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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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What is your native language? As it stands, it is not very clear what is being asked. Maybe you can post in your native language and someone can translate. –  Pedro Tamaroff Sep 29 '12 at 22:01
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@PeterTamaroff I would say that Medan is asking if the matrix definition of the exponential extends to linear maps, and the answer would be yes. –  Fly by Night Sep 29 '12 at 22:05
    
@FlybyNight Oh, well. Good then. But this is still in risk of uncalled downvotes. –  Pedro Tamaroff Sep 29 '12 at 22:07
    
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. –  Medan Sep 30 '12 at 3:26
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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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Yes, this is important. My operator looks like $A:=\frac{\partial}{\partial x}+\frac{\partial^2}{\partial x^2}$. If I remember correctly it can be bounded or unbounded depending what is the space of functions. However, I am looking at the "splitting methods" for the odes and pde and in order to prove the splitting error they just do expansions for the exponential. –  Medan Sep 30 '12 at 13:17
    
And what is your space? I am not sure about your note, differential operators tend to by unbounded... Anyway, you can not use the series, you will loose differentiability in the process. –  kalvotom Sep 30 '12 at 16:14
    
Assume the space is such the linear operator $A$ is unbounded. And assume it is a Laplacian operator on the interval for $x$. Then, I can't represent $e^A$, however, I can discretize the operator and get the approximation matrix of it $A_h$(using three points, for example). Then I can do $e^{A_h}$. Is that correct? Sounds like cheating, just by doing a consistent discretization it avoids all unboundedness problems? –  Medan Sep 30 '12 at 19:26
    
I do not think so. It depends on what you are trying to do. Even if you are on an interval, than there are many operators which act as a Laplacian but with a different domain. Their exponential is then different also. Your discretiazation might correspond to one of those operators, but I am not sure about that. On the other hand, I am positive that you can write down an explicit expression for the exponential of any of those operators. It will act as some integral operator. –  kalvotom Sep 30 '12 at 20:41
    
Let me tell you where my question comes from. Consider the pde $$u_t=Au$$ with $A:=\frac{\partial}{\partial x},u(0)=u_0$. Then the solution can be written as $$u(t)=u_0e^{At}$$So this is where I get stuck, if $A$ is a matrix I can do the matrix expansion and get $u(t)$ as a series, otherwise I can't. I want to keep it as an operator and not specify how I would discretize that, but I run into trouble working with unbounded operators? –  Medan Sep 30 '12 at 21:50
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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.

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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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