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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) onto 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

If T was a matrix, the condition for convergence is that all of the eigenvalues of T - I have absolute value smaller than 1. For the derivative operator T applied to a function f, and under some conditions that guarantee convergence, 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, especially since the coefficients can also be complex numbers. These functions usually have a Taylor series expansion too. However, it does not apply to functions such as polynomials, due to lack of convergence of the formula, in that case.

In short, we have found a formula to compute the integral of a function, based solely on the function itself and its successive derivatives. The same technique can be used to invert more complicated linear operators, such as Laplace transforms.

Exercise

Apply the derivative operator to the pseudo-integral of a function f, using the above formula for the pseudo-integral. The result should be equal to f. This is the case if f belongs to the same family of functions as described above. Can you identify functions not belonging to that family of functions, for which the theory is still valid? Hint: try f(x) = exp(b x^2) or f(x) = x exp(b x), where b is a parameter.

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.

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Comment by scottedwards2000 on February 20, 2019 at 4:23am

Not sure i follow this part:

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:

Could you please elaborate a bit? 

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