Say z is a number much smaller than 1 (written z << 1, or for absolute values |z| << 1). Then by the identity of section M-4|
(1 – z)(1 + z) = 1 – z2
Since z2 is much smaller than 1 or z, we can write, using the symbol ~ for "approximately equal"
(1 – z)(1 + z) ~ 1
and dividing both sides by (1 – z)
(1+z) ~ 1/(1– z)
(Many texts use the symbol ~ not alone but placed above an equal sign; however, that combination is not available for web documents). For example (check with your calculator)
which is close enough to (1+z) for many purposes.
The basic rule is: one may neglect small quantities like z, z2, z3 etc. when they are added to (or subtracted from) something much bigger. One may not do so if they are just multiplied or divided, because then, if they are removed, nothing is left of the expression containing them.
Here z can be either positive or negative. If we write z = – y, where y is a small number of opposite sign, we get
(1– y) ~ 1/(1+y)
which is another useful result, valid for any small number. If that small number is again renamed and is now called z (not the same z as before, of course), we get
(1– z) ~ 1/(1+z)
which can also be obtained from the earlier equation
(1 – z)(1 + z) ~ 1
by dividing both sides by (1 + z).
In section (34a) where the distance to the Lagrangian point L1 is derived, it turns out necessary to approximate 1/[1– z]3. You start from (1+z) ~ 1/(1– z) and raise both sides to their 3rd powers:
(1+z)3 ~ 1/(1– z)3
Multiply out the left side:(1 + z)3 = (1+z)(1+z)(1+z) = (1 + 3z + 3z2 + z3)
However, if z2 and z3 are much smaller than z, then dropping the terms containing them only increases the error slightly, leaving
1/(1– z)3 ~ 1 + 3z
The next section is optional.
A step beyond: The Binomial Theorem
Formally 1/(1–z)3 is (1– z) to the power -3. It suggests that more generally, for small z and for any value of a
(1–z)a ~ 1 – az
1/(1– z)a ~ 1 + az
(1 + z)a ~ 1 + az
1/(1 + z)a ~ 1 – az
(these are the same formula, for positive and negative z and a). That in fact is true, and a may be positive, negative or fractional. It is the consequence of a result first proved by Newton, his so-called binomial theorem. For those interested, that formula states
(1 + z)a = 1 + az + [a(a-1)/2] z2 + [a(a-1)(a-2)/6] z3 + ...
where the denominator of the fraction preceding any power zn is obtained by multiplying together the whole numbers (1,2,3... n), a number generally denoted as n! and called "n factorial."
If a is a positive whole number, the sequence a, (a-1), (a-2)... ultimately reaches zero, and the term where that first happens itself equals zero, as do all the ones that follow, all of whom contain a multiplier ("factor") zero. The series of powers of z then ends with za and we get formulas like the one derived earlier for a=3:
(1 + z)3 = (1 + 3z + 3z2 + z3)
Those cases of the binomial theorem were in fact known before Newton. What he showed that the theorem also held for negative and fractional values of a, where the series on the right side goes on to higher and still higher powers of z, without end. If z is small these powers quickly become negligibly small, and it is no great error to leave them out and write (for z of either sign)
1/(1 + z)a ~ 1/(1 + az) ~ 1 – az