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Emil Jeřábek
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Let $F : \mathbb{R}[X] \rightarrow \mathbb R[X]$ be a linear map and let $H \in \mathbb{R}[u,x,y,z]$ be a polynomial. Suppose that

$$ F(P \times Q) = H(F(P),F(Q),P,Q)$$$$ F(P \cdot Q) = H(F(P),F(Q),P,Q)$$

for all $P, Q \in \mathbb{R}[X]$. Two obvious solutions for $F$ and $H$ are

  • $F = I$, the identity function, and $H(u,x,y,z) = \alpha u x + \beta u z + \gamma yx + \delta yz$ where $\alpha,\beta,\gamma,\delta \in \mathbb{R}$ and $\alpha +\beta+\gamma+\delta = 1$;

  • $F = D$, the derivative, and $H(u,x,y,z) = u z + x y$, from the Leibniz rule.

Do there exist solutions $F$ and $H$ in which $F$ is not a linear combination of $I$ and $D$?

Let $F : \mathbb{R}[X] \rightarrow \mathbb R[X]$ be a linear map and let $H \in \mathbb{R}[u,x,y,z]$ be a polynomial. Suppose that

$$ F(P \times Q) = H(F(P),F(Q),P,Q)$$

for all $P, Q \in \mathbb{R}[X]$. Two obvious solutions for $F$ and $H$ are

  • $F = I$, the identity function, and $H(u,x,y,z) = \alpha u x + \beta u z + \gamma yx + \delta yz$ where $\alpha,\beta,\gamma,\delta \in \mathbb{R}$ and $\alpha +\beta+\gamma+\delta = 1$;

  • $F = D$, the derivative, and $H(u,x,y,z) = u z + x y$, from the Leibniz rule.

Do there exist solutions $F$ and $H$ in which $F$ is not a linear combination of $I$ and $D$?

Let $F : \mathbb{R}[X] \rightarrow \mathbb R[X]$ be a linear map and let $H \in \mathbb{R}[u,x,y,z]$ be a polynomial. Suppose that

$$ F(P \cdot Q) = H(F(P),F(Q),P,Q)$$

for all $P, Q \in \mathbb{R}[X]$. Two obvious solutions for $F$ and $H$ are

  • $F = I$, the identity function, and $H(u,x,y,z) = \alpha u x + \beta u z + \gamma yx + \delta yz$ where $\alpha,\beta,\gamma,\delta \in \mathbb{R}$ and $\alpha +\beta+\gamma+\delta = 1$;

  • $F = D$, the derivative, and $H(u,x,y,z) = u z + x y$, from the Leibniz rule.

Do there exist solutions $F$ and $H$ in which $F$ is not a linear combination of $I$ and $D$?

edited tags; edited tags; edited tags
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Federico Poloni
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I rewrote the question.
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Mark Wildon
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Does there exist another form of derivatethe derivative for polynomials?

Consider linear $F: \mathbb R[x] \rightarrow \mathbb R[x]$ withLet : $F(P\times Q)=H(F(P),F(Q),P,Q)$ for some$F : \mathbb{R}[X] \rightarrow \mathbb R[X]$ be a linear map and let $H$$H \in \mathbb{R}[u,x,y,z]$ be a polynomial. Suppose that

A possible solution for $H$ is $H(u,x,y,z)=u\times z+x \times y$, where $F$ is the classical derivate $D$.$$ F(P \times Q) = H(F(P),F(Q),P,Q)$$

Does there exist any other non-trivial* solutions for $H$all ?

*:$P, Q \in \mathbb{R}[X]$. Two obvious solutions for $F$ isn't a linear combinaison of $Id$ and $D$.$H$ are

  • $F = I$, the identity function, and $H(u,x,y,z) = \alpha u x + \beta u z + \gamma yx + \delta yz$ where $\alpha,\beta,\gamma,\delta \in \mathbb{R}$ and $\alpha +\beta+\gamma+\delta = 1$;

  • $F = D$, the derivative, and $H(u,x,y,z) = u z + x y$, from the Leibniz rule.

Do there exist solutions $F$ and $H$ in which $F$ is not a linear combination of $I$ and $D$?

Does there exist another form of derivate for polynomials?

Consider linear $F: \mathbb R[x] \rightarrow \mathbb R[x]$ with : $F(P\times Q)=H(F(P),F(Q),P,Q)$ for some $H$.

A possible solution for $H$ is $H(u,x,y,z)=u\times z+x \times y$, where $F$ is the classical derivate $D$.

Does there exist any other non-trivial* solutions for $H$ ?

*: $F$ isn't a linear combinaison of $Id$ and $D$.

Does there exist another form of the derivative for polynomials?

Let $F : \mathbb{R}[X] \rightarrow \mathbb R[X]$ be a linear map and let $H \in \mathbb{R}[u,x,y,z]$ be a polynomial. Suppose that

$$ F(P \times Q) = H(F(P),F(Q),P,Q)$$

for all $P, Q \in \mathbb{R}[X]$. Two obvious solutions for $F$ and $H$ are

  • $F = I$, the identity function, and $H(u,x,y,z) = \alpha u x + \beta u z + \gamma yx + \delta yz$ where $\alpha,\beta,\gamma,\delta \in \mathbb{R}$ and $\alpha +\beta+\gamma+\delta = 1$;

  • $F = D$, the derivative, and $H(u,x,y,z) = u z + x y$, from the Leibniz rule.

Do there exist solutions $F$ and $H$ in which $F$ is not a linear combination of $I$ and $D$?

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Dattier
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YCor
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YCor
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