# link  d_Qndfs  'economics, markets/Puetz - Universal Waves Series, model derivations.txt'
# www.BillHowell.ca  20Apr2020 initial taken fromn 18Apr2020 'Puetz - Universal Waves Series, header.ndf'
# view this file in a text editor, with [constant width font, tab = 3 spaces], no line-wrap




#****************************
# List of derivations, generated with :
# $ cat  "$d_Qndfs""Puetz - Universal Waves Series, model derivations.txt"  |  grep  "^#]"  |  sed 's/^#\]/  /'
#


#***********
#] References
# the [formulae, data] come from :
[1]	Stephen Puetz, Glenn Borchardt 2011  "Universal Cycle Theory, Neomechanics of the hierarchically infinite universe" www.OutskirtsPress.com, ISBN 978-1-4327-8133-0  http://www.uct-news.com 
# see also : 
[2]	Stephen Puetz 2009 "The Unified Cycle Theory, How cycles dominate the structure of the universe and influence life on earth" www.OutSkirtsPress.com, 489pp, ISBN 978-1-4327-1216-7 http://www.uct-news.com 
# the following table is copied-over from "$d_Qndfs""Puetz - Universal Waves Series.ods"

# tables of cycle parameters
#		see p565-566 Tables C3.2.1-.6 for 		  UWS constants
#		see p567		 Tables C3.3.1-.3 for Double-UWS constants


#*************************
#]  Derivation of t_lag_cycle(n) from θ(n)
20Apr2020 initial

COMMENT :  time lags are NOT the same as phase angles!
blah, blah, blah


# page 564h0.7 Equation A1.3.1, rename t_lag to t_lag_app
(1)	y(i) =  A*sin((2*pi*(t(i) + t_scale 				  + t_lag_app)/lambda(n)) +  theta(n))

I will use [t_lag_cycle, t_lag_app]
		y(i) =  A*sin((2*pi*(t(i) + t_scale + t_lag_cycle + t_lag_app)/lambda(n)))
therefore  
		2*pi*(t(i) + t_scale + t_lag_app)				  /lambda(n)  +  theta(n)
	=	2*pi*(t(i) + t_scale + t_lag_cycle + t_lag_app)/lambda(n)
...
	theta(n) =  	2*pi*(t(i) + t_scale + t_lag_cycle + t_lag_app)/lambda(n)  
					-  2*pi*(t(i) + t_scale 				  + t_lag_app)/lambda(n)		 
...
	theta(n)*lambda(n)/(2*pi)
	=     (t(i) + t_scale + t_lag_cycle + t_lag_app)
		-  (t(i) + t_scale 					+ t_lag_app)
	=  t_lag_cycle
or ...
	t_lag_cycle  =  theta(n)*lambda(n)/(2*pi)




#*************************
#]  Derivation of θn (initial attempts - not used)

18Apr2020
# page 564h0.7 Equation A1.3.1
(1)	y(i) =  A*sin((2*pi*(t(i) + t_scale + t_lag)/lambda(n)) +  theta(n))

Example theta_1st_max(n), p465h0.8, use Table A1.4.2, p465h0.8 : 
	y(i)/A = 1
	t_lag		= 0, is the lag time that depends on the process being investigate (here no time lag)
	t_scale	= -2000, for BP/AP with year 0 Gregorian as basis
	t(i)		= "present" = 2010
	t_inc		= 1/(3*365), time increment down column
	n			= 3, for 28.7 day period
	lambda(n)= 0.0028970478728148
	theta(n)	= 3.391242799
Example : 
	y(i)		=  A*sin((2*pi()*(t(i) + t_scale + t_lag)/lambda(n)) +  theta(n))
	y(i)/A	=    sin((2*pi()*(t(i) + t_scale + t_lag)/lambda(n)) +  theta(n))
				=    sin( 2*pi()*(2010 - 2000		+ 0	 )/0.0028970478728148 + 3.391242799)
	OK - this worked just as in the text.  I made many mistakes getting there!
	

Derive formula for theta(n) :
	theta(n) 
	=  arcsin(y(i)/A) - { 2*pi*(t(i) + t_scale + t_lag) /lambda(n) }  - not a function as it's many-valued
	=  arcsin(y(i)/A) -   2*pi*(t(i) + t_scale + t_lag) /lambda(n)

theta(n) : is an angle, so it can vary from +-infinity, 
	but p565h0.6 Table C3.2.1 varies from +-2*pi, so it takes the first max? 
	we typically want just the first max, for which arcsin(y(i)/A) = 1 : 
		arcsin(y(i)/A) = arcsin(1) =  pi/2
		so take the modulus (remainder) after dividing by lambda(n)

therefore : 
	theta_1st_max(n) 
	=  modulus[pi()/2 - 2*pi()*(t(i) + t_scale + t_lag)/lambda(n), 2*pi()]

Assumptions - follow the same parameters as for the example above, but use the updated table
	y(i)/A 	= 1
	t_lag		= 0, is the lag time that depends on the process being investigate (here no time lag)
	t_scale	= -2000, for BP/AP with year 0 Gregorian as basis
	t(i) 		= "present" = 2010
	n			= formula will be use to calculate all n
	lambda(n)= as calculated from n=0 (multiples of 3)
then : 
	theta_1st_max(n) 
	=  modulus[pi()/2 - 2*pi()*(t(i) + t_scale + t_lag)/lambda(n), 2*pi()]
	=  modulus[pi()/2 - 2*pi()*(2010	- 2000	 + 0	  )/lambda(n), 2*pi()]
	=  modulus[pi()/2 - 2*pi()*10/lambda(n), 2*pi()]
This does NOT work :
	theta_1st_max(n) :
		given  = 1.910577326
		result = 2.892167090
		large n give constant theta_1st_max(n) = 1.570796327, which is wrong! 
			but it makes sense given my assumptions!
		
Try 2 Assumptions :
	y(i)/A 	= 1
	t_lag		= 0, is the lag time that depends on the process being investigate (here no time lag)
	t_scale	= 0, for BP/AP with year 2000 as basis
	t(i) 		= "present" = 2010
	n			= formula will be use to calculate all n
	lambda(n)= as calculated from n=0 (multiples of 3)
then : 
	theta_1st_max(n)
	=  modulus[pi()/2 - 2*pi()*(t(i) + t_scale + t_lag)/lambda(n), 2*pi()]
	=  modulus[pi()/2 - 2*pi()*(2010	+ 0		 + 0	  )/lambda(n), 2*pi()]
	=  3.272536818 vs 1.910577326 std

NO - I must solve problem at higher n
	this seems to mean that the time basis has changed - look at abbreviated
	divide by the [day, yr, kyr, My, Gy, Ty] - yes, as per section titles
	OK - this works
	
Now try to find t_lags to make it work
	find a t_lag for each section that works, these must align UWS to actual situation?  
	Perhaps can't calculate? 
	what is t(i) for each section? eg year 2000?
	
NYET - A better approach would be to assume that he manually set the θn, so manually input all of them




# enddoc