Hello! First post, over here, hopefully it'll pique some interest, and awesome answers, but any and all answers are welcome!(adsbygoogle = window.adsbygoogle || []).push({});

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 [itex]f(x)[/itex] and defining [tex]\int f = \int_0^x f(t) dt[/tex] we wish to find the function [itex]f(x)[/itex] such that [tex]\int f = f-1[/tex] Is the case. Then, after rearranging, we get:

[tex]1=f-\int f[/tex]

We "factor" and receive:

[tex]1=\left(1-\int\right)f[/tex]

Then, taking the "inverse":

[tex]f=\left(1-\int\right)^{-1}1[/tex]

We, then (somehow, magically, as I see it) "apply" the geometric series to receive:

[tex]f=\left(1+\int+\int^2+\int^3+\cdots\right)1[/tex]

[tex]f=1+\left(\int 1\right)+\left(\int^2 1\right)+\left(\int^3 1\right)+\cdots[/tex]

[tex]f=1+x+\left(\int x\right)+\left(\int^2 x\right)+\cdots[/tex]

[tex]f=1+x+\frac{x^2}{2}+\left(\int \frac{x^2}{2}\right)+\cdots[/tex]

[tex]f=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots[/tex]

[tex]f=e^x[/tex]

Which is, very obviously, the correct answer. Anyways, could anyone explain what isactuallygoing 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Playing with Analytical Operators

**Physics Forums | Science Articles, Homework Help, Discussion**