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edited Oct 27, 2019 at 17:43

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I would not recommend "Taylor $k$-differentiable", because the name "Taylor" always refers to the standard polynomial made by successive derivatives at $x_0$, and because as far as we know this generalization was not in the mind of Taylor. A quite standard and self-explanatory name is: $f$ has a polynomial expansion of order $k$ at $x_0$, where "order $k$" refers to the form of the remainder, $o(|x-x_0|^k)$. If you like a shorter name linked to a mathematician, I'd go for Peano k-differentiable, which I think is also customary, since I think Peano (I'll hopefully add a reference later) made quite an extensive and thorough study of this objectsubject. ItIn particular, the implication you mention and the form of the remainder $o(x-x_0)^k$ is due to Peano (it's a one-page note, Une nouvelle forme du reste dans la formule de Taylor, 1889). This polynomial expansion has a nice calculus, analogous to the calculus of derivatives, though, as soon as you want deeper results, such as e.g. the inverse mapping theorem, you need to assume the existence of this polynomial expansion everywhere, with continuously depending coefficients: but this is equivalent to being a standard $C^k$ function.

I would not recommend "Taylor $k$-differentiable", because the name "Taylor" always refers to the standard polynomial made by successive derivatives at $x_0$, and because as far as we know this generalization was not in the mind of Taylor. A quite standard and self-explanatory name is: $f$ has a polynomial expansion of order $k$ at $x_0$, where "order $k$" refers to the form of the remainder, $o(|x-x_0|^k)$. If you like a shorter name linked to a mathematician, I'd go for Peano k-differentiable, which I think is also customary, since Peano (I'll hopefully add a reference later) made quite an extensive and thorough study of this object. It has a nice calculus, analogous to the calculus of derivatives, though, as soon as you want deeper results, such as e.g. the inverse mapping theorem, you need to assume the existence of this polynomial expansion everywhere, with continuously depending coefficients: but this is equivalent to being a standard $C^k$ function.

I would not recommend "Taylor $k$-differentiable", because the name "Taylor" always refers to the standard polynomial made by successive derivatives at $x_0$, and because as far as we know this generalization was not in the mind of Taylor. A quite standard and self-explanatory name is: $f$ has a polynomial expansion of order $k$ at $x_0$, where "order $k$" refers to the form of the remainder, $o(|x-x_0|^k)$. If you like a shorter name linked to a mathematician, I'd go for Peano k-differentiable, which I think is also customary, since I think Peano (I'll hopefully add a reference later) made quite an extensive and thorough study of this subject. In particular, the implication you mention and the form of the remainder $o(x-x_0)^k$ is due to Peano (it's a one-page note, Une nouvelle forme du reste dans la formule de Taylor, 1889). This polynomial expansion has a nice calculus, analogous to the calculus of derivatives, though, as soon as you want deeper results, such as e.g. the inverse mapping theorem, you need to assume the existence of this polynomial expansion everywhere, with continuously depending coefficients: but this is equivalent to being a standard $C^k$ function.

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created Oct 27, 2019 at 17:28

Pietro Majer

60.5k
4
122
269

I would not recommend "Taylor $k$-differentiable", because the name "Taylor" always refers to the standard polynomial made by successive derivatives at $x_0$, and because as far as we know this generalization was not in the mind of Taylor. A quite standard and self-explanatory name is: $f$ has a polynomial expansion of order $k$ at $x_0$, where "order $k$" refers to the form of the remainder, $o(|x-x_0|^k)$. If you like a shorter name linked to a mathematician, I'd go for Peano k-differentiable, which I think is also customary, since Peano (I'll hopefully add a reference later) made quite an extensive and thorough study of this object. It has a nice calculus, analogous to the calculus of derivatives, though, as soon as you want deeper results, such as e.g. the inverse mapping theorem, you need to assume the existence of this polynomial expansion everywhere, with continuously depending coefficients: but this is equivalent to being a standard $C^k$ function.