Built on Facts

Sunday Function

Posted by Matt Springer on April 28, 2009
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Yes, it’s Tuesday. Busy weekend, including a Relay for Life (you should find your local one and donate!). Today we’ll get back on schedule a bit and make this one a rather more utilitarian than usual Sunday Function, playing into the last step we’ll need in order to complete the Bose-Einstein condensation discussion.

In the course of our journey we will meet this somewhat alarming creature:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

It’s a function that takes two parameters: a more traditional variable z, and a number (lowercase Greek letter nu) denoting the particular subspecies of function g. The capital Greek letter gamma is itself a function.

We’re only going to have to deal with this for nu = 3/2, so we may as well plot the stuff under the integral sign. Taking z = 1/2 just for the random heck of it:

i-7b7142f1efa09b47c3b8242f89bcb9f4-graph.png

As z changes that curve will change shape and correspondingly the area under the curve will change as well. Is there some z that will maximize that area? Sometimes you can save yourself trouble if you just plot the graph, and you can save the rigorous proof work for later. We’ll do that. This is the actual graph of g(z):

i-3fdee7de4197c08986e3407853fa09fa-graph2.png

It attains a maximum at z = 1, and we shall see that in our case the physical meaning of z requires that it not be able to go above 1 anyway.

Mystery upon mystery at the moment, I know. Unless you’ve done this before, of course. But we’ll put it all together shortly and I think it’ll be worth the wait.

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Comments

  1. #1 Eric Lund
    April 28, 2009

    The apparent reason for the restriction on z is obvious: for z > 1 you have a simple pole in that integrand. Does the function have an analytic continuation, or is it completely undefined?

    The z = 1 case also has a restriction to ν > 1, in order to make the singularity integrable. Without seeing the physical setting in which this function arises, I can’t say what that physically means, but you say that we will get the specific version of this function with ν = 3/2.

  2. #2 Just being dumb here
    April 28, 2009

    I plugged in z= 1/2 and nu = 3/2 and did the integral and found g(1/2) = .624. After doing the integral the dummy variable x disappears. SO you write that you are using z = 1/2 to plot this but isn’t z the variable here?
    For z = 1/2, g(z) = a constant.
    What exactly are you plotting g vs x? or g vs z?
    How command do i use to plot this mathematica?

  3. #3 just being dumb here
    April 28, 2009

    Just to make it clear. I understand how you get the second graph where g->1 as z->infinity.
    But I just don’t understand the first plot.
    The integral of g(z) when nu = 3/2 is
    PolyLog[3/2,z] and plotting this from 0 to 5 gives the second graph. But what about the first? when z = 1/2 g(z) should be a constant and you get a lognormal-type graph.

  4. #4 Matt Springer
    April 28, 2009

    We integrate a function of x and evaluate that function with the appropriate z values to get g(z). So what I’ve plotted in the first graph is just the function we’re integrating to get g(z). This, in other words:

    1/Gamma[3/2] x^((3/2) – 1)/((1/2)^-1 (Exp[x] – 1))

  5. #5 Uncle Al
    April 28, 2009

    You have rediscovered social engineering! The first 40% is interpolated linear, the next 30% is interpolated linear with a slightly larger slope. The last 30% extrapolated, legislated, and enforced by jackbooted State compassion is observed as disaster. Economics wll espy your exposition and weep tears of self-realization, as will Baby Boomers expecting Social Security and Medicare.

    Like infinity, centrally administered social engineering gets real exciting toward the end.

  6. #6 Anonymous
    April 28, 2009

    g(z) looks a lot like the riemann zeta function (in one of its forms). Is that just a coincidence, or is there something more significant in that it looks like zeta(nu) (with a 1/z in the denominator of the integrand)? And is there a name for g(z)?

  7. #7 Kobra
    April 29, 2009

    By any chance does g(1) = e?

  8. #8 Anonymous
    May 1, 2009

    I think g(1)=zeta( nu ), where zeta is the Riemann zeta function (http://en.wikipedia.org/wiki/Riemann_zeta_function#Mellin_transform)

    so if nu=3/2=1.5, g(1)=zeta(1.5)=2.61237534868549… it’s pretty close to e though