7
$\begingroup$

$$ \newcommand{\qa}{P} \newcommand{\qb}{Q} \newcommand{\da}{dP} \newcommand{\db}{dQ} \newcommand{\positiverealnumbers}{\mathbb{R}_+} \newcommand{\realnumbers}{\mathbb{R}} \newcommand{\naturalnumbers}{\mathbb{N}} \newcommand{\positiveintegers}{\mathbb{Z}_+} $$

We frequently solve geometrical and physical problems by obtaining an approximate expression for differential $\da$ in terms of differential $\db$ and then integrating $\da$ to obtain $\qa$. We assume that the expression for $\qa$ is exact even though we used an approximate formula for $\da$. This is justified by saying that the differentials are infinitely small quantities. For example, when we derive an expression for the area of a circular disc (see example 1) we set $dA = 2 \pi r dr$ which is an approximate expression when the diffentials are interpreted as real numbers. In this article we try to define a method for computing $\qa$ so that we don't need approximate expressions in the derivation.

Theorem 1

Let $a,b \in \realnumbers$ and $a < b$. Suppose that $$ \Delta f = g(x) \Delta x + h(x, \Delta x) $$ for all $x \in \realnumbers$ and $\Delta x \in \positiverealnumbers$ for which $x, x + \Delta x \in [a, b]$. Suppose also that $g$ is Riemann integrable and for all $\varepsilon \in \positiverealnumbers$ there exists $R \in \positiverealnumbers$ so that $$ \left\vert \frac{h(x, \Delta x)}{\Delta x} \right\vert < \varepsilon \tag{1} $$ for all $x, x + \Delta x \in [a, b]$, $0 < \vert \Delta x \vert < R$. Let $t \in [a, b]$. Let $n \in \positiveintegers$, and define $\Delta' x := \frac{t - a}{n}$ and $x_i := a + \frac{i}{n} \Delta' x$ $i \in \naturalnumbers$, $i \leq n$. Define also $\Delta f_i := f(x_{i+1}) - f(x_i)$ where $i \in \naturalnumbers$, $i < n$. Define $$ f(t) := \lim_{n \to \infty} \sum_{i=0}^{n-1} \Delta f_i $$ Now $$ f(t) = \int_a^t g(x) dx $$ and $df = g(x) dx$.

Proof

Let $$ f(t) := \lim_{n \to \infty} \sum_{i=0}^{n-1} \Delta f_i = \lim_{n \to \infty} \sum_{i=0}^{n-1} g(x_i) \Delta'x + \lim_{n \to \infty} \sum_{i=0}^{n-1} h(x_i, \Delta'x). $$ Now $$ \lim_{n \to \infty} \sum_{i=0}^{n-1} g(x_i) \Delta'x = \int_a^t g(x) dx $$ by the definition of the Riemann integral.

Let $\varepsilon \in \positiverealnumbers$. Choose $R_1 \in \positiverealnumbers$ so that $$ \left\vert \frac{h(x, \Delta x)}{\Delta x} \right\vert < \frac{\varepsilon}{t-a} $$ for all $x, x + \Delta x \in [a, b]$, $0 < \vert \Delta x \vert < R_1$. Let $n \in \positiveintegers$ so that $$ n > \frac{t-a}{R_1} . $$ Now \begin{align*} \left\vert \sum_{i=0}^{n-1} h(x_i, \Delta'x) \right\vert & < \sum_{i=0}^{n-1} \frac{\varepsilon}{t-a} \vert \Delta'x \vert = n \frac{\varepsilon}{t-a} \frac{t-a}{n} = \varepsilon . \end{align*} Hence $$ \lim_{n \to \infty} \sum_{i=0}^{n-1} h(x_i, \Delta'x) = 0 . $$

$$\tag*{$\blacksquare$}$$

Theorem 2

A sufficient condition for inequality (1) is that there exist $S, C \in \positiverealnumbers$ so that $$ \vert h(x, \Delta x) \vert < C \vert \Delta x \vert^2 $$ for all $x, x + \Delta x \in [a, b]$ and $0 < \vert \Delta x \vert < S$.

Proof

Set $R := \min \{ S, \varepsilon / C \}$. Now $$ \left\vert \frac{h(x, \Delta x)}{\Delta x} \right\vert < C \vert \Delta x \vert < \varepsilon . $$

$$\tag*{$\blacksquare$}$$

Note that we can't prove Theorem 1 directly by the Fundamental Theorem of Calculus because we would need to \textit{assume} that there exists a function $f : [a,b] \to \realnumbers$ for which $f(x + \Delta x) - f(x) = \Delta f(x, \Delta x)$ in order to use it.

Example 1

Derive an expression for the area of a disc whose inner radius is $r_a$ and outer radius $r_b$.

The area to be computed in example 1

Solution

Define $\Delta A$ to be the area of a disc with inner radius $r$ and width $\Delta r$. We have $$ 2 \pi r \Delta r \leq \Delta A \leq 2 \pi (r + \Delta r) \Delta r $$ By setting $g(r) := 2 \pi r$ and $h(r, \Delta r) := 2 \pi (\Delta r)^2$ we get $A = \pi r_b^2 - \pi r_a^2$ by Theorems 1 and 2.

Example 2

Suppose that a particle is moving under influence of a constant force $F = ma$ for time $T$ and the particle is initially at rest. Derive an expression for the kinetic energy of the particle. Assume that the work done by a constant force $F$ is $W = F s$ where $s$ is the distance that the particle moves in the direction of the force. Assume also that the kinetic energy of a particle at rest is $0$.

Solution

We define $\Delta s$ to be the distance that the particle moves in the time interval $[t, t + \Delta t]$. We have $v = at$, $$ a t \Delta t \leq \Delta s \leq a (t + \Delta t) \Delta t , $$ and $$ a (t + \Delta t) \Delta t = a t \Delta t + a (\Delta t)^2 . $$ Set $g(t) := a t$ and $h(t, \Delta t) := a (\Delta t)^2$ and it follows from Theorems 1 and 2 that the distance that the particle moves in time $T$ is $$ s = \int_0^T a t dt = \frac{1}{2} a T^2 $$ By setting $v_f = a T$ we obtain $$ E_k = W = \frac{1}{2} F a T^2 = \frac{1}{2} m a^2 T^2 = \frac{1}{2} m v_f^2 . $$


Do you find this formalism useful?

Tommi Höynälänmaa

$\endgroup$
3
  • 7
    $\begingroup$ I'm curious about the \newcommands you have in the beginning of your post. Do you really find it more convenient to type \qa than to type P? And is \Bbb R_+ (or even \mathbb R_+) actually simplified by \positiverealnumbers? $\endgroup$
    – Arthur
    Feb 16 '19 at 9:20
  • $\begingroup$ I view $d$ as an operator, almost the same as $\frac{d}{dx}$. Both operations act on functions, and the resulting $df$ and $\frac{d}{dx}f$ are also functions. $df := \frac{d}{dx}f \Delta x$. So the definition of $df$ requires you to know what $\frac{d}{dx}f$ means. The two are essentially the same operation. There is one function where the differential of it, is just an increment. consider $g: x \mapsto x$, then $dg = 1 \Delta x$. When you see $\Delta x$, it's often written as $dx$ . Or independent variable $\Delta r$ often as $dr$. When you turn the increment of an independent variable $\endgroup$
    – DWade64
    Feb 16 '19 at 19:42
  • $\begingroup$ into a differential like $dx$ or $dr$, you are viewing the independent variable as an identity function (you can only apply $d$ to functions). $\int$ means anti-derivative, it can also mean anti-differential. That's how I view differentials. I don't add anything new to it besides what a derivative is (though it still is slightly different: tangent height as opposed to tangent slope) $\endgroup$
    – DWade64
    Feb 16 '19 at 19:46
1
$\begingroup$

I am not finding anything quite new in your formalism to bypass the use of differentials , in the realm of standard analysis.

As for the applicability of theorem 1 in example 1, the formal definition of area in a plane it can be proven quite formally (For reference see Apostol calculus volume 1 the very first chapter on integral calulus), so it seems nothing new.

For the application in example 2, What are taking the formal definition of work? that your assuming monotonicity in the very first inequality appearing in the solution of example 2 ? If you are taking somewhat an axiomatic definition of work as given in Apostol Calculus volume 1 chapter on Application of Integral calculus then the question comes why to remember such set of axioms rather than taking the definition of work as the integral $\large\int_a^{a'} \vec F.d\vec s$ in general, when they are equivalent in the special case of real line?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.