- 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 csc^{2}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) = (4Ã^{3}(z) - g_{2}Ã(z) - g_{3})^{-1/2}and dz/d sin z = (1 - sin^{2 }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 - t^{2})^{-1/2}dt [Edw1, back cover, formula 103].