3 ms·
> A dot product is nothing of the sort. Its nearly the same thing, isn't it? If you denote by Tx the left-shift operator defined by (Tx f)(y) = f(y+x), then t
by enriquto 6y ago
> A dot product is nothing of the sort.
Its nearly the same thing, isn't it? If you denote by Tx the left-shift operator defined by (Tx f)(y) = f(y+x), then the correlation of f and g evaluated at x is precisely the dot product of f and Tx g. If you evaluate your function at a certain point, you obtain a scalar product.
- rualca 6y ago> (...) then the correlation of f and g evaluated at x It really isn't the same, and oddly enough you unknowingly show that off, by mentioning that convolution is the function that maps input functions to the output function, but the dot product is at best a single point evaluated with the output function.