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Problem of the Week #17 - July 23rd, 2012

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Chris L T521

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Jan 26, 2012
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Thanks to those who participated in last week's POTW!! Here's this week's problem.

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Problem: Recall that the convolution of $f$ and $g$ is defined by the integral

\[(f\ast g)(t) = \int_0^{t}f(t-\tau)g(\tau)\,d\tau.\]

Establish the commutative, distributive, and associative properties of convolution, i.e.

(1) $f\ast g = g\ast f$
(2) $f\ast (g_1 + g_2) = f\ast g_1 + f\ast g_2$
(3) $f\ast(g\ast h) = (f\ast g)\ast h$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's question was correctly answered by Sudharaka and Badhi. You can find their solutions below.

Sudharaka's solution:

\[(f*g)(t)=\int_{0}^{t}f(t-\tau)g(\tau)\,d\tau\]

Substituting \(u=t-\tau\) we get,


\begin{eqnarray}


(f*g)(t)&=&-\int_{t}^{0}f(u)g(t-u)\,du\\


&=&\int_{0}^{t}f(u)g(t-u)\,du\\


&=&(g*f)(t)


\end{eqnarray}


\[\therefore f*g=g*f~~~~~~~~~~~~~(1)\]


\begin{eqnarray}


[f*(g_{1}+g_{2})](t)&=&\int_{0}^{t}f(t-\tau)(g_{1}+g_{2})(\tau)\,d\tau\\


&=&\int_{0}^{t}f(t-\tau)[g_{1}(\tau)+g_{2}(\tau)]\,d\tau\\


&=&\int_{0}^{t}f(t-\tau)g_{1}(\tau)\,d\tau+\int_{0}^{t}f(t-\tau)g_{2}(\tau)\,d\tau\\


&=&(f*g_{1})(t)+(f*g_{2})(t)


\end{eqnarray}


\[\therefore f*(g_{1}+g_{2})=(f*g_{1})+(f*g_{2})~~~~~~~~~~~~~(2)\]


\begin{eqnarray}


[(f*g)*h](t)&=&\int_{0}^{t}(f*g)(t-\tau)h(\tau)\,d\tau\\


&=&\int_{0}^{t}\left[\int_{0}^{t-\tau}f(t-\tau-u)g(u)\,du\right]h(\tau)\,d\tau\\


\end{eqnarray}


Substituting \(w=u+\tau\) we get,


\[[(f*g)*h](t)=\int_{0}^{t}\left[\int_{\tau}^{t}f(t-w)g(w-\tau)\,dw\right]h(\tau)\,d\tau\]


Changing the order of integration,


\begin{eqnarray}


[(f*g)*h](t)&=&\int_{0}^{t}\left[\int_{0}^{w}f(t-w)g(w-\tau)h(\tau)\,d\tau\right]\,dw\\


&=&\int_{0}^{t}f(t-w)\left[\int_{0}^{w}g(w-\tau)h(\tau)\,d\tau\right]\,dw\\


&=&\int_{0}^{t}f(t-w)(g*h)(w)\,dw\\


&=&[f*(g*h)](t)


\end{eqnarray}


\[\therefore (f*g)*h=f*(g*h)~~~~~~~~~~~~~(3)\]

BAdhi's solution:

from the definition,



$$


(f\ast g)(t)=\int_0^tf(t-\tau)g(\tau)\,d \tau


$$


(1) Proving that $f\ast g=g\ast f$


$f\ast g =\int_0^tf(t-\tau )g(\tau )\,d \tau$


with the substitution,
$$\begin{align*}t-\tau=v\\
\,dv=-\,d\tau\\
\end{align*}$$


then,
$$\tau=0 \implies v=t, \qquad \tau=t \implies v=0$$
$\begin{align*}
f\ast g &=\int_0^tf(t-\tau )g(\tau )\,d \tau\\
&=-\int_t^0f(v)g(t-v)\,dv\\
\end{align*}
$


Since the definite integral is independent from the variable, we can substitute $v$ with $\tau$


then the equation becomes,


$\begin{align*}f\ast g &=-\int_t^0f(\tau )g(t-\tau )\, d\tau\\
&=\int_0^tf(\tau )g(t-\tau )\, d\tau\\
&=\underbrace{\int_0^tg(t-\tau )f(\tau )\, d\tau }_{g\ast f}\\
\end{align*}$


thus proves the commutative property,
$$f\ast g=g\ast f$$


(2) proving, $f\ast (g_1+g_2)=f\ast g_1 +f\ast g_2$


$\begin{align*}
f\ast(g_1+g_2)&=\int_0^tf(t-\tau )\left[g_1(\tau )+g_2(\tau )\right]\, d\tau \\
&=\int_0^t f(t-\tau )g_1(\tau ) + f(t-\tau)g_2(\tau )\, d\tau \\
&=\underbrace{\int_0^t f(t-\tau )g_1(\tau )\, d\tau }_{f\ast g_1} + \underbrace{\int_0^t f(t-\tau )g_2(\tau )\, d\tau }_{f\ast g_2}\\
\end{align*}
$


thus proves the distributive property,
$$f\ast (g_1+g_2)=f\ast g_1 +f\ast g_2$$


(3) proving that, $f\ast (g\ast h)=(f\ast g)\ast h$


$


\begin{align*}
f\ast (g\ast h) &= (g\ast h)\ast f \qquad \qquad \left(from \;proof: (1)\right)\\
&=\int_0^t\left[ (g\ast h)(t-\tau )\right] f(\tau ) \,d\tau \\
&=\int_0^tf(\tau )\left[ \int_0^{t-\tau} g(t- \tau -u)h(u) \,du\right]\,d\tau\\
&=\int_0^t\int_0^{t-\tau}f(\tau )g(t-\tau -u)h(u)\,du\,d\tau \\
\end{align*}$


by changing the order of the integral, the change of boundaries of $\tau$ and $u$ are,


$0<\tau<t,\: 0<u<(t-\tau) \longrightarrow 0<u<t, \: 0<\tau<(t-u) $


then,


$
\begin{align*}
f\ast (g\ast h) &=\int_0^t\int_0^{t-u}f(\tau )g(t-\tau -u)h(u)\,d\tau\,du\\
\end{align*}$


since $h(u)$ is independent from $\tau$,


$\begin{align*}
f\ast (g\ast h) &=\int_0^th(u)\underbrace{\left[ \int_0^{t-u}f(\tau )g(t-u-\tau )\,d\tau\right]}_{(g\ast f)(t-u)} \,du\\
&=\int_0^th(u)[(g\ast f)(t-u)]\,du\\
&=(g\ast f)\ast h
\end{align*}$


from $proof : (1) $,


$f\ast (g\ast h)=(f\ast g)\ast h$


Thus proves the associative property,


$$f\ast (g\ast h)=(f\ast g)\ast h$$
 
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