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.
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Published at Mon, 04 Feb 2019 00:30:00 +0000