Re:Abusing the algebra of algebraic data types – why does th...
by dannymi on Monday October 10, 2022 @08:28PM
(#33158029)
Indeed the signature of the function on the lhs is f(x) and of the one on the rhs is f(u,v).
On the rhs, it very much is not lambda x: f(u(x), v(x)) that the differential operator ∂/∂u (or ∂/∂v) is applied on—because the former would be a function of x.
(I use only x rather than (x,y) for simplicity so you can see where a big problem would lie)
Be aware that the notation in ∂f/∂x = ∂f/∂u ∂u/∂x + ∂f/∂v ∂v/∂x is already very simplified and the actual whole thing would be (in modern notation, as opposed to the one used by SICP):
Where (as is usual) the (...) denotes function calls, and the ∂/∂x f0 makes a new function, here of x. Note that the function call is done on that new function (this order of precedence is the usual in mathematics)
Also, f0(x) = f(u(x),v(x))
Understanding what exactly is going on where in here is already the basis for fluid dynamics.
On the rhs, it very much is not lambda x: f(u(x), v(x)) that the differential operator ∂/∂u (or ∂/∂v) is applied on—because the former would be a function of x.
(I use only x rather than (x,y) for simplicity so you can see where a big problem would lie)
Be aware that the notation in ∂f/∂x = ∂f/∂u ∂u/∂x + ∂f/∂v ∂v/∂x is already very simplified and the actual whole thing would be (in modern notation, as opposed to the one used by SICP):
∂/∂x f0(x) = ∂/∂u f(u(x),v(x)) ∂/∂x u(x) + ∂/∂v f(u,v) ∂/∂x v(x)
Where (as is usual) the (...) denotes function calls, and the ∂/∂x f0 makes a new function, here of x. Note that the function call is done on that new function (this order of precedence is the usual in mathematics)
Also, f0(x) = f(u(x),v(x))
Understanding what exactly is going on where in here is already the basis for fluid dynamics.