# Thread: compact support gaussian function

1. ## compact support gaussian function

Hi, I am not mathematician. (I am computer science)
I need your help to construct a compact support gaussian function

It is ok for an interval [a, b]?
[tex]f(x) = \exp(-0.5\frac{(x-\mu)^2}{\sigma^2})[x/tex]
$\displaystyle f(x) = \exp(-0.5\frac{(x-\mu)^2}{\sigma^2})$
if x \in (a, b)
and
$\displaystyle f(x) = 0$ otherwise

And how I can to prove this

2. ## Re: compact support gaussian function

Originally Posted by jorjasso
And how I can to prove this
Prove what?

3. ## Re: compact support gaussian function

is it ok that definition for compact support gaussian function?

4. ## Re: compact support gaussian function

I am not familiar with "compact support gaussian function" considered as one term. I found only one occurrence of this phrase on the web. Strictly speaking, this is not a Gaussian function because the latter has infinite support. You can give any definition you like; the question is in the properties of the defined object. Do you need to guarantee any particular properties of this function?

5. ## Re: compact support gaussian function

Yes Emakarov, I want to use the gaussian function as membership function of fuzzy numbers.
An definition of fuzzy numbers from [1] is that
fuzzy numbers are functions [tex] f:R\to [0,1] [\tex]
that satisfy four conditions, one condition is that the support has to be closed and bounded (compact).

[1] Salih Aytar, Serpil Pehlivan, Musa A. Mammadov, The core of a sequence of fuzzy numbers, Fuzzy Sets and Systems, Volume 159, Issue 24, 16 December 2008, Pages 3369-3379, ISSN 0165-0114

6. ## Re: compact support gaussian function

Depending on the definition of support, you may need to change the condition x ∈ (a, b) to x ∈ [a, b] because (a, b) is not closed.

Originally Posted by jorjasso
An definition of fuzzy numbers from [1] is that
fuzzy numbers are functions [tex] f:R\to [0,1] [\tex]
that satisfy four conditions, one condition is that the support has to be closed and bounded (compact).
Then your function definitely satisfies this condition. Note, however, that if $\displaystyle f(x)=e^{- { (x-\mu)^2 \over 2 \sigma^2 } }$, then $\displaystyle \int_{-\infty}^{\infty}f(x)\,dx=\sqrt{2\pi}\sigma$, but $\displaystyle \int_{a}^{b}f(x)\,dx<\sqrt{2\pi}\sigma$, so $\displaystyle f(x)/(\sqrt{2\pi}\sigma)$ may serve as a probability density on $\displaystyle \mathbb{R}$, but not on [a, b].

Thanks,