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The matrix exponential is a well know thing but when I see online it is provided for matrices. Does it the same expansion for a linear operator? That is if $A$ is a linear operator then $e^A=I+A+\frac{1}{2}A^2+\cdots+\frac{1}{k!}A^k+\cdots$

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    sorry, I had to read it before posting. Fly by Night had a right guess. I wanted to make sure I can extend matrix exponential to the case of linear operators.2012-09-30

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Yes, you can define an exponential of any linear BOUNDED operator by this series. If the operator is unbounded then it is not always possible.

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    May be this is what confuses me: you are saying "In the present case the exponential $e^{tA}$ is a linear bounded operator". Well, if $A$ is the second derivative operator and thus unbounded, but $e^{tA}$ is bounded? I just wanted to say that assume $A$ is such that it is unbounded, whatever one s.t. you can't write close solution. Then, you can approximate that with a matrix. In the first case you can't write $e^{tA}$ as a series but once you approximate that and have a matrix you can. Is that correct? My goal is to be able to write $e^{tA}$ as a series.2012-10-06
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The exponential series has a remarkably "ubiquitiuos" convergence. As soon as you have a $\mathbb Q$-algebra $M$ with a norm such that $||X Y||\le c\cdot ||X||\cdot ||Y||$ for some $c$, then $\exp(A)$ converges for all $A$ with respect to this norm. Hence if $M$ is complete, you indeed obtain an element of $M$. Moreover, if $AB=BA$ then $\exp(A+B)=\exp(A)\exp(B)$ holds.

There are even cases when the exponential series is useful even when division by $k!$ is undefined. One just has to be careful that $A$ must be nilpotent enough (i.e. $A^k=0$ for all $k$ for which divison by $k!$ is undefined)

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As you have suggested, if $A$ is a linear operator then:

$\exp A = I + A + \frac{1}{2}A^2 + \cdots + \frac{1}{k!}A^k + \cdots \, . $

These are very common in physics. Here is a link to a PDF file.