/*	Howell - do binomial expansion of : 
/$		(1 - β^2*sin(θ)^2)^(-3/2)  
  Binomial series - Kreyszig1972 p52h0.0 Equation (2) 
/$		(1 + z)^(-m) = sum(n = 0 to ∞: choose( - m,n)*z^n)
/*  where choose() is the choose operation 
/$		(1 + z)^(-m) 
		= 1 - m*z 
			 + m*(m + 1)/2!*z^2 
			 - m*(m + 1)*(m + 2)/3!*z^3 
			 + ...

/*  Substituting for z= (-β^2*sin^2(O)), m=3/2 
/$		 (1 -       β^2*sin(θ)^2)^(-3/2)  
		= 1 - 3/2*(-β^2*sin(θ)^2) +  3/2*(3/2 + 1)/2!*(-β^2*sin(θ)^2)^2 -   3/2*(3/2 + 1)*(3/2 + 2)/3!*(-β^2*sin(θ)^2)^3 + ...
		= 1 + 3/2  *β^2*sin(θ)^2  - 15/8             *  β^4*sin(θ)^4    + 105/48                      *  β^6*sin(θ)^6    + ...
		= 1 + 3/2  *β^2*sin(θ)^2  - 15/8             *  β^4*sin(θ)^4    +  35/16                      *  β^6*sin(θ)^6    + ...

/* 20Aug2019 correction - Above I had put in Lucas's factors, NOT my binomial series results!
	 Corrected version : 
	Substituting for z= (-β^2*sin^2(O)), m=3/2 
/$		 (1 -       β^2*sin(θ)^2)^(-3/2)  
		= 1 + 3/2*β^2*sin(θ)^2 + 3/8*β^4*sin(θ)^4 - 1/16*β^6*sin(θ)^6 + 3/128*β^8*sin(θ)^8 - 3/256*β^10*sin(θ)^10 + ...


/*++++++++++++++++++++++++++++++++++++++
/*add_eqn  "no_issue   
   04_38
   Binomial_expansion_for_E0_terms 

/$L		(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 + ...  
/%H		(1 - β^2*sin(θ)^2)^(-3/2)  = 1 + 3/2*β^2*sin(θ)^2 +  3/8*β^4*sin(θ)^4 -  1/16*β^6*sin(θ)^6 + 3/128*β^8*sin(θ)^8 - 3/256*β^10*sin(θ)^10 + ...

/*	WRONG! -  Lucas has the wrong coefifficients for a binomial series!