MHB Write f(x) in terms of the unit step function u(x)

alexmahone
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Write f(x) in terms of the unit step function u(x).

$u(x)=\left\{ \begin{array}{rcl} 1\ &\text{if}& \ x\geq 0 \\ 0\ &\text{if}& \ x<0\end{array} \right.$

$f(x)=\left\{ \begin{array}{rcl} 1\ &\text{if}& \ 2n\le x\le 2n+1 \\ 0\ &\text{elsewhere}\end{array} \right.$
 
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A more useful definition of 'Haeviside Step Function' or 'Unit Step Function' is...

$\mathcal{u}(x)=\begin{cases}1 &\text{if}\ x>0\\ \frac{1}{2} &\text{if}\ x=0\\ 0 &\text{if}\ x<0\end{cases}$ (1)

... and the function...

$f(x)=\begin{cases} 1 &\text{if}\ 2n<x<2n+1\\ \frac{1}{2} &\text{if}\ x=n\\ 0 &\text{elsewhere}\end{cases}$ (2)

... can be written as...

$\displaystyle f(x)= \sum_{n=0}^{\infty} (-1)^{n} \mathcal{u}(x-n)$ (3)

Kind regards

$\chi$ $\sigma$
 
chisigma said:
A more useful definition of 'Haeviside Step Function' or 'Unit Step Function' is...

But I need to use the definition in post #1.

I get $\displaystyle f(x)=\sum_{-\infty}^\infty u(x-2n)u(2n+1-x)$
 
chisigma said:
A more useful definition of 'Haeviside Step Function' or 'Unit Step Function' is...

$\mathcal{u}(x)=\begin{cases}1 &\text{if}\ x>0\\ \frac{1}{2} &\text{if}\ x=0\\ 0 &\text{if}\ x<0\end{cases}$ (1)

... and the function...

$f(x)=\begin{cases} 1 &\text{if}\ 2n<x<2n+1\\ \frac{1}{2} &\text{if}\ x=n\\ 0 &\text{elsewhere}\end{cases}$ (2)

... can be written as...

$\displaystyle f(x)= \sum_{n=0}^{\infty} (-1)^{n} \mathcal{u}(x-n)$ (3)

Kind regards

$\chi$ $\sigma$

It is curious the fact that the function...

$\displaystyle f_{a}(x) =\begin{cases} 1 &\text{if}\ 2n \le x \le 2n+1\\ 0 &\text{elsewhere}\end{cases}$ (1)

... has Laplace Transform...

$\displaystyle \mathcal{L}\{f_{a}(x)\}= F(s)= \frac{1}{s\ (1+e^-s)}$ (2)

... and the (2) has Inverse Laplace Transform...

$\displaystyle \mathcal{L}^{-1}\{F(s)\}= f_{b}(x)=\sum_{n=0}^{\infty} (-1)^{n}\ \mathcal{u}(x-n)$ (3)

... so that for the unicity of the inverse L-transform $f_{a}(x)$ and $f_{b}(x)$ are 'pratically' the same function... where 'pratically' means that the difference between then is a 'null function'...

Kind regards

$\chi$ $\sigma$
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...

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