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@@ -0,0 +1,6 @@
+{
+ "cells": [],
+ "metadata": {},
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
diff --git a/15_challenger-logistic/15_logistic-regression_challenger.ipynb b/15_challenger-logistic/15_logistic-regression_challenger.ipynb
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--- /dev/null
+++ b/15_challenger-logistic/15_logistic-regression_challenger.ipynb
@@ -0,0 +1,2197 @@
+{
+ "cells": [
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "setdefaults"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "%plot --format svg"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 115,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans =\n",
+ "\n",
+ " 0.081014\n",
+ " 0.317493\n",
+ " 0.690279\n",
+ " 1.169170\n",
+ " 1.715370\n",
+ " 2.284630\n",
+ " 2.830830\n",
+ " 3.309721\n",
+ " 3.682507\n",
+ " 3.918986\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "N=10;\n",
+ "A_beam=diag(ones(N,1))*2+diag(ones(N-1,1)*-1,-1)+diag(ones(N-1,1)*-1,1);\n",
+ "[v,e]=eig(A_beam);\n",
+ "diag(e)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Nonlinear Regression\n",
+ "\n",
+ "We can define any function and minimize the sum of squares error even if the constants cannot be separated.\n",
+ "\n",
+ "$S_{r}=\\left[y-f(z_{1},z_{2},...)\\right]^{2}$\n",
+ "\n",
+ "Consider the function, \n",
+ "\n",
+ "$f(x) = a_{0}(1-e^{a_{1}x})$\n",
+ "\n",
+ "We can define the sum of squares error as a function of $a_{0}$ and $a_{1}$:\n",
+ "\n",
+ "$f_{SSE}(a_{0},a_{1})=\\sum_{i=1}^{n}\\left[y- a_{0}(1-e^{a_{1}x})\\right]^{2}$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "function [SSE,yhat] = sse_nonlin_exp(a,x,y)\n",
+ " % This is a sum of squares error function based on \n",
+ " % the two input constants a0 and a1 where a=[a0,a1]\n",
+ " % and the data is x (independent), y (dependent)\n",
+ " % and yhat is the model with the given a0 and a1 values\n",
+ " a0=a(1);\n",
+ " a1=a(2);\n",
+ " yhat=a0*(1-exp(a1*x));\n",
+ " SSE=sum((y-a0*(1-exp(a1*x))).^2);\n",
+ "end"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Where the data we are fitting is:\n",
+ "\n",
+ "| x | y |\n",
+ "|---|---|\n",
+ " | 0.0 | 0.41213|\n",
+ " | 1.0 | -2.65190|\n",
+ " | 2.0 | 15.04049|\n",
+ " | 3.0 | 5.19368|\n",
+ " | 4.0 | -0.71086|\n",
+ " | 5.0 | 12.69008|\n",
+ " | 6.0 | 29.20309|\n",
+ " | 7.0 | 58.68879|\n",
+ " | 8.0 | 91.61117|\n",
+ " | 9.0 | 173.75492|\n",
+ " | 10.0 | 259.04083|"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "data=[\n",
+ " 0.00000 0.41213\n",
+ " 1.00000 -2.65190\n",
+ " 2.00000 15.04049\n",
+ " 3.00000 5.19368\n",
+ " 4.00000 -0.71086\n",
+ " 5.00000 12.69008\n",
+ " 6.00000 29.20309\n",
+ " 7.00000 58.68879\n",
+ " 8.00000 91.61117\n",
+ " 9.00000 173.75492\n",
+ " 10.00000 259.04083\n",
+ "\n",
+ "];\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 116,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "the sum of squares for a0=-2.00 and a1=0.20 is 98118.4\n"
+ ]
+ },
+ {
+ "data": {
+ "image/svg+xml": [
+ ""
+ ],
+ "text/plain": [
+ ""
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "[SSE,yhat]=sse_nonlin_exp([-2,0.2],data(:,1),data(:,2));\n",
+ "fprintf('the sum of squares for a0=%1.2f and a1=%1.2f is %1.1f',...\n",
+ "-2,0.2,SSE)\n",
+ "plot(data(:,1),data(:,2),'o',data(:,1),yhat)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 127,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "a =\n",
+ "\n",
+ " -1.71891 0.50449\n",
+ "\n",
+ "fsse = 633.70\n"
+ ]
+ }
+ ],
+ "source": [
+ "[a,fsse]=fminsearch(@(a) sse_nonlin_exp(a,data(:,1),data(:,2)),[-2,0.2])"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 128,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "data": {
+ "image/svg+xml": [
+ ""
+ ],
+ "text/plain": [
+ ""
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "[sse,yhat]=sse_nonlin_exp(a,data(:,1),data(:,2));\n",
+ "plot(data(:,1),data(:,2),'o',data(:,1),yhat)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 130,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "data": {
+ "image/svg+xml": [
+ ""
+ ],
+ "text/plain": [
+ ""
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "[sse,yhat]=sse_nonlin_exp(a,data(:,1),data(:,2));\n",
+ "plot(data(:,1),data(:,2)-yhat)\n",
+ "title('residuals of function')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Case Study: Logistic Regression\n",
+ "\n",
+ "Many times the variable you predict is a binary (or discrete) value, such as pass/fail, broken/not-broken, etc. \n",
+ "\n",
+ "One method to fit this type of data is called [**logistic regression**](https://en.wikipedia.org/wiki/Logistic_regression).\n",
+ "\n",
+ "[Logistic Regression link 2](http://www.holehouse.org/mlclass/06_Logistic_Regression.html)\n",
+ "\n",
+ "We use a function that varies from 0 to 1 called a logistic function:\n",
+ "\n",
+ "$\\sigma(t)=\\frac{1}{1+e^{-t}}$\n",
+ "\n",
+ "We can use this function to describe the likelihood of failure (1) or success (0). When t=0, the probability of failure is 50%. "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 131,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "data": {
+ "image/svg+xml": [
+ ""
+ ],
+ "text/plain": [
+ ""
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "t=linspace(-10,10);\n",
+ "sigma=@(t) 1./(1+exp(-t));\n",
+ "plot(t,sigma(t))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Now we make the assumption that we can predict the boundary between the pass-fail criteria with a function of our independent variable e.g.\n",
+ "\n",
+ "$y=\\left\\{\\begin{array}{cc} \n",
+ "1 & a_{0}+a_{1}x +\\epsilon >0 \\\\\n",
+ "0 & else \\end{array} \\right\\}$\n",
+ "\n",
+ "so the logistic function is now:\n",
+ "\n",
+ "$\\sigma(x)=\\frac{1}{1+e^{-(a_{0}+a_{1}x)}}$\n",
+ "\n",
+ "Here, there is not a direct sum of squares error, so we minimize a cost function: \n",
+ "\n",
+ "$J(a_{0},a_{1})=\\sum_{i=1}^{n}\\left[-y_{i}\\log(\\sigma(x_{i}))-(1-y_{i})\\log((1-\\sigma(x_{i})))\\right]$\n",
+ "\n",
+ "y=0,1 \n",
+ "\n",
+ "So the cost function either sums the "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "### Example: Challenger O-ring failures\n",
+ "\n",
+ "The O-rings on the Challenger shuttles had problems when temperatures became low. We can look at the conditions when damage was observed to determine the likelihood of failure. \n",
+ "\n",
+ "[Challenger O-ring data powerpoint](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwjZvL7jkP3SAhUp04MKHXkXDkMQFggcMAA&url=http%3A%2F%2Fwww.stat.ufl.edu%2F~winner%2Fcases%2Fchallenger.ppt&usg=AFQjCNFyjwT7NmRthDkDEgch75Fc5dc66w&sig2=_qeteX6-ZEBwPW8SZN1mIA)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 132,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "oring =\n",
+ "\n",
+ " 1.00000 53.00000 1.00000\n",
+ " 2.00000 57.00000 1.00000\n",
+ " 3.00000 58.00000 1.00000\n",
+ " 4.00000 63.00000 1.00000\n",
+ " 5.00000 66.00000 0.00000\n",
+ " 6.00000 66.80000 0.00000\n",
+ " 7.00000 67.00000 0.00000\n",
+ " 8.00000 67.20000 0.00000\n",
+ " 9.00000 68.00000 0.00000\n",
+ " 10.00000 69.00000 0.00000\n",
+ " 11.00000 69.80000 1.00000\n",
+ " 12.00000 69.80000 0.00000\n",
+ " 13.00000 70.20000 1.00000\n",
+ " 14.00000 70.20000 0.00000\n",
+ " 15.00000 72.00000 0.00000\n",
+ " 16.00000 73.00000 0.00000\n",
+ " 17.00000 75.00000 0.00000\n",
+ " 18.00000 75.00000 1.00000\n",
+ " 19.00000 75.80000 0.00000\n",
+ " 20.00000 76.20000 0.00000\n",
+ " 21.00000 78.00000 0.00000\n",
+ " 22.00000 79.00000 0.00000\n",
+ " 23.00000 81.00000 0.00000\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "% read data from csv file \n",
+ "% col 1 = index\n",
+ "% col 2 = temperature\n",
+ "% col 3 = 1 if damaged, 0 if undamaged\n",
+ "oring=dlmread('challenger_oring.csv',',',1,0)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 134,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "data": {
+ "image/svg+xml": [
+ ""
+ ],
+ "text/plain": [
+ ""
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "plot(oring(:,2),oring(:,3),'o')\n",
+ "xlabel('Temp (F)')\n",
+ "ylabel('failure (1)/ pass (0)')"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 135,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "function J=sse_logistic(a,x,y)\n",
+ " % Create function to calculate cost of logistic function\n",
+ " % t = a0+a1*x\n",
+ " % sigma(t) = 1./(1+e^(-t))\n",
+ " sigma=@(t) 1./(1+exp(-t));\n",
+ " a0=a(1);\n",
+ " a1=a(2);\n",
+ " t=a0+a1*x;\n",
+ " J = 1/length(x)*sum(-y.*log(sigma(t))-(1-y).*log(1-sigma(t)));\n",
+ "end"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 142,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "J = 0.88822\n",
+ "a =\n",
+ "\n",
+ " 15.03501 -0.23205\n",
+ "\n"
+ ]
+ },
+ {
+ "data": {
+ "image/svg+xml": [
+ ""
+ ],
+ "text/plain": [
+ ""
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "J=sse_logistic([10,-0.2],oring(:,2),oring(:,3))\n",
+ "a=fminsearch(@(a) sse_logistic(a,oring(:,2),oring(:,3)),[0,-3])\n",
+ "\n",
+ "T=linspace(50,85);\n",
+ "plot(oring(:,2),oring(:,3),'o',T,sigma(a(1)+a(2)*T),T,a(1)+a(2)*T)\n",
+ "axis([50,85,-0.1,1.2])"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 139,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "probability of failure when 70 degrees is 23.00% \n",
+ "probability of failure when 60 degrees is 75.25%\n",
+ "probability of failure when 36 degrees is 99.87%\n"
+ ]
+ }
+ ],
+ "source": [
+ "fprintf('probability of failure when 70 degrees is %1.2f%% ',100*sigma(a(1)+a(2)*70))\n",
+ "fprintf('probability of failure when 60 degrees is %1.2f%%',100*sigma(a(1)+a(2)*60))\n",
+ "fprintf('probability of failure when 36 degrees is %1.2f%%',100*sigma(a(1)+a(2)*36))\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Interpolation\n",
+ "\n",
+ "Using regression (linear and nonlinear) you are faced with the problem, that you have lots of noisy data and you want to fit a physical model to it. \n",
+ "\n",
+ "You can use interpolation to solve the opposite problem, you have a little data with very little noise.\n",
+ "\n",
+ "## Linear interpolation\n",
+ "\n",
+ "If you are trying to find the value of f(x) for x between $x_{1}$ and $x_{2}$, then you can match the slopes\n",
+ "\n",
+ "$\\frac{f(x)-f(x_{1})}{x-x_{1}}=\\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$\n",
+ "\n",
+ "or\n",
+ "\n",
+ "$f(x)=f(x_{1})+(x-x_{1})\\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$\n",
+ "\n",
+ "### Example: Logarithms\n",
+ "\n",
+ "Engineers used to have to use interpolation in logarithm tables for calculations. Find ln(2) from \n",
+ "\n",
+ "a. ln(1) and ln(6)\n",
+ "\n",
+ "b. ln(1) and ln(4)\n",
+ "\n",
+ "c. just calculate it as ln(2)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 149,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ln(2)~0.358352\n",
+ "ln(2)~0.462098\n",
+ "ln(2)~0.549306\n",
+ "ln(2)=0.693147\n"
+ ]
+ }
+ ],
+ "source": [
+ "ln2_16=log(1)+(log(6)-log(1))/(6-1)*(2-1);\n",
+ "fprintf('ln(2)~%f\\n',ln2_16)\n",
+ "ln2_14=log(1)+(log(4)-log(1))/(4-1)*(2-1);\n",
+ "ln2_13=log(1)+(log(3)-log(1))/(3-1)*(2-1);\n",
+ "fprintf('ln(2)~%f\\n',ln2_14)\n",
+ "fprintf('ln(2)~%f\\n',ln2_13)\n",
+ "ln2=log(2);\n",
+ "fprintf('ln(2)=%f\\n',ln2)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 147,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "data": {
+ "image/svg+xml": [
+ ""
+ ],
+ "text/plain": [
+ ""
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "x=linspace(1,6);\n",
+ "plot(x,log(x),2,log(2),'*',...\n",
+ "[1,2,6],[log(1),ln2_16,log(6)],'o-',...\n",
+ "[1,2,4],[log(1),ln2_14,log(4)],'s-')\n",
+ "ylabel('ln(x)')\n",
+ "xlabel('x')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Quadratic interpolation (intro curvature)\n",
+ "\n",
+ "Assume function is parabola between 3 points. The function is can be written as:\n",
+ "\n",
+ "$f_{2}(x)=b_{1}+b_{2}(x-x_{1})+b_{3}(x-x_{1})(x-x_{2})$\n",
+ "\n",
+ "When $x=x_{1}$\n",
+ "\n",
+ "$f(x_{1})=b_{1}$\n",
+ "\n",
+ "when $x=x_{2}$\n",
+ "\n",
+ "$b_{2}=\\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$\n",
+ "\n",
+ "when $x=x_{3}$\n",
+ "\n",
+ "$b_{3}=\\frac{\\frac{f(x_{3})-f(x_{2})}{x_{3}-x_{2}}\n",
+ "-\\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}}{x_{3}-x_{1}}$\n",
+ "\n",
+ "#### Reexamining the ln(2) with ln(1), ln(4), and ln(6):"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 154,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Z =\n",
+ "\n",
+ " 1 1 1\n",
+ " 1 4 16\n",
+ " 1 600 360000\n",
+ "\n",
+ "ans = 5.1766e+05\n",
+ "ans =\n",
+ "\n",
+ " -4.6513e-01\n",
+ " 4.6589e-01\n",
+ " -7.5741e-04\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "x=[1,4,600]';\n",
+ "Z=[x.^0,x.^1,x.^2]\n",
+ "cond(Z)\n",
+ "Z\\log(x)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 155,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "b1 = 0\n",
+ "b2 = 0.46210\n",
+ "b3 = -0.051873\n"
+ ]
+ }
+ ],
+ "source": [
+ "x1=1;\n",
+ "x2=4;\n",
+ "x3=6;\n",
+ "f1=log(x1);\n",
+ "f2=log(x2);\n",
+ "f3=log(x3);\n",
+ "\n",
+ "b1=f1\n",
+ "b2=(f2-b1)/(x2-x1)\n",
+ "b3=(f3-f2)/(x3-x2)-b2;\n",
+ "b3=b3/(x3-x1)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 157,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ans = 0.56584\r\n"
+ ]
+ },
+ {
+ "data": {
+ "image/svg+xml": [
+ ""
+ ],
+ "text/plain": [
+ ""
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "x=linspace(1,6);\n",
+ "f=@(x) b1+b2*(x-x1)+b3*(x-x1).*(x-x2);\n",
+ "plot(x,log(x),2,log(2),'*',...\n",
+ "[1,4,6],[log(1),log(4),log(6)],'ro',...\n",
+ "x,f(x),'r-',2,f(2),'s')\n",
+ "f(2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "### Newton's Interpolating Polynomials\n",
+ "\n",
+ "For n-data points, we can fit an (n-1)th-polynomial\n",
+ "\n",
+ "$f_{n-1}(x)=b_{1}+b_{2}(x-x_{1})+\\cdots+b_{n}(x-x_{1})(x-x_{2})\\cdots(x-x_{n})$\n",
+ "\n",
+ "where \n",
+ "\n",
+ "$b_{1}=f(x_{1})$\n",
+ "\n",
+ "$b_{2}=\\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$\n",
+ "\n",
+ "$b_{3}=\\frac{\\frac{f(x_{3})-f(x_{2})}{x_{3}-x_{2}}\n",
+ "-b_{2}}{x_{3}-x_{1}}$\n",
+ "\n",
+ "$\\vdots$\n",
+ "\n",
+ "$b_{n}=f(x_{n},x_{n-1},...,x_{2},x_{1})\n",
+ "=\\frac{f(x_{n},x_{n-1},...x_{2})-f(x_{n-1},x_{n-2},...,x_{1})}{x_{n}-x_{1}}$\n",
+ "\n",
+ "**e.g. for 4 data points:**\n",
+ "\n",
+ ""
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 160,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "b =\n",
+ "\n",
+ " 0.00000 0.54931 -0.08721 0.01178\n",
+ " 1.09861 0.28768 -0.02832 0.00000\n",
+ " 1.38629 0.20273 0.00000 0.00000\n",
+ " 1.79176 0.00000 0.00000 0.00000\n",
+ "\n",
+ "ans = 0.66007\n",
+ "ans =\n",
+ "\n",
+ " 0.00000\n",
+ " 1.09861\n",
+ " 1.38629\n",
+ " 1.79176\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "Newtint([1,3,4,6],log([1,3,4,6]),2)\n",
+ "log([1,3,4,6]')"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ln(2)=0.693147\n",
+ "ln(2)~0.693147\n"
+ ]
+ },
+ {
+ "data": {
+ "image/svg+xml": [
+ ""
+ ],
+ "text/plain": [
+ ""
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "\n",
+ "\n",
+ "x=[0.2,1,2,3,4]; % define independent var's\n",
+ "y=log(x); % define dependent var's\n",
+ "xx=linspace(min(x),max(x)+1);\n",
+ "yy=zeros(size(xx));\n",
+ "for i=1:length(xx)\n",
+ " yy(i)=Newtint(x,y,xx(i));\n",
+ "end\n",
+ "plot(xx,log(xx),2,log(2),'*',...\n",
+ "x,y,'ro',...\n",
+ "xx,yy,'r-')\n",
+ "\n",
+ "fprintf('ln(2)=%f',log(2))\n",
+ "fprintf('ln(2)~%f',Newtint(x,y,2))"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Octave",
+ "language": "octave",
+ "name": "octave"
+ },
+ "language_info": {
+ "file_extension": ".m",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "octave",
+ "version": "0.19.14"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
diff --git a/15_challenger-logistic/Newtint.m b/15_challenger-logistic/Newtint.m
new file mode 100644
index 0000000..e4c6c83
--- /dev/null
+++ b/15_challenger-logistic/Newtint.m
@@ -0,0 +1,34 @@
+function yint = Newtint(x,y,xx)
+% Newtint: Newton interpolating polynomial
+% yint = Newtint(x,y,xx): Uses an (n - 1)-order Newton
+% interpolating polynomial based on n data points (x, y)
+% to determine a value of the dependent variable (yint)
+% at a given value of the independent variable, xx.
+% input:
+% x = independent variable
+% y = dependent variable
+% xx = value of independent variable at which
+% interpolation is calculated
+% output:
+% yint = interpolated value of dependent variable
+
+% compute the finite divided differences in the form of a
+% difference table
+n = length(x);
+if length(y)~=n, error('x and y must be same length'); end
+b = zeros(n,n);
+% assign dependent variables to the first column of b.
+b(:,1) = y(:); % the (:) ensures that y is a column vector.
+for j = 2:n
+ for i = 1:n-j+1
+ b(i,j) = (b(i+1,j-1)-b(i,j-1))/(x(i+j-1)-x(i));
+ end
+end
+%b
+% use the finite divided differences to interpolate
+xt = 1;
+yint = b(1,1);
+for j = 1:n-1
+ xt = xt*(xx-x(j));
+ yint = yint+b(1,j+1)*xt;
+end
diff --git a/15_challenger-logistic/challenger_oring.csv b/15_challenger-logistic/challenger_oring.csv
new file mode 100644
index 0000000..11d647e
--- /dev/null
+++ b/15_challenger-logistic/challenger_oring.csv
@@ -0,0 +1,24 @@
+Flight#,Temp,O-Ring Problem
+1,53,1
+2,57,1
+3,58,1
+4,63,1
+5,66,0
+6,66.8,0
+7,67,0
+8,67.2,0
+9,68,0
+10,69,0
+11,69.8,1
+12,69.8,0
+13,70.2,1
+14,70.2,0
+15,72,0
+16,73,0
+17,75,0
+18,75,1
+19,75.8,0
+20,76.2,0
+21,78,0
+22,79,0
+23,81,0
diff --git a/15_challenger-logistic/newton_interpolation.png b/15_challenger-logistic/newton_interpolation.png
new file mode 100644
index 0000000..5990cb5
Binary files /dev/null and b/15_challenger-logistic/newton_interpolation.png differ
diff --git a/15_challenger-logistic/octave-workspace b/15_challenger-logistic/octave-workspace
new file mode 100644
index 0000000..8c437bb
Binary files /dev/null and b/15_challenger-logistic/octave-workspace differ