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Is there a useful (see below to understand bit more what is useful) function with the properties:

f(x)=0 \quad x< k and $f(x)=1 \quad x\geq k$ to $k \in \mathbb{R}$

But it couldn't be a piecewise function like

$f(x)=$ \begin{cases} 0 & x< k \\ 1 & x\geq k \end{cases}

because this type of function can't help so much to manipulate formulas, for use in integration, derivative, transformations, hiperbolic and so on. This kind of function is calculated by a comparision, and I need a calculation using calculus, products, sums, etc.

This kind of function could be very useful and interesting to validate two different formulas. For example, if I have a formula (g1(x)) that holds to a problem when x and another that holds when $x\geq k$ (g2(x)), I could use this function (f(x)) to wrote something like:

$h(x)=f(x)g1(x)+(1-f(x))g2(x)$

and this give to me a function $h(x)$ that will hold to all $x\in \mathbb{R}$.

To functions in integers we can write something like

$h(x)=\frac{(-1)^{n-1}+1}{2}g1(x)+\frac{(-1)^n+1}{2}g2(x)$

and the $g1(x)$ will hold to all $x\equiv 1 \mod 2$ and $g2(x)$ to all $x\equiv 0 \mod 2$.

Pratice:

Here are one example to understand a little about this strange $f$ function.

To $k=0$, $f(x)=0$ to x<0 and $f(x)=1$ to $x\geq 0$

Se we can define $h(x)=f(x)x-(1-f(x))x=xf(x)-x+xf(x)=2xf(x)-x=x(2f(x)-1)$

and this function, $x(2f(x)-1)$ is the modulus of the absolute value of a real number $x$.

Test this to $x=0$, $x=-1$ (or negative) and $x=1$ (or positive) and you will see.

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    @PeterT.off, I'm not familiated with this function so I will take a look first, but I really hope this function will be *useful*.2012-04-13

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A large number of functions like this exist, for instance: $\lim_{n\rightarrow\infty}\frac{1}{2}+\frac{1}{\pi}\arctan(nx)$

But there is really no reason not to use the Heaviside step function as it is popularly called, you can write the derivative as a dirac delta function.

More analytic function that suite your problem can be found here as well.

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    I think is this what I'm looking for $H(x+k)$ holds. Thx.2012-04-13