Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $g*g*g$?

share|improve this question
1  
What about looking at the Laplace transform of $f(t)$ raised to the $n^{\mathrm{th}}$ power? –  AnonSubmitter85 May 8 at 14:38
    
like turning $1*1$ into $(1/s)^2$ first ? –  user100523 May 8 at 14:44

3 Answers 3

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, $\displaystyle \underbrace{1*1*\cdots*1}_{n\ \mbox{times}}=\frac{t^{n-1}}{(n-1)!}$

share|improve this answer

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*}

share|improve this answer

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 $$1*1=x \\ 1*x=\frac{x^2}{2} \\ \vdots \\ 1*\cdots *1=\frac{x^{n-1}}{(n-1)!}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.