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Area under Curve for Unbounded Guassian & Laurentzian Distributions

Hi - I believe (maybe mistakenly) that there are mathematical formulas for obtaining (unbounded) area under curves for Guassian & Laurentzian distributions that do NOT involve integration.  I cannot seem to find those formulas.  Does anyone have them or point me to a link that might.

Thank you in advance.

Sincerely,

Don
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Hey Don,

There are a few VIs that uses integration to find the area under the curve. As for your case, it seems you will need to write your own algorithm to do this.

Here is an example for Calculating the Area with Respect to the Y-Axis on an XY Graph in LabVIEW. This should give you an idea as to what you'll need to do.

Hopefully this helps!!

Aashish M
CEO
TransferFi
www.transferfi.com
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No, I am looking for something that does not involve summations or integrations as stated in my original message.  I went thru a few mathematical handbooks and could not find what I was looking for.  I may be mistaken that this is actually possible to do without integration, but I thought that there might be an area calculation just using amplitude, center, and full-with half-max that characterizes the guassian curve.
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If you are searching for a closed form solution for the Gaussian there isn't a straight forward solution only approximations.  The integral can be broken down into an incomplete gamma function or a maclaurin series, so you would just get an approximation at best.  If I am not mistaken that is a very sought after problem, the closed form solution to the Normal Distr. integral, but I could be mistaken.
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I would be interested to see any further information you know about this. 

Thanks,

Don

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I got this from Wolfram MathWorld real fast, I wish I had some of my old notes from class but here it is...  You are still going to be doing some summation.
 
Phi(z) = 1/(sqrt(2pi))int_0^ze^(-x^2/2)dx
(3)
= 1/2erf(z/(sqrt(2))),
(4)

where erf is the error function.  Notice this is the positive reals.

This is the Maclaurin Series breakdown of the funciton

erf(z) = 2/(sqrt(pi))sum_(n=0)^(infty)((-1)^nz^(2n+1))/(n!(2n+1))
(6)
= 2/(sqrt(pi))(z-1/3z^3+1/(10)z^5-1/(42)z^7+1/(216)z^9+...)

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Incomplete Gamma Function

 erf(z)=pi^(-1/2)gamma(1/2,z^2),

where the gamma(.5,z^2) is the incomplete gamma function;

The "Upper Incomplete Gamma Function"

 Gamma(a,x)=int_x^inftyt^(a-1)e^(-t)dt.
(1)

For a an integer n

Gamma(n,x) = (n-1)!e^(-x)sum_(k=0)^(n-1)(x^k)/(k!)
(2)
= (n-1)!e^(-x)e_(n-1)(x),
(3)

 Gamma(a,x)=int_x^inftyt^(a-1)e^(-t)dt.
(1)

For a an integer n

Gamma(n,x) = (n-1)!e^(-x)sum_(k=0)^(n-1)(x^k)/(k!)
(2)
= (n-1)!e^(-x)e_(n-1)(x),
(3)

 

For other

The cumulative distribution function, which gives the probability that a variate will assume a value <=x, is then the integral of the normal distribution,

D(x) = int_(-infty)^xP(x^')dx^'
(8)
= 1/(sigmasqrt(2pi))int_(-infty)^xe^(-(x^'-mu)^2/(2sigma^2))dx^'
(9)
= 1/2[1+erf((x-mu)/(sigmasqrt(2)))],
(10)

It is a bit "thrown together", I hope this helps some...

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This IS starting to look like what I saw in the Math Handbook I looked at. Thanks for taking the time.  Not sure if it will get me to where I need but let me study it further....Don


Message Edited by DonRoth on 05-21-2008 01:36 PM
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Don,

I came across an article in IEEE Instrumentation and Measurement Magazine, December 2007, by Stephen Dyer and Justin Dyer, titled "Approximations to Error Functions." They discuss three approximations, particularly looking at the magnitudes of the errors and the computational load. They suggest that the approximations require much less calculation effort.

Lynn
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Hello again and sorry for the delays. I spoke with a physicist here and he has stated that one can estimate the guassian and laurentzian areas using the following formulas.  I have not validated this by comparing against integration or other approximation methods (yet).  He does a lot of work with fitting Raman Spectra.

Laurentzian: peak area = (pi/2) * fwhm * peak amplitude

Guassian: [sqrt(pi)/(2*sqrt(ln2))] * fwhm * peak amplitude


fwhm = full width half max

Any comments?  I am tied up with a few things and may not get to validate immediately.  I asked him for a reference for this but did not get one.


Sincerely,

Don

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