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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.2012-09-29
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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.2012-09-29
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    @FlybyNight Oh, well. Good then. But this is still in risk of uncalled downvotes.2012-09-29
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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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    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.2012-09-30
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    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.2012-09-30
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    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?2012-09-30
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    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.2012-09-30
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    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?2012-09-30
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    Well, in this case you have $(e^{At}u_0)(x) = u_0(x+t)$. I am assuming that $u_0$ is defined on $\mathbb{R}$. The case of a bounded interval might be complicated.2012-09-30
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    the same question then for $A:=\frac{\partial^2 }{\partial x^2}$?2012-09-30
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    In this case you have $$ (e^{tA} u_0)(x) = \int_\mathbb{R} \frac{1}{\sqrt{4\pi t}} e^{-(x-y)^2/4t} u_0(y) \,\mathrm{d}y. $$ You can derive this formula employing the Fourier transform. See also "heat kernel" on wikipedia.2012-10-01
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    I know it has a closed form solution but I want to represent the solution at the new time $t$ as a series, which is supposed to converge to the integral you have written. However, it is not going to converge if the operator is unbounded, right? The point is to explore what is $e^{tA}$2012-10-02
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    I do not see the purpose of this. In the present case the exponential $e^{tA}$ is a linear bounded operator in $L^2(\mathbb{R})$. In general, it is defined as a solution of the initial value problem $u_t = Au$, $u(0) = u_0$, that is $(e^{tA}u_0) := u(t)$.2012-10-02
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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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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)