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\begin{document}

\title{Psuedo-Convolution}



\author[1]{Benedict Irwin}%
\affil[1]{University of Cambridge}%


\vspace{-1em}



  
  \date{May 08, 2024}


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\sloppy


If we have a function which is defined as the convolution

\begin{equation}
h(z) = \mathrm{conv}(f,g) = \int_{-\infty}^\infty f(t)g(z-t)\;dt
\end{equation}

we might consider the case where f and g have series expansions of the
form

\par\null

\begin{equation}
f(x)=\sum_{k=0}^\infty a_k x^k
\end{equation}
\begin{equation}
g(x)=\sum_{k=0}^\infty b_k x^k
\end{equation}

If we write these expansions into the integral, the linear property of
the integral would allow us to integrate term by term, but each term
individually may diverge because of the infinite support. Instead, it
might be interesting to consider a new definition of the convolution,
which I will call the pseudo-convolution. This will use the relationship
between Laplace transforms and convolutions, and we could define

\par\null

\begin{equation}
\textrm{pseudoconv}(f,g)=\mathcal{L}^{-1}[\mathcal{L}[f](s)\mathcal{L}[g](s)](x)
\end{equation}

Where L is a Laplace transform. For this definition we have things like

\par\null

\begin{equation}
\mathrm{pseudoconv}(x,x)=\frac{x^3}{6}
\end{equation}

If we do the term by term expansion by taking the Cauchy product of the
series for f and g and integrating, then defining the integrals as
pseudo-convolutions, we arrive at an expansion (under this definition)

\par\null

\begin{equation}
h(z)=\sum_{k=1}^\infty \frac{t^k}{k!}\sum_{l=0}^{k-1} l! (k-l-1)! a_l b_{k-l-1}
\end{equation}

This is probably not an equality, but defines a transform mapping two
functions to one new function. We get results like

\par\null

\begin{equation}
\mathrm{pseduoconvolve}(e^{-ax},e^{-bx})=\frac{e^{-bx}-e^{-ax}}{a-b}
\end{equation}

If we write recursions such as

\par\null

\begin{align}
a_0=\mathrm{pc}(1,1)\\
a_1=\mathrm{pc}(1,a_0+1)\\
\cdots \\
a_k=\mathrm{pc}(1,a_{k-1}+1)\\
\end{align}
then
\begin{equation}
a_\infty = x \exp(x)
\end{equation}



Text

where `pc' means `pseudo-convolve'. Similarly

\begin{align}
a_0=\mathrm{pc}(1,1)\\
a_1=\mathrm{pc}(1,2a_0)\\
\cdots \\
a_k=\mathrm{pc}(1,2a_{k-1})\\
\end{align}
then
\begin{equation}
\sum_{k=0}^\infty a_k = \frac{e^{2x}-1}{2}
\end{equation}



It seems that if
\begin{align}
a_0=\mathrm{pc}(1,1)\\
a_1=\mathrm{pc}(1,na_0)\\
\cdots \\
a_k=\mathrm{pc}(1,na_{k-1})\\
\end{align}
then
\begin{equation}
\sum_{k=0}^\infty a_k = \frac{e^{nx}-1}{n}
\end{equation}



Other definitions exist such as:
\begin{align}
a_0=\mathrm{pc}(1,1)=x\\
a_k=\mathrm{pc}(1,ka_{k-1})=\frac{x^{k+1}}{k+1}\\
\sum_{k=0}^\infty a_k= -\log(1-x)
\end{align}
and
\begin{align}
a_0=\mathrm{pc}(1,1)=x\\
a_k=\mathrm{pc}(1,k^2 a_{k-1})=\frac{k!x^{k+1}}{k+1}
\end{align}
and even
and
\begin{align}
a_0=\mathrm{pc}(1,1)=x\\
a_k=\mathrm{pc}(1,a_{k-1}/k)=\frac{x^{k+1}}{(k+1)k!k!}\\
\sum_{k=0}^\infty a_k= \sqrt{x}I_1(2\sqrt{x})
\end{align}
for Bessel $I_1(x)$ function. To shorten this notation somewhat we can define the notation
\begin{equation}
\Sigma_{\mathrm{pc},x}(1,a_{k-1}/k) \to \sqrt{x}I_1(2\sqrt{x})
\end{equation}
where the subscript $x$ means, the first term is $x$. Then we can write \begin{align}
\Sigma_{\mathrm{pc},x}(-x,a_{k-1}) \to \sin(x) \\
\Sigma_{\mathrm{pc},1}(-x,a_{k-1}) \to \cos(x) \\
\Sigma_{\mathrm{pc},x}(x,a_{k-1}) \to \sinh(x) \\
\Sigma_{\mathrm{pc},1}(x,a_{k-1}) \to \cosh(x) \\
\Sigma_{\mathrm{pc},1}(-1,a_{k-1}) \to \exp(-x) \\
\Sigma_{\mathrm{pc},1}((-1)^{k+1},a_{k-1}) \to \sin(x)+\cos(x)\\
\Sigma_{\mathrm{pc},1}\left(-\frac{(2k-1)x}{2k},a_{k-1}\right) \to J_0(x)\\
\Sigma_{\mathrm{pc},1}\left(\frac{(2k-1)x}{2k},a_{k-1}\right) \to I_0(x)\\
\end{align}
Very interestingly it seems that
\begin{align}
\Sigma_{\mathrm{pc},x}(-4x,a_{k-1}) \to \sin(x)\cos(x) \\
\Sigma_{\mathrm{pc},x}(4x,a_{k-1}) \to \sinh(x)\cosh(x) \\
\Sigma_{\mathrm{pc},x^2}(-16x,a_{k-1}) \to \sin^2(x)\cos^2(x)\\
\Sigma_{\mathrm{pc},x}\left(\frac{k^2}{(k+1)},a_{k-1}\right) \to \mathrm{Li}_2(x) \\
\Sigma_{\mathrm{pc},x}\left(\frac{k^3}{(k+1)^2},a_{k-1}\right) \to \mathrm{Li}_3(x) \\
\Sigma_{\mathrm{pc},x}\left(f_{\theta}(k)x^{2k},a_{k-1}\right) \to \frac{1}{2}(1+\theta_3(0,x)) \\
\end{align}
where $f_{\theta}(k)$ goes like $1,12,630,80080,18386775,\cdots$. It is not clear what this sequence is. It is not in the OEIS on it's own.

\begin{align}
\Sigma_{\mathrm{pc},x}(-1,a_{k-1}) \to \log(x+1) \\
\Sigma_{\mathrm{pc},x}\left(\frac{-k}{k+1},a_{k-1}\right) \to \gamma + \Gamma(0,x) + \log(x) \\
\Sigma_{\mathrm{pc},x}\left(\frac{-k}{k+2},a_{k-1}\right) \to 2\left(\gamma + \Gamma(0,x) + \log(x) + \frac{1-x-\cosh(x)+\sinh(x)}{x}\right)
\end{align}

If you want to recreate the generating function for the partitions powers $x+x^2+x^3+x^5+x^7+x^{11}+x^{15}+x^{22}+\cdots$, then the powers of $x$ inside the recursion follow a related sequence $0,0,1,1,3,3,6,7,11,13,20,\cdots$ which may be A083751.
\begin{align}
\Sigma_{\mathrm{pc},1}(k,a_{k-1}) \to \frac{1}{1-x} \\
\Sigma_{\mathrm{pc},1}(2k(2k-1)x,a_{k-1}) \to \frac{1}{1-x^2} \\
\Sigma_{\mathrm{pc},1}(3k(3k-1)(3k-2)x^2/2,a_{k-1}) \to \frac{1}{1-x^3} \\
\Sigma_{\mathrm{pc},1}((2k-1)!/(k-1)! x^{k-1},a_{k-1}) \to \frac{\theta_2(0,\sqrt{x})}{2x^{1/8}} \\
\end{align}

One of my favourite results so far is the following

\par\null

\begin{align}
a_{-1}=1 \\
a_{0}=1 \\
a_k = \mathrm{pc}((F(k+1)-1),a_{k-1}a_{k-2})
\end{align}
where $F(k)$ are the Fibbonacci numbers, $F(1)=1,F(2)=2,F(3)=3,F(4)=5,\cdots$

The sum is the function \begin{equation}
\sum a_k = 1 + x + x^2 + x^4 + x^7 + x^{12} + x^{20} + x^{33} + \cdots
\end{equation}
so this is the function whose internal recursion coefficients are the same as the powers in the output! There is also the function
\begin{align}
a_{0}=1 \\
a_{1}=\mathrm{pc}(1,a_0) \\
a_{2}=\mathrm{pc}(2,a_0a_1) \\
a_{3}=\mathrm{pc}(4,a_0a_1a_2) \\
a_{4}=\mathrm{pc}(8,a_0a_1a_2a_3) \\
\cdots \\
\sum_{k=0}^\infty a_k = 1 + x + x^2 + x^4 + x^8 + x^{16} + x^{32} + \cdots
\end{align}
and similarly
\begin{align}
a_{0}=1 \\
a_{1}=\mathrm{pc}(1,a_0) \\
a_{2}=\mathrm{pc}(3,a_0^2a_1) \\
a_{3}=\mathrm{pc}(12,a_0^3a_1^2a_2) \\
a_{4}=\mathrm{pc}(60,a_0^4a_1^3a_2^2a_3) \\
\cdots \\
\sum_{k=0}^\infty a_k = 1 + x + x^3 + x^{12} + x^{60} + x^{360} + x^{2520} + \cdots
\end{align}
where the powers are likely A001710. We also have 
\begin{align}
a_{0}=1 \\
a_{1}=\mathrm{pc}(1,a_0) \\
a_{2}=\mathrm{pc}(3,a_0^2) \\
a_{3}=\mathrm{pc}(10,a_0^3) \\
a_{4}=\mathrm{pc}(41,a_0^4) \\
\cdots \\
\sum_{k=0}^\infty a_k = 1 + x + x^3 + x^{10} + x^{41} + x^{206} + x^{1237} + \cdots
\end{align}
where th powers are most likely A002627.
\begin{align}
a_{0}=1 \\
a_{k}=\mathrm{pc}(2^k-1,a_{k-1}^2) \\
\cdots \\
\sum_{k=0}^\infty a_k = \sum_{k=1}^\infty x^{2^k -1}
\end{align}



\begin{align}
\Sigma_{\mathrm{pc},x^{1/4}}((8k-3)(8k+1)x/16,a_{k-1}) \to -\frac{x^{1/4}}{x^2-1} \\
\end{align}

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