least squares function 

	i varies from 1 to N.

	L2 = ∑[ yi - f(yi, pi)]2	(1)
	∂[L2]/∂pi = 0	(2)

	For a linear function f(yi, pi) = a xi + b 
	
	Thus:
	L2 = ∑[ yi - (a xi + b) ]2 =
	∑ [ yi2 - 2 yi(a xi + b) + (a xi + b) 2]=
	∑ [yi2 - 2 a yi xi - 2b yi + a2 xi2 + b2 + 2ab xi] 


	The condition (2) gives:

	∂[L2]/∂a = 0
	∂[L2]/∂b = 0
	
	The first leads to:
	∑ [a xi2 + b xi - yixi] = 0
	The second to:
	∑ [a xi - yi + b] = 0 
	
	Then:

	a ∑xi2 + b ∑xi - ∑yixi = 0
	a ∑xi - ∑yi + N b = 0 

	The latter gives:
	b = (1/N) [∑yi - a ∑xi]
	This value, substituted to the former leads to:

	a ∑xi2 + [(1/N) [∑yi - a ∑xi]] ∑xi  = ∑yixi. 
	
	Thus:

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	a = [N ∑xiyi - ∑xi ∑yi]/[N ∑xi2 - (∑xi)2]
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	and 
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	b = [∑yi ∑xi2 - ∑xi ∑xiyi]/[N ∑xi2 - (∑xi)2]
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