Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following Gaussian function:

$$\rho(r) = q_i (\alpha/\pi)^{3/2} \exp(-\alpha r^2)$$

Qualitatively, the "width" of this Gaussian is related to $\frac{1}{\alpha}$: the larger the value of $\alpha$, the smaller the "width" of the Gaussian.

This Wikipedia article uses this definition of a Gaussian function:

$$f(x) = a\exp \left(-\frac{(x-b)^2}{2c^2}\right)$$

and says that one way to define the width is to consider the full width at half maximum:

$$\text{FWHM} = 2 \sqrt{2 \ln 2} c \approx 2.35482c$$

In other words, the width of $f(x)$ is proportional to $c$. But, in my function $\rho(r)$, $\alpha$ appears in two places: in the exponential and as a coefficient of the exponential. How should I define the width of $\rho(r)$?

share|improve this question
1  
A multiplicative factor provides a vertical stretch, that do not change the FWHM value. –  enzotib Aug 11 '12 at 20:00
1  
In your example the width of the Gaussian should be related to $\alpha^{-1/2}$ not to $\alpha^{-1}$. –  Fabian Aug 11 '12 at 21:25
add comment

2 Answers 2

up vote 5 down vote accepted
+50

The proportionality constant in a Gaussian function does not affect its width. Consider the general form of a Gaussian function (may not be a probability density function):

$f(x) = a \exp\Big(-\frac{(x-b)^2}{c^2}\Big)$

The maxima occurs at $x=b$, and its value is $a$. Hence, the value of $f(x)$ at half maximum is $0.5a$. You can now find the value of $x$ at which $f(x) = 0.5a$:

$f(x^*) = a\exp\Big(-\frac{(x^*-b)^2}{c^2}\Big) = 0.5a$

The proportionality constant $a$ cancels out from both sides, and we end up with $(x^*-b) = \pm c\sqrt{\ln 2}$. The $FWHM$ thus becomes $2c\sqrt{\ln 2}$. Even if you were to define width using any other definition, say full width at $\alpha$ maximum (where $\alpha < 1$), the proportionality constant would always cancel out. Further, the center ($b$) would not matter as well since the Gaussian function is symmetric.

share|improve this answer
add comment

In the one-dimensional case, the width of $f_c:x\mapsto(2\pi c^2)^ {-1/2}\exp(-(x-b)^2/(2c^2))$ is defined as follows.

One notes that the measure $f_c(x)\mathrm dx$ coincides with $g(z)\mathrm dz$ for some function $g$ independent of $c$, with $z=(x-b)/c$, precisely for the standard gaussian density $g:z\mapsto(2\pi)^ {-1/2}\exp(-z^2/2)$. Since $x=cz+b$, the argument of $f_c$ scales like $c$ hence the width of $f_c$ is proportional to $c$ (the exact coefficient of proportionality, whether one chooses $2\sqrt{2\ln2}$ or another value, being essentially irrelevant).

Likewise, in the three-dimensional case, $\rho_\alpha(r)=C\alpha^{3/2}\exp(-\alpha r^2)$ hence the radial part of the volume measure is $\rho_\alpha(r)r^2\mathrm dr=\sigma(s)\mathrm ds$ for some function $\sigma$ independent of $\alpha$, with $s=\alpha^{1/2}r$, precisely for $\sigma(s)=Cs^2\exp(-s^2)$. Since $r=\alpha^{-1/2}s$, the argument of $\rho_\alpha$ scales like $\alpha^{-1/2}$ hence the width of $\rho_\alpha$ is proportional to $\alpha^{-1/2}$ (the exact coefficient of proportionality being irrelevant).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.