Find the function f(x) satisfy: x=0,-> f(x) =0, x<>0, f(x)=1

[Viet Tu]

My problem is simple, but I couldn't solve it:
This is: Find a function f(x) of a variable x, x is in real field. the F(x) satisfy:
When x = 0, f(x) = 0
When x<>0, f(x) =1.

Anyone can help me?

Thank you.

 

[haruspex]

My problem is simple, but I couldn't solve it:
This is: Find a function f(x) of a variable x, x is in real field. the F(x) satisfy:
When x = 0, f(x) = 0
When x<>0, f(x) =1.

Anyone can help me?

Thank you.

That appears to be a complete definition of the function. What more is there to find?
Are you really asking whether it can be expressed in terms of standard functions?
Is this homework?

 

[Viet Tu]

Thanks for reply.
Yes, it exactly is express form of function. It's not homework but created by myself when I solve a problem in my job.

 

[jedishrfu]

Perhaps you could describe the problem that came up at work.

My guess is you're trying to represent your function definition using common math functions in software where instead you could simply define a function to return a zero with x/=0 and a 1 when x=0.

This could be related to a Dirac Delta function (which really isn't a function):

https://en.wikipedia.org/wiki/Dirac_delta_function

where you could write it as something like this: $f(x) = 1 - \delta(x)$

but of course this wouldn't work because $\delta(x)$ is infinite at x=0

 

[Mark44]

Perhaps you could describe the problem that came up at work.

Yes, we would like to see the scenario that this function describes.

My guess is you're trying to represent your function definition using common math functions in software where instead you could simply define a function to return a zero with x/=0 and a 1 when x=0.

This could be related to a Dirac Delta function (which really isn't a function):

https://en.wikipedia.org/wiki/Dirac_delta_function

where you could write it as something like this: $f(x) = 1 - \delta(x)$

but of course this wouldn't work because $\delta(x)$ is infinite at x=0

Alternatively, the function could be defined piecewise.
$f(x) = \begin{cases} 0 & \text{if }x = 0 \\ 1 & \text{if }x \ne 0 \end{cases}$

If this function is the result of a spreadsheet calculation, it's easy to build logic into a spreadsheet cell for this.

 

[haruspex]

Thanks for reply.
Yes, it exactly is express form of function. It's not homework but created by myself when I solve a problem in my job.

Express it in what context? Are you writing a computer program? A spreadsheet formula? Or just a technical paper?

 

[Viet Tu]

Express it in what context? Are you writing a computer program? A spreadsheet formula? Or just a technical paper?

Express as polynomial form. This is in spreadsheet formula. I have alternative solution, using "if" command. However, I'm trying to find a solution in math.

 

[haruspex]

Express as polynomial form. This is in spreadsheet formula. I have alternative solution, using "if" command. However, I'm trying to find a solution in math.

No chance of polynomial form, or any "school" algebra. But don't think of math as limited to that. Math allows functions to be defined as you wrote it in post #1, and the appropriate way to express it in a spreadsheet is with IF.

 

[Viet Tu]

No chance of polynomial form, or any "school" algebra. But don't think of math as limited to that. Math allows functions to be defined as you wrote it in post #1, and the appropriate way to express it in a spreadsheet is with IF.

Thanks for your response.

 

[WWGD]

How about $1- \chi_0$ , where $\chi_0$ is the characteristic function of 0 , i.e. $\chi_a$ is 1 at a and 0 elsewhere ? EDIT : In my experience, most of the time, it is $\chi_A$ , where $A$ is a set, but then you can just use $A=$ {$a$}

 

[Viet Tu]

How about $1- \chi_0$ , where $\chi_0$ is the characteristic function of 0 , i.e. $\chi_a$ is 1 at a and 0 elsewhere ? EDIT : In my experience, most of the time, it is $\chi_A$ , where $A$ is a set, but then you can just use $A=$ {$a$}

I think it is a good solution. Nonetheless, it's too complicated for my case. Thank you so much.