TITLE: Why do we need to reverse a function in the convolution operaton?
QUESTION [7 upvotes]: As I understand, convolution is one way to describe how 2 functions correlate to each other.
According to the wikipedia,

The convolution of $f$ and $g$ is written $f∗g$, using an asterisk or star.
  It is defined as the integral of the product of the two functions
  after one is reversed and shifted.

I can somehow accpet the shift operation. But why do we need to reverse one of the function? If we just want to make the 2 funcitons collide, I think shifting is enough.
ADD 1
An interesting article recommended by my lecturer:
Understanding Convolution

REPLY [9 votes]: Here is one way to discover the discrete convolution operation.  Let $S: \mathbb R^n \to \mathbb R^n$ be the circular shift operator defined by
$$
S \begin{bmatrix} x_0 \\ x_1 \\ \vdots \\ x_{n-1} \end{bmatrix}
= \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_{n-1} \\ x_0 \end{bmatrix}.
$$
Assume that the linear operator $A:\mathbb R^n \to \mathbb R^n$ is "shift-invariant" in the sense that $A(Sx) = S(Ax)$ for all $x$.  In other words, if you shift the input, the output is shifted the same way.  We can easily check that $A S^n x = S^n A x$ for all integers $n$, positive or negative.  Shift-invariant linear operators are of fundamental importance, and for example they are useful in signal processing and numerical differentiation.
Key idea: Let's find the solution to $Ax = b$, assuming that we have already found a "fundamental solution" $x_0$ which satisfies 
$$
Ax_0 = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = e_0.
$$  
We are off to a good start if we notice that 
\begin{align}
A S^{-1} x_0 = \begin{bmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = e_1
\end{align}
and in fact
$$
A S^{-j} x_0 = e_j \quad \text{for } j = 0,\ldots,n-1.
$$ 
It follows that
\begin{align}
A(b_0 x_0 + b_1 S^{-1} x_0 + b_2 S^{-2} x_0 + \cdots + b_{n-1} S^{n-1} x_0) &= b_0 e_0 + e_1 b_1 + \cdots + b_{n-1} e_{n-1} \\&=b.
\end{align}
We have solved $Ax = b$.
The solution to $Ax = b$ is a particular combination of $x_0$ and $b$, and this combination of $x_0$ and $b$ is called the "convolution" of $x_0$ and $b$, and is denoted $x_0 \ast b$.  We have discovered the convolution operation.  This explains why we care about convolution, and why convolution is defined in the way that it is.
We can discover the convolution of functions using a similar line of reasoning, with the delta function $\delta$ in place of $e_0$.