Symbolic Integration part 2: Some theory

By ajb on May 30, 2007 1:48 PM | Permalink | Comments (0) | TrackBacks (0)

The second part in a series. You may wish to read part 1 first.

In this action-packed episode, we go through some of the theory that we need to understand integration algebraically.

Derivatives are actually defined in terms of infinitessimals and limits, but once you've got the basic operations sorted out, it's rare that you ever need to do this in practice.

More likely, you apply a simple recursive procedure to the structure of the function to be differentiated. Some of the rules look like this, for example:

$\begin{align*} D(a+b) & = Da + Db \\ D(ab) & = (Da) b + a (Db) \\ D(1/a) & = - \frac{Da}{a^2} \end{align*}$

There are several reasons why there's no analogous simple recursive procedure for anti-differentiation.

One problem seems to be the product rule (the second one in the examples above); it means that there is no simple inverse for the derivative of a product.

Another problem is that even the simplest formulas have odd exceptions. For example (ignoring the constants of integration):

$\begin{align*} \int x^{-3} dx & = - \frac{1}{3} x^{-2} \\ \int x^0 dx & = x \\ \int x^1 dx & = \frac{1}{2} x^2 \\ \int x^2 dx & = \frac{1}{3} x^3 \\ \int x^{\frac{1}{2}} dx & = \frac{2}{3} x^{\frac{3}{2}} \end{align*}$

And in general:

$\int x^k dx = \frac{1}{k+1} x^{k+1}$

Except for:

$\int x^{-1} dx = \log x$

It's not obvious why the rule breaks down in this case.

OK, let's go over some theory, so we can try to get a handle on all of this stuff.

The basic format of the rest of this entry will be a bunch of definitions. I'll follow the trickier definitions with some intuition on how to understand it.

Definition: A field has a fairly well-understood mathematical definition. To briefly recap:

A field has two binary operations (addition and multiplication).
There are at least two different elements, called 0 and 1, which are identities for addition and multiplication respectively.
Both addition and multiplication are associative and commutative.
Every element of the field has an additive inverse, and every element except 0 has a multiplicative inverse.
Multiplication distributes over addition (i.e. a(b+c) = ab + ac).

Intuition: Fields are mathematical structures that support the four basic operations (addition, subtraction, multiplication and division), which work in the obvious way.

For our purposes, the most interesting fields, the ones that we're doing calculus on, are those which include the rational numbers. (For those who know a bit of field theory, we're working with fields of characteristic zero.) This gives us a bunch of nice properties, not the least of which is that we have all of the integers at our disposal.

We also need to formalise the notion of what it means for a field to "include" another:

Definition: $F$ is a subfield of $K$ (written $F \subseteq K$ ) if $F$ is a field for which every element of $F$ is also an element of $K$ .

Intuition: One example is that the field of rational numbers is a subfield of the field real numbers, which is in turn a subfield of the field of complex numbers.

However, we can also extend a field by adding one or a few new elements, as well as everything that's "closed" under the field operations. For example, we can extend the rational numbers $\Q$ by "adding" the element $\sqrt{2}$ , and every number that can be constructed from $\sqrt{2}$ and the rational numbers using addition, subtraction, multiplication and division. We denote this new field by $\Q [\sqrt{2}]$ .

Definition: A differential field $(K,D)$ is a field $K$ together with a map $D$ from $K$ to $K$ which satisfies the following three laws:

$\begin{align*} D(a+b) & = Da + Db \\ D(ab) & = (Da) b + a (Db) \\ D(a/b) & = \frac{g Df - f Dg}{g^2} \end{align*}$

Such a map is called a derivation on $K$ .

Intuition: These are the familiar rules for computing derivatives.

Definition: A constant $a$ is an element of $K$ for which $Da = 0$ .

Intuition: This is how constants behave. Note that for any differential field, 0 and 1 must be constants. You may like to prove that the constants form a subfield, and hence that for any field which has the rationals as a subfield, all of the rationals are constants.

As an exercise, try proving that differential fields obey the power rule for n>0:

$D(f^n) = n f^{n-1} Df$

Definition: A differential field $(F,D)$ is a differential extension of $(K,D)$ if $K$ is a subfield of $F$ and the derivation on $F$ extends the one on $K$ .

Intuition: This simply means that a differential extension is a field extension where the derivation means the same thing on the common subfield.

The way we'll be doing this is by extending some field with a new element $t$ (and its closure under the field operations), and then just defining what $Dt$ means.

One example is adding constants to a field. Extending $(K,D)$ with a new constant $c$ gives a new differential field $(K(c),D)$ where $Dc=0$ . This is what it means to "add constants" to a differential field.

Definition: Let $(E,D)$ be a differential extension of $(K,D)$ . An element $t \in E$ is a monomial over $K$ if:

$t$ is transcendental over $K$ , and
$Dt \in K[t]$ .

Intuition: A monomial is a special kind of field extension which behaves in a nice way.

What it means for $t$ to be transcendental over $K$ is that it is not the root of a polynomial built from coefficients from $K$ . So $\sqrt{2}$ is not transcendental over $\Q$ , but $\pi$ is.

One of the most important monomials that we're going to use is $x$ , the variable of integration. It's defined as a differential extension of the base field, with $Dx=1$ . (This is, obviously, what happens when you take the derivative of $x$ .)

The differential field $(\Q[x],D)$ has as elements, the rational functions in $x$ . Every element is one polynomial in $x$ divided by another polynomial in $x$ , and $D$ takes the derivative with respect to $x$ .

The reason why we use monomials is that the fancy functions that we're interested in integrating are expressible as monomials. For example:

$\begin{align*} \theta_1 = e^x & \Rightarrow D\theta_1 = \theta_1 \\ \theta_2 = \log x & \Rightarrow D\theta_2 = \frac{1}{x} \\ \theta_3 = \tan x & \Rightarrow D\theta_3 = 1 + \theta_3^2 \end{align*}$

There's a slight catch with sin and cos, which can't be expressed as monomials, but we can cheat a bit. You can extend a field with sin and cos together, using a pair of coupled monomials. If $c = \cos x$ and $s = \sin x$ , then:

$\begin{align*} Dc & = -s \\ Ds & = c \end{align*}$

Of course, if your field $F$ contains $c = \cos x$ and $s = \sin x$ , then $t = \tan x$ is not a monomial, because it's not transcendental over $F$ . (Obviously, $t=s/c$ .) This is an important point which we'll deal with in a bit more detail later.

The neat thing is that we've completely abstracted away any details about what cos, exp etc mean. We can, in a sense, deal with them entirely in terms of their representation as monomials.

But the details will have to wait until next time. Stay tuned...

On to part 3...

0 TrackBacks

Listed below are links to blogs that reference this entry: Symbolic Integration part 2: Some theory.

TrackBack URL for this entry: http://andrew.bromage.org/mt/mt-tb.cgi/70

Symbolic Integration part 2: Some theory

Categories

Tags

0 TrackBacks

Leave a comment

Search

About this Entry