Re: Evaluating a gradient function with vectors
I think your approach is wrong for the denominator. My understanding is that w*x is a inner product that makes a scalar, so the only vector part of the gradient is the yn*xn on the top of the fraction.
(In particular, all 3 w and x terms would be used in the exponential in the denominator for each of the 3 values in the gradient. But your numerator would use the 3 different x's for the 3 different terms.)
