hw6 fix

HW6/README.md

@@ -34,12 +34,14 @@ variable is failure (1=fail, 0=pass). Create a function called `cost_logistic.m`
 takes the vector `a`, and independent variable `x` and dependent variable `y`. Use the
 function, $\sigma(t)=\frac{1}{1+e^{-t}}$ where $t=a_{0}+a_{1}x$. Use the cost function,
 
-    $J(a_{0},a_{1})=\sum_{i=1}^{n}\left[-y_{i}\log(\sigma(t_{i}))-(1-y_{i})\log((1-\sigma(t_{i})))\right]$
+    $J(a_{0},a_{1})=1/m\sum_{i=1}^{n}\left[-y_{i}\log(\sigma(t_{i}))-(1-y_{i})\log((1-\sigma(t_{i})))\right]$
 
     and gradient
 
    $\frac{\partial J}{\partial a_{i}}=
-   1/m\sum_{k=1}^{N}\left(\sigma(t_{k})-y_{k}\right)t_{k}$
+   1/m\sum_{k=1}^{N}\left(\sigma(t_{k})-y_{k}\right)x_{k}^{i}$
+
+   where $x_{k}^{i} is the k-th value of temperature raised to the i-th power (0, and 1)
 
    a. edit `cost_logistic.m` so that the output is `[J,grad]` or [cost, gradient]
 

HW6/problem_1_data.m

@@ -1,7 +1,7 @@
 
 % part a
 xa=[1 2 3 4 5]';
-yb=[2.2 2.8 3.6 4.5 5.5]';
+ya=[2.2 2.8 3.6 4.5 5.5]';
 
 % part b
 
@@ -10,6 +10,6 @@
 
 % part c
 
-xc=[0.5 1 2 3 4 5 6 7 9];
-yc=[6 4.4 3.2 2.7 2.2 1.9 1.7 1.4 1.1];
+xc=[0.5 1 2 3 4 5 6 7 9]';
+yc=[6 4.4 3.2 2.7 2.2 1.9 1.7 1.4 1.1]';
 

lecture_21/.ipynb_checkpoints/lecture_21-checkpoint.ipynb

@@ -0,0 +1,6 @@
+{
+ "cells": [],
+ "metadata": {},
+ "nbformat": 4,
+ "nbformat_minor": 2
+}