Confusion on Delta Rule and Error












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I'm currently reading Mitchell's book for Machine Learning, and he just started gradient descent. There's one part that's really confusing me.



At one point, he gives this equation for the error of a perceptron over a set of training examples.



enter image description here



$O_d$ is the actual output of $ vec{W} cdot vec{X}$, where $ vec{X}$ is the input vector and $vec{W}$ is the weights vector.



$t_d$ is the target output, what we want to get.



The sum over all the $D$ means we sum over every single $vec{X}$ we can input.



Okay, so far so good, I understand that.



However, he then gives this example:



enter image description here



But that is just not true!!!! That equation for the error does NOT give us a single minimum!!!



According to his previous rule, if we're considering the error for a single weight vector and a single training vector, the equation for the error would be:



$E(vec{w}) = frac{1}{2} * (t_d - (w_0 * x_0 + w_1 * x_1))^2$



Which has an infinite number of minimums!!! Every time $(w_0 * x_0 + w_1 * x_1) = t_d$



I graphed it here to show you:



enter image description here



In that picture, $x$ and $y$ are the two rows of the weight vector $vec{w}$.



Please help! I've been confused about this for the last three hours!



Thanks






machine-learning neural-network training gradient-descent perceptron






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    0












    $begingroup$


    I'm currently reading Mitchell's book for Machine Learning, and he just started gradient descent. There's one part that's really confusing me.



    At one point, he gives this equation for the error of a perceptron over a set of training examples.



    enter image description here



    $O_d$ is the actual output of $ vec{W} cdot vec{X}$, where $ vec{X}$ is the input vector and $vec{W}$ is the weights vector.



    $t_d$ is the target output, what we want to get.



    The sum over all the $D$ means we sum over every single $vec{X}$ we can input.



    Okay, so far so good, I understand that.



    However, he then gives this example:



    enter image description here



    But that is just not true!!!! That equation for the error does NOT give us a single minimum!!!



    According to his previous rule, if we're considering the error for a single weight vector and a single training vector, the equation for the error would be:



    $E(vec{w}) = frac{1}{2} * (t_d - (w_0 * x_0 + w_1 * x_1))^2$



    Which has an infinite number of minimums!!! Every time $(w_0 * x_0 + w_1 * x_1) = t_d$



    I graphed it here to show you:



    enter image description here



    In that picture, $x$ and $y$ are the two rows of the weight vector $vec{w}$.



    Please help! I've been confused about this for the last three hours!



    Thanks






    machine-learning neural-network training gradient-descent perceptron






    share









    $endgroup$















      0












      0








      0





      $begingroup$


      I'm currently reading Mitchell's book for Machine Learning, and he just started gradient descent. There's one part that's really confusing me.



      At one point, he gives this equation for the error of a perceptron over a set of training examples.



      enter image description here



      $O_d$ is the actual output of $ vec{W} cdot vec{X}$, where $ vec{X}$ is the input vector and $vec{W}$ is the weights vector.



      $t_d$ is the target output, what we want to get.



      The sum over all the $D$ means we sum over every single $vec{X}$ we can input.



      Okay, so far so good, I understand that.



      However, he then gives this example:



      enter image description here



      But that is just not true!!!! That equation for the error does NOT give us a single minimum!!!



      According to his previous rule, if we're considering the error for a single weight vector and a single training vector, the equation for the error would be:



      $E(vec{w}) = frac{1}{2} * (t_d - (w_0 * x_0 + w_1 * x_1))^2$



      Which has an infinite number of minimums!!! Every time $(w_0 * x_0 + w_1 * x_1) = t_d$



      I graphed it here to show you:



      enter image description here



      In that picture, $x$ and $y$ are the two rows of the weight vector $vec{w}$.



      Please help! I've been confused about this for the last three hours!



      Thanks






      machine-learning neural-network training gradient-descent perceptron






      share









      $endgroup$




      I'm currently reading Mitchell's book for Machine Learning, and he just started gradient descent. There's one part that's really confusing me.



      At one point, he gives this equation for the error of a perceptron over a set of training examples.



      enter image description here



      $O_d$ is the actual output of $ vec{W} cdot vec{X}$, where $ vec{X}$ is the input vector and $vec{W}$ is the weights vector.



      $t_d$ is the target output, what we want to get.



      The sum over all the $D$ means we sum over every single $vec{X}$ we can input.



      Okay, so far so good, I understand that.



      However, he then gives this example:



      enter image description here



      But that is just not true!!!! That equation for the error does NOT give us a single minimum!!!



      According to his previous rule, if we're considering the error for a single weight vector and a single training vector, the equation for the error would be:



      $E(vec{w}) = frac{1}{2} * (t_d - (w_0 * x_0 + w_1 * x_1))^2$



      Which has an infinite number of minimums!!! Every time $(w_0 * x_0 + w_1 * x_1) = t_d$



      I graphed it here to show you:



      enter image description here



      In that picture, $x$ and $y$ are the two rows of the weight vector $vec{w}$.



      Please help! I've been confused about this for the last three hours!



      Thanks






      machine-learning neural-network training gradient-descent perceptron




      machine-learning neural-network training gradient-descent perceptron





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      asked 4 mins ago









      Joshua RonisJoshua Ronis

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