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Find $$1*1*1*\cdots*1\quad n\,\,\text{ factors}$$ that is, a function $f(t)=1$ convolution with itself for a total of $n$ factors.

Would anyone mind helping me? I have no idea what I should do.

Also, if we define $g(t) = t$, what is $g*g*g$?

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What about looking at the Laplace transform of $f(t)$ raised to the $n^{\mathrm{th}}$ power? – AnonSubmitter85 May 8 '14 at 14:38
like turning $1*1$ into $(1/s)^2$ first ? – user100523 May 8 '14 at 14:44
up vote 2 down vote accepted

For two functions $f(t), g(t)$ $$\mathcal{L}((f*g)(t))=\mathcal{L}(f(t))\mathcal{L}(g(t))$$Since $\displaystyle \mathcal{L}(1(t))=\frac{1}{s}$ it follows that $$\mathcal{L}(\underbrace{1*1*\cdots*1}_{n \ \mbox{times}})=\frac{1}{s^n}$$ Now, $$\mathcal{L}(t^{n-1})=\frac{(n-1)!}{s^n}$$ hence comparing, $\displaystyle \underbrace{1*1*\cdots*1}_{n\ \mbox{times}}=\frac{t^{n-1}}{(n-1)!}$

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Working from the definition of convolution, $(f*g)(t):=\int_0^t f(s)g(t-s)\,ds$, and using the fact that $*$ is associative, we see \begin{align*} 1*1&=\int_0^t 1\cdot 1\,ds=t\\ 1*1*1&=(1*1)*1=t*1=\int_0^t s\,ds={t^2\over 2}\\ 1*1*1*1&=(1*1*1)*1={t^2\over 2}*1=\int_0^t {s^2\over 2}\,ds={t^3\over 3\cdot 2}\\ 1*1*1*1*1&=(1*1*1*1)*1={t^3\over 3\cdot 2}*1=\int_0^t {s^3\over 3\cdot 2}\,ds={t^4\over 4\cdot3\cdot 2}\\ \vdots\\ \underbrace{(1*1*\cdots*1)}_{n\text{ times}}&={t^{n-1}\over (n-1)!}. \end{align*}

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The Laplace Transform is $$\mathcal{L}(1)^n=\frac{1}{s^n}=\frac{(n-1)!}{s^n}/(n-1)!=\mathcal{L}(x^{n-1})/(n-1)!$$ so it would be $$1*1*\ldots*1=\frac{x^{n-1}}{(n-1)!}$$

Alternatively, note that $$1*(cx^a)=\int_0^x (1)ct^a\,\mathrm{d}t=c\frac{x^{a+1}}{a+1}$$ so by induction you would get $$1*1=x \\ 1*x=\frac{x^2}{2} \\ \vdots \\ 1*\cdots *1=\frac{x^{n-1}}{(n-1)!}$$

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