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See this notethis note describing the algebraic tangent space of a $C^k$ manifold. In particular, on $\mathbb{R}$, there exists a basis of the cotangent space $\mathfrak{m}_p\, / {\mathfrak{m}_p}^2$ containing the functions $|x-p|^{k+\alpha}$ for all $\alpha \in (0, 1)$.

We can therefore define a derivation at $p$ which maps $|x-p|^{k+\frac{1}{2}}$ to $1$ and every other basis element to $0$.

See this note describing the algebraic tangent space of a $C^k$ manifold. In particular, on $\mathbb{R}$, there exists a basis of the cotangent space $\mathfrak{m}_p\, / {\mathfrak{m}_p}^2$ containing the functions $|x-p|^{k+\alpha}$ for all $\alpha \in (0, 1)$.

We can therefore define a derivation at $p$ which maps $|x-p|^{k+\frac{1}{2}}$ to $1$ and every other basis element to $0$.

See this note describing the algebraic tangent space of a $C^k$ manifold. In particular, on $\mathbb{R}$, there exists a basis of the cotangent space $\mathfrak{m}_p\, / {\mathfrak{m}_p}^2$ containing the functions $|x-p|^{k+\alpha}$ for all $\alpha \in (0, 1)$.

We can therefore define a derivation at $p$ which maps $|x-p|^{k+\frac{1}{2}}$ to $1$ and every other basis element to $0$.

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user113407

See this note describing the algebraic tangent space of a $C^k$ manifold. OnIn particular, on $\mathbb{R}$, wethere exists a basis of the cotangent space $\mathfrak{m}_p\, / {\mathfrak{m}_p}^2$ containing the functions $|x-p|^{k+\alpha}$ for all $\alpha \in (0, 1)$.

We can therefore define a derivation at $p$ which maps $|x-p|^{k+\frac{1}{2}}$ to $1$ and every other basis element to $0$.

See this note describing the algebraic tangent space of a $C^k$ manifold. On $\mathbb{R}$, we can therefore define a derivation at $p$ which maps $|x-p|^{k+\frac{1}{2}}$ to $1$ and every other basis element to $0$.

See this note describing the algebraic tangent space of a $C^k$ manifold. In particular, on $\mathbb{R}$, there exists a basis of the cotangent space $\mathfrak{m}_p\, / {\mathfrak{m}_p}^2$ containing the functions $|x-p|^{k+\alpha}$ for all $\alpha \in (0, 1)$.

We can therefore define a derivation at $p$ which maps $|x-p|^{k+\frac{1}{2}}$ to $1$ and every other basis element to $0$.

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user113407

LetSee $M = \mathbb{R}$ and check thatthis note describing the following map isalgebraic tangent space of a $C^k$ manifold. On $\mathbb{R}$, we can therefore define a derivation at $p \in \mathbb{R}$: $$ D(f) = \begin{cases} \lim\limits_{x \to p} \dfrac{f(x) - f(p)}{\;|x-p|^{r + \frac{1}{2}}} &\text{if the limit exists} \\[1ex] 0 &\text{otherwise} \end{cases}. $$$p$ which maps $|x-p|^{k+\frac{1}{2}}$ to $1$ and every other basis element to $0$.

Let $M = \mathbb{R}$ and check that the following map is a derivation at $p \in \mathbb{R}$: $$ D(f) = \begin{cases} \lim\limits_{x \to p} \dfrac{f(x) - f(p)}{\;|x-p|^{r + \frac{1}{2}}} &\text{if the limit exists} \\[1ex] 0 &\text{otherwise} \end{cases}. $$

See this note describing the algebraic tangent space of a $C^k$ manifold. On $\mathbb{R}$, we can therefore define a derivation at $p$ which maps $|x-p|^{k+\frac{1}{2}}$ to $1$ and every other basis element to $0$.

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