Roots of the normal distribution

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SUMMARY

The discussion centers on the roots of the normal distribution function, defined as $$f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{\frac{-(x-\mu)^2}{2 \sigma ^{2}}}$$. Participants clarify that the function does not have any roots since it never reaches zero; instead, it approaches zero as x approaches positive or negative infinity. The correct interpretation is that the roots are non-existent, and the limit of the function as x approaches infinity is zero.

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Homework Statement


$$f:\mathbb{R} \rightarrow \mathbb{R},$$

$$ f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{\frac{-(x-\mu)^2}{2 \sigma ^{2}}}$$

What are the roots of this equation?

Homework Equations

The Attempt at a Solution



The roots of an equation are the values of x such that f(x) = 0. This is the first time I have seen a question like this and am still getting my head around the normal distribution, but as far as I'm aware the curve never does reach f(x) = 0 so I want to express the idea that the roots of this equation are +/- \infty but I don't know how to do this...

lim_{x \rightarrow +/- \infty} f(x) = 0

I'd appreciate some guidance,

thanks :)
 
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It doesn't look to me like you need much guidance on this. You have it exactly correct. But I wouldn't say the roots are ##\pm \infty##. Just say it has no roots but the limit is 0 as you have stated.
 
LCKurtz said:
It doesn't look to me like you need much guidance on this. You have it exactly correct. But I wouldn't say the roots are ##\pm \infty##. Just say it has no roots but the limit is 0 as you have stated.

Well that is good news, thanks!
 

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