Attempting the NEW oblate Q integral

Attempting the NEW oblate Q integral PhL 2.2.10 As usual, consideration of a relatively simple integral has led to massive misunderstandings of basic things at the kindergarten level. 1. "Today's problem" 1 2. Stupid Paradox. 1 3. Lars Kindergarten Revisited. 1 4. A Closer Look at Eigenfunctions 2 4. Back to our integral. 6 1. "Today's problem" is evaluation of this integral, where m is a negative integer: ( umν, umν') = !Syntax Error, Idζ Qνm(iζ)* Qν'm(iζ) = Kνm δν,ν' where ν = |m| – N ν' = |m| – N' N and N' are positive, odd, and ≠ 0. This integral is purportedly an orthogonality integral for a certain ODE system described in "when are ODE solutions orthogonal.doc". 4. A Closer Look at Eigenfunctions Let's go back to this again. Recall that our spectrum is ν = |m|- N with N = 1,3,5.... but ν > -1/2. Therefore, 1+ν-|m| = 1 + |m|- N - |m| = 1-N = 0, -2, -4 and so on. Now look at our Q form given page 133 (36). The "a" argument of F(a,b,c,z) is then always a negative integer, which means the series truncates. We have: Qνm(iζ) = 2ν [ Γ(1+ν) Γ(1+ν+m)/ Γ(2+2ν) ] (iζ +1)m/2-ν-1 (iζ -1)-m/2 F(1+ν-m, 1+ν, 2+2ν; 2/(1+ iζ) ) Assume for the moment that N = 1 so we have 1+ν-|m| = 0. Then ( μ = m, either symbol same) F(1+ν-μ, 1+ν, 2+2ν; 2/(1+ iζ) ) = F(0,1+ν,2+2ν; 2/(1+ iζ) ) But we know that in this case, F = 1 + abz/c + ... = 1 so then (m = μ) Qνm(iζ) = 2ν [ Γ(1+ν) Γ(1+ν+m)/ Γ(2+2ν) ] (iζ +1)m/2-ν-1 (iζ -1)-m/2 [1 - (1+ iζ)-1 ] Q|m|-1m(z) = 2ν [ Γ(1+ν) Γ(1+ν+μ)/ Γ(2+2ν) ] (z +1)μ/2-ν-1 (z -1)-μ/2 [1 ] Let's try this for some simple case, say m = -2 : Q1-2(z) = 21 [ Γ(1+1) Γ(1+1-2)/ Γ(2+2*1) ] (z +1)-2/2-1-1 (z -1)-(-2)/2 [1 ] = 21 [ Γ(2) Γ(0)/ Γ(4) ] (z +1)-3 (z -1)1 [1 ] STOP! Why are we getting a pole here? and indeed there is no log. My formula above gives Qνμ(z) = 2ν [ Γ(1+ν) Γ(1+ν+μ)/ Γ(2+2ν) ] (z +1)μ/2-ν-1 (z -1)-μ/2 [1 - (1+ z)-1 ] Q02(z) = 20 [ Γ(1+0) Γ(1+0+2)/ Γ(2+2*0) ] (z +1)2/2-0-1 (z -1)-2/2 [1 - (1+ z)-1 ] = [ 2! /1 ] (z -1)-1 [1 - (1+ z)-1 ] = 2(z -1)-1 [1 - 1/(1+ z) ] = 2(z -1)-1 [ 1+z - 1]/(z+1) = 2(z -1)-1 z/(z+1) = 2z/ [(z+1)(z-1)] // agrees! I would like to see a table of such things. Maxima cannot do this kind of parameter combination (but it can differentiate one it can so) ! I can make Maple do it like this: so I could make my own table. Multiple derivatives: g := diff(f,x$n) n can be a variable. So, back to our general formula for our "eigenfunctions" ! Qνμ(z) = 2ν [ Γ(1+ν) Γ(1+ν+μ)/ Γ(2+2ν) ] (z +1)μ/2-ν-1 (z -1)-μ/2 F(1+ν-μ, 1+ν, 2+2ν; 2/(1+ z) ) Leave μ, but set in ν = μ-N, N = 2,4... > -1/2 Qμ-Nμ(z) = 2μ-N [ Γ(1+ μ-N) Γ(1+ μ-N +μ)/ Γ(2+2[μ-N]) ] (z +1)μ/2-[μ-N]-1 (z -1)-μ/2 F(1+(μ-N)-μ, 1+μ-N, 2+2(μ-N); 2/(1+ z) ) = 2μ-N [ Γ(1+μ-N) Γ(1+ 2μ-N)/ Γ(2+2μ-2N]) ] (z +1)-μ/2+N-1 (z -1)-μ/2 F(1-N, 1+μ-N, 2+2μ-2N); 2/(1+ z) ) Let's try this differently using μ = ν + N Qνμ(z) = 2ν [ Γ(1+ν) Γ(1+ν+μ)/ Γ(2+2ν) ] (z +1)μ/2-ν-1 (z -1)-μ/2 F(1+ν-μ, 1+ν, 2+2ν; 2/(1+ z) ) Qνν+N(z) = 2ν [ Γ(1+ν) Γ(1+ν+ν+N)/ Γ(2+2ν) ] (z +1)(ν+N)/2-ν-1 (z -1)-(ν+N)/2 F(1-N, 1+ν, 2+2ν; 2/(1+ z) ) Qνν+N(z) = 2ν [ Γ(1+ν) Γ(1+2ν+N)/ Γ(2+2ν) ] (z +1)(N-ν)/2-1 (z -1)-(N+ν)/2 F(1-N, 1+ν, 2+2ν; 2/(1+ z) ) Well, μ is really pre-determined by the φ problem in a separation scenario, so better the other way: Qμ-Nμ(z) == 2μ-N [ Γ(1+μ-N) Γ(1+ 2μ-N)/ Γ(2+2μ-2N) ] (z +1)-μ/2+N-1 (z -1)-μ/2 F(1-N, 1+μ-N, 2+2μ-2N); 2/(1+z) ) Notice that Γ(2+2μ-2N) / Γ(1+μ-N) = Γ(2x)/Γ(x) x = 1+μ-N = (1/) 22x-1Γ(x+1/2) = (1/) 22(1+μ-N)-1Γ(3/2+μ-N) = (1/) 22(μ-N)+1Γ(3/2+μ-N) so we could replace: Γ(1+μ-N)/ Γ(2+2μ-2N) = 2-2(μ-N)+1 1/ Γ(3/2+μ-N) to get Qμ-Nμ(z) ==2μ-N [ 2-2(μ-N)+1 Γ(1+ 2μ-N)/ Γ(3/2+μ-N) ] (z +1)-μ/2+N-1 (z -1)-μ/2 F(1-N, 1+μ-N, 2+2μ-2N); 2/(1+z) ) = 2-μ+N+1 [Γ(1+ 2μ-N)/ Γ(3/2+μ-N) ] (z +1)-μ/2+N-1 (z -1)-μ/2 F(1-N, 1+μ-N, 2+2μ-2N); 2/(1+z) ) which is perhaps a little more compact. This IS our eigenfunction set for N = 2,4...Nmax. Let's go back to our earlier (safer) formula: Qμ-Nμ(z) == 2μ-N [ Γ(1+μ-N) Γ(1+ 2μ-N)/ Γ(2+2μ-2N) ] (z +1)-μ/2+N-1 (z -1)-μ/2 F(1-N, 1+μ-N, 2+2μ-2N); 2/(1+z) ) and define an ancillary parameter κ ≡ 1+μ-N = 1+ν. Then our formula reads Qμ-Nμ(z) = 2μ-N [ Γ(κ) Γ(κ+μ)/ Γ(2κ) ] (z +1)-μ/2+N-1 (z -1)-μ/2 F(1-N, κ, 2κ; 2/(1+z) ) We can then evaluate: N=2 F(-1, κ, 2κ; ψ) = 1 + (-1)(κ)/(2κ) ψ = 1-(κ/2)ψ N=4 F(-3, κ, 2κ; ψ) = 1 + (-3)(κ)/(2κ) ψ + (-3) (-2) (κ) (κ+1)/ [2!(2κ) (2κ+1)] ψ2 = 1 - (3/2)ψ + (3/2) (κ+1)/(2κ+1) ψ2 + a ψ3 term Statement: The eigenfunctions are all of this form which shows only powers, no logs: Qμ-Nμ(z) = AμN (z +1)-μ/2+N-1 (z -1)-μ/2 Σn=1N-1 anμ ψn ψ = 2/(1+z) If μ= m is a positive integer, this formula is still correct, but we have a simpler way to compute things: Qm-Nm(z) = (z2-1)m/2∂mQm-N(z) For example, if m = 2 we have Q2-N2(z) = (z2-1) ∂2Q2-N(z) and this is the only allowed value of N for m=2: // if N = 2, get ν = -2, no good! N=2: Q02(z) = (z2-1)2∂2 Q0(z) = 2z/(z2-1) as Maple verified If m = 4, we then have Q4-N4(z) = (z2-1)2 ∂4Q4-N(z) Then we have several cases: N=0: Q44(z) = (z2-1)2 ∂4Q4(z) + ?? // not a legal N! N=2: Q24(z) = (z2-1)2 ∂4Q2(z) = 48z/(z2-1)2 Maple AND Maxima! N=4: Q04(z) = (z2-1)2 ∂4Q0(z) = 24z(z2+1)/ (z2-1)2 Maple Notice that the N=2 and N=4 legal EF's vanish at z = 0, as our homo BC requires. I skipped m = 3, but it will have ν = 3-2 = 1 as the sole eigenfunction. From Maxima I find that Q44(z) = { ln[ (z+1)/(z-1)] * 105( z8 - 4z6 +6z4 - 4z2 + 1) + 2(-105z7 + 385z5 -511 z3 + 279 z) } / [ 2(z2-1)2 ] This is illegal because it is not in my spectrum, and also you can see it does not vanish at z = 0 so it violates the homo BC (which is why it is not in the spectrum). 4. Back to our integral. Now we can at least look at the examples I did in "when are ODE solutions orthogonal.doc". First, we should find that: I = !Syntax Error, Idζ Q24(iζ)* Q04(iζ) = 0 Now we know exactly how to do this integral: I = !Syntax Error, Idζ [48z/(z2-1)2]* [24z(z2+1)/ (z2-1)2] = !Syntax Error, Idζ [48(iζ)/(-ζ2-1)2]* [24(iζ)(-ζ2+1)/ (-ζ2-1)2] = !Syntax Error, Idζ [48(-iζ)/(-ζ2-1)2] [24(iζ)(-ζ2+1)/ (-ζ2-1)2] = 48*24 !Syntax Error, Idζ ζ2(1-ζ2) / (1+ζ2)4 = 48*24 !Syntax Error, Idx x2(1-x2) / (1+x2)4 = 48*24 [ !Syntax Error, Idx x2/ (1+x2)4 - !Syntax Error, Idx x4/ (1+x2)4 ] and Maple tells us that: so amazingly enough, we really do get 0, to full accuracy. Now let's try this integral !Syntax Error, Idζ Q24(iζ)* Q24(iζ) = = !Syntax Error, Idζ 48z*/(z2-1)2 48z/(z2-1)2 = 482 !Syntax Error, Idζ ζ2 / (-ζ2-1)4 = 482 !Syntax Error, Idx x2 / (x2+1)4 = 482  π/32 = (3/2)*48*π = 72π This agrees exactly with the numeric integration in "when are ODE solutions orthog"