# Sum of Integrals, Derivatives

The M and P functions that I mentioned in “Higher Order Operators and Functions” are a lot more interesting that I expected.

First, a recap on the definition:

M(f(x)) = f(x) + $\int f(x) + \int \int f(x)$ + … (general antiderivatives here)
P(f(x)) = f(x) + f'(x) + f”(x) + f”'(x) +…
Originally I guessed that there is no function such that M(f(x)) and P(f(x)) converge, but I have found out how wrong I was.
For the simple function f(x) = x,
$M(f(x)) = \sum_{k=1}^{\infty} \frac{x^n}{n!}$ which is the Taylor series expansion for $e^x$ except the constant term.
P(x) = x + 0 + 0 … = x
So x is doubly-convergent.
For a general monomial a x^b we multiply the monomial by a term k so that it has the form:
$\frac{1}{b!} x^b$ which matches the taylor series form.$k = \frac{1}{a*b!}$
$M(a x^b) = k (\sum_{i = 0}^{\infty} x^i - \sum_{i = 0}^{b-1} x^i)$
$M(a x^b) = k e^x - k (\sum_{i = 0}^{b-1} x^i)$
Which is convergent. And we showed before that all polynomials are P convergent.
Integration is distributive over addition, so we can add the above expression for each term of a polynomial together to get a combined M(p(x)) for any polynomial. All polynomials are doubly convergent then.
Because integration and differentiation are distributive over addition, the M and P functions are also distributive (modulo some technical difficulties that may occur re: convergence).
An example polynomial:
$M(x + x^3) = M(x) + M(x^3) = e^x + 6 M(\frac{1}{6} x^3)$
$M(x + x^3) = 7 * e^x - 3 x^2 - 6 x - 6$
Rationals:
f(x) = 1 / x is P convergent, but the value and convergence status of  M(1 / x) is unknown.
$P(\frac{1}{x}) = \sum_{k=0}^{\infty} \frac{-1^k}{x^(k+1)}$
$P(\frac{1}{x}) = \frac{e^(- \frac{1}{x})}{x}$