By adding factors one-by-one, we may understand the effects of each factor
and thereby gain the ability to adjust our model to better simulate reality.
Example 1. [Arn1, §1.6]®[Arn1,
§1.7] ([Arn1, Fig. 9]®[Arn1,
Fig.11 (p.23, l.!11)]).
Example 2. [Arn1, §1.6]®[Arn1,
§1.8]®[Arn1,
§1.9]®[Arn1,
§1.10] ([Arn1, Fig.13 (Notice the
point of divergence [p.25, l.19])]®[Arn1,
Fig.14]®[Arn1, Fig.15]).
The process of mathematical modeling [Edw, p.4, l.-7-p.5,
l.-9].
We must be flexible when effectively establishing a model. Our goal is to
use a simple model to solve sophisticated problems. We should not use a sophisticated model
to solve simple problems.
[Wat1, §20.222] uses csc2 z to establish a model so that we may follow the pattern to derive the differential equation given in [Wat1, p.437, l.2].
In [Wat1, p.439, l.9-l.10], Watson proves c = 0 by using the power series of f(z)
and f '(z). He sticks with a strictly axiomatic approach by considering f(z) as
only a series. It is unnecessary to use this complicated method to establish a
simple model. We can let z = p/2 to prove c = 0. In
addition, [Wat1, p.436, l.-1] has nothing to do with
the above complicated method. Thus, Watson's complicated method involving
negative powers only creates a
huge obstacle for us to establish a simple model.
It would be difficult to understand that the development given in [Wat1,
§20.6] is based on standardization if we fail to recognize the
primitive model upon which the development is based. Since dz/dÃ(z)
= (4Ã3(z)
- g2Ã(z)
- g3)-1/2
and dz/d sin z = (1 - sin2
z)-1/2
are similar, z = ò{f(t)}-1/2
dt [Wat1, p.453, l.5] can be considered the corresponding counterpart of z =
ò(2at - t2)-1/2
dt [Edw1, back cover, formula 103].