why are p and q SCRAPS

This scrap was at the end of the Appendix: Comments: So we start with our Legendre ODE and we end up with these solutions w±(z) = Pn±m(z) = (z+1)±m/2 (z-1)∓m/2 F(-n,n+1; 1∓m; [1-z]/2) as the two independent solutions for m ≠ integer. The expansion here is around z = 1, not around z = 0. If you want an expansion around z = 0, you can use p 126 P (22) where the first term is even in z and the second is odd in z, so in general Pn±m(z) are neither even nor odd. So I am trying to set a foundation for the functions P(z) before thinking about p and q and why they are even only. So compare the ODEs Lz = –(1-z2) ∂z2 + 2z ∂z + m2/(1-z2) // modified from Leg prop doc Lzu(z) = n(n+1)u(z) Lz= Lζ = +(1+ζ2) ∂ζ2 - 2ζ ∂ζ + m2/(1+ζ2) Lζv(ζ) = n(n+1)v(ζ) I think something must change in the ODE theory when your poles are off the real axis. Well, we think that for m ≠ integer, these should still be the two independent solutions: v±(ζ) = Pn±m(iζ) but we know if we use P (22) we get something like this: v+(ζ) = Pnm(iζ) = eiδ { F(ζ2) + iζ G(ζ2) } so for sure this thing is complex. But we could do this: v+(ζ)* = Pnm(iζ)* = e-iδ { F(ζ2) - iζ G(ζ2) } Now, since Lζ is real, both v and v* should be Legendre solutions! [ Wrong! ] So we are cutting new ground here. Then we can say: f(ζ) = e-iδ v+(ζ) + eiδ v+(ζ)* = 2 F(ζ2) = real even solution g(ζ) = (1/i) [ e-iδ v+(ζ) – eiδ v+(ζ)*] = 2 ζ G(ζ2) = real odd solution I think f(ζ) is my p function, and here I have come up with an odd real function! So finally we are getting somewhere. Are v+(ζ) and v+(ζ)* independent? I cannot write one as a multiple of the other, so I guess so. And this is the main scrap stuff from my earlier attempts at the "why are p and q" document. I think it can all be thrown out, but I will keep it for a while. General Fact: In Bateman's tables for P and Q forms, there are usually two terms. In general, you can define the P and Q functions as a certain one-term expression (which solves the Legendre ODE). All the other forms are then obtained by the Kummer ui definitions and triple-relations on pp 105-108. But all the six basic functions are solutions of the HGE, so each triple relation always replaces a solution with a linear combination of two solutions. This is reflected in the HGE world, so that each term of a two term form is always itself a solution of the Legendre ODE. I have not conclusively proven this claim, but I am 99% sure it is true. We have seen above an explicit example of the claim. And I showed if for Q (40) in a separate doc "A Maple study of Bateman Q (40).doc". [ I tried again to prove this in recent doc " Deriving the Bateman P and Q forms.doc" but I did not succeed. I still think it is true. ] Given the above fact, we may generalize our conclusions above to general z, not just real x, and say: qnm(z)even = (1-z2)-m/2 2A' F'(z2) m and n real qnm(z)odd = (1-z2)-m/2 2 B' z G'(z2) or, recalling that (z2-1)-m/2 = (∓i)2m (1-z2)-m/2, qnm(z)even = (±i)2m (z2-1)-m/2 2A' F'(z2) m and n real qnm(z)odd = (±i)2m (z2-1)-m/2 2 B' x G'(z2) just to stay consistent. So these functions are now even and odd in general complex z, not just x in (-1,1). We can use the first form and consider these results when z = iζ = in the imaginary z axis. then we have qnm(iζ)even = (1+ζ2)-m/2 2A' F'(-ζ2) m and n real (-i)qnm(iζ)odd = (1+ζ2)-m/2 2 B' ζ G'(-ζ2) where we added the (-i) factor to kill of the i from z = iζ on the right side linear factor. So let's then make up some new names here (by adding primes). q'nm(ζ)even = (1+ζ2)-m/2 2A' F'(-ζ2) m and n real q'nm(ζ)odd = (1+ζ2)-m/2 2 B' ζ G'(-ζ2) Example: Let's try to compute these even and odd functions for a simple case n = 1 and m = 0 : q10(x)even = 2 (1/2) [Γ(1)/Γ(3/2)] F(-1/2,1,1/2; z2) = 2 F(-1/2,1,1/2; z2) = 2 times function shown below: Notice that this f2 function is manifestly even because the (...) quantity is odd. The odd function is q10(x)odd = 2 x F(0,3/2,3/2; x2) = 2 x // F = 1 We know that P1(x) = x, so x is a good solution. And it is easy to show that the log term f2 is a solution: Compare our qeven result to the even q(ζ) function obtained in the P and Q doc. I will start here by quoting some text from far above: " Start with Q (40) in z2, appropriate for pondering symmetry near z = 0, and we find: e-iπm Qnm (z) = = (z2-1)-m/2 (±i)m-n {∓i A' F'(z2) + B' z G'(z2)} where the primed factors may be read from (40). Note that we have already extracted the phase factors. One thing is (now) obvious just looking at the form: You won't be able to isolate either term in {...} by making a linear combination of Q(z) and Q(-z). The reason is this Q(+z) = f1(z) {∓i A' F'(z2) + B' z G'(z2)} Q(- z) = f2(z){±i A' F'(z2) – B' z G'(z2)} = – f2(z) {∓i A' F'(z2) + B' z G'(z2)} = f3(z) {∓i A' F'(z2) + B' z G'(z2)} So we have to give up that idea at once." Now let's reconsider what happens if z = iζ with ζ real. We then have: e-iπm Qnm (iζ) = [(∓i)m (1+ζ2)-m/2] (±i)m-n {∓i A' F'(-ζ2) + B' iζ G'(-ζ2)} = i [(∓i)m (1+ζ2)-m/2] (±i)m-n {∓A' F'(-ζ2) + B' ζ G'(-ζ2)} where now the upper/lower sign refers to the sign of ζ, and where we use these results to deal with the power factor, (z2-1)-m/2 = (1-z2)-m/2 e±iπ(-m/2) // page 123 (12) rule so that (-ζ2-1)-m/2 = (1+ζ2)-m/2 e±iπ(-m/2) = (1+ζ2)-m/2 e∓iπm/2 = (∓i)m (1+ζ2)-m/2 Our conclusion is now different! We will be able to isolate the terms in {}. Rewrite the above as the first line below, then take ζ → -ζ to get the second line e-iπm Qnm (iζ) = i (1+ζ2)-m/2 (±i)-n {∓A' F'(-ζ2) + B' ζ G'(-ζ2)} e-iπm Qnm (-iζ) = i (1+ζ2)-m/2 (∓i)-n {±A' F'(-ζ2) - B' ζ G'(-ζ2)} Oops! things just collapsed! In fact, my entire P and Q real functions document has just crashed into nothingness (that I have been working on for 2 weeks) because I made a sign error! Look more closely. e-iπm Qnm (iζ) = i (1+ζ2)-m/2 (+i)-n {–A' F'(-ζ2) + B' ζ G'(-ζ2)} // ζ > 0 e-iπm Qnm (iζ) = i (1+ζ2)-m/2 (–i)-n {+A' F'(-ζ2) + B' ζ G'(-ζ2)} // ζ < 0 Recall way back where we wrote e-iπm Qnm (z) = (z2-1)-m/2 { (±i)m-n-1 A' F'(z2) + (±i)m-n B' z G'(z2)} = (z2-1)-m/2 (±i)m-n { (±i)-1 A' F'(z2) + B' z G'(z2)} = (z2-1)-m/2 (±i)m-n {∓i A' F'(z2) + B' z G'(z2)} ******************************* Comment: Our problem above was to consider the Legendre ODE limiting to the interval z in (-1,1) where we call it x. In retrospect, finding purely real functions f(x) which solve the Legendre ODE and which have definite parity in the range x in (-1,1) was "no big deal" (though it was a very big deal when I started and didn't know how to do it!). Now I want to repeat the Q work, but instead I want to be talking about the flipped ζ Legendre ODE and I want to find real definite-parity solutions of that ODE on ζ in (-∞,∞). This is the same as the regular Legendre ODE on the interval z = pure imaginary = iζ . *********************************************************************** Finally I am seeing the light. Start with P (22) and Q (40) in z2, appropriate for pondering symmetry near z = 0, and we find: Pnm(z) = (z2-1)-m/2 { A F(z2) + B z G(z2)} e-iπm Qnm (z) = (z2-1)-m/2 { (±i)m-n-1 A' F'(z2) + (±i)m-n B' z G'(z2)} = (z2-1)-m/2 (±i)m-n { (±i)-1 A' F'(z2) + B' z G'(z2)} = (z2-1)-m/2 (±i)m-n {∓i A' F'(z2) + B' z G'(z2)} where A,B,F,G and primed versions are all real when n, m and z are real. Before continuing, we have to understand the meaning of our first factor f(z) = " (z2-1)-m/2 " = (z+1)-m/2 (z-1)-m/2 I put the function in quotes as a reminder that it is a "misnomer" because the quoted complex function differs from what is shown on the right. We then have f(-z) = " ([-z]2-1)-m/2 " = (-z+1)-m/2 (-z-1)-m/2 = (1-z)-m/2 (-z-1)-m/2 = [ e∓iπ(-m/2) (z-1)-m/2 ] [ e∓iπ(-m/2) (z+ 1)-m/2 ] = e∓iπ(-m) (z-1)-m/2(z+ 1)-m/2 = e±iπm " (z2-1)-m/2 " = (±i)2m " (z2-1)-m/2 " Therefore, when we take z→-z in our two equations, we take this into account and write Pnm(+z) = (z2-1)-m/2 { A F(z2) + B z G(z2)} Pnm(–z) = (±i)2m (z2-1)-m/2 { A F(z2) – B z G(z2)} e-iπm Qnm (+z) = (z2-1)-m/2 (±i)m-n {∓i A' F'(z2) + B' z G'(z2)} e-iπm Qnm (–z) = (±i)2m (z2-1)-m/2 (∓i)m-n {±i A' F'(z2) – B' z G'(z2)} = (±i)2m (z2-1)-m/2 (±i)n-m {±i A' F'(z2) – B' z G'(z2)} = (z2-1)-m/2 (±i)n+m {±i A' F'(z2) – B' z G'(z2)} = (z2-1)-m/2 (±i)m+n {±i A' F'(z2) – B' z G'(z2)} So our two Q's are these: e-iπm Qnm (+z) = (z2-1)-m/2 (±i)m-n {∓i A' F'(z2) + B' z G'(z2)} e-iπm Qnm (–z) = (z2-1)-m/2 (±i)m+n {±i A' F'(z2) – B' z G'(z2)} Let's do Q first since it is harder. In order to construct an even function , we need to cancel the second terms. So consider: (∓i)m-n e-iπm Qnm (+z) + (∓i)m+n e-iπm Qnm (–z) = (z2-1)-m/2 {∓i A' F'(z2) + B' z G'(z2)} + (z2-1)-m/2 {±i A' F'(z2) – B' z G'(z2)} = 0 For n, m, z real, these two functions have mixed symmetry, they are neither even nor odd functions of z. And the function P is real but Q is complex. We could construct some even and odd functions like this, AND have them all be real : [ the sum of any two ODE solutions is also an ODE solution ] Pnm(z) + Pnm(-z) = (z2-1)-m/2 2A F(z2) even Pnm(z) – Pnm(-z) = (z2-1)-m/2 2B z G (z2) odd For Q, things are so messy now that I have to write it all out longhand. Here we go: e-iπm [e∓iσ Qnm(z) + e±iσ Qnm(-z)] = = e∓iσ { (z2-1)-m/2 { e±iδ A' F'(z2) + e±iσ B' z G'(z2)} + e±iσ { (z2-1)-m/2 { e∓iδ A' F'(z2) - e∓iσ B' z G'(z2)} = { (z2-1)-m/2 { e±i(δ-σ) A' F'(z2) + B' z G'(z2)} + { (z2-1)-m/2 { e∓i(δ-σ) A' F'(z2) - B' z G'(z2)} = { (z2-1)-m/2 { e±i(δ-σ) A' F'(z2)} + { (z2-1)-m/2 { e∓i(δ-σ) A' F'(z2)} = { (z2-1)-m/2 A' F'(z2){ e±i(δ-σ) + e∓i(δ-σ)} = { (z2-1)-m/2 A' F'(z2){ 2cos(δ-σ)} = { (z2-1)-m/2 A' F'(z2){ 2cos(-π/2)} = { (z2-1)-m/2 A' F'(z2){ 0)} = 0 = (z2-1)-m/2 [e±i(δ-σ) + e∓i(δ-σ)] A' F'(z2) = (z2-1)-m/2 cos(δ-σ) 2A' F'(z2) = 0 even (i) e-iπm [ e∓iδ Qnm(z) – e±iδ Qnm(-z)] = (z2-1)-m/2 B' z G'(z2) [e∓i(δ-σ) - e±i(δ-σ)](i) odd = (z2-1)-m/2 sin(δ-σ) 2A' F'(z2) = – (z2-1)-m/2 2A' F'(z2 To summarize: pe(z) ≡ Pnm(z) + Pnm(-z) = (z2-1)-m/2 2A F(z2) even po(z) ≡ Pnm(z) – Pnm(-z) = (z2-1)-m/2 2B z G (z2) odd qe(z) ≡ e-iπm [Qnm(z) + Qnm(-z)] = (z2-1)-m/2 cos(δ-σ) 2A' F'(z2) even qo(z) ≡ (i) e-iπm [ e∓iδ Qnm(z) – e±iδ Qnm(-z)] = (z2-1)-m/2 sin(δ-σ) B' z G'(z2) odd In this way, we can generate four real solutions of the Legendre ODE which have definite parity. For general m, any two solutions ( or our 6 shown above) can be taken as the two independent functions. For integer m, you have to be more careful about which pairs of the 6 solutions are really independent. I cannot remember ever seeing these four new solutions mentioned. Even and odd P and Q functions of ζ. Let's just rewrite the four equations above as pe(iζ) ≡ Pnm(iζ) + Pnm(-iζ) = (-ζ2-1)-m/2 2A F(-ζ2) even po(iζ) ≡ Pnm(iζ) – Pnm(-iζ) = (-ζ2-1)-m/2 2B iζ G (-ζ2) odd qe(iζ) ≡ e-iπm [Qnm(iζ) + Qnm(-iζ)] = (-ζ2-1)-m/2 cos(δ) 2A' F'(-ζ2) even qo(iζ) ≡ (i) e-iπm [ e∓iδ Qnm(iζ) – e±iδ Qnm(-iζ)] = (-ζ2-1)-m/2 sin(δ) B' iζ G'(-ζ2) odd where now ± refers to the sign of ζ. Now we can back up and write (z2-1)-m/2 = (1-z2)-m/2 e±iπ(-m/2) // page 123 (12) rule so that (-ζ2-1)-m/2 = (1+ζ2)-m/2 e±iπ(-m/2) = (1+ζ2)-m/2 e∓iπm/2 = (∓i)m (1+ζ2)-m/2 We install this change, AND we left divide the 2nd and 4th equations by i , AND then we multiply all four equations by (±i)m ; pe(iζ) ≡ Pnm(iζ) + Pnm(-iζ) = (∓i)m (1+ζ2)-m/2 2A F(-ζ2) even po(iζ)/i ≡ [Pnm(iζ) – Pnm(-iζ)]/i = (∓i)m (1+ζ2)-m/2 2B ζ G (-ζ2) odd qe(iζ) ≡ e-iπm [Qnm(iζ) + Qnm(-iζ)] = (∓i)m (1+ζ2)-m/2 cos(δ) 2A' F'(-ζ2) even qo(iζ)/i ≡ e-iπm [ e∓iδ Qnm(iζ) – e±iδ Qnm(-iζ)] = (∓i)m (1+ζ2)-m/2 sin(δ) B' ζ G'(-ζ2) odd Start with P (22) and Q (40) in z2, appropriate for pondering symmetry near z = 0, and we find: Pnm(z) = (z2-1)-m/2 { A F(z2) + B z G(z2)} e-iπm Qnm (z) = (z2-1)-m/2 { e±iδ A' F'(z2) + e±iσ B' z G'(z2)} δ= (π/2)(m-n-1) Now install z = iζ [ note that A' and B' are here defined to include the usual Q function e+iπm factor ] Pnm(iζ) = (-ζ2-1)-m/2 { A F(-ζ2) + B iζ G(-ζ2)} e-iπm Qnm (iζ) = (-ζ2-1)-m/2 { e±iδ A' F'(-ζ2) + e±iσ B' iζ G'(-ζ2)} δ= (π/2)(m-n-1) where now ± refer to the sign of ζ , before it was sign of Im(z). Now we can back up and write (z2-1)-m/2 = (1-z2)-m/2 e±iπ(-m/2) // page 123 (12) rule so that (-ζ2-1)-m/2 = (1+ζ2)-m/2 e±iπ(-m/2) = (1+ζ2)-m/2 e∓iπm/2 = (∓i)m (1+ζ2)-m/2 and we get these improved results Pnm(iζ) = (∓i)m (1+ζ2)-m/2 { A F(-ζ2) + B iζ G(-ζ2)} e-iπm Qnm (iζ) = (∓i)m (1+ζ2)-m/2 { e±iδ A' F'(-ζ2) + e±iσ B' iζ G'(-ζ2)} δ= (π/2)(m-n-1) or (±i)m Pnm(iζ) = (1+ζ2)-m/2 { A F(-ζ2) + B iζ G(-ζ2)} (**) e-iπm (±i)m Qnm (iζ) = (1+ζ2)-m/2 { e±iδ A' F'(-ζ2) + e±iσ B' iζ G'(-ζ2)} δ= (π/2)(m-n-1) which the reader should compare to Pnm(z) = (z2-1)-m/2 { A F(z2) + B z G(z2)} (*) e-iπm Qnm (z) = (z2-1)-m/2 { e±iδ A' F'(z2) + e±iσ B' z G'(z2)} δ= (π/2)(m-n-1) Now recall how from these two equations (*), we obtained the following four: pe(z) ≡ Pnm(z) + Pnm(-z) = (z2-1)-m/2 2A F(z2) even po(z) ≡ Pnm(z) – Pnm(-z) = (z2-1)-m/2 2B z G (z2) odd qe(z) ≡ e-iπm [Qnm(z) + Qnm(-z)] = (z2-1)-m/2 cos(δ) 2A' F'(z2) even qo(z) ≡ (i) e-iπm [ e∓iδ Qnm(z) – e±iδ Qnm(-z)] = (z2-1)-m/2 sin(δ) B' z G'(z2) odd In similar fashion, using the same steps, we start from (**) and create the following four: p'e(ζ) ≡ (±i)m Pnm(iζ) + (∓i)m Pnm(-iζ) = (1+ζ2)-m/2 2A F(-ζ2) even p'o(ζ) ≡ (±i)m Pnm(iζ) – (∓i)m Pnm(-iζ) = (1+ζ2)-m/2 2B iζ G (-ζ2) odd q'e(ζ) ≡ e-iπm [(±i)m Qnm(iζ) + (∓i)m Qnm(-iζ)] = (1+ζ2)-m/2 cos(δ-σ) 2A' F'(-ζ2) even q'o(ζ) ≡ (i) e-iπm [ e∓iδ (±i)m Qnm(iζ) – e±iδ (∓i)m Qnm(-iζ)] = (1+ζ2)-m/2 sin(δ-σ) B' iζ G'(-ζ2) odd Notice that, when we take ζ to -ζ, the term (±i)m Pnm(iζ) becomes (∓i)m Pnm(-iζ) , and so on, because now the signs refer to sign of ζ, upper sign if ζ > 0. We are not quite done, however, because we need to divide the 2nd and 4th equations by i to real RHS's. So p'e(ζ) ≡ (±i)m Pnm(iζ) + (∓i)m Pnm(-iζ) = (1+ζ2)-m/2 2A F(-ζ2) even p'o(ζ)/i ≡ (±i)m [Pnm(iζ) – (∓i)m Pnm(-iζ)]/i = (1+ζ2)-m/2 2B iζ G (-ζ2) odd q'e(ζ) ≡ e-iπm [(±i)m Qnm(iζ) + (∓i)m Qnm(-iζ)] = (1+ζ2)-m/2 cos(δ-σ) 2A' F'(-ζ2) even q'o(ζ)/i ≡ e-iπm [ e∓iδ (±i)m Qnm(iζ) – e±iδ (∓i)m Qnm(-iζ)] = (1+ζ2)-m/2 sin(δ-σ) B' iζ G'(-ζ2) odd So here our four real functions of ζ that satisfy the flipped Legendre equation! We can simplify a bit as follows: δ = (π/2)(m-n-1) => e±iδ = e±i(π/2)(2δ/π) = (±i)(2δ/π) = (±i)m-n-1 Then we have e±iδ(∓i)m = (±i)m-n-1(∓i)m = (±i)-n-1 But these functions are complex. Now just as before, we can construct four real functions of definite symmetry as follows: (±i)m Pnm(iζ) + (∓i)m Pnm(-iζ) = (1+ζ2)-m/2 2A F(-ζ2) even (1/i)[ (±i)m Pnm(iζ) – (∓i)m Pnm(-iζ)] = 2B ζ G(-ζ2) odd (±i)m Qnm(iζ) + (∓i)m Qnm(-iζ) = (1+ζ2)-m/2 { e±iδ + e∓iδ } A' F'(-ζ2) = (1+ζ2)-m/2 cos(δ) 2 A' F'(-ζ2) (±i)m e∓iδ Qnm(iζ) – (∓i)m e±iδ Qnm(-iζ) = (1+ζ2)-m/2 { e∓iδ + e±iδ A' F'(-ζ2)