Proving Bounds of γ'' & β'' in [T''] Matrix

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[ T'' ] = [ T ] [ T' ]

[ T'' ] =

[ ( γ'' ) , ( - γ'' β'' ) ]

[ ( - γ'' β'' ) , ( γ'' ) ]

γ'' = γ γ' ( 1 + β β' ) → no problem, proven to be b/w [ 1 , ∞ )

β'' = ( β + β' ) / ( 1 + β β' ) → can't figure out how to prove it's b/w [ 0 , 1 )
 
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Each of the individual betas ought to be less than 1, right?
 


Muphrid said:
Each of the individual betas ought to be less than 1, right?

Yes, for each β:

0 ≤ β < 1
 


So in the limiting case of each beta going to 1, what do you get for the final result?
 


Hint: Assuming both β and β' are in [0,1), what happens to β'' if β (or β') increases?
What about the limit β=β'=1?

Can you combine both to a formal proof?
 


Perhaps yet another way to go from here is to note that that looks awfully like the tangent addition formula ... and then making a substitution.
 


I've figured it out.

( β + β' ) / ( 1 + β β' ) - 1 + 1

[ ( β + β' ) - ( 1 + β β' ) ] / ( 1 + β β' ) + 1

[ - ( 1 - β ) ( 1 - β' ) ] / ( 1 + β β' ) + 1

1 ≤ ( 1 + β β' ) < 2

0 < ( 1 - β ) ≤ 1

0 < ( 1 - β ) / ( 1 + β β' ) ≤ 1

0 < ( 1 - β ) ( 1 - β' ) / ( 1 + β β' ) ≤ 1

-1 ≤ - ( 1 - β ) ( 1 - β' ) / ( 1 + β β' ) < 0

0 ≤ - ( 1 - β ) ( 1 - β' ) / ( 1 + β β' ) + 1 < 1
 
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