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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 $x<k$ 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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    Your first and second paragraphs each define a function; in fact, exactly the same function. Therefore, what you are asking seems to be, is there a function $f$ which is not the same as the same function $f$.2012-04-13
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    Your question makes no sense, because the function is defined by **what it does**, not by how you *express* what it does.2012-04-13
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    The answer is: Yes, it is a step function and is indeed very useful in math and electrical engineering. See for example the Heaveside Step Function.2012-04-13
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    Rahul Narain is right: you specify the function in your first paragraph, and then ask for a different function with the same values. Strictly speaking, that makes no sense. Maybe what you mean is: "Is there a way to _describe_ this function without defining it piecewise?" nubis gives an answer to this question, by describing it as the limit of smooth functions2012-04-13
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    Engineers might have a bone to pick with you if you claim the Heaviside step function is not useful...2012-04-13
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    @RahulNarain, the truth is the latex that I tried to express the piecewise function in the second paragraph didn't work, so I had removed it.2012-04-13
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    @ZevChonoles, well, this question really have sense to me. Several times I need something like this. But, is a little hard define this function precisely but I think the answer below will help me and I will take a look.2012-04-13
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    @you, you're right, I was searching a way to describe it withou the piecewise. The truth, describe in a useful way.2012-04-13
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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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    Strictly, there is exactly *one* such function :)2012-04-13
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    @MarianoSuárez-Alvarez Would you be interested in reading a ""paper"" in Spanish I want to present to a professor in the UBA? (it is not long at all, and I'm guessing you'll find it interesting).2012-04-13
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    @MarianoSuárez-Alvarez - There are more than one, as you can see in the Wikipedia link. The logistic function is but one.2012-04-13
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    What you mean is that there are *many formulas which give that function* but the function is *one*.2012-04-13
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    @MarianoSuárez-Alvarez - There are many functions whose limits converge point-wise to the Heavyside step function.2012-04-13
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    For example: let $H$ be the Heavyside step function, take $g:\mathbb R\to\mathbb R$ to be any function *whatsoever* and consider the sequence $f_n=H+\frac1n g$, $n\geq1$. Then $f_n\to H$ pointwise.2012-04-13
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    @PeterT.off, send me an email.2012-04-13
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    @Mariano Where to?2012-04-13
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    Google will find my homepage, which has my address :)2012-04-13
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    @MarianoSuárez-Alvarez I have sent it. Let me know!2012-04-13
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    I think is this what I'm looking for $H(x+k)$ holds. Thx.2012-04-13