5 ms·
Can you please elaborate on "you never get the expression"? When you input the variable values into a symbolic derivative you just get a value at the end. d/dx
by laplacesdemon48 6y ago
Can you please elaborate on "you never get the expression"?
When you input the variable values into a symbolic derivative you just get a value at the end. d/dx x^2 = 2x. If x = 0.5 then d/dx = 1. The same is true for symbolic integrals. For most practical applications, we don't really care about the full symbolic expression. We just want the answer, or at least a good approximation. This post uses a specific example of the difference between two Beta distributions. We want to get that 0.71. It is very hard to "automatically" make that happen.
- adamnemecek 6y agoAre you familiar with automatic differentiation? https://blog.demofox.org/2014/12/30/dual-numbers-automatic-differentiation/ https://blog.demofox.org/2014/12/30/dual-numbers-automatic-d... note that you get f(x) and f'(x) for some x without getting the derived function.
- justinpombrio 6y agoOoh, that's a neat technique. I think you can get an expression for the derivative from it. Say we want the derivative of f(x) = x^3 - 2x^2 + 5. That becomes: (x+e)^3 - 2(x+e)^2 + 5 = x^3 + 3x^2e - 2x^2 -4xe + 5 = (x^3 - 2x^2 + 5) + (3x^2 - 4x)e The term in front of 'e' is "3x^2 - 4x", which is f'(x).
- judofyr 6y agoJust to clarify: This result follows directly from the definition of the derivative: f'(x) = lim_{e->0} (f(x+e) - f(x))/e If you're able to express f(x+e) on the form (f(x) + y e) then it follows that y is the derivative. It also should be noted that auto-diff doesn't let you skip the rules for derivation and you're using the same calculation as you would to show that e.g. f'(x^n)=n*x^{n-1}.
- foxes 6y agoFor algebraic expressions you get an expression if you expand in dual numbers. But what about something like cos(x). Either you can lazily evaluate the power series or you know its sin(x).