Lucas
	Gausss electrostatic Law from (4-11)()  
/$		ET(r,v)   
		=  E0(r) + Ei(r,v) 
		=  E0(r)*(1 - λ(v))/(1 - β^2*sin(θ)^2)^(3/2) 

/*--+
  First approach - using Lucas's (4-37)
  Howell - Lucas's result for 4-37 Er_second_iteration 
/$		ET(r,v)   
		=   E0(r)     *(1 + 3/2*β^2*sin(θ)^2 + 15/8*β^4*sin(θ)^4 + 35/16*β^6*sin(θ)^6 + ...)  
		  - E0(r)*λ(v)*(1 + 3/2*β^2*sin(θ)^2 + 15/8*β^4*sin(θ)^4 + 35/16*β^6*sin(θ)^6 + ...)  
/*	Using the binomial expansion result of (4-38) : 
	04_38   Binomial_expansion_for_E0_terms 
/$		(1 - β^2*sin(θ)^2)^(-3/2)  = 1 + 3/2*β^2*sin(θ)^2 + 15/8*β^4*sin(θ)^4 + 35/16*β^6*sin(θ)^6 + ...  
		ET(r,v)   
		=   E0(r)     *(1 - β^2*sin(θ)^2)^(-3/2)  
		  - E0(r)*λ(v)*(1 - β^2*sin(θ)^2)^(-3/2)  
		ET(r,v) = (1 - λ(v))*E0(r)/(1 - β^2*sin(θ)^2)^(3/2)  
/* so this works well starting with Lucas's (4-37), but the latter may have incorrectly retained the E0*lambda term!?

/*----+
	Second approach - using Howells (4-37 iteration4 26Jun2016)
	I did NOT get the same kind of result as Lucas - mine is more like Newtons binomial expansion, than the binomial series that is used in the text.
	Isaac Newton  :
/$		(1 - x^2)^(3/2) = 1 - 3/2*x^2 + 3/8*x^4 + 1/16*x^6 ...
/*	Howell 4th iteration : 
/$		ET = + E0     *(1 + 3/2*(β*sin(θ))^2 +  3/4 *(β*sin(θ))^4 +  1/4  *(β*sin(θ))^6 +  1/8 *(β*sin(θ))^8 +  3/32 *(β*sin(θ))^10 )
			  - E0*λ(v)*(                                    +  1/6  *(β*sin(θ))^6 +  1/8 *(β*sin(θ))^8 )

/*	26Jun2016  Further iterations will PERHAPS reduce the value of coefficents beyond the "betaSin^4" term, but it appears that the coefficient of the "betaSin^4" term itself WONT reduce with further iterations (?).
	Binomial expansion of 
/$		(1 - (β*sin(θ))^2)^(3/2) 
/*	would give : 
/%		ETods(POIo,t=0)  
		= + E0ods(POIo,t=0)*(1 + 3/2*β*sin(Aθoc(POIo))^2 +  3/8 *β*sin(Aθoc(POIo))^4 -  1/16 *β*sin(Aθoc(POIo))^6 +  3/128*β*sin(Aθoc(POIo))^8 - 3/256*β*sin(Aθoc(POIo))^10   +  7/1024*β*sin(Aθoc(POIo))^12 + ...  )
		= + E0ods(POIo,t=0)*(1 - 3/2*β^2*sin(Aθpc(POIo(t),t=0))^2)^(3/2) 

/*	I assume that the E0*L(v)* term disappears with iterations, and that Howells 4th iteration is "immature" and probably harbours errors.
	Therefore, at this time, just take the binomial expansion instead : 
/%		ETods(POIo,t=0) 
		= E0ods(POIo,t=0)*(1 - 3/2*β^2*sin(Aθpc(POIo(t),t=0))^2)^(3/2) 
		= E0ods(POIo,t=0)*(1 + 3/2*β*sin(Aθoc(POIo))^2 +  3/8 *β*sin(Aθoc(POIo))^4 -  1/16 *β*sin(Aθoc(POIo))^6 +  3/128*β*sin(Aθoc(POIo))^8 - 3/256*β*sin(Aθoc(POIo))^10   +  7/1024*β*sin(Aθoc(POIo))^12 + ...  )



/*++++++++++++++++++++++++++++++++++++++
/*add_eqn  "no_issue   
   04_39
/*	E(r,v) for constant velocity, non-point charge, observer reference frame  
/$L		ET(r,v) = (1 - λ(v))*E0(r)/(1 - β^2*sin(θ)^2)^(3/2)  
/$H		ET(r,v) = (1 - λ(v))*E0(r)/(1 - β^2*sin(θ)^2)^(3/2)  
/*	WARNING : simple using Lucas's (4-37), but is this incorrect? Second approach using Newtons expansion shown as well.   
	In the second approach : (1 - L(v)) does NOT appear!!  
   
/%		ETods(POIo,t=0) = E0ods(POIo,t=0)*(1 - 3/2*β^2*sin(Aθpc(POIo(t),t=0))^2)^(3/2)