legendre my problem vol 2

Legendre my problem: Volume 2 PhL 11.1.09 Note: Part I contains various unmarked incorrect statements which are corrected in Part II. I decided to keep Part I in the doc as part of "the record". Overview of Part I: Original Contents of this Doc (2 pages 9.23.10) 1 Overview of Part II: Slippery When Wet (+ Theorem 1A) ( 2 pages 9.24.10) 3 Part I: Original Contents of this Doc 5 1. Review of Past Efforts on the half sphere problem. 5 2. Let's look for a moment at the whole sphere problem. 6 2a. Same as 2 in bra ket notation. 7 3. Resume Half Sphere. 8 Dependence and Independence of Vectors: notion of a Basis (118) 9 4. Integration Range. 10 5. Force I(x) = 0 on lower half sphere. 11 6. More General Situation. 11 7. Theorem 1: 12 8. Review of attempts to solve: 12 Plan C Reviewed 12 Plan B Reviewed 14 Summary: 15 Task Still Remaining: 16 Part II: Slippery When Wet 9.22.10. 16 1. Full (a,b) interval series. 16 2. Enclosed (A,B) view of series. 16 3. Aside: 16 4. Extension by 0 of the enclosed interval function. 16 5. Bowl problem applying ∫dΩs Ynm(Ωs) to both sides of V(s) equation. 17 6. Bowl problem applying ∫b dΩs Ynm(Ωs) to both sides of V(s) equation. 19 7. The V(s) equation without applying any ∫dΩs Ynm(Ωs) . 21 8. A quick review and statement of equations in Stakgold Units. 24 9. Proof that !Syntax Error, Idx f(x) Pn(x) = δn0 has no solution f(x). 25 ______________________________________________________________________________ Overview of Part I: Original Contents of this Doc (2 pages 9.23.10) In Section 1 I quickly review how starting with ∫σ E(s|ξ)I(ξ)dSξ = 1 for half sphere, we get to: Σn Pn(y) In = 2 (*) where In ≡ !Syntax Error, I dx Pn(x) I(x) !Syntax Error, I dx K(y,x) I(x) = 2 (**) where K(y,x) ≡ Σn Pn(y) Pn(x) I later show in Part II Section 7 how to derive these from scratch. It is the Σn Pn(y) In series whose existence will be coming into question below. [ equation (**) is never further used in this doc. ] In Section 2 I define ψn(x) = (n+1/2)1/2 Pn(x) to be the orthonormal Legendres and I make sure the full charged sphere problem works out right with these. I am planning to ponder Hilbert Space once again, and I guess these little ψn are going to be the normalized basis vectors in a HS. In Section 2a I rewrite everything in Section 2 using bra-ket notation where now ψn(x) = <x|n> and then these |n> are the normalized base states spanning the HS. I don't really know where I was going with this idea, but see next section maybe. In Section 3 I resume the flow of Section 1. I restate the two equations seen above, then I review my own Stakgold notes on spanning sets with the big example being the powers xn. A spanning set "spans" a Hilbert Space H and L2 on the interval (0,1) is such a space and the powers xn form such a spanning set. The set of points in H which you can reach using a spanning set -- call this set S -- is "dense" in the HS, but does not equal the HS. In fact H = , the closure of S. Any spanning set that is orthogonal on the interval is a basis. The classic problem is that as you add more terms in a series to approach some desired point f in H (a function which is not a polynomial, and assume spanning set xn) , the coefficients of the lower terms MAY "move around" and never stabilize. If this happens, the series with an infinite number of terms which is supposed to add up to f does not exist. I then applied these facts to the set { Pn(x) } on the interval (0,1) and concluded that the series Σn=0∞ Pn(x) In does not exist for x on (0,1) because the Pn are not orthogonal on this interval and therefore do not form a basis, just a spanning set. In Part II I will change this conclusion, but that lies ahead. Continuing on the (incorrect) assumption that the series does not exist, I concluded that my development of the half sphere solution was "wrong" going between these two steps: 2 (1/4π) ∫σ dΩξ { Σn N0n Σm (1/Nmn) Ynm(Ωξ) Y*nm(Ωs) } I(Ωξ) = 1 σ = half sphere 3 (1/4π) Σn N0n Σm (1/Nmn) Y*nm(Ωs) ∫σ dΩξ Ynm(Ωξ) I(Ωξ) = 1 σ = half sphere I felt line 2 was OK, but line 3 then creates this non-existent series. Things are a little clouded here by the presence of index m which Part II shows can be set to 0 so we are then talking Pn functions only. I then restate my conclusion this way 2 = !Syntax Error, I dx Σn Pn(y) Pn(x) I(x) ≠ Σn Pn(y) !Syntax Error, I dx Pn(x) I(x) ≡ Σn Pn(y) In where the ≠ shows where the non-existent series is "generated" by the sum/integral order interchange. In Section 4 I then make some vague comments (but dismiss them) about the fact that on an enclosing interval (-1,1), the Pn do form a basis. I will return to this with more clarity in Part II. In Section 5 I repeat the comments of Section 4. I can see now that I was doubting this conclusion that the series does not exist and was trying to get more traction on the idea. In Section 6 I repeat the comments yet again in a more general framework, !Syntax Error, Idx {Σn fn(y) gn(x)} I(x) ≠ Σn fn(y) !Syntax Error, Idx gn(x) I(x) ≡ Σn fn(y) In where the ≠ applies if fn(y) is not a basis on (a,b). In retrospect, if fn(y) were powers yn, then the series in the left term Σn fn(y) gn(x) would not exist either! I will be countermanding these conclusions later if it is the case that the fn form a basis on an interval which encloses (a,b). In Section 7 I restate the conclusion still again, this time as "Theorem 1". This theorem needs to be modified and I will do that in Part II. It is Theorem 1A and it appears in the Part II Overview! In Section 8 I "pretend" that the series in question does exist (despite all my work above), and I review some of my attempts to solve the half-sphere problem using this series. It all begins with ∫σ E(s|ξ)I(ξ)dSξ = 1/2 which is the Stak "integral equation" we know and love. This is correctly converted to Σn=0∞ Pn(y) In= 1 (assuming the series does exist) with the definition In ≡ !Syntax Error, I dx Pn(x) I(x). I then (wrongly) assume that In = δn,0 is the one and only possible solution of this little equation on (0,1), and is therefore the correct solution to it for the half sphere problem. This leads to the system of equations !Syntax Error, I dx Pn(x) I(x) = δn,0. It then became the subject of my two "legendre my problem" docs to solve this equation for I(x). I then review my Plan C attempt (from my previous "leg my prob" doc) to solve for I(x). This numerical work suggested (correctly) that no solution I(x) exists! This Plan C is also reviewed in my previous doc's overview, so I won't review it again here. The review given below is also pretty good. Next, I review something I call Plan B, but which is really subsection 8 of my first "potential of half-spherical shell" doc. This Plan also led to the same conclusion that no I(x) exists. In retrospect, it is OK that no I(x) solution exists to a system of equations which are an incorrect representation of my half sphere problem of interest. __________________________________________________________________________________ Overview of Part II: Slippery When Wet (+ Theorem 1A) ( 2 pages 9.24.10) In Section 1 I state the expansion/projection transform associated with a set of functions φn complete on an interval (a,b). The expansion is a series which "exists" in the sense of stable coefficients. I don't care really whether the high end of the series diverges or not, which is a distinct meaning of "existence". I am here only concerned with the first meaning. In Section 2 I comment that the expansion series certainly continues to "exist" if you decide to only look at it on an enclosed interval (A,B) within (a,b). The coefficients (projections) involve full (a,b) integrals still in this case. This is the "peephole" idea. In Section 3 I wander off briefly on something not relevant or useful. In Section 4 I observe that, if f(x) exists on (A,B), you can extend it to the full (a,b) by adding 0 pieces, call this fext(x). Then you can expand fext(x) = Σn cn φn(x) on (a,b) [ series exists ]. The peephole view of this series on (A,B) is then f(x) = Σn cn φn(x) which series must then also exist. Moreover, the cn involve only (A,B) integrals of f(x). By this process, I have proven the following theorem: Theorem 1A. Let (a,b) enclose (A,B), let φn be complete on (a,b), and let f(x) be defined only on (A,B). Then the series f(x) = Σn cn φn(x) on (A,B) exists, and cn = !Syntax Error, Idx f(x) φn(x). Conversely, suppose f(x) = Σn cn φn(x) on (A,B) where there exists no enclosing interval (a,b) on which the φn are complete. In this case, the φn are only a spanning set on all (a,b) enclosing intervals. It is likely then that the series Σn cn φn(x) which attempts to equal f(x) will not exist, because the coefficients cn will keep moving as the series is lengthened. An example would be φn = xn. For any given finite last term N, you would determine the optimal cn by a least squares fit. There is no formula for cn as in the previous case. Of course if f(x) in this case happens to be a polynomial, the series will exist (and be finite). In Section 5 I apply a full 4π ∫dΩs Ynm(Ωs) to both sides of the V(s) = Σ q/R equation and this takes me down a wrong alley and all I conclude is the following, which does not contain my "series of interest", ∫dzs Pn(zs) V(zs) = 4πa (2n+1)-1∫b dzξ Pn(zξ) σ(zξ) This equation is pretty useless since it just relates integrals of two unknown functions. In Section 6 I apply just a bowl integral ∫b dΩs Ynm(Ωs) to both sides of the V(s) = Σ q/R equation and this takes me down another wrong alley which is nevertheless mildly interesting. In place of the result quoted above, I get V0 ∫b dzs Pn(zs) = 2πa Σn" { ∫b dzs Pn(zs) Pn"(zs) } [ ∫b dzξ Pn"(zξ) σ(zξ) ] I then give the names kn(b), hnn"(b) and σn"(b) to the three integrals shown to get kn(b) V0 = 2πa Σn" hnn"(b) σn"(b) I then rescale the vector k by defining K(b)n = kn(b) V0/(2πa) and write the above in matrix form: K(b) = H(b) σ(b) => σ(b) = H(b)-1 K(b) This seems to be a new formal method of solving the half sphere problem. You pick some N and compute the H and K objects since they are just known integrals of Pn functions, and get the σn(b) from which you can get σ(z) which is what you want -- the charge density on the half sphere. As you increase N, I think the solution would converge. I could try this in Maple some rainy day. However, this is not really relevant to the discussion at hand. In Section 7 I finally get around to deriving the equation which has the series of interest in it. In fact I obtain two related "charged bowl" equations each having a series: V(zs) = 2πa Σn Pn(zs) σn(b) valid for all zs in (-1,1) V0 = 2πa Σn Pn(zs) σn(b) valid for zs in (β,1) // σn(b) ≡ ∫b dzξ Pn(zξ) σ(zξ) The second equation is of course just the first with zs located "on the bowl", and it contains our "series of interest". We ask: does this series exist or not? My answer is yes, by the following argument: there is no issue with the first equation's series existing since the Pn are complete on (-1,1). The second equation's series is just a peephole view of the first equation's series, so it also exists. Tempest in a teapot. I also consider another series expansion σ(z) = Σn σ'n Pn(z) on (β,1) and show this exists as well since the Pn are complete on an enclosing interval (-1,1). In Section 8 I review the conclusions above, admit that my "series does not exist" claim of Part I was incorrect, and convert a few equations from green-Jackson units to Stakgold units. In Section 9 I give a simple proof that !Syntax Error, Idx f(x) Pn(x) = δn0 has no solution f(x). Had I been aware of this simple proof at the start, I never would have written these two docs! ______________________________________________________________________________ Part I: Original Contents of this Doc 0. Sometimes taking a personal voyage to a place helps in your understanding of that place, compared to just reading about that place. I think the place involved here is the subject of "infinite dimensional Hilbert Spaces", but I am not yet quite sure. This was just another boring chapter of Stakgold Volume I that one has to slog through. But now I think I have been drawn into that place "in my face". 1. Review of Past Efforts on the half sphere problem. Change scene now to our problem of charge self-distribution on a hemispherical shell, or on any spherical cap smaller than a whole sphere. We start with the self-consistency integral equation ∫σ E(s|ξ)I(ξ)dSξ = 1 σ = half sphere We insert an expansion for 1/R in E, assume I has no azimuthal dependence, and we arrive here, where we define y = zs = cos(θs) and x = zξ = cos(θξ) where dSξ = 12dΩξ : [ will show wrong below! ] [ but then I later show it is right !! ] Σn Pn(y) !Syntax Error, I dx Pn(x) I(x) = 2 half sphere valid (0,1) for xs = cos(θs) ≡ y where the 2 comes from E = 1/(4πR) and azimuthal integral → 2π. This can be rewritten in two ways: Σn Pn(y) In = 2 (*) where In ≡ !Syntax Error, I dx Pn(x) I(x) !Syntax Error, I dx K(y,x) I(x) = 2 (**) where K(y,x) ≡ Σn Pn(y) Pn(x) In either form, we see the stringency of the condition: the RHS is magically independent of y, whereas the LHS depends on y. The second form appears in my "canonical structures" PDF book by VSV that I luckily downloaded yesterday. For our purposes here, this simply confirms that I have not made a "math error" in developing the two lines shown above. 2. Let's look for a moment at the whole sphere problem. We can use normalized basis functions defined as follows: ψn(x) = (n+1/2)1/2 Pn(x) and then we can write orthogonality and completeness in this way !Syntax Error, Idy ψm(y)ψn(y) = δm,n δ(x-y) = Σn ψn(x)ψn(y) Since I am extremely prone to errors, I will verify this left equation: !Syntax Error, Idy ψm(y)ψn(y) = !Syntax Error, Idy [(m+1/2)1/2 Pm(x)] [(n+1/2)1/2 Pn(x)] = = (m+1/2)1/2(n+1/2)1/2!Syntax Error, Idy Pm(x) Pn(x) = (m+1/2)1/2(n+1/2)1/2 δnm (n+1/2)-1 = (n+1/2)1 δnm (n+1/2)-1 = δnm OK and from this we can write the following expansion theorem: I(x) = Σn In ψn(x) In = !Syntax Error, Idx ψn(x)I(x) Now we can rewrite (*) for the whole sphere [ which is Σn Pn(y) In = 2] in this way Σn ψn(y) Jn = 2 where Jn = (n+1/2)-1/2 In In = (n+1/2)1/2 Jn and now let's make sure we did this right: Σn ψn(y) Jn = Σn [ (n+1/2)1/2 Pn(x) ][ (n+1/2)-1/2In ] = Σn Pn(x) In We can then apply !Syntax Error, Idy ψm(y) to both sides of our equation on the last line, we get !Syntax Error, Idy ψm(y)Σn ψn(y) Jn = !Syntax Error, Idy ψm(y) 2 Σn Jn!Syntax Error, Idy ψm(y)ψn(y) = 2 !Syntax Error, Idy ψm(y) Σn Jn δm,n = 2 δm,0 Jm = 2 δm,0 In = (n+1/2)+1/2 Jn = (n+1/2)+1/2 2 δn,0 = (1/2)1/2 2 δn,0 = 2-1/2 2 δn,0 = δn,0 Then we use our expansion to find I(x) = Σ In ψn(x) = ψ0(x) = [ (0+1/2)1/2 P0(x) ] = 2-1/2 1 = 1 which agrees with our solution elsewhere of this full sphere problem. 2a. Same as 2 in bra ket notation. Now let's rewrite some of our whole sphere equations in Hilbert Space notation: !Syntax Error, Idy ψm(y)ψn(y) = δm,n δ(x-y) = Σn ψn(x)ψn(y) <m|n> = δm,n 1 = Σn |n><n | δ(x-y) = <x|y> __________________________________________________________________________ I(x) = Σn In ψn(x) In = !Syntax Error, Idx ψn(x)I(x) <x|I> = Σn <x|n><n|I> <n|I> = !Syntax Error, Idx <n|x><x|I> __________________________________________________________________________ Σn ψn(y) Jn = 2 where Jn = (n+1/2)-1/2 In In = (n+1/2)1/2 Jn Σn <y|n><n|J> = 2 = <y|J> = J(y) so J(y) = 2, but we want to know I(y) : __________________________________________________________________________ !Syntax Error, Idy ψm(y)Σn ψn(y) Jn = !Syntax Error, Idy ψm(y) 2 !Syntax Error, Idy <m|y>Σn <y|n>Jn = 2 !Syntax Error, Idy <m|y><y|0> Σn Jn <m [ !Syntax Error, Idy |y> <y| ] n> = 2 <m|0> Σn Jn<m| n> = 2 <m|0> Jm = 2 δm,0 Im = (m+1/2)1/2 Jm = 2/ δm,0 = δm,0 I(x) = Σn In ψn(x) ie, <x|I> = Σn <x|n><n|I> I(x) = Σn δn,0 ψn(x) = ψ0(x) = (0+1/2)1/2 P0(x) = 1 __________________________________________________________________________ 3. Resume Half Sphere. Now having done all of the above for the whole sphere, we return to our half sphere situation: Σn Pn(y) In = 2 (*) where In ≡ !Syntax Error, I dx Pn(x) I(x) !Syntax Error, I dx K(y,x) I(x) = 2 (**) where K(y,x) ≡ Σn Pn(y) Pn(x) The functions Pn(x) do not form a complete set of basis functions on (0,1), so we cannot "construct" our infinite dimensional Hilbert Space method used in 2 and 2a above. The functions Pn(x) could just as well be xn , another set of functions on (0,1) which do not form a basis on that interval. I want now to look back at Stakgold Chapter 2 on this subject and recall what was taught there. Here are some quotes from my raw notes. David Hilbert (1862-1943, German) was a very big gun, see wiki. Did physics math support among other things, for QM and gen rel, stated the 23 problems, some still unsolved. Godel friend, etc. definition: "a spanning set" . Let H = some Hilbert space. Suppose we can find a countable set of vectors f1, f2 ... (such as xn) whose linear combinations form a space which is dense in H. If such a set can be found, then those f's are set to form a spanning set (which is not necessarily a basis, see later). An immediate example of such a countable set would be the powers tn or some polynomials of degree n in a Hilbert space consisting of functions of a real variable going to same. Here n = 1,2,3.... is where the countable comes in. We have already seen how such polynomials are dense in L2. So our Hilbert Space is L2 on (0,1). The functions Pn(x) or xn on this interval are countable and no doubt are dense in L2 so the set of Pn(x) or xn do form a spanning set for our problem of the half sphere. definition: "separable Hilbert Space" : one which has at least one spanning set. So without doubt L2 on (0,1) is a separable Hilbert Space, I just listed two different spanning sets. Dependence and Independence of Vectors: notion of a Basis (118) Detail on the spanning set: a spanning set f1, f2 ... means that you can get arbitrarily close to an element x in H by doing a finite sum x n akfk. As you require a lower , you in general need more terms, that is, you have to increase n. If, when you do this all the way to the dimensionality of the vector space N, the ak don't change for earlier terms in the series, then the fk also form a basis because then you have written x = nmax akfk . For N = finite, a spanning set is always a basis because once you reach N terms in the sum, you have your ak and you are done and the fk must be a basis. For N = infinite, it is easy to find cases where the ak don't stay put as you crunch down on your closeness requirement. I think, for example, that in L2 , the set of powers tk = fk forms a spanning set, but not a basis. However, any orthogonal set of spanning set functions fk (perhaps made by G-S) do always form a basis, and this is formally stated as Theorem 1 a few pages down in these notes. Notice that, even if fk are not a basis, a spanning set spans a space that is dense in H since you can get arbitrarily close. Call this space S. If you add all the points that you cannot quite reach, that is the closure , and then you have = H. I think the closure of any space S that is dense in H is H. So consider our situation above Σn Pn(y) In = 2 (*) where In ≡ !Syntax Error, I dx Pn(x) I(x) Here Pn(x) form a spanning set but not a basis for the interval (0,1). We are thinking about f(y) = 2 = Σn=0∞ Pn(y) an f(y) = 2 ≈ Σn=0N Pn(y) an For any given N < ∞, we can find a set of an that make this "fit" as good as possible. But as we increase N, the ak keep moving, no matter how large we make N. The conclusion is this. Suppose you write f(y) = Σn=0∞ Pn(y) an on (0,1) You can write these symbols on paper, but such a series DOES NOT EXIST ! There is no set of an that makes this series true! For finite N you can find a set of ak that make a best fit. Notice that we are not saying the series does not converge, we are saying it does not exist! If something does not exist, you cannot even talk about whether it converges or not. Now let's go back and examine our developmental steps from the integral equation for I. 1 ∫σ E(s|ξ)I(ξ)dSξ = 1 σ = half sphere 2 (1/4π) ∫σ dΩξ { Σn N0n Σm (1/Nmn) Ynm(Ωξ) Y*nm(Ωs) } I(Ωξ) = 1 σ = half sphere 3 (1/4π) Σn N0n Σm (1/Nmn) Y*nm(Ωs) ∫σ dΩξ Ynm(Ωξ) I(Ωξ) = 1 σ = half 4. (1/4π) Σn N0n(1/N0n) Y*n0(Ωs) ∫σ dΩξ Yn0(Ωξ) I(Ωξ) = 1 σ = half 5. (1/2) Σn Pn(zs) !Syntax Error, I dzξ Pn(zξ) I(zξ) = 1 half sphere valid (0,1) for zs 6. (1/2) Σn Pn(zs) In = 1 half sphere valid (0,1) for zs We know step 1 is valid, and we know that step 6 has a series which does not exist! There is no set of coefficients In which make equation 6 true (sums are all n=0,∞) . Therefore equation 5 is also not valid, because the integral there is In. The problem presumably has some solution I(x), so these integrals which we call In would then be a set of actual numbers we could compute for all n. But that would imply that equation 6 is valid, which we know it is not. Again, equation 5 is not valid. Equation 2 is valid because we have just inserted a classical expansion for 1/R. If that expansion did not exist, we would not be seeing it in the literature! This expansion is a double series which exists and which converges for any points ξ and s on the sphere. If equation 3 were valid, then equation 4 would follow from it, undeniably because we know the azimuthal integral is 0 unless m = 0, Conclusion: The error is made going from equation 2 to equation 3. This is a place I always knew was a potential problem, but thought is was OK here. This is an interchange of order of summation and integration. Doing this interchange creates a series that does not exist! 4. Integration Range. Look again at our claim: 2 = !Syntax Error, I dx Σn Pn(y) Pn(x) I(x) ≠ Σn Pn(y) !Syntax Error, I dx Pn(x) I(x) ≡ Σn Pn(y) In Why does the integration range have to relate to our interval of interest (a,b)? Suppose (0,1) is a subset of (a,b), for example, (a,b) = (-1,1). Then on this full interval, the series on the right DOES exist, and there ARE some In which satisfy 2 = Σn=0∞ Pn(y) In . That is OK, but then we have to regard the equation 2 = !Syntax Error, I dx Σn Pn(y) Pn(x) I(x) = Σn Pn(y) In as being valid on y in all of (-1,1). But that is like saying ∫σ E(s|ξ)I(ξ)dSξ = 1 where σ is the half sphere, but we allow s to go onto the lower half sphere. This is then the WRONG integral equation for charge self consistency. 5. Force I(x) = 0 on lower half sphere. Suppose we insist that I(ξ) = 0 on the lower half sphere and consider ∫σ E(s|ξ)I(ξ)dSξ = 1 where now σ is the entire sphere. This equation then applies for ξ (within the integration) anywhere on the whole sphere, but s only on the upper half sphere. We then write our integral equation and posited sum in either of these manners, 2 = !Syntax Error, I dx Σn Pn(y) Pn(x) I(x) = Σn Pn(y) In In = !Syntax Error, I dx Pn(x) I(x) or 2 = !Syntax Error, I dx Σn Pn(y) Pn(x) I(x) = Σn Pn(y) In In = !Syntax Error, I dx Pn(x) I(x) BUT either of these equations is only valid on y ε (0,1). The equation is known to be invalid outside of this y range, in fact, because the potential is not going to be "1" on the lower half sphere. So we have to restrict the equation to being valid on L2 over just (0,1). But in this case, the series shown does not exist. We could consider a different integral equation which is true on y ε (-1,1), and in that case the series DOES exist and In = 2 δn,0 and I(x) = 1.But this equation is the whole sphere problem, and our assumption that I(x) = 0 on the lower half sphere is no good. 6. More General Situation. Assume then that we have a Fredholm equation of this form: !Syntax Error, Idx {Σn fn(y) gn(x)} I(x) = h(y) where we have written the kernel as an infinite diagonal sum which presumably exists and the f and g are just some functions. The whole theory of Fredholm equations is based on this equation applying for y on the range (a,b) which appear as endpoints (the Volterra equation has an upper endpoint y instead of b). So we assume that the equation is meant to be true on all of (a,b) and we have L2 on (a,b) as our HS. Then we try to process the LHS as follows !Syntax Error, Idx {Σn fn(y) gn(x)} I(x) =?= Σn fn(y) !Syntax Error, Idx gn(x) I(x) ≡ Σn fn(y) In If the functions fn(y) do not form a basis on (a,b), then this series does not exist, and we have to replace the symbol =?= with ≠ !!! So in this situation, interchanging the order of integration and summation is not allowed !!! . There exists no set of numbers In such that h(y) = Σn=0∞ fn(y) In on (a,b). This was exactly our situation where (a,b) = (0,1) and fn = gn = Pn . 7. Theorem 1: Consider the expansion h(y) = Σn=0∞ fn(y) In . Whether this series exists or not depends on the range over which you mean the expansion to be true. Suppose fn(y) form a basis on (a,b). Then this series exists only on the range (a,b). For any other range, the series does NOT exist. If we know that h(y) is a periodic function on the range (a,b+N(b-a))for N=integer, then in that special case the series would also exist, but not for general h(y). So the default situation really is that the above series does NOT exist. For it to exist, the functions have to form a basis on the interval in question. So you really need to think about this any time you do a change of integration order. The interchange is not valid any time you create a series that does not exist! 8. Review of attempts to solve: !Syntax Error, I dx Pn(x)I(x) = δn,0 I erroneously converted the half-sphere integral equation into the following series form, with the intended region of validity y in (0,1). [Here I rescale the RHS to make things simpler, which would correspond to integral equation ∫σ E(s|ξ)I(ξ)dSξ = 1/2, except of course it is invalid ] Σn=0∞ Pn(y) In = 1 (*) where In ≡ !Syntax Error, I dx Pn(x) I(x) I now know that the series does not exist on (0,1), so there is no point in continuing. But what happens if you assume it does exist? Then you are led down mysterious alleyways that go nowhere, to wit: Plan C Reviewed The first thing you do, assuming the expansion (*) of 1 exists, is to claim that you have an immediate solution for the coefficients In (of the equation Σn=0∞ Pn(y) In =1 ). In = δn,0 [ In retrospect, it is true that this is a solution of the equation Σn=0∞ Pn(y) In =1 for y on (0,1), but it is not the solution which is correct for the half sphere. But I did not know that when writing this doc. ] This then leads you to try to find a function I(x) on (0,1) which has these properties !Syntax Error, I dx Pn(x)I(x) = δn,0 (*) [ This equation comes from our definition that In ≡ !Syntax Error, I dzξ Pn(zξ) I(zξ) to which we add our knowledge that I(zξ) = 0 for the lower sphere, which tells is we can also write In = !Syntax Error, I dzξ Pn(zξ) I(zξ). So we are then combining the definition of I(n), knowledge that I=0 on the bottom, and our incorrect solution that In = δn,0 to get the equation (*) shown above. We then try to solve this incorrectly derived equation. ] In Section 6 (Plan C) of my document "legendre my problem.doc", I tried to solve these equations in the following manner. I found a set of complete orthonormal basis functions φn(x) = (2n+1)1/2 Pn(-1+2x) on the interval (0,1), and I tried expanding both I(x) and Pn(x) onto these functions, certainly allowable. My expansions were [ I call the coefficients of I(x) fn to not confuse them with In shown above ] I(x) = Σm fm φm(x) fm = !Syntax Error, Idx φn(x)f(x) = (2n+1)1/2!Syntax Error, I Pn(-1+2x) f(x) Pn(x) = Σk Gnk φk(x) Gnk = !Syntax Error, Idx Pn(x) φk(x) = known numbers Insertion of these expansions into (*) and use of orthonormality then yielded Σm Gnm fm = δn,0 or G f = 1 I considered this a square matrix equation, and figured maybe I could invert it and find a solution for the vector f whose components are fm θ f = G-1 1 => fn = Σm [ G-1]nm δm,0 = [ G-1] n0 = (first column of matrix G-1) I had Maple compute the matrix G and G-1 for various finite summation end values N. This was an exact calculation, since Gnk are well-defined numbers. What I found was that the first column of G-1 has numbers which "don't move" -- you just get a longer list of numbers as you increase N. For N = 6 for example we got this for fn I then used these resulting values for fn and installed them into my valid expansion above to get I(x), and then I plotted I(x). I even then recomputed !Syntax Error, I dx Pn(x)I(x) to make sure I was getting δn,0, and this all worked out fine. BUT, what I seemed to find was that I(x) so-computed was not approaching any stable limit. For any finite N, I was able to obtain an exact function I(x) which satisfied my requirement !Syntax Error, I dx Pn(x)I(x) = δn,0 where solution is: I(x) = Σm fm φm(x) Now this series for I(x) on (0,1) is "legal series", but I was finding that it diverges! The fm coefficients just keep getting larger and larger as m increases. I had a sequence of functions, call it this IN(x) = Σm=0N fm φm(x) How do you test to see whether a sequence converges or not? You really need to know something about what the fm do in size, and my method here makes that pretty hard to see. But all my numerical evidence was that the fm were getting large without limit, and it certainly seemed that IN(x) was not approaching a limit. Each N made a new function that was different from the previous value of N. I thought maybe this sequence was approaching some kind of distribution, if not a function. So this is where this little effort stopped. The conclusion seemed to be this: no function I(x) exists such that !Syntax Error, I dx Pn(x)I(x) = δn,0 on the interval (0,1). As you increase N, you force I(x) to bend more ways to maintain all previous orthogonalities and have one new one. I was very disturbed by this apparent fact, since it seemed to say that my half sphere shell had no solution for its charge density I(x). But now I know that the set of conditions !Syntax Error, I dx Pn(x)I(x) = δn,0 is not valid for this half sphere problem. Thus the fact that these conditions are satisfied by no function is less bothersome. And later I learned that the half sphere solution is I(x) = (π/2) 1/, and this very obviously does not satisfy !Syntax Error, I dx Pn(x)I(x) = δn,0 as Maple happily shows for various n. Plan B Reviewed I came back for more of the same I think, a slight variation of the above. I start with In = !Syntax Error, Idx Pn(x)I(x) = δn,0 and then activate those same φn(x) on (0,1) as above. This time we try this φn(x) = Σm=0∞ knm Pm(x) (*) which we now know is a non-existent sum on (0,1) [ say for some particular value of n ] . The reason is that the Pm are not a complete set on (0,1) so the knm for given n should "keep moving" as we increase N and do best bit to φn(x). We can "extend" the range of both sides to (-1,1). This gets us involved with values like Pn(-3) on the LHS, but never mind. After extension, the sum does exist at least, and we can solve for knm , knm = (1/2) (2m+1) (2n+1)1/2 !Syntax Error, I dx Pm(x) Pn(-1+2x) = (1/2) (2m+1) !Syntax Error, I dx Pm(x) φn(x) kn0 = (1/2) (2n+1)1/2 !Syntax Error, I dx Pn(-1+2x) // special case for use below which are a well defined set of numbers for finite n and m. Now we go back to our (0,1) world where we have the expansion theorem fn = !Syntax Error, Idx φn(x)I(x) I(x) = Σn=0∞fnφn(x) x = cosθ If we insert (*) into the projection here we get fn = !Syntax Error, Idx { Σm=0∞ knm Pm(x)I(x) } = Σm=0∞ knm!Syntax Error, Idx Pm(x)I(x) = Σm=0∞ knm δm0 = kn0 Notice that these expansion coefficients fn are the same as those used in Plan A above. There we only knew that fn = [ G-1] n0, but now we have a nice expression for fn as the integral kn0. So the rest of the story comes out exactly as in Plan A above. That is to say, we have our candidate series for I(x) on (0,1) I(x) = Σn=0∞fnφn(x) = Σn=0∞ kn0 φn(x) Maple gives us these values for the first 11 kn0 coefficients > with(orthopoly); > phi := (2*n+1)^(1/2)*P(n,-1+2*x): > kn0 := (1/2)*(2*n+1)^(1/2) * int(P(n,-1+2*x), x=-1..1): > for n from 0 to 10 do kn0 od; which you see agree with the smaller fn list shown in Plan A above. Our series diverges! Summary: we attempt to solve !Syntax Error, Idx Pn(x)I(x) = δn,0 for I(x). This is an equation which we earlier found had no solution, but we try anyway. Our candidate solution is I(x) = Σn=0∞ kn0 φn(x) where kn0 = (1/2) (2n+1)1/2!Syntax Error, I dx Pn(-1+2x) Although this series at least "exists" on (0,1) since φn(x) form a complete basis on (0,1), in order to make the series solve our condition !Syntax Error, Idx Pn(x)I(x) = δn,0 we find that the series severely diverges. So again our conclusion is that there is no solution I(x) that meets all these conditions. For any finite N we can find I(x) that meets N conditions. Each time we increment N by 1, the solution I(x) changes violently everywhere in the range. The sequence IN(x) defined as partial sums does not converge. And I must hasten to add: this is no longer bothersome because !Syntax Error, Idx Pn(x)I(x) = δn,0 is now understood to be an invalid encapsulation of our surface charge integral equation for the half sphere. Task Still Remaining: Solve ∫σ E(s|ξ)I(ξ)dSξ = 1 for I(ξ) ! [ I much later solved this integral equation using a Polyanin lookup. ] Part II: Slippery When Wet 9.22.10. I still have not nailed down this series existence issue, it keeps squirming around. Let's try it once again. In what follows, Σn always means Σn=0∞. 1. Full (a,b) interval series. Suppose you have f(x) = Σn fnφn(x) on interval (a,b) where φn are complete and orthonormal on that interval. This series exists, all is well, and we know that fn = !Syntax Error, If(x) φn(x) are the coefficients. We assume f(x) is some reasonable function so fn exists and the series converges. 2. Enclosed (A,B) view of series. The series of course also is fine if we examine it just on some subinterval (A,B) which is contained in (a,b). On (A,B) the series exists and converges. The coefficients are the same fn shown above. They are NOT Fn ≡!Syntax Error, If(x) φn(x), although the Fn can surely be computed. 3. Aside: Since we can compute these Fn , we can at least ask about the possible series Σn Fnφn(x). Since the Fn differ from the fn, we know that Σn Fnφn(x) ≠ f(x) on all of (a,b) because the fn are unique for a given f(x). On the interval (a,b), since φn(x) are complete, it is possible that Σn Fnφn(x) = g(x), ie, the series converges to some well defined function. Since φn are complete on (a,b), the only malfunction here would be if this series failed to converge. If it converges, then the series is good. In this case, we would conclude that we can write Fn = !Syntax Error, Ig(x) φn(x). Then we have Fn = !Syntax Error, Ig(x) φn(x) = !Syntax Error, If(x) φn(x). 4. Extension by 0 of the enclosed interval function. Here is one possibly useful concept. We are handed f(x) on the internal interval (A,B) on which the φn are not a complete set. We can "extend" f(x) to the larger interval (a,b) [ on which φn are a complete set] by making it be 0 outside (A,B): In this case we can expand the extended function on the complete set φn so that f(x) = Σn Fnφn(x) where Fn =!Syntax Error, If(x) φn(x) [ = !Syntax Error, If(x) φn(x) = fn ] as above. In this situation, the series f(x) = Σn Fnφn(x) certainly "exists" despite the fact that a counter theorem wants to make it not exist. The counter theorem is that on (A,B), the φn do not form a complete set, they are just a spanning set, and therefore Σn Fnφn(x) on (A,B) does not exist as would be the case if the φn were just powers. But we see that the counter theorem is "overridden" if (1) the φn are a complete set on some enclosing interval, and (2) on the enclosing interval we extend f(x) by making it be 0 outside the enclosed interval. 5. Bowl problem applying ∫dΩs Ynm(Ωs) to both sides of V(s) equation. Let's now try to apply this idea to our half sphere situation. Start with this general statement, where b means an integral over the bowl, and where we use the Stak version of the 1/R expansion, but let's change units so E = 1/R so we are in green Jackson units. Then V(s) = ∫b dSξ E(s|ξ)σ(ξ) = ∫b dSξ 1/R σ(ξ) = a2 ∫b dΩξ 1/R σ(ξ) = a2 ∫b dΩξ { (1/a) Σn" N0n" Σm" (1/Nm"n") Yn"m"(Ωξ) Y*n"m"(Ωs)} σ(ξ) = a Σn" N0n" Σm" (1/Nm"n") Y*n"m"(Ωs) [ ∫b dΩξ Yn"m"(Ωξ) σ(ξ) ] The object [...] is analogous to !Syntax Error, If(x) φn(x) of our last section, an integral over an enclosed interval so to speak. Now apply ∫dΩs Ynm(Ωs) to both sides where we are doing a full 4π integration. We get ∫dΩs Ynm(Ωs) V(s) = a Σn" N0n" Σm" (1/Nm"n") {∫dΩs Ynm(Ωs) Y*n"m"(Ωs )} [ ∫b dΩξ Yn"m"(Ωξ) σ(ξ) ] Using the Stak p 395 result we have ∫dΩs Ynm(Ωs) Y*n"m"(Ωs ) = δnn"δmm" Nnm so we get ∫dΩs Ynm(Ωs) V(s) = a Σn" N0n" Σm" (1/Nm"n") { δnn"δmm" Nnm } [ ∫b dΩξ Yn"m"(Ωξ) σ(ξ) ] = a N0n (1/Nmn) { Nnm } [ ∫b dΩξ Ynm(Ωξ) σ(ξ) ] = a N0n [ ∫b dΩξ Ynm(Ωξ) σ(ξ) ] = a {4π/(2n+1) } [ ∫b dΩξ Ynm(Ωξ) σ(ξ) ] = 4πa (2n+1)-1 [ ∫b dΩξ Ynm(Ωξ) σ(ξ) ] Meanwhile, we know that V(s) is really V(θs) only. So the LHS becomes ∫dΩs Ynm(Ωs) V(s) = δm0 ∫dΩs Yn0(Ωs) V(s) = 2πδm0 ∫dzs Pn(zs) V(zs) We can make the same argument on the [...] integral so say that ∫b dΩξ Ynm(Ωξ) σ(ξ) = 2πδm0 ∫b dzξ Pn(zξ) σ(zξ) Our overall equation then becomes: 2πδm0 ∫dzs Pn(zs) V(zs) = 4πa (2n+1)-1 2πδm0 ∫b dzξ Pn(zξ) σ(zξ) If m ≠ 0, this says 0 = 0 which is fine. So only real content is when m = 0 and we have ∫dzs Pn(zs) V(zs) = 4πa (2n+1)-1∫b dzξ Pn(zξ) σ(zξ) where on the LHS he have the enclosing region integral (-1,1), and on the RHS an enclosed region (0,1) for the half shell bowl. If we had a full sphere at this point, then V(zs) = V0 for all zs in (-1,1) and we then get ∫dzs Pn(zs) V(zs) = 4πa (2n+1)-1∫b dzξ Pn(zξ) σ(zξ) 2 δn0 V0 = 4π a(2n+1)-1∫b dzξ Pn(zξ) σ(zξ) 2 V0 = 4π a∫b dzξ σ(zξ) = a 2 σ0 => V0 = 4π a σo Q would then be Q = 4πa2σ0 = 4πa2(V0/4πa) = V0a, so C = Q/V0 = a. So where has this left us? For the bowl situation, we end up with this general equation: ∫dzs Pn(zs) V(zs) = 4πa (2n+1)-1∫b dzξ Pn(zξ) σ(zξ) where we know V(zs) = V0 on the bowl, but we don't know it beyond the bowl edge. And we don't know σ(zs) on the bowl. So both sides of this equation contain unknown functions. Unfortunately, I don't see how this equation can be connected with our discussion of section 4 above. I have managed to get rid of all azimuthal stuff and just have the Legendre functions appear. 6. Bowl problem applying ∫b dΩs Ynm(Ωs) to both sides of V(s) equation. Suppose we were to apply ∫b dΩs Ynm(Ωs) to both sides instead of a full integral. Then looking at the above, we would have: ∫b dΩs Ynm(Ωs) V(s) = a Σn" N0n" Σm" (1/Nm"n") {∫b dΩs Ynm(Ωs) Y*n"m"(Ωs )} [ ∫b dΩξ Yn"m"(Ωξ) σ(ξ) ] We could right at this point make our two replacements: ∫b dΩs Ynm(Ωs) V(s) = δm0 ∫bdΩs Yn0(Ωs) V(s) = 2πδm0 ∫b dzs Pn(zs) V(zs) ∫b dΩξ Yn"m"(Ωξ) σ(ξ) = 2πδm"0 ∫b dzξ Pn"(zξ) σ(zξ) and rewrite the above as 2πδm0 ∫b dzs Pn(zs) V(zs) = a Σn" N0n" Σm" (1/Nm"n") {∫b dΩs Ynm(Ωs) Y*n"m"(Ωs )} [ 2πδm"0 ∫b dzξ Pn"(zξ) σ(zξ) ] = 2πa Σn" N0n" (1/N0n") {∫b dΩs Ynm(Ωs) Y*n"0(Ωs )} [ ∫b dzξ Pn"(zξ) σ(zξ) ] = 2πa Σn" {∫b dΩs Ynm(Ωs) Y*n"0(Ωs )} [ ∫b dzξ Pn"(zξ) σ(zξ) ] But of course we know that ∫b dΩs Ynm(Ωs) Y*n"0(Ωs ) = δm0 ∫b dΩs Yn0(Ωs) Y*n"0(Ωs ) = 2π δm0 ∫b dzs Pn(zs) Pn"(zs) so we then get 2πδm0 ∫b dzs Pn(zs) V(zs) = 2πa Σn" {2π δm0 ∫b dzs Pn(zs) Pn"(zs) } [ ∫b dzξ Pn"(zξ) σ(zξ) ] Again if m ≠ 0 we get 0 = 0, so we take m = 0 to get ∫b dzs Pn(zs) V(zs) = 2πa Σn" { ∫b dzs Pn(zs) Pn"(zs) } [ ∫b dzξ Pn"(zξ) σ(zξ) ] But we know V(zs) = V0 for the entire LHS integration, so we can rewrite as V0 ∫b dzs Pn(zs) = 2πa Σn" { ∫b dzs Pn(zs) Pn"(zs) } [ ∫b dzξ Pn"(zξ) σ(zξ) ] We might then define some known integrals ∫b dzs Pn(zs) ≡ kn(b) ∫b dzξ Pn(zξ) Pn"(zξ) = hnn"(b) Then our equation above becomes kn(b) V0 = 2π a Σn" hnn"(b) [ ∫b dzξ Pn"(zξ) σ(zξ) ] Finally, let's define our "enclosed integral" projection this way σn"(b) ≡ ∫b dzξ Pn"(zξ) σ(zξ) Then we have kn(b) V0 = 2πa Σn" hnn"(b) σn"(b) So this is what happened to our early statement that ∫σ E(s|ξ)I(ξ)dSξ = V0 . It is a matrix equation K(b) = H(b) σ(b) K(b)n = kn(b) V0/(2πa) We can solve this if H can be inverted, so get σ(b) = H(b)-1 K(b) So this is all "well and good", but again, I see no connection to section 4 above. It does suggest a new method of solving the bowl problem. You might use this last line to find that values of σn"(b). We do know that σn"(b) ≡ ∫b dzξ Pn"(zξ) σ(zξ) = ∫ dzξ Pn"(zξ) σ(zξ) because we know σ(zξ) = 0 beyond the bowl. Therefore we can invert to get σ(zξ) in terms of σn"(b). Here then we are "extending" the function σ(zξ) by having it be 0 outside our enclosed interval. In this last step we are then in effect adding the second dual integral equation boundary condition. 7. The V(s) equation without applying any ∫dΩs Ynm(Ωs) . I was hoping in doing the above work to arrive at an equation of this form: (1/2) Σn Pn(zs) !Syntax Error, I dzξ Pn(zξ) σ(zξ) = 1 half sphere valid (0,1) for zs and then I could talk about whether this series exists or not. But neither of the two threads I just traced out in sections 5 or 6 led to such an equation! So now I have to go study Part C of "potential of half-spherical shell.doc" to see how this arises. In both my threads above, I applied some sort of ∫dΩs Ynm(Ωs) to both sides. In case 5 it was the full integral, in case 6 the bowl only integral. But in Part C we don't apply any operator like this. So let's back up to the start of section 5 above where we had V(s) = ∫b dSξ E(s|ξ)σ(ξ) = ∫b dSξ 1/R σ(ξ) = a2 ∫b dΩξ 1/R σ(ξ) = a2 ∫b dΩξ { (1/a) Σn" N0n" Σm" (1/Nm"n") Yn"m"(Ωξ) Y*n"m"(Ωs)} σ(ξ) = a Σn" N0n" Σm" (1/Nm"n") Y*n"m"(Ωs) [ ∫b dΩξ Yn"m"(Ωξ) σ(ξ) ] [ This is in Jackson units, and I can get back to Part C Stak units by multiplying V by 4π. ] Right at this point, let us use this fact: ∫b dΩξ Yn"m"(Ωξ) σ(ξ) = 2πδm"0 ∫b dzξ Pn"(zξ) σ(zξ) to get V(s) = a Σn" N0n" Σm" (1/Nm"n") Y*n"m"(Ωs) [ ∫b dΩξ Yn"m"(Ωξ) σ(ξ) ] = a Σn" N0n" Σm" (1/Nm"n") Y*n"m"(Ωs) [ 2πδm"0 ∫b dzξ Pn"(zξ) σ(zξ) ] = a Σn" N0n" (1/N0n") Y*n"0(Ωs) [2π ∫b dzξ Pn"(zξ) σ(zξ) ] = 2πa Σn Pn(zs) [ ∫b dzξ Pn(zξ) σ(zξ) ] [ Aside: we can at this point, if we like, write this as V(s) = 2πa ∫b dzξ σ(zξ) { Σn Pn(zs) Pn(zξ) } ] As at the end of section 6 we define then σn(b) ≡ ∫b dzξ Pn(zξ) σ(zξ) and we end up with V(zs) = 2πa Σn Pn(zs) σn(b) This equation (assuming series exists) is valid for all zs in (-1,1). We know that when zs lies on the bowl, this equation becomes: V0 = 2πa Σn Pn(zs) σn(b) half sphere valid (0,1) for zs and THIS then is the equation I have been looking for. So, does this series exist or not? series f(zs) ≡ Σn Pn(zs) σn(b) valid for zs in (β,1) β = 0 for half sphere where σn(b) ≡ ∫b dzξ Pn(zξ) σ(zξ) // = Fn Well, I think the answer is quite simple. The series V0 = 2πa Σn Pn(zs) σn(b) is just a "peephole view" looking at the enclosed interval (β,1) of the following series V(zs) = 2πa Σn Pn(zs) σn(b) which is defined on the full enclosing interval (-1,1). On this interval the Pn are a complete set of functions, and the series exists and all is well. If the enclosing series exists, then of course if you examine it just on the enclosed interval, it still exists. Conclusion: The series on the RHS of this equation -- which equation is only true for limited zs -- V0 = 2πa Σn Pn(zs) σn(b) half sphere valid (0,1) for zs does in fact exist, Now, what about the idea of "extending" the function σ(zξ) to the full range by making it be 0 outside the bowl. Physically we know we can do this, because we know σ really is 0 outside the bowl. Using this method, we can show that a slightly different series also exists. Consider what the charge density looks like in our bowl problem, where we regard the full curve as the extended σ or σext(z) , σext(z) exists on the full (-1,1), and matches σ(z) on (β,1). We know we can expand σext(z) on the Pn(z) in an "existent" series with some coefficients which we write this way σext(z) = Σn σ'n Pn(z) where kn σ'n = !Syntax Error, I dz Pn(z) σext(z) = !Syntax Error, I dz Pn(z) σ(z). where kn = 1/(n+1/2) is a normalizing factor since the Pn are not orthonormal, recall fn = !Syntax Error, Idx f(x) Pn(x) f(x) = (1/2)Σn(2n+1)fn Pn(x) If we now look in the peep-hole bowl region, we can write this as σ(z) = Σn σ'n Pn(z) where kn σ'n = !Syntax Error, I dz Pn(z) σext(z) = !Syntax Error, I dz Pn(z) σ(z). So this is another "series" that we can consider being valid on (β,1) which is not the full (-1,1) but the series definitely exists because it is a peep-hole view of another series we know exists. 8. A quick review and statement of equations in Stakgold Units. (a) We derived this equation, where I now suppress the parameter b , V(zs) = 2πa Σn Pn(zs) σn where σn ≡ ∫b dzξ Pn(zξ) σ(zξ) and the ∫b came from our original integral equation " ∫b E(s|ξ)I(ξ)dSξ = 1 ". This equation "knows about" the fact that σ = 0 outside the bowl. If σ were not zero outside the bowl, this equation as stated would not be true because Σ q/R would have to include this extra σ. (b) The equation is valid on the entire (-1,1) for zs , on which the Pn are complete, so the series which appears, namely Σn Pn(zs) σn, does not have "existence issues" in the sense that all the coefficients keep moving because Pn is only a spanning set. If we take a peep-hole view of the above equation, we get V0 = 2πa Σn Pn(zs) σn for zs in (β,1) σn ≡ ∫b dzξ Pn(zξ) σ(zξ) so this series also exists since it is a peephole view of the previous series. (c) As a separate subject, we can consider a series expansion of σ(z) and we find on all of (-1,1) that σext(z) = Σn σ'n Pn(z) where kn σ'n = !Syntax Error, I dz Pn(z) σext(z) = !Syntax Error, I dz Pn(z) σ(z). In this case, the peephole view is σ(z) = Σn σ'n Pn(z) where kn σ'n = !Syntax Error, I dz Pn(z) σext(z) = !Syntax Error, I dz Pn(z) σ(z). The first series is defined on all of (-1,1) and thus has no existence issues. The second equation is a peephole view of the first, and therefore also has no existence issues. (d) The coefficients are different for the two series considered in (b) and (c), but involve the exact same integrals. The difference is the normalizing factor kn which involves (2n+1). (e) So, my earlier conclusion that the change of order of integration and summation created a non-existent series was WRONG !!! I now have to backtrack and fix that up wherever that conclusion is used. (f) If we want to convert our last several equations to Stak units, multiply V by 4π. Thus: 2V(zs) = a Σn Pn(zs) σn(b) 2V0 = a Σn Pn(zs) σn(b) half sphere valid (0,1) for zs and of course I was using V0 =1 and sphere radius a = 1, so these then become 2V(zs) = Σn Pn(zs) σn(b) 2 = Σn Pn(zs) σn(b) half sphere valid (0,1) for zs Our "kernel form" equation above was V(s) = 2πa ∫b dzξ σ(zξ) { Σn Pn(zs) Pn(zξ) } ] in Jackson, so we can convert it as well to Stak to get 2 V(s) = a ∫b dzξ σ(zξ) { Σn Pn(zs) Pn(zξ) } and again with a = 1 and s on the bowl and V0 = 1 this becomes 2 = ∫b dzξ σ(zξ) { Σn Pn(zs) Pn(zξ) } = ∫b dzξ σ(zξ) K(zs, zξ) 9. Proof that !Syntax Error, Idx f(x) Pn(x) = δn0 has no solution f(x). I now realize that I had a proof sitting there in my first doc [ end of Section 4, "how do we resolve this mystery"] . Here I will formalize it, motivated a bit by what happened in this second doc. We start with our equation to be solved !Syntax Error, Idx f(x) Pn(x) = δn0 n = 0,1...∞ Assume that some solution f(x) exists. Extend that solution to the full (-1,1) interval by adding a zero piece to the left, call this fext(x). Surely we can do that. Then we have fext(x) = f(x) θ(x) on the interval (-1,1). We can certainly "work with" this function. Here is the Legendre expansion and projection for this function, on the interval (-1,1) for which the Pn are a complete set: cn = !Syntax Error, Idx fext(x) Pn(x) fext(x) = Σn=0∞(n+1/2)cn Pn(x) But of course we know then that cn = !Syntax Error, Idx fext(x) Pn(x) = !Syntax Error, Idx f(x) θ(x) Pn(x) = !Syntax Error, Idx f(x) Pn(x) = δn0 The expansion then says: fext(x) = Σn=0∞(n+1/2)cn Pn(x) = Σn=0∞(n+1/2) δn0 Pn(x) = (1/2) We then end up with the following statement on (-1,1): f(x) θ(x) = 1/2 For x = -1/2, say, this says 0 = 1/2 But that is a contradiction. Therefore our "assumption" that a solution f(x) exists must be false. QED. In simple terms, we have a Legendre expansion on (-1,1) and we are given cn = δn0 right at the start. This completely determines fext(x) to be fext(x) = 1/2 on the interval (-1,1). We cannot on the one hand accept the data cn = δn0 and then on the other hand try to insist that the solution have the form fext(x) = θ(x)f(x) on (-1,1). This is too many conditions on the problem and there is no solution f(x). In our two docs, we learned the hard way that f(x) does not exist. Restate once again. We imagine a function fext(x) that is 0 on (-1,0) and is some f(x) on (0,1). Can the Legendre expansion coefficients be δn0 for this function fext(x)? Well, those coefficients imply that the function is fext(x) =(1/2) on (-1,1). But we assumed fext(x) is 0 on (-1,0). So the answer is: No, δn0 cannot be the coefficients of any such function fext(x). In other words, no such function fext(x) exists which has δn0 as coefficients. In turn, this means that f(x) does not exist where fext(x) = θ(x)f(x). That means that our infinite set of integral equations !Syntax Error, Idx f(x) Pn(x) = δn0 has no solution f(x).