The theory of operators is an entire area of which I only know some basics.
However, even basic knowledge about it can simplify things quite a bit and sometimes makes impossible things possible. I will just discuss one nice cute application.
However, even basic knowledge about it can simplify things quite a bit and sometimes makes impossible things possible. I will just discuss one nice cute application.
Suppose we want to compute
\int xe^x dx
This is very simple: we just do integration by part,
\int xe^x dx = \int x de^x = xe^x -\int e^x dx = xe^x -e^x +C
However, what if we want to compute
\int P(x)e^x dx
where P(x) is a polynomial of degree n?
After a moment reflection, you might want to say that you can repeat integration by part n times and each time the deg of the polynomial inside the integral is reduced by one. Yes, but if you want to do this by hand, this will cost you a big piece of paper and xxx minutes before you find out how the terms evolve in each iteration and conclude the result by induction.
However, with the help of operators, it is just a piece of cake:
Let us us D be the differential operator, i.e. Df = f'.
Then 1/D is the integral operator.
It is a known fact that for any integer k,
D^k (e^{ax}y)=e^{ax}(D+a)^k y
Therefore,
\int P(x)e^{ax} dx = \frac{1}{D} e^{ax} P(x)= e^{ax} \frac{1}{D+a} P(x)= \frac{e^{ax}}{a} \frac{1}{1+D/a} P(x) = \frac{e^{ax}}{a} (1-D/a +D^2/a^2 - D^3/a^3 \ldots)P(x)= \frac{e^{ax}}{a} (P(x)-P'(x)/a +P''/a^2 - P^{(3)}/a^3 \ldots)
Notice that we only need the terms up to D^n ( since taking n+1 times differentiation of a poly of degree n
will result in 0).
Now, we can see every term is very simple to compute.
Actually, this integral appears quite often in various contexts, for eg. Laplace transform, Fourier analysis, moment generating function, where we often need to convolute functions with exponentials..
