






A Preliminary DerivationGiven a fraction a/b, one may multiply or divide its top and bottom ("numerator and denominator") by the same number c:
where (remember?) the two letters ac stand for "a times c" and similarly for bc. That is so because (c/c) = 1, no matter what the value of c is (except of course zero: "Thou Shalt not Divide by Zero") and multiplying anything by 1 does not change its value. In multiplying fractions, the rule is to multiply top with top, bottom with bottom, so we get
As for dividing top and bottom by the same number d
it follows at once from the preceding, if we choose the number c to equal 1/d.
Working with Small QuantitiesSome equations, identities or formulas contain small quantities, and these can be made much simpler and easier to use by sacrificing a little accuracy. In fact, some equations which have no simple solution at all (like Kepler's equation in section (12a)) can yield in this way an approximate solution, often good enough for most uses, or else open to further improvement.Many such calculations make use of the following observation. When we derive squares, 3rd powers, 4th powers etc. of numbers larger than 1, the results are always bigger, while for numbers smaller than 1, the results are always smaller. For example:

power  More than 1  Less than 1 
number  10  0.1 
square  100  0.01 
3rd power  1000  0.001 
4th power  10,000  0.0001 
The above also holds for negative numbers, if one understands "bigger" and "smaller" to refer to the absolute value (the value without sign). For instance:

power  More than 1  Less than 1 
number  – 10  – 0.1 
square  100  0.01 
3rd power  – 1000  – 0.001 
4th power  10,000  0.0001 
5th power  – 100,000  – 0.00001 
Say z is a number much smaller than 1 (written z << 1, or for absolute values z << 1). Then by the identity near the end of of section M4
Since z^{2} is much smaller than 1 or z, we can write, using the symbol ~ for "approximately equal"
and dividing both sides by (1 – z)
(Many texts use the symbol ~ not alone but placed above an equal sign). For example (check with your calculator)
then 1/(1– z) = 1/0.99 = 1.010101... The basic rule is: one may neglect small quantities like z, z^{2}, z^{3} 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
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
which can also be obtained from the earlier equation
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:
Multiply out the left side: However, if z^{2} and z^{3} are much smaller than z, then dropping the terms containing them only increases the error slightly, leaving
The next section is optional. A step beyond: The Binomial TheoremFormally 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
and similarly (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 socalled binomial theorem. For those interested, that formula states
where the denominator of the fraction preceding any power z^{n} is obtained by multiplying together the whole numbers up to n, giving the number generally denoted as n! and called "n factorial." For example If a is a positive whole number, the sequence a, (a1), (a2)... 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 z^{a} and we get formulas like the one derived earlier for a=3:

Note: Why not divide by zero? It does not work. There exists no number like 1/0 (except maybe infinity, which is not a regular number), and use of expressions like 0/0 can lead to contradictions such as 2 = 3. 
Next Stop: #M6 The Theorem of Pythagoras
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last edited 27 October 2016