# Recursive Integration over Piecewise Polynomials: Closed form?

Is there a closed form to the following recursive integration? $$f_0(x) = \begin{cases} 1/2 & |x|<1 \\ 0 & |x|\geq1 \end{cases} \\ f_n(x) = 2\int_{-1}^x(f_{n-1}(2t+1)-f_{n-1}(2t-1))\mathrm{d}t$$ It's very clear that this converges against some function and that quite rapidly, as seen in this image, showing the first 8 terms:

Furthermore, the derivatives of it have some very special properties.
Note how the (renormalized) derivatives consist of repeated and rescaled functions of the previous degree which is obviously a result of the definition of the recursive integral:

EDIT
Basically by trying around, I found the following probably more usable Fouriertransform of the expression above. I do not have a formal proof but it holds for all terms I tried it with (first 11). $$\mathcal{F}_x\left[f_n(x)\right](t)=\frac{\sin \left(2^{-n} t\right) \left(\prod _{k=1}^n \frac{2^{k+1} \sin \left(2^{-k} t\right)}{t}\right)}{\sqrt{2 \pi } t}$$
Here an image of how that looks like (first 10 terms in Interval $[-8\pi,8\pi]$):

With this, my question becomes:
"What, if there is one, is the closed form inverse fourier transform of $\mathcal{F}_x\left[f_n(x)\right](t)=\frac{\sin \left(2^{-n} t\right) \left(\prod _{k=1}^n \frac{2^{k+1} \sin \left(2^{-k} t\right)}{t}\right)}{\sqrt{2 \pi } t}$, especially for the case $n\rightarrow\infty$?"

On an unrelated note: This is my first question here, so I hope I did pose it decently. I searched through mathstack, to my best knowledge using terms of the involved math, without success.

(minor Edit: it seems like the image sharing service I'm using tends to be somewhat unreliable. I'm sorry for that. If some or all of the images don't show up, that's why.)

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2
The integral has unbalanced parentheses and no limits; the right-hand side isn't a well-defined function of $x$. Also, in mathematics, when defining functions by cases one doesn't usually follow the programming convention of implying a conjunction with earlier conditions in later conditions, but states mutually exclusive conditions. – joriki Nov 19 '12 at 16:46
@joriki: I tried to improve what you said. Is that correct now? I'm not sure what I'm supposed to do with the conditions. Can you clarify, please? – kram1032 Nov 19 '12 at 17:03

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The limits are OK now; I would have added an opening parenthesis instead of removing the closing one, to clarify the scope of the integral; about the conditions, I'd write $\begin{cases}1/2&|x|\lt1\;,\\0&|x|\ge1\;.\end{cases}$ – joriki Nov 19 '12 at 17:13
@joriki thank you for your input! Is this how you'd like it? – kram1032 Nov 19 '12 at 17:17

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nathang 19:50 Note f_n(0) = 1 for n>0, so (f_n)'(-1/2) = 4 for n>1. So it can't be the box function. – Peter Sheldrick Nov 20 '12 at 21:52
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## 2 Answers

Here is a formula for $f_n$:

$$f_n(x) = \sum_{j=0}^{2^n} \left( \frac{c_n(j) - c_n(j-1)}{2}\frac{\left(2^n x + 2^n - 2j\right)^n H\left(2^nx + 2^n - 2j\right)} {n!2^{n(n-1)/2}} \right).$$

Here $H$ is the Heaviside step function, $c_n$ is defined by $$c_n(j) = \begin{cases} 0 & \text{if $j<0$}\\ (-1)^{s(j)} & \text{if $0\leq j < 2^n$} \\ 0 & \text{if $j\geq 2^n$} \end{cases}$$ and $s(j)$ is the sum of the digits of the binary representation of $j$. (For example $s(13) = s(0\text{b}1101) = 3$.)

While the Heaviside function is crucial to deriving the formula, it can be removed from the final result using the floor function (denoted $\lfloor \cdot \rfloor$):

$$f_n(x) = \sum_{j=0}^{\lfloor2^{n-1}(x+1)\rfloor} \left( \frac{c_n(j) - c_n(j-1)}{2}\frac{\left(2^n x + 2^n - 2j\right)^n} {n!2^{n(n-1)/2}} \right).$$

Here is a plot of $f_{15}$ using this formula:

# Deriving the formula

First, separate the definition into two integrals and change variables, $2t+1 \mapsto t$ in the first, and $2t-1\mapsto t$ in the second, giving

$$f_{n+1}(x) = \int_{-1}^{2x+1} f_n(t)\ dt - \int_{-3}^{2x-1}f_n(t)\ dt$$

Of course, we can change the -3 to -1 and combine these to a single integral:

$$f_{n+1}(x) = \int_{2x-1}^{2x+1} f_n(t)\ dt$$

Then rewrite $f_0=(1/2)(H(t+1) - H(t-1))$. Note that the integral of $H(t)$ is $tH(t)$, whose integral is $(t^2/2)H(t)$, and so forth. Now we can write $f_n$ as a single iterated integral, for example

$$f_3(x) = \frac12 \int_{2x-1}^{2x+1} \int_{2y-1}^{2y+1} \int_{2z-1}^{2z+1} (H(t-1) - H(t+1))dt\ dz\ dy$$

Each integration can be done doing several different changes of variables. This gives rise to the powers of 2 in the denominator.

# Notes

Each $f_n$ is symmetric. The part from -1 to -0.5 is repeated four times. Due to the way that Heaviside functions work, it is computationally easiest to compute values for $f_n(x)$ for $x$ closer to -1.

# Code

Here is some Python code to compute $f_n(x)$.

````from __future__ import division
from math import factorial

def c(j, n):
    if j < 0 or j >= 2**n:
        return 0
    else:
        return (-1)**bin(j).count("1")

def f(x, n):
    numerator = 0
    for j in xrange(int(2**(n-1) * (x+1))):
        numerator += (c(j, n) - c(j-1, n)) * (2**n * x + 2**n - 2*j)**n
    denominator = 2 * 2**(n*(n-1)/2) * factorial(n)
    return numerator/denominator

print f(-0.75, 10)
````
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That is awesome! Definitely upvoted. I'll wait with accepting for now, since you said it's WIP. – kram1032 Nov 22 '12 at 9:37

Suppose $f$ is a fixed point of the iterations. Then $$f(x) = 2\int_{-1}^x\big(f(2t+1)-f(2t-1)\big)\,\mathrm{d}t,$$ which, upon differentiating both sides by $x$, implies that $$f'(x) = 2\big(f(2x+1)-f(2x-1)\big).$$ I'll assume that $f$ vanishes outside $[-1,1]$, which you can presumably prove from the initial conditions. Then we get $$f'(x) = \begin{cases} 2f(2x+1) & \text{if }x\le0, \\ -2f(2x-1) & \text{if }x>0. \end{cases}$$ This is pretty close to the definition of the Fabius function. In fact, your function would be $\frac{\text{Fb}'(\frac{x}{2}+1)}{2}$

The Fabius function is smooth but nowhere analytic, so there isn't going to be a nice closed form for your function.

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– Rahul Narain Nov 24 '12 at 16:32
Ah, nice. So my type of function now has a name. Good to know. - It's always hard to research stuff you have no idea how it's called. I already suspected this kind of thing to be tried. - It's too simple an idea to be new. – kram1032 Nov 24 '12 at 17:16
– Rahul Narain Nov 24 '12 at 17:31
Nice, so I guess our two questions are related. – kram1032 Nov 24 '12 at 17:35
– kram1032 Nov 25 '12 at 0:43
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