By D. E. Rutherford

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1 Defining the model We can begin by defining our model as a function which maps our input attributes, in this case Olympic year, to our output or target values – winning time. g. g. years since the first modern games, 1980 − 1896 = 84) that would make no real difference to the underlying assumptions. There are many functions that could be used to define this mapping. In general, this function will take as an input x (the Olympic year) and will return t (the winning time in seconds). In other words, t is a function of x.

In these cases, we will omit the size subscript. A key property of the identity matrix is that any vector or matrix multiplied by a suitably sized identity matrix is equal to the original matrix or vector. For example, if y = [y1 , . . , yD ]T and ID is the D × D identity matrix, yT ID = y, Iy = y. Similarly, for an N × M matrix, a11 a21 A= . .. aN 1 a12 a22 .. aN 2 ... .. ... a1M a2M .. aN M AIM = A, IN A = A. Multiplying a scalar by an identity results in a matrix with the scalar value on each diagonal element.

1). 10. Consider augmenting our data matrix X with an additional column, x2n : 1 x1 x21 1 1 x2 x22 xn = xn , X = . .. .. 2 . xn 1 xN x2N and adding an extra parameter to w: w0 w = w1 , w2 resulting in f (x; w) = wT x = w0 + w1 x + w2 x2 . 16 to find w but the function we are fitting is quadratic in the data. m). Also shown is the function we get if we try and fit our original linear (in the data) model (dashed line, t = w0 + w1 x). It is clear from the quality of the fit to the data that the quadratic model is a more appropriate model for this data.