# Derivatives of sums and products “isomorphic” to powers of products and sums

Let $f$ and $g$ be smooth Real functions.

From basic algebra, any power of a product is a product of powers:

$$(f\cdot g)^n = f^n\cdot g^n,$$

and a power of a sum is given by the binomial expansion

$$(f+g)^n = \sum_{k=0}^n \binom nk f^{n-k}g^k = \sum_{k=0}^n \frac{n!}{(n-k)!k!}f^{n-k}g^k.$$

And from calculus, we also have the derivative of a sum

$$\frac{d^n}{dx^n}(f+g) = f^{(n)}+g^{(n)}$$

and the derivative of a product

$$\frac{d^n}{dx^n}(f\cdot g) = \sum_{k=0}^n \binom nk f^{(n-k)}g^{(k)} = \sum_{k=0}^n \frac{n!}{(n-k)!k!}f^{(n-k)}g^{(k)},$$

where the superscript parentheses $g^{(k)}$ denote the $k$th derivative of $g$.

So it seems that addition, multiplication, and exponentiation are "isomorphic", in some sense, to multiplication, addition, and differentiation.

$$\Big(C^\infty\mathbb R,\;+\;,\;\cdot\;,\;\wedge n\Big)\sim\Big(C^\infty\mathbb R,\;\cdot\;,\;+\;,\;\frac{d^n}{dx^n}\Big)$$

I know it's not an isomorphism in the strict sense, but I don't know what other word to use. "Relation" is too weak.

Is this useful? Has it been studied before? What exactly is this relation?

These other questions are similar, but they focus on the binomial stuff. This question is focused on the duality between addition and multiplication.

The exponential function looks like it might be useful here; it maps addition to multiplication, but it maps multiplication to exponentiation, not back to addition.

$$\exp(f+g) = \exp(f)\cdot\exp(g)$$

$$\exp(f\cdot g) \neq \exp(f)+\exp(g)$$

(I have no idea how to tag this question.)

• Uhm... Careful there: the algebraic correlations between sum and product are not symmetric. Namely, multiplication is distributive over addition, not the other way around. – Saucy O'Path Jun 8 '18 at 23:24
• It's a very nice observation. In Fourier analysis there's a duality between the operations $f\mapsto x^nf$ and $f\mapsto f^{(n)}$, it might have something to do with it. – Berci Jun 8 '18 at 23:25
• @SaucyO'Path -- I know. That's one of the reasons it's not truly an isomorphism. – mr_e_man Jun 8 '18 at 23:26