 # We know that, f:RR to RR satisfies f′(x)=f(x) for all x in RR then f(x)=A e^x What is a function f:RR^d to RR^d, analogous to the exponential function, if x in RR^d was a d dimensional vector? lexi13xoxla 2022-08-13 Answered
We know that, $f:\mathbb{R}\to \mathbb{R}$ satisfies ${f}^{\prime }\left(x\right)=f\left(x\right)$ for all $x\in \mathbb{R}$ then $f\left(x\right)=A{e}^{x}$
What is a function $f:{\mathbb{R}}^{d}\to {\mathbb{R}}^{d}$, analogous to the exponential function, if $\mathbf{x}\in {\mathbb{R}}^{d}$ was a d dimensional vector?
Is there a name given to such functions whose derivatives are the same as the function itself?
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There is a problem with defining a derivative of a map as a map between the same spaces as the original map. Given a map $f:{\mathbb{R}}^{m}\to {\mathbb{R}}^{n}$ its differential at a point $p\in {\mathbb{R}}^{m}$ (if it exists) is a linear map $Df\left(p\right):T{\mathbb{R}}^{m}\to T{\mathbb{R}}^{n}$ of tangent spaces. If m=1 then f is a vector-valued function and its differential is also a vector-valued function. In this case we can define
${f}^{\prime }\left(t\right)=\left({f}_{1}^{\prime }\left(t\right),...,{f}_{n}^{\prime }\left(t\right)\right)$
which is in fact the differential at the point t. Note that here we identify the space ${\mathbb{R}}^{n}$ with the space of linear maps $T\mathbb{R}\to T{\mathbb{R}}^{n}$ which is (luckily) n-dimensional. But if m>1 then this space of linear maps is no more n-dimensional and we can not define the derivative of a map in the same way we do in single variable calculus.