<?xml version="1.0" encoding="utf-8"?><!DOCTYPE article  PUBLIC '-//OASIS//DTD DocBook XML V4.4//EN'  'http://www.docbook.org/xml/4.4/docbookx.dtd'><article><articleinfo><title>FAQ/polyeqn</title><revhistory><revision><revnumber>16</revnumber><date>2015-04-15 14:41:49</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>15</revnumber><date>2014-10-14 14:12:59</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>14</revnumber><date>2014-10-14 14:11:20</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>13</revnumber><date>2014-10-08 15:45:53</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>12</revnumber><date>2013-03-08 10:17:44</date><authorinitials>localhost</authorinitials><revremark>converted to 1.6 markup</revremark></revision><revision><revnumber>11</revnumber><date>2012-10-22 12:32:26</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>10</revnumber><date>2012-10-22 12:31:35</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>9</revnumber><date>2012-10-22 12:30:20</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>8</revnumber><date>2012-08-15 12:18:42</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>7</revnumber><date>2012-08-15 12:15:41</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>6</revnumber><date>2012-08-15 12:14:45</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>5</revnumber><date>2012-08-15 12:14:08</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>4</revnumber><date>2012-08-15 12:10:29</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>3</revnumber><date>2012-08-15 12:09:28</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>2</revnumber><date>2012-08-15 12:09:13</date><authorinitials>PeterWatson</authorinitials></revision><revision><revnumber>1</revnumber><date>2012-08-15 12:05:52</date><authorinitials>PeterWatson</authorinitials></revision></revhistory></articleinfo><section><title>Summary measures for learning curves</title><para>Suppose we wish to look at change in processing success (y) with ease of task (x). </para><para>We fit a cubic polynomial and obtain using multiple regression a best fitting cubic of form -0.000005x<superscript>3 </superscript> + 0.0003 x<superscript>2 </superscript> + 0.0026x + 0.7508. This is the curve in red in this <ulink url="https://lsr-wiki-02.mrc-cbu.cam.ac.uk/statswiki/FAQ/polyeqn/statswiki/FAQ/polyeqn?action=AttachFile&amp;do=get&amp;target=plot.jpeg">plot.</ulink> </para><para>The asymptote occurs at a turning point when its first derivative (a quadratic) equals zero. For the curve in red this quadratic, obtained from differentiating the above cubic, is of form -0.000015x^2 + 0.0006x - 0.0026 =0. This quadratic has a turning point in the sampled interval [10,60] at 43.94 obtained by inputting the regression coefficients into the <ulink url="http://www.mathsisfun.com/quadratic-equation-solver.html">quadratic root solver here.</ulink> </para><para>The predicted learning slope, from baseline to asymptote, is obtained by subtracting the predicted response, obtained using the regression coefficients in the cubic, at the start (x=10) from that at the asymptote(x=43.94) of the learning line and dividing by the change in x (43.94-10),  </para><para>So for the red curve the change from baseline to asymptote = (1.02 – 0.80)/(43.94-10)=0.0064. </para><para>Note that Howell (1995) shows that a quadratic curve with a linear component is sufficient to yield an asymptotic model. In this instance one would obtain the asymptote by differentiating the quadratic to get a linear equation with the asymptote equal to its root or, equivalently, by solving the quadratic equation and averaging the (real) roots (the vertex method). </para><para>Similarly if we anticipate having three turning points we can fit predictors of form x^4,  x^3,  x^2 and x, obtain their regression coefficients together with that of the intercept term and differentiate this quartic to obtain a cubic curve. The roots of this cubic (obtained using a cubic root solver calculator such as <ulink url="http://easycalculation.com/algebra/cubic-equation.php">here</ulink>) will give the location of the three turning points. </para><para><emphasis role="underline">Reference</emphasis> </para><para>Howell DC (1997). Statistical methods for psychology. Fourth Edition. Wadsworth, Belmont, CA. </para></section></article>