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I have $f(x)$=$(2x,e^x)$ what does this notation mean? Notation: $Df(\frac{∂}{∂x})$

Certainly $Df(x)$=$(2,e^x)$ but how can I replace $x$ with $\frac{∂}{∂x}$?

Particularly, how can I make sense of $e^{\frac{∂}{∂x}}$

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  • $\begingroup$ This is from a Differential Geometry course, but has no geometry so far. So it is possible that I actually need to invoke some functional calculus. @Potato $\endgroup$ – Faust Nov 5 '13 at 2:52
  • $\begingroup$ I could guess that you can use the Taylor series of $\exp$. The result is a valid linear operator on some class of functions. $\endgroup$ – Tunococ Nov 5 '13 at 2:55
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    $\begingroup$ There are two common meanings/usages for $\frac{\partial}{\partial x}$. It can either mean taking partial derivatives w.r.t $x$ or refer to the "tangent vector" associated with the coordinate curve for $x$. In your case, it is the second definition they are using. Remember, $Df$ is not a function from $\mathbb{R}$ to $\mathbb{R}^2$. Instead, it is a function between the corresponding tangent bundles. To avoid confusion, I will probably use the expression $DF_{x}(\frac{\partial}{\partial x})$ instead. $\endgroup$ – achille hui Nov 5 '13 at 2:58
  • $\begingroup$ @Potato I guess hui is getting the point $\endgroup$ – Faust Nov 5 '13 at 3:02
  • $\begingroup$ @achillehui I think you are right. $\endgroup$ – Faust Nov 5 '13 at 3:04
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In your case $f$ is a map from $\mathbb{R}^1$ to $\mathbb{R}^2$, so at any point $x\in \mathbb{R}^1$ $Df$ is a linear map from $T\mathbb{R}^1_x$ to $T\mathbb{R}^2_{f(x)}$, and is given by the $1\times 2$ matrix $(2,e^x)^T$. To clarify the notation, $\frac{\partial}{\partial x}$ is notation for the unit tangent vector in the tangent space at $x$ (pointing in the positive direction), so if $u$ and $v$ are your coordinates on $\mathbb{R}^2$ we have $Df\left( \frac{\partial}{\partial x}\right)=2\frac{\partial}{\partial u}+e^x\frac{\partial}{\partial v}$ (where $\frac{\partial}{\partial u}$ and $\frac{\partial}{\partial v}$ are the unit tangent vectors in the positive $u$ and $v$ directions, which are a basis for the tangent space at $f(x)$).

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  • $\begingroup$ I think this is the final solution to my naive question... Thanks all that helped. $\endgroup$ – Faust Nov 5 '13 at 3:37
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The keyword is Operational Calculus (let Google be your friend).
The following reference is an introductory exposure of the subject:

Re: Why exp(-st) in the Laplace Transform?

About the second part of your question. Any differentiable function can be developed into a Taylor series: $$ f(x+a) = f(x) + a.f'(x) + \frac{1}{2} a^2.f''(x) + \frac{1}{3!} a^3 f'''(x) + \cdots $$ Write as follows: $$ f(x + a) = \left[ 1 + a\frac{d}{dx} + \frac{1}{2} a^2\frac{d^2}{dx^2} + \frac{1}{3!} a^3\frac{d^3}{dx^3} +\cdots \right] f(x) $$ $$ = \left[ 1 + \left(a\frac{d}{dx}\right) + \frac{1}{2} \left(a\frac{d}{dx}\right)^2 + \frac{1}{3!} \left(a\frac{d}{dx}\right)^3 +\cdots \right] f(x) $$ The series expansion of $e^x$ is recognized in the expression between the square brackets. Therefore we write symbolically: $$ f(x+a) = e^{a\frac{d}{dx}} f(x) $$

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    $\begingroup$ This isn't right. The OP was very vague, but he was actually asking about differential geometry, not the functional calculus. $\endgroup$ – Potato Nov 6 '13 at 3:21
  • $\begingroup$ @Potato: Then I don't see where his $e^{\frac{\partial}{\partial x}}$ comes from. $\endgroup$ – Han de Bruijn Nov 6 '13 at 15:12
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    $\begingroup$ It's viewed as a map on the tangent bundle of $\mathbb R$ at $x$. (The notation the OP uses is really confusing!) $\endgroup$ – Potato Nov 6 '13 at 17:35

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