Day: May 23, 2022

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202205231502 Exercise 1.3 A (Q1)

Exercises 1-9, let \mathrm{d}x be an infinitesimal and prove the given formula.

1. (\mathrm{d}x+1)^2=2\,\mathrm{d}x+1

Extracted from Michael Corral. (2020). Elementary Calculus.

Background. (Infinitesimal)

A number \delta is an infinitesimal if the conditions (a) – (d) hold: (a) \delta\neq 0; (b) If \delta >0 then \delta is smaller than any positive real number; (c) If \delta <0 then \delta is larger than any negative real number; (d) \delta^2=0 (and hence all higher powers of \delta, such as \delta^3 and \delta^4, are also 0) N.b. Any infinitesimal multiplied by a nonzero real number is also an infinitesimal, while 0 times an infinitesimal is 0.

Proof.

Suppose the contrary is true:

(\mathrm{d}x+1)^2\neq 2\,\mathrm{d}x+1.

\begin{aligned} \textrm{LHS} & = (\mathrm{d}x+1)^2 \\ & = (\mathrm{d}x)^2+2(\mathrm{d}x)(1)+(1)^2 \\ & \stackrel{(\textrm{d})}{=} 0+2\,\mathrm{d}x+1 \\ & = 2\,\mathrm{d}x+1 \\ & = \textrm{RHS} \qquad \perp\\ \end{aligned}

Thus converse is the case.