From Infinite Matrices to New Integration Formula
This is another interesting problem, off-the-beaten-path. It ends up with a formula to compute the integral of a function, based on its derivatives solely.
For simplicity, I’ll start with some notations used in the context of matrix theory, familiar to everyone: T(f) = g, where f and g are vectors, and T a square matrix. The notation T(f) represents the product between the matrix T, and the vector f. Now, imagine that the dimensions are infinite, with f being a vector whose entries represent all the real numbers in some peculiar order.
In mathematical analysis, T is called an operator, mapping all real numbers (represented by the vector f) into another infinite vector g. In other words, f and g can be viewed as real-valued functions, and T transforms the function f into a new function g. A simple case is when T is the derivative operator, transforming any function f into its derivative g = df/dx. We define the powers of T as T^0 = I (the identity operator, with I(f) = f), T^2(f) = T(T(f)), T^3(f) = T(T^2(f)) and so on, just like the powers of a square matrix. Now let the fun begins.
Exponential of the Derivative Operator
We assume here that T is the derivative operator. Using the same notation as above, we have the same formula as if T was a matrix:
Applied to a function f, we have:
This is a simple application of Taylor series. So the exponential of the derivative operator is a shift operator.
Inverse of the Derivative Operator
Likewise, as for matrices, we can define the inverse of T as
Applied to a function f, and under some conditions that guarantee converge, it is easy to show that
The coefficients (for instance 1, -4, 6, -4, 1 in the last term displayed above) are just the binomial coefficients, with alternating signs.
We call the inverse of the derivative operator, the pseudo-integral operator. It is easy to prove that the pseudo-integral operator (as defined above), applied to the exponential function, yields the exponential function itself. So the exponential function is a fixed point (the only continuous one) of the pseudo-integral operator. More interestingly, in this case, the pseudo-integral operator is just the standard integral operator: they are both the same. Is this always the case regardless of the function f? It turns out that this is true for any function f that can be written as
This covers a large class of functions. In short, we have found a formula to compute the integral of a function, based solely on the function itself and its successive derivatives.
To not miss this type of content in the future, subscribe to our newsletter. For related articles from the same author, click here or visit www.VincentGranville.com. Follow me on on LinkedIn, or visit my old web page here.
Published at Mon, 04 Feb 2019 00:30:00 +0000
Leave a Reply