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

1. Mar 10, 2014

### 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!

2. Mar 10, 2014

### jfizzix

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

3. Mar 10, 2014

### 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?

Last edited: Mar 10, 2014
4. Mar 10, 2014

### DrewD

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 $e$.

5. Mar 11, 2014

### Number Nine

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.

6. Mar 11, 2014

### 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.

7. Mar 11, 2014

### 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.

8. Mar 11, 2014

### DrewD

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

9. Mar 12, 2014

### johnqwertyful

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.

10. Mar 12, 2014

### DrewD

Agreed.

11. Mar 12, 2014

### paisiello2

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

f(x) = 0^x

12. Mar 12, 2014

### micromass

Staff Emeritus
How would that be defined for negative real numbers?

13. Mar 12, 2014

### paisiello2

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

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

14. Mar 12, 2014

### DrewD

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

15. Mar 12, 2014

### micromass

Staff Emeritus
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.

Last edited: Mar 13, 2014