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(M-5) Deriving Approximate Results

A Preliminary Derivation

  Given a fraction a/b, one may multiply or divide its top and bottom ("numerator and denominator") by the same number c:

(a/b)   =   (ac)/(bc)

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

(a/b) (c/c)   =   (ac)/(bc)

As for dividing top and bottom by the same number d

(a/b)   =   [a/d]/ [b/d]

it follows at once from the preceding, if we choose the number c to equal 1/d.

Working with Small Quantities

  Some 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 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)

    If           z = 0.01,     (1+z) = 1.01,      (1– z) = 0.99,

    then      1/(1– z) = 1/0.99 = 1.010101...

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

and similarly

(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

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:   #M-6    The Theorem of Pythagoras

Author and Curator:   Dr. David P. Stern
     Mail to Dr.Stern:   audavstern("at" symbol)erols.com.