Geometric sum as functional inverse (2): the backstory

My latest contribution about using geometric sum as inverses is surprisingly well-received, so I decided to give it a sequel.

The immediate question is: are we reinventing Maclaurin series? Is this even possible without differentiation? Well, not really. If we decompose $f$ into power series then we will see precisely what happened. If $f = \sum a_k x^k$ then $a_i = D^i[\sum a_k x^k](0)/i!$ where $D$ is the differential operator since I am too lazy to write fractions right? We can recover this by calculating $M^i[D^i[\sum a_k x^k](0)]$:

$M^i[D^i[\sum a_k x^k](0)] = M_i[i!a_i] = a_ix^i$.

In other words, we are just representing the $x^i$-term in forms of $a_i'x^i$ instead of constant $a_i$. 

Let us recall the definition of the (properly defined) operator $M$: $Mf(x) = \int ^x_0 f(u)du$.

The miracle of the exponential series actually happens because $M^i[1]$ is precisely the $x^i$-term of the exponential series. Clearly this is not guaranteed because $M^i[g] = a_ix^i$ for all $i$ iff $g$ is a constant. Sadly $g$ is pre-determined by the function $f$ we use.

Consider $f = \log (1+x)$. It looks bad further away from zero but it should be fine in an neighborhood around zero. Suppose we wave our hand say from $(I-M)f = g$ we obtain $f = (\sum M^k)g$, we want to check what's $M^kg$ exactly.

First we compute $g = (I-M)f = x(1-\log(1+x))$. Not good-looking but okay. Next we compute $Mg$:

$Mg = \frac{1}{2}(x^2+\frac{1}{2}x(x-2)-(x^2-1)\log(1+x))$

according to Wolframalpha. Soon you realized the fact that $M^kg$ will never be the k-th term of the Maclaurin series because the log term will never go away. 

Everything we argued in $L^{\infty}$ still works, even for this nasty $f$! Each $M^kg$ is a proper function and the sum indeed converges to $f$ uniformly if we restrict the domain to $[-\varepsilon, \varepsilon]$ for some small $\varepsilon \in (0,1)$. However, it simply does not coincides with the Maclaurin series because $g$ is not a constant. When is $g$ a constant, or rather, when is $(I-M)f$ a constant? Differential equation $y-y' = 0$ has solution $ce^x$ and is the only instance that fits all our speculation. 

What a pity.

You then asked: what about the trig example at the end? Why are we recovering the Maclaurin series like that even when the theory already crumbled?

To clarify let us repeat the above again: $g = (I-M)[\sin] = \sin x + \cos x -1$. 

$Mg = \sin x - \cos x + 1 -x$.

$M^2g = -\sin x - \cos x + 1 + x - x^2/2$.

Well...an oscillation? Nope. This is clear if you plot them on a graph.

Red is $g$, blue is $Mg$ and green is $M^2g$. As predicted by the norm of $M$, the $L^{\infty}$-norm do shrink exponentially which gives us the convergence. 

We can also prove convergence in analytically: note that when k is 3 mod 4, all the trigonometric terms in $(I+\ldots + M^k)g$ vanished, leaving the Maclaurin series of $\sin x$ up to degree of $4k-1$. For other k's can be sandwiched using the fact that $\| M^k g\|_{\infty}$ shrinks exponentially within the interval [-1,1] (or something slightly better than (-1,1) -- since $g$ is not a constant but that is no big deal). Again, we wanted more that convergence, and $\sin x$ is not good enough for that because $g$ is not a constant.

The last hope lingers on the fact that $(\sum (M^4)^k)[x - x^3/6] = \sin x$. Anything special about the trig functions that allows such representation? I will truly leave that to the readers this time. Hint: what is the relation between $\sin x$ and $\exp x$?