veneciasp

Answered

2022-07-03

I am currently studying Hardy's Pure Course of Mathematics and am on chapter 2, section 24: Rational Functions. In this chapter, Hardy defines a rational function as the quotient of two polynomials such that: $R(x)={\displaystyle \frac{P(x)}{Q(x)}}$

Hardy mentions that P(x) and Q(x) must have no common factor, though the removal of common factors changes the function. He provides the following example:

$\frac{({\displaystyle \frac{1}{x+1}}+{\displaystyle \frac{1}{x-1}})}{({\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{x-2}})}$

which may be reduced to:

$\frac{{x}^{2}(x-2)}{(x-1{)}^{2}(x+1)}$

The first equation is not defined for values x=-1,1,0,2 and the second is not defined for values x=-1,1

So my questions are:

1) Do all simplifications of rational functions lead to different functions?

2) When manipulating equations, are reductions by common factors actually "mathematical errors?"

3) And what are the implications of what Hardy had described?

Hardy mentions that P(x) and Q(x) must have no common factor, though the removal of common factors changes the function. He provides the following example:

$\frac{({\displaystyle \frac{1}{x+1}}+{\displaystyle \frac{1}{x-1}})}{({\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{x-2}})}$

which may be reduced to:

$\frac{{x}^{2}(x-2)}{(x-1{)}^{2}(x+1)}$

The first equation is not defined for values x=-1,1,0,2 and the second is not defined for values x=-1,1

So my questions are:

1) Do all simplifications of rational functions lead to different functions?

2) When manipulating equations, are reductions by common factors actually "mathematical errors?"

3) And what are the implications of what Hardy had described?

Answer & Explanation

Perman7z

Expert

2022-07-04Added 13 answers

One way to look at what a function is, and a way that is not usually introduced until much later in your math education, is to say that rather than being a rule, a function is a set. In particular, a function f which maps a domain X to a codomain Y is some set of ordered pairs (x,y) where x comes from X and y comes from Y, and where if both (a,b) and (a,c) are in f then b=c.

That might seem complicated, but here is an example. We can define a function $f:\mathbb{N}\to \mathbb{N}$ (that is to say, f is a function where the domain of f is the natural numbers, and the codomain is also the natural numbers) to be

$f(n)=n+1$

Now we can look at this function as the set

$\{(0,1),(1,2),(2,3),\dots \}$

The algebraic expression "n+1" just gives us a convenient way to specify the elements of the set f. You've of course looked at functions in kind of this way probably, by forming a table of values:

x | f(x)

0 | 1

1 | 2

2 | 3

..| ..

If you were able to keep writing forever and ever, you could write down every single pair, and that infinitely long table would uniquely specify the function f.

The question that you're asking is this: when do we say that two functions are equal? Well if we look at a function in a more general way, as set of ordered pairs rather than just as an algebraic rule, then it seems easy enough to say when two functions are equal. When things are equal, they are identical. So if two functions g and h are equal, then they better be made up of the exact same set of pairs.

So let's look at $f(x)={x}^{3}/{x}^{2}$ and $g(x)={x}^{2}/x$. Since we can't divide by 0, it's standard to assume that 0 is not in the domain of either of these functions—otherwise the definitions would be meaningless. So we can say that they are defined for all real numbers except 0. Now since for any number except 0, ${x}^{3}/{x}^{2}={x}^{2}/x=x$, f and g have all the exact same points. Since a function is nothing but its points, that means that the two functions are identical.

But now if we cancel all the common factors we end up with another function that looks like h(x)=x. Is it true that f=g=h? Well, if we don't specify anything about the domain of h then no, because h contains the point (0,0), and neither f nor g do, so they aren't identical functions. But if we specify that the domain of h is all real numbers except 0, then yes, because we've taken the point (0,0) out of h.

So basically, cancelling common factors does not change a function as long as you keep track of the domain.

What are the implications of this? Well off the top of my head, consider the following problem, with which type you may be familiar: find the maximum of the function

$f(x)=\frac{{x}^{2}(2-x)}{x}+2x+5$

if it exists.

So the first thing you might like to try is to cancel the common factors and write f as

$x(2-x)+2x+5$

Now, you know what that is. It's an upside down parabola, and its maximum occurs at x=0. But wait! There is no x=0 in f; we infer from the context that f is defined on all the real numbers except for 0. So the answer to this is actually that f has no maximum.

That might seem complicated, but here is an example. We can define a function $f:\mathbb{N}\to \mathbb{N}$ (that is to say, f is a function where the domain of f is the natural numbers, and the codomain is also the natural numbers) to be

$f(n)=n+1$

Now we can look at this function as the set

$\{(0,1),(1,2),(2,3),\dots \}$

The algebraic expression "n+1" just gives us a convenient way to specify the elements of the set f. You've of course looked at functions in kind of this way probably, by forming a table of values:

x | f(x)

0 | 1

1 | 2

2 | 3

..| ..

If you were able to keep writing forever and ever, you could write down every single pair, and that infinitely long table would uniquely specify the function f.

The question that you're asking is this: when do we say that two functions are equal? Well if we look at a function in a more general way, as set of ordered pairs rather than just as an algebraic rule, then it seems easy enough to say when two functions are equal. When things are equal, they are identical. So if two functions g and h are equal, then they better be made up of the exact same set of pairs.

So let's look at $f(x)={x}^{3}/{x}^{2}$ and $g(x)={x}^{2}/x$. Since we can't divide by 0, it's standard to assume that 0 is not in the domain of either of these functions—otherwise the definitions would be meaningless. So we can say that they are defined for all real numbers except 0. Now since for any number except 0, ${x}^{3}/{x}^{2}={x}^{2}/x=x$, f and g have all the exact same points. Since a function is nothing but its points, that means that the two functions are identical.

But now if we cancel all the common factors we end up with another function that looks like h(x)=x. Is it true that f=g=h? Well, if we don't specify anything about the domain of h then no, because h contains the point (0,0), and neither f nor g do, so they aren't identical functions. But if we specify that the domain of h is all real numbers except 0, then yes, because we've taken the point (0,0) out of h.

So basically, cancelling common factors does not change a function as long as you keep track of the domain.

What are the implications of this? Well off the top of my head, consider the following problem, with which type you may be familiar: find the maximum of the function

$f(x)=\frac{{x}^{2}(2-x)}{x}+2x+5$

if it exists.

So the first thing you might like to try is to cancel the common factors and write f as

$x(2-x)+2x+5$

Now, you know what that is. It's an upside down parabola, and its maximum occurs at x=0. But wait! There is no x=0 in f; we infer from the context that f is defined on all the real numbers except for 0. So the answer to this is actually that f has no maximum.

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