# Matrix Expression of Convolution and Its Generalized Continuous Form

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## Abstract

**:**

## 1. Introduction

- (1)
- If the matrix representing the function (image) f is A and the matrix representing the function g is B, then the convolution $f\ast g$ is represented by the sum of all elements of $A\circ B$ and this is the same as $tr\left(A{B}^{T}\right)$ where ∘ is array multiplication, T is the transpose, and $tr$ is the trace. Thus, the convolution in artificial intelligence (AI) is the same as $tr\left(A{B}^{T}\right)$.
- (2)
- The generalized continuous form of the convolution in AI can be represented as$$V\left(f\right)=\Phi \left(u\right){\int}_{0}^{\infty}{e}^{-t\Delta}f\left(t\right)\phantom{\rule{3.33333pt}{0ex}}dt,$$$$\Delta =\Delta (\delta ,u)=\frac{ln[1+\frac{\delta -1}{u}]}{\delta -1}.$$

## 2. Matrix Expression of Convolution in Convolutional Neural Network (CNN)

**Example**

**1.**

**Definition**

**1.**

**Example**

**2.**

## 3. Generalized Continuous Form of Matrix Expression of Convolution

**Lemma**

**1.**

**Theorem**

**1.**

- (1)
- (Duality with Laplace transform) If $\pounds \left(f\right)={F}_{\ast}\left(s\right)$ is the Laplace transform of a function $f\left(t\right)$, then it satisfies the relation of $V\left(f\right)=\Phi \left(u\right)\xb7{F}_{\ast}(\Delta )$.
- (2)
- (Shifting theorem) If $f\left(t\right)$ has the transform $F\left(u\right)$, then ${e}^{at}f\left(t\right)$ has the transform $\Phi \left(u\right)\xb7F(\Delta -a)$. That is,$$V\left[\phantom{\rule{3.33333pt}{0ex}}{e}^{at}f\left(t\right)\right]=\Phi \left(u\right)\xb7F(\Delta -a).$$Moreover, If $f\left(t\right)$ has the transform $F\left(u\right)$, then the shifted function $f(t-a)h(t-a)$ has the transform ${e}^{-a\Delta}\xb7\Phi \left(u\right)F(\Delta )$. In formula,$$V\phantom{\rule{3.33333pt}{0ex}}\left[f(t-a)h(t-a)\right]={e}^{-a\Delta}\xb7\Phi \left(u\right)F(\Delta )$$for $h(t-a)$ is Heaviside function (We write h since we need u to denote u-space).
- (3)
- (Linearity) Let $V\left(f\right)$ be the variant of Laplace-type transform. Then $V\left(f\right)$ is a linear operation.
- (4)
- (Existence) If $f\left(t\right)$ is defined, piecewise continuous on every finite interval on the semi-axis $t\ge 0$ and satisfies$$\left|f\left(t\right)\right|\le M{e}^{kt}$$for all $t\ge 0$ and some constants M and k, then the variant of Laplace-type transform $V\left(f\right)$ exists for all $\Delta >k$.
- (5)
- (Uniqueness) If the variant of Laplace-type transform of a given function exists, then it is uniquely determined.
- (6)
- (Heaviside function)$$vi\left[\phantom{\rule{3.33333pt}{0ex}}h(t-a)\right]={\int}_{0}^{\infty}{e}^{-t\Delta}h(t-a)\phantom{\rule{3.33333pt}{0ex}}dt={\int}_{a}^{\infty}{e}^{-t\Delta}\xb71\phantom{\rule{3.33333pt}{0ex}}dt$$$$={e}^{-a\Delta}/\Delta ,$$where h is Heaviside function.
- (7)
- (Dirac’s delta function) We consider the function$${f}_{k}(t-a)=\left\{\begin{array}{cc}1/k\hfill & if\phantom{\rule{3.33333pt}{0ex}}a\le t\le a+k\hfill \\ 0\hfill & otherwise.\hfill \end{array}\right.$$In a similar way to Heaviside, taking the integral of Laplace-type transform, we get$$vi\left[{f}_{k}(t-a)\right]={\int}_{0}^{\infty}{e}^{-t\Delta}{f}_{k}(t-a)\phantom{\rule{3.33333pt}{0ex}}dt=-\frac{1}{k\Delta}{\left[{e}^{-t\Delta}\right]}_{a}^{a+k}$$$$=-\frac{1}{k\Delta}({e}^{-(a+k)\Delta}-{e}^{-a\Delta})=-\frac{1}{k\Delta}{e}^{-a\Delta}({e}^{-k\Delta}-1).$$If we denote the limit of ${f}_{k}$ as $\delta (t-a)$, then$$vi\left(\delta (t-a)\right)=\underset{k\to 0}{lim}\phantom{\rule{3.33333pt}{0ex}}\left(vi\right)\left[{f}_{k}(t-a)\right]={e}^{-a\Delta}.$$
- (8)
- (Shifted data problems) For a given differential equation ${y}^{\u2033}+a{y}^{\prime}+by=r\left(t\right)$ subject to $y\left({t}_{0}\right)={c}_{0}$ and ${y}^{\prime}\left({t}_{0}\right)={c}_{1}$, where ${t}_{0}\ne 0$ and a and b are constant, we can set $t={t}_{1}+{t}_{0}$. Then $t={t}_{0}$ gives ${t}_{1}=0$ and so, we have$${y}_{1}^{\u2033}+a{y}_{1}^{\prime}+b{y}_{1}=r({t}_{1}+{t}_{0}),\phantom{\rule{3.33333pt}{0ex}}{y}_{1}\left(0\right)={c}_{0},\phantom{\rule{3.33333pt}{0ex}}{y}_{1}^{\prime}\left(0\right)={c}_{1}$$for input $r\left(t\right)$. Taking the variant, we can obtain the output $y\left(t\right)$.
- (9)
- (Transforms of derivatives and integrals) Let a function f is n-th differentiable and integrable, and let us consider the fraction Δ as an operator. Then $V\left(f\right)$ of the n-th derivatives of $f\left(t\right)$ satisfies$$vi\left({f}^{\left(n\right)}\right)={\Delta}^{n}vi\left(f\right)-\sum _{k=1}^{n}{\Delta}^{n-k}{f}^{(k-1)}\left(0\right)\phantom{\rule{2.em}{0ex}}$$and$$V\phantom{\rule{3.33333pt}{0ex}}\left[{\int}_{0}^{t}f\left(\tau \right)\phantom{\rule{3.33333pt}{0ex}}d\tau \right]=\Phi \left(u\right)\xb7\frac{1}{\Delta}V\left(f\right).$$
- (10)
- (Convolution) If two functions f and g are integrable for * is the convolution, then $V(f\ast g)$ satisfies$$V(f\ast g)=\Phi \left(u\right)\xb7F(\Delta )G(\Delta )$$for $V\left(f\right)=F(\Delta )$.

**Proof.**

**Example**

**3.**

**Solution.**

- (1)
- Since this equation is $y+y\ast t=1$, taking the integral of Laplace-type transform on both sides, we have$$Y+(Y\xb7\frac{1}{{\Delta}^{2}})=\frac{1}{\Delta}$$$$Y=\frac{\Delta}{{\Delta}^{2}+1}$$Let us do the check by expansion. Expanding, we get ${y}^{\u2033}\left(t\right)+y\left(t\right)=0$. Since ${\int}_{a}^{a}f=0$, we get $y\left(0\right)=1$ and ${y}^{\prime}\left(0\right)=0$. Thus, we obtain $y=cost$.
- (2)
- This is rewritten as a convolution$$y\left(t\right)-y\ast sint=t.$$Taking the integral of Laplace-type transform, we have$$Y\left(u\right)-Y\left(u\right)\frac{1}{{\Delta}^{2}+1}=Y\left(u\right)(1-\frac{1}{{\Delta}^{2}+1})=\frac{1}{{\Delta}^{2}}$$$$Y\left(u\right)=\frac{1}{{\Delta}^{2}}+\frac{1}{{\Delta}^{4}}$$$$y\left(t\right)=t+\frac{1}{6}{t}^{3}.$$
- (3)
- Note that the equation is the same as $y-(1+t)\ast y=1-sinht$. Taking the transform, we get$$Y-(\frac{1}{\Delta}+\frac{1}{{\Delta}^{2}})Y=\frac{1}{\Delta}-\frac{1}{{\Delta}^{2}-1},$$$$Y\phantom{\rule{3.33333pt}{0ex}}(1-\frac{1}{\Delta}-\frac{1}{{\Delta}^{2}})=\frac{{\Delta}^{2}-\Delta -1}{\Delta ({\Delta}^{2}-1)}.$$Simplification gives$$Y=\frac{\Delta}{{\Delta}^{2}-1},$$$$y\left(t\right)=cosht$$

**Theorem**

**2.**

**Proof.**

**Example**

**4.**

- (a)
- The string is initially at rest on the x-axis from $x=0$ to ∞.
- (b)
- For $t>0$ the left end of the string is moved in a given fashion, namely, according to a single sine wave$$w(0,t)=f\left(t\right)=\left\{\begin{array}{cc}sin\phantom{\rule{3.33333pt}{0ex}}2t\hfill & if\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}0\le t\le \pi \hfill \\ 0\hfill & otherwise.\hfill \end{array}\right.$$
- (c)
- Furthermore, $w(x,t)\to 0$ as $x\to \infty $ for $t\ge 0$.

**Example**

**5.**

**Solution.**Taking the integral of Laplace-type transform on both sides of ${w}_{t}={c}^{2}{w}_{xx}$, we have

## 4. Conclusions

## Author Contributions

## Funding

## Acknowledgments

## Conflicts of Interest

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Geum, Y.H.; Rathie, A.K.; Kim, H.
Matrix Expression of Convolution and Its Generalized Continuous Form. *Symmetry* **2020**, *12*, 1791.
https://doi.org/10.3390/sym12111791

**AMA Style**

Geum YH, Rathie AK, Kim H.
Matrix Expression of Convolution and Its Generalized Continuous Form. *Symmetry*. 2020; 12(11):1791.
https://doi.org/10.3390/sym12111791

**Chicago/Turabian Style**

Geum, Young Hee, Arjun Kumar Rathie, and Hwajoon Kim.
2020. "Matrix Expression of Convolution and Its Generalized Continuous Form" *Symmetry* 12, no. 11: 1791.
https://doi.org/10.3390/sym12111791