[regression] expandend exercise according to feedback from Alex

regression/exercises/expdecaydescent.m

@@ -23,12 +23,3 @@ function [tau, taus, mses] = expdecaydescent(t, x, tau0, epsilon, threshold)
       count = count + 1;
   end
 end
-
-function mse = expdecaymse(t, x, tau)
-  mse = mean((x - expdecay(t, tau)).^2);
-end
-
-function gradient = expdecaygradient(t, x, tau)
-  h = 1e-7;   % stepsize for derivative
-  gradient = (expdecaymse(t, x, tau+h) - expdecaymse(t, x, tau))/h;
-end

regression/exercises/expdecaygradient.m

@@ -0,0 +1,11 @@
+function gradient = expdecaygradient(t, x, tau)
+% Gradient of MSE for a decaying exponential.
+%
+% Arguments: t, vector of time points.
+%            x, vector of the corresponding measured data values.
+%            tau, value for the time constant. 
+%
+% Returns: gradient: the derivative of the MSE with respect to tau.
+  h = 1e-7;   % stepsize for derivative
+  gradient = (expdecaymse(t, x, tau+h) - expdecaymse(t, x, tau))/h;
+end

regression/exercises/expdecaygradientplot.m

@@ -0,0 +1,15 @@
+...
+
+gradients = zeros(length(taus), 1);
+for i = 1:length(taus)
+  gradients(i) = expdecaygradient(time, voltage, taus(i));
+end
+
+subplot(2, 1, 2)
+plot(taus, gradients);
+xlabel('tau [ms]')
+ylabel('gradient [mV^2/ms]')
+savefigpdf(gcf, 'expdecaygradientplot.pdf', 10, 10);
+
+
+

regression/exercises/expdecaymse.m

@@ -0,0 +1,10 @@
+function mse = expdecaymse(t, x, tau)
+% Mean squared error for a decaying exponential.
+%
+% Arguments: t, vector of time points.
+%            x, vector of the corresponding measured data values.
+%            tau, value for the time constant. 
+%
+% Returns:   mse: mean squared error
+  mse = mean((x - expdecay(t, tau)).^2);
+end

regression/exercises/expdecaymseplot.m

@@ -0,0 +1,13 @@
+expdecaydata;                             % generate data
+close all;
+
+taus = 0.0:0.1:20.0;
+mses = zeros(length(taus), 1);
+for i = 1:length(taus)
+  mses(i) = expdecaymse(time, voltage, taus(i));
+end
+
+subplot(2, 1, 1)
+plot(taus, mses);
+xlabel('tau [ms]')
+ylabel('mean squared error [mV^2]')

regression/exercises/expdecayplot.m

@@ -11,7 +11,7 @@ plot(taus, '-o');
 plot([1, length(taus)], [tau, tau], 'k'); % line indicating true tau
 hold off;
 xlabel('Iteration');
-ylabel('tau');
+ylabel('tau [ms]');
 subplot(2, 2, 3);                         % bottom left panel
 plot(mses, '-o');
 xlabel('Iteration steps');

regression/exercises/gradientdescent-1.tex

@@ -56,6 +56,29 @@
       \lstinputlisting{expdecaydata.m}
     \end{solution}
 
+    \part Write a function that returns the mean squared error.  The
+    function takes as arguments the measured data (time points and
+    corresponding voltage measurements) and a value for the membrane
+    time constant at which the mean squred error should be
+    evaluated. Plot the mean squared error as a function of $\tau$.
+    \begin{solution}
+      \lstinputlisting{expdecaymse.m}
+      \lstinputlisting{expdecaymseplot.m}
+    \end{solution}
+
+    \part Write a function that returns the gradient of the mean
+    squared error, i.e. its derivative with respect to the $\tau$
+    parameter.  The function takes as arguments the measured data
+    (time points and corresponding voltage measurements) and a value
+    for the membrane time constant at which the gradient should be
+    evaluated. Plot the gradient as a function of $\tau$.
+    \begin{solution}
+      \lstinputlisting{expdecaygradient.m}
+      \newsolutionpage
+      \lstinputlisting{expdecaygradientplot.m}
+      \includegraphics{expdecaygradientplot}
+    \end{solution}
+
     \part Implement the gradient descent algorithm for finding the
     least squares for the exponential function \eqref{expfunc}. The
     function takes as arguments the measured data, an initial value
@@ -78,7 +101,6 @@
     for each iteration step, (ii) the mean squared error for each
     iteration step, and (iii) the measured data and the fitted
     exponential function.
-    \newsolutionpage
     \begin{solution}
       \lstinputlisting{expdecayplot.m}
       \includegraphics{expdecayplot}
@@ -140,9 +162,16 @@
     \part Plot the path of the gradient descent algorithm into the
     error surface.
 
-    Why appears (potentially) the path of the gradient descent
+    Why does (potentially) the path of the gradient descent algorithm
-    algorithm not perpendicular to the contour lines of the error
+    appear to be not perpendicular to the contour lines of the error
     surface?
+    \begin{solution}
+      Because on the plot the $c$ and $\tau$ axes are (in general) not
+      equally scaled. That is a unit step in $c$ direction covers a
+      physical distance on paper or the screen than a unit stpe in
+      $\tau$ direction. Use \code{axis equal} to force the plot to
+      have equally scaled $c$ and $\tau$ axes.
+    \end{solution}
 
     \part Plot the data together with the fitted exponential function.