by unrolling them into a single vector. After you have completed the code to compute the gradients, the next cell run a gradient check\n",
"(available in `utils.checkCostFunction`) to numerically check the implementation of your gradients (this is similar to the numerical check that you used in the neural networks exercise. If your implementation is correct, you should find that the analytical and numerical gradients match up closely.\n", "(available in `utils.checkCostFunction`) to numerically check the implementation of your gradients (this is similar to the numerical check that you used in the neural networks exercise. If your implementation is correct, you should find that the analytical and numerical gradients match up closely.\n",
@@ -809,7 +809,7 @@
"\n", "\n",
"$$ \\frac{\\partial J}{\\partial x_k^{(i)}} = \\sum_{j:r(i,j)=1} \\left( \\left(\\theta^{(j)}\\right)^T x^{(i)} - y^{(i,j)} \\right) \\theta_k^{(j)} + \\lambda x_k^{(i)} $$\n", "$$ \\frac{\\partial J}{\\partial x_k^{(i)}} = \\sum_{j:r(i,j)=1} \\left( \\left(\\theta^{(j)}\\right)^T x^{(i)} - y^{(i,j)} \\right) \\theta_k^{(j)} + \\lambda x_k^{(i)} $$\n",
"\n", "\n",
"$$ \\frac{\\partial J}{\\partial \\theta_k^{(j)}} = \\sum_{i:r(i,j)=1} \\left( \\left(\\theta^{(j)}\\right)^T x^{(i)}- y^{(i,j)} \\right) x_k^{(j)} + \\lambda \\theta_k^{(j)} $$\n", "$$ \\frac{\\partial J}{\\partial \\theta_k^{(j)}} = \\sum_{i:r(i,j)=1} \\left( \\left(\\theta^{(j)}\\right)^T x^{(i)}- y^{(i,j)} \\right) x_k^{(i)} + \\lambda \\theta_k^{(j)} $$\n",
"\n", "\n",
"This means that you just need to add $\\lambda x^{(i)}$ to the `X_grad[i,:]` variable described earlier, and add $\\lambda \\theta^{(j)}$ to the `Theta_grad[j, :]` variable described earlier.\n", "This means that you just need to add $\\lambda x^{(i)}$ to the `X_grad[i,:]` variable described earlier, and add $\\lambda \\theta^{(j)}$ to the `Theta_grad[j, :]` variable described earlier.\n",
"\n", "\n",