Let $\mathbb{F}$ be a field, and consider $\mathbb{F}^\mathbb{F}$ as an algebra over $\mathbb{F}$ with the standard function multiplication. Let $D$ be a derivation on a subalgebra of $\mathbb{F}^\mathbb{F}$ closed under function composition that takes the identity function to the function that sends everything to $1,$ as in the answers to these two questions. Does $D$ necessarily satisfy the chain rule for arbitrary $\mathbb{F}$?
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1Are you asking this: In a universe where the only functions are polynomials, do we need the chain rule? – Ted Shifrin Dec 20 '22 at 02:22
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3That is not what I'm asking. – mathlander Dec 20 '22 at 02:22
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I'm asking if the chain rule is implied for any derivation that takes the identity function to 1. – mathlander Dec 20 '22 at 02:23
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In that case, how would you differentiate $f(\sin x)$? – Ted Shifrin Dec 20 '22 at 02:27
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1Why is that important? – mathlander Dec 20 '22 at 02:28
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2I don’t see how transcendental functions are dealt with in this discussion. But perhaps I didn’t wade through enough pages. In that case, you should write a coherent brief discussion. – Ted Shifrin Dec 20 '22 at 02:34
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4It's much easier to prove that a chain rule gives a product rule than the other way around. I expect it's a stronger statement than the existence of a derivation since function composition is generally more complicated than mere multiplication. – CyclotomicField Dec 20 '22 at 02:58
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2The linked answer does not show this for an arbitrary derivation satisfying your assumptions when $\mathbb{F}=\mathbb{R}$; it only shows it under the additional assumptions (7) and (8) (in particular, (8) is a very strong assumption). – Eric Wofsey Dec 21 '22 at 04:44
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Thanks for informing me. – mathlander Dec 21 '22 at 04:45
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I believe the answer is no, the product rule does not imply the chain rule, although counterexamples seem tedious to construct. Let $F = \mathbb{R}$ or $\mathbb{C}$ and let $A$ be the smallest $F$-subalgebra of $F^F$ closed under composition and containing the functions $x$ and $\exp(x)$. I believe that there exists a derivation $D$ on $A$ with the following properties:
- $D(x) = 1$
- $D(\exp(x)) = \exp(x)$
- $D(\exp^{\circ k}(x)) = 0$ for all $k \ge 2$.
If so then $D$ does not satisfy the chain rule. To motivate the existence of such a derivation let's see what constrains the possible values of $D$. Every element of $A$ is a sum of terms of the form $x^k \exp(a)$ where $a \in A$. By applying the sum-to-product exponent rule to $\exp(a)$ we can write every element of $A$ as a sum of products of copies of either $x$ or $\exp(a)$ where $a$ itself has the form $x^k \exp(a')$ for some $a' \in A$. $D$ being a derivation then means that the value of $D$ is determined by $D(x)$ and by the values $D(\exp(x^k \exp(a))$.
Now the point here is that being a derivation does not constrain the value of $D(\exp(x^k \exp(a))$ at all, except in the trivial case $D(\exp(c)) = 0$ where $c$ is a scalar. In fact we ought to be able to show that $A$ is basically a free algebra, albeit on a very large set of generators (if so, we can freely specify the value of $D$ on every generator). So we have a great deal of freedom to assign values to expressions of the form $D(\exp(-))$, although again it seems tedious to pin down whether we have enough freedom for the construction we want. Hopefully somebody else can do this!
What is true is that if $D$ is a derivation (so satisfies linearity and the product rule) then $D(p(f)) = p'(f) D(f)$ where $p$ is a polynomial and $p'$ is its ordinary derivative.
mathlander
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Qiaochu Yuan
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5@mathlander: it means iterated composition, e.g. $\exp^{\circ 2}(x) = \exp(\exp(x))$ and so forth. – Qiaochu Yuan Dec 20 '22 at 05:47
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1Strictly speaking, if $\Bbb F=\Bbb R$, you need summands of the form $\pm x^k\exp(a)$ - or switch to linear combinations instead of sums to begin with – Hagen von Eitzen Jan 29 '23 at 18:05