# Playing with Analytical Operators

1. Aug 16, 2012

### LolWolf

Hello! First post, over here, hopefully it'll pique some interest, and awesome answers, but any and all answers are welcome!

So, point is, we were given a guest lecture over how to 'jokingly' find answers to specific problems using methods that appear to not be justified; now, after the lecture I asked a question regarding such, and I was told that such operations are actually justified, what is not is the notation. Anyways, enough storytelling, the problem and 'derivation' is as follows:

Given some function $f(x)$ and defining $$\int f = \int_0^x f(t) dt$$ we wish to find the function $f(x)$ such that $$\int f = f-1$$ Is the case. Then, after rearranging, we get:
$$1=f-\int f$$
$$1=\left(1-\int\right)f$$
Then, taking the "inverse":
$$f=\left(1-\int\right)^{-1}1$$
We, then (somehow, magically, as I see it) "apply" the geometric series to receive:
$$f=\left(1+\int+\int^2+\int^3+\cdots\right)1$$
$$f=1+\left(\int 1\right)+\left(\int^2 1\right)+\left(\int^3 1\right)+\cdots$$
$$f=1+x+\left(\int x\right)+\left(\int^2 x\right)+\cdots$$
$$f=1+x+\frac{x^2}{2}+\left(\int \frac{x^2}{2}\right)+\cdots$$
$$f=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$$
$$f=e^x$$

Which is, very obviously, the correct answer. Anyways, could anyone explain what is actually going on here? I know that the integrals can be dealt with as operators and such... but the notation confuses absolutely anything I have been able to come up with.

Last edited: Aug 16, 2012
2. Aug 16, 2012

### Muphrid

This is rooted in the theory of linear operators on functions and is pretty well justified; it's different notation, though. I would just do something like, define

$$\underline S(f) = \int_0^x f(t) \; dt$$

When given a particular $x$, this is a functional, but I expect the intention is to treat this as an operator of a function, so as such, I've suppressed the dependence on $x$ in the left-hand side.

Now, all you need is an identity operator. Call it $\underline I(f) = f$. From here, everything that follows is pretty clear. From a physicist's perspective, it all seems very natural to relate to Hilbert spaces and how that's used in quantum mechanics. You could probably express all of the above in terms of Dirac notation. Linear operators on functions have inverses and can be expanded in terms of power series just as operators on vectors can be.

You get something like

$$(\underline I - \underline S)(f) = \underline I(f) - \underline S(f) = 1$$

It should be clear at this point that $\underline I - \underline S$ is an operator itself, and barring some strange stuff, that can be inverted, and the power series method works just fine for operators (most clearly seen with matrices, but still).

3. Aug 17, 2012

### LolWolf

Oh, that's pretty cool, thanks for the explanation. And, yeah, I was told it was justified, although this was confusing (albeit hysterical) notation.

Anyways, how is it expressible using bra-ket notation? Just out of curiosity, since I only have an understanding of it in terms of states/probability amplitudes.

4. Aug 17, 2012

### Muphrid

I would say something like,

$$\langle x | \hat S | f \rangle = \int_0^x f(t) \; dt$$

It may need a little more justifying glue than that, but it works for me. Dirac notation doesn't care what your functions represent (whether they're probability amplitudes in QM or something else). The basic ideas about what it all represents are the same.

5. Aug 17, 2012

### Millennial

Here's the way I'd solve it:
$$\int_{0}^{t}f(t)\,dt=f(t)-1$$
Take Laplace transform:
$$\frac{F(s)}{s}=F(s)-\frac{1}{s}$$
$$F(s)=sF(s)-1$$
$$(1-s)F(s)=-1$$
$$(s-1)F(s)=1$$
$$F(s)=\frac{1}{s-1}$$
Taking inverse Laplace here gives $f(t)=e^t$, which was the desired result.

6. Aug 17, 2012

### chiro

Take a look at the operator theory for differential operators and make the note that the integral is 1/D, where D is a differential operator with respect to some variable (and is a linear operator).

A good text on differential equations should cover this and the theory should exist somewhere (it has been proven).

7. Aug 17, 2012

### Mute

Just to add to this, in order to expand inverse operators like$(\underline I - \underline S)^{-1}$ in a power series, the operator norm $||\underline S||$ must be less than 1 (in this particular case).

8. Aug 18, 2012

### LolWolf

Oh, awesome, thanks for the replies! All right, now it comes to make more sense, esp. since we are defining the unity within the first expression to be the identity function. Which makes it actually valid. Cool, thanks for the help.
As for the power series, yeah, that makes sense, didn't even know operators had norms, but I think I can grasp the concept, and perhaps apply it to simple examples.