Automatic differentiation using dual numbers

[Albert Chan]

Perhaps it is better to think of dual number as a container, holding symbolic variables.

That's why even with complicated expressions, derivative only has slight rounding errors,
due to symbolic variables really stored numerically.

This is why variable as {x, 1}: f(x) = x → f'(x) = 1
Plain constant simply {c, 0}:   f(x) = c → f'(x) = 0

(06-20-2022 05:31 PM)Thomas Klemm Wrote:  \(
\begin{align}
(2 + 3 \varepsilon)(5 - \varepsilon) = 10 + 13 \varepsilon
\end{align}
\)

2 ENTER 3 COMPLEX
5 ENTER -1 COMPLEX
XEQ "*"

10 i13

Think of this as getting {f*g, (f*g)'}, forget about ε

(f*g) = 2*5 = 10
(f*g)' = f'*g + f*g' = 3*5 + 2*-1 = 13

Again, do not think of them as numbers; think of them as symbolic, stored numerically.

---

If we wanted to extend container to hold more derivatives, ε is a bad idea.
Example, we want to carry 2 derivatives, instead of 1.

CAS> expand((1+2*ε+3*ε^2) * (3+ε+4*ε^2))

3 + 7*ε + 15*ε^2 + 11*ε^3 + 12*ε^4

Do this symbolically (recommended), we get correct answer.

{f, f', f''} = {1,2,3}
{g,g',g''} = {3,1,4}

(f*g) = 1*3 = 3
(f*g)' = f'*g + f*g' = 2*3 + 1*1 = 7
(f*g)'' = f''*g + f'*g' + f'*g' + f*g'' = 3*3 + 2*(2*1) + 1*4 = 17