square root of sum vs. sum of quare roots for a certain form

Question

Given my original formula $\sqrt { \left( 1-a-b \right) \left( 1+c+d \right) } $, i notice that it is approximately equal to $\sqrt { \left( 1-a \right) \left( 1+c \right) } +\sqrt { \left( 1-b \right) \left( 1+d \right) } -1$ under many conditions. this is useful since each of the a,b,c,d can be brought out into its own square root. I wish to retian the individual sqrt terms as shown in the approximation, however, i was looking for a way to improve the match to the original formula, perhaps by adding or multiplying by a correction factor, or finding a better formula for breaking the square root of sums into sums of square roots, and then fixing it. is this possible? I also wish to keep extending it, as in approximating $\sqrt { \left( 1-a-b-c \right) \left( 1+d+e+f \right) } $ by something like

$\sqrt { \left( 1-a \right) \left( 1+d \right) } +\sqrt { \left( 1-b \right) \left( 1+e \right) } +\sqrt { \left( 1-c \right) \left( 1+f \right) } -2$

Thanks greatly, -bob.

Answer 1

Well, I am a programmer and not a mathematical wizard. I looked at your question and thought to myself, "Is that really true?" As a practical, hands-on sort of guy, I figured "Heck - let's just code it up and see..."

Python is a practical language - you don't have to understand the syntax, but here's a couple of computer functions to evaluate your formulas:

>>> import math, random
>>> def f1(a, b, c, d):
...   return math.sqrt((1-a-b) * (1+c+d))
...
>>> def f2(a, b, c, d):
...   t1 = math.sqrt((1-a) * (1+c))
...   t2 = math.sqrt((1-b) * (1 + d))
...   return t1 + t2 - 1
...

If you don't want to deal with imaginary numbers, then I guess you want (1-a-b) to be positive, as well as (1-a).

So, here's an evaluation with a,b,c,d all between 0 and 1 (that just happens to not go negative in (1-a-b) or (1-a)):

>>> a,b,c,d = [random.uniform(0,1) for x in range(4)]
>>> a,b,c,d
(0.3200221451898907, 0.5040920089635285, 0.6495066218935408, 0.8141228038413111)

>>> f1(a,b,c,d)
0.6582685966973517
>>> f2(a,b,c,d)
1.0075618762189977
>>>

I'm not sure what your idea of "under many conditions" really looks like, but is 0.658 approximately equal to 1.007? I guess that sort of depends on your problem domain.

I figured, well, maybe if c and d are bigger (since they don't cause the radical to go negative)?

>>> a,b,c,d = (0.51, 0.31415, 42, 21)
>>> f1(a,b,c,d)
3.354757815401881
>>> f2(a,b,c,d)
7.4746245552735076

I don't know... I guess I am unconvinced of some of the premises here and am not seeing why your initial two formulas would be considered approximately equal under "many conditions"?