All, I'm trying to simplifiy the following equation for the zero'th order modified Bessel function of the first kind. I'm using the approximation for the Bessel function as Io(z) = exp(z)/sqrt(2*pi*z) I would like to have the natural logarithm of this function log(Io(z)) = log(exp(z) / sqrt(2*pi*z)) = z + log(1/sqrt(2*pi*z)) but i don't want to have to calculate the log of a sqrt function. What I'd like is a simplification that gives me an easy way to calc the log(Io(z)) for each value of z. Any ideas? CW

# Bessel function simplification problem

Started by ●March 4, 2005

Reply by ●March 4, 20052005-03-04

Hello CW, If Io(z) approx exp(z)/sqrt(2*pi*z) then log(Io(z)) approx z-0.5*log(2*pi*z) = z-log(z)/2 -A where constant A = log(2pi)/2 IHTH, Clay

Reply by ●March 4, 20052005-03-04

"CW" <prada_white@yahoo.ca> wrote in message news:8c58d246.0503040759.2d9d6f60@posting.google.com...> All, > > I'm trying to simplifiy the following equation for the zero'th order > modified Bessel function of the first kind. I'm using the > approximation for the Bessel function as > > Io(z) = exp(z)/sqrt(2*pi*z) > > I would like to have the natural logarithm of this function > > log(Io(z)) = log(exp(z) / sqrt(2*pi*z)) > > = z + log(1/sqrt(2*pi*z)) > > but i don't want to have to calculate the log of a sqrt function. > What I'd like is a simplification that gives me an easy way to calc > the log(Io(z)) for each value of z. Any ideas? > > CWHi CW loq(sqrt(x)) = log(x)/2 (I think) so you are trying to simplify z-log(z)/2-log(2*pi). good luck - Mike

Reply by ●March 4, 20052005-03-04

Sorry, i should have added that z is not bounded -1 < z < 1 so I'm not sure that the Mclaurin series will help me. what i can say about z is that it will be a real non-negative number. Does that shed any more light? CW

Reply by ●March 4, 20052005-03-04

"CW" <prada_white@yahoo.ca> wrote in message news:8c58d246.0503040759.2d9d6f60@posting.google.com...> All, > > I'm trying to simplifiy the following equation for the zero'th order > modified Bessel function of the first kind. I'm using the > approximation for the Bessel function as > > Io(z) = exp(z)/sqrt(2*pi*z) > > I would like to have the natural logarithm of this function > > log(Io(z)) = log(exp(z) / sqrt(2*pi*z)) > > = z + log(1/sqrt(2*pi*z)) > > but i don't want to have to calculate the log of a sqrt function. > What I'd like is a simplification that gives me an easy way to calc > the log(Io(z)) for each value of z. Any ideas? > > CWHi CW again : I should have said z-log(z)/2 -log(2*pi)/2 sorry. Mike

Reply by ●March 4, 20052005-03-04

CW wrote:> All, > > I'm trying to simplifiy the following equation for the zero'th order > modified Bessel function of the first kind. I'm using the > approximation for the Bessel function as > > Io(z) = exp(z)/sqrt(2*pi*z) > > I would like to have the natural logarithm of this function > > log(Io(z)) = log(exp(z) / sqrt(2*pi*z)) > > = z + log(1/sqrt(2*pi*z)) > > but i don't want to have to calculate the log of a sqrt function. > What I'd like is a simplification that gives me an easy way to calc > the log(Io(z)) for each value of z. Any ideas? > > CWMy first idea is that you've forgotten some basic concepts about logarithms. log(a*b) = log(a) + log(b). log(a^x) = x * log(a). 1/sqrt(2*pi*z) = (2*pi*z)^(-1/2) therefore log(1/sqrt(2*pi*z)) = -(log(2*pi) + log(z))/2. You still have to calculate the log, but you now don't have to calculate the square root as well. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com

Reply by ●March 4, 20052005-03-04

Reply by ●March 4, 20052005-03-04

Thanks for all the quick responses Clay, Mike and Tim. So I guess i'm down to calculating z-log(z)/2 upon each iteration, as the other term can be precomputed. So is the calculation of a log term computationally expensive? That was my main motivation for trying to remove it. CW

Reply by ●March 4, 20052005-03-04

"CW" <prada_white@yahoo.ca> wrote in message news:1109954279.983132.154090@g14g2000cwa.googlegroups.com...> Thanks for all the quick responses Clay, Mike and Tim. > > So I guess i'm down to calculating z-log(z)/2 upon each iteration, as > the other term can be precomputed. So is the calculation of a log term > computationally expensive? That was my main motivation for trying to > remove it. > > CW >Hi CW - in general yes. There are a whole load of tricks to simplify it though most of which seem to have been discussed in this newsgroup so you ought to find something to suit you in the archives if you are using fixed point math. Also I believe that Ray Andraka published a trick on www.dpsguru.com if you are using floating point. best of luck - mike

Reply by ●March 4, 20052005-03-04

CW wrote: