a study of Q1(z) using Bateman Q (40)
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Short working note by Phil dated 2.23.10, using Maple code from batemanpq.mws. He evaluates the two terms of Bateman's Q (40) separately and then sums them, using Bateman p 123 (12) to handle the (1-z) phase and recover (z/2) ln[(z+1)/(z-1)] - 1. Comments note the required order of Maple convert commands and that each term separately solves the Legendre ODE.
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A study of Q1(z) using Bateman Q (40) PhL 2.23.10
It is interesting and useful to see how the various Bateman forms reproduce known simple functions such as Q1(z).
The Maple code is located in batemanpq.mws. The first order of business is to enter the expressions. I had Q (43) in this file and edited it to become Q (40).
Notice that the (z2-1) powers are written out the right way (otherwise we get in trouble!). The g1 and g2 factors appear in various places so I have them separated. I include the usual extra e+iπm phase of the Bateman Q function only in the last summing expression. I am going to evaluate the two terms separately here, and then the sum. Here is the first term,
which I could write as (z/2) ln[(z+1)/(1-z)] - 1 . This is almost the Q1 function, except for the (1-z) in the denominator. Now let's get the second term:
which I could write as (∓i)πz/2 . So our total result is this:
(z/2) ln[(z+1)/(1-z)] - 1 (∓i)πz/2
Now we use Bateman p 123 (12) to write
(1-z) = e∓iπ(z-1)
Then we resolve the result like this:
(z/2) ln[(z+1)/{ e∓iπ(z-1)}] - 1 (∓i)πz/2
= (z/2) ln[(z+1)/{(z-1)}] + (±iπ)(z/2) - 1 (∓i)πz/2
= (z/2) ln[(z+1)/(z-1)] - 1
which at last is the correct result.
Comment #1: Notice how you have to use the two "convert" commands exactly in the sequence I show above. We get results first in terms of a tanh-1 function, then we convert that to log.
Comment #2: The "log part" is contained entirely in the first term.
Comment #3: We know that the function f(z) = z solves the ODE. So here we find that each of the two terms separately solves the Legendre ODE. This is something I have wondered about. I think in general that the "two terms" are how you add two independent solutions of the ODE to construct the official Q function, so I would speculate that it is a general fact that each term is separately a solution in all those Bateman forms.
// I show this in a general way for Q (40) in my recent doc "A Maple study of Bateman Q (40).doc" . Maple cannot do it for general arguments n,m,z but if you pick some random numbers, you can show it is true.