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* = d*f/*d*x*. 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.

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