Function that equals 1 at x=0, but 0 everywhere else?

[Kepler_]

Is there a function that equals 1 at x=0 and equals 0 when x isn't 0 without using a piecewise function? I've been experimenting with limits and derivatives but haven't made much progress.

The closest thing I was able to think of is 1-x/x, which is indeterminate at x=0, and 0 everywhere else.

Thanks!

 

[jfizzix]

How about the limit as a approaches \infty of f(x) = e^{-a x^{2}}?

 

[Kepler_]

That'll do it :P
It seems like any real constant greater than 1 works in place of e. Is there a reason why you chose e?

 

[DrewD]

That'll do it :P
It seems like any non-zero real constant works in place of e. Is there a reason why you chose e?

Out of infinitely many possible bases to choose from, $e$ is the only one that makes sense most of the time. In this case it doesn't matter, but why would you choose something else? If you want, you could rewrite this as
$\lim_{a\rightarrow\infty}\left(\frac{1}{e^a}\right)^{x^2}$in which case it is the same as
$\lim_{b\rightarrow0}\left( b\right)^{x^2}$
What matters is the $x^2$ exponent and the continuous, decreasing function as a base. Most mathematicians are used to $e$ being part of that base.

Also, this is sometimes thought of as a limit of normal distributions. These are defined using $e$.

 

[Number Nine]

without using a piecewise function?

Is there a reason for this requirement? "Piecewise" is just a notation for describing a function; it isn't actually a "kind" of function. Even your description of your function is piecewise: 1 at zero, and 0 everywhere else.

 

[PeroK]

Two functions are equal if they take the same value at every point. So, regardless of whether you describe a function piecewise or as the limit of a sequence of functions, it's the same function in the end.

 

[johnqwertyful]

I think this demonstrates a fundamental misunderstanding of functions. Functions are a rule associating two sets. That's it. Really, "piecewise function" doesn't really make sense. It's just a way of writing things.

 

[DrewD]

I think this demonstrates a fundamental misunderstanding of functions. Functions are a rule associating two sets. That's it. Really, "piecewise function" doesn't really make sense. It's just a way of writing things.

But sometimes piecewise makes calculus hard. I think this is a good point to make though.

 

[johnqwertyful]

But sometimes piecewise makes calculus hard. I think this is a good point to make though.

But "Piecewise" is only a way of describing a function. There is no such thing as a "piecewise function". f(x)=x if x>0, -x if x≤0. Or f(x)=|x|. Both are the same function. One is written "piecewise" the other isn't.

 

[DrewD]

But "Piecewise" is only a way of describing a function. There is no such thing as a "piecewise function". f(x)=x if x>0, -x if x≤0. Or f(x)=|x|. Both are the same function. One is written "piecewise" the other isn't.

Agreed.

 

[paisiello2]

Here's another function that would satisfy the requirement but no limits involved:

f(x) = 0^x

 

[micromass]

Here's another function that would satisfy the requirement but no limits involved:

f(x) = 0^x

How would that be defined for negative real numbers?

 

[paisiello2]

Ooops, you're right, I forgot to modify it:

f(x) = 0^(x^2)

 

[DrewD]

Everytime I have encountered it, $0^0$ has been considered an indeterminate form and not 1.

 

[micromass]

Everytime I have encountered it, $0^0$ has been considered an indeterminate form and not 1.

I agree, but let's not open that can of worms here. Paisiello2 clearly meant $0^0=1$, so let's keep it at that, even if it's nonstandard.

mod note: not a $0^0$ argument please. All posts on this will be deleted.[/color]