Beyond The Boundary

   
Recent changes
Table of contents
Links to this page
FRONT PAGE / INDEX

Subscribe!
@ColinTheMathmo

My latest posts can be found here:
Previous blog posts:
Additionally, some earlier writings:

Beyond the Boundary - 2014/05/23

So when we were talking about how to Fill In The Gaps we saw that for the function $f(x)=sin(x)/x$ we didn't have a value for x=0. That's clear from the formula, because when we substitute the value 0 for x we get $0/0,$ and that's undefined.

Actually f is defined on all complex numbers except 0, but for the moment we are concentrating on the reals.
So it brought us to the idea that when we define a function we also need to say what its domain is. We need to say what the values are that we are allowed to feed to the function. In this case we see that f is only defined on the real numbers without 0.

But we can see just by plotting $\frac{sin(x)}{x}$ that it makes sense to define f(0)=1. That fills the gap, the resulting curve flows smoothly, everything seems to be OK.

In fact, we can show that this expression:

  • $1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+...$

We need to be very careful here. Just because we can write something down, that doesn't mean it necessarily makes sense! In this case we really should ask "What does the 'dot dot dot' mean, and does it make sense?" In fact we can assign a meaning to it, and provided we are careful, it does make sense. A little more about that later, but a complete treatment is overkill for this article.
has the same value as $f(x)=sin(x)/x$ everywhere that f is defined, and it also has the value f(0)=1. So augmenting $f(x)=sin(x)/x$ with the value f(0)=1 seems to make sense.

There doesn't seem to be much that's controversial about all that. More, we can make precise what we mean when we say that a particular augmentation of a function "makes sense." There's a process called "Analytic Continuation" that basically means "Extend the existing definition in a way that keeps everything smooth and well-behaved."

Let me show you an example.

Consider the expression:

  • $g(x)=1+x+x^2+x^3+x^4+x^5+...$

In this case we need to be careful about exactly what we mean by the "dot dot dot" here. Here's how we do that.

Think of any value for x - say $\frac{1}{2}.$ Put that in and we get an infinite sum:

$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...$

If we cut that off early we find that we get the sequence of partial sums:

$1,\quad\frac{3}{2},\quad\frac{7}{4},\quad\frac{15}{8},\quad...$

Technically, let's suppose there's a special location, and every time you give me a permitted error, there comes a point after which all the elements of the sequence are closer. Then if you become more demanding, I just need to go further to find that all subsequent elements satisfy your demands. When that happens, the sequence is said to approach that special location as a "limit."
This sequence gets relentlessly closer to 2, as close as we like. The sequence has 2 as a limit. In this sense we say that the infinite sum converges, and that it makes sense to talk about a value of the infinite sum.

In short, when x is $\frac{1}{2},$ we say that the infinite sum is equal to 2.

You can substitute any value of x between (but not including) -1 and 1, and you find that the infinite sum converges to a value. Because of this we say that the expression:

$g(x)=1+x+x^2+x^3+x^4+x^5+...$

is well-defined for x on the domain (-1,1) (the round brackets (as opposed to square brackets) showing that we do not include the endpoints) and all is well.

For x=1 and bigger, the infinite sum does not converge. It does not make sense to talk about "the value" of g(x) when $x\le-1$ or $x\ge{1}.$ In particular, when x=2 we get the expression:

1+2+4+8+16+...

and clearly the partial sums don't converge to a limit, they disappear off into the distance.

But now comes the interesting part. Consider the function $h(x)=\frac{1}{1-x}$ which is defined for all x except x=1. When $-1\lt{x}\lt{1}$ we find that the value of g(x) is the same as the value of h(x). In other words, everywhere that g(x) is defined and has a reasonable value, that value is the same as the function h(x).

But h(x) is defined in places where g(x) is not defined. In particular, if we look at x=2 we get:

  • $g(x)=1+2+4+8+16+...$
  • $h(x)=\frac{1}{1-2}=-1$

The first of these does not make sense - g(x) is not defined outside of the range $-1\lt{x}\lt{1}$ - but h(x) is defined and has a perfectly sensible value.

We say that the function h(x) extends the function g(x).

Of course, it's easy to take an original function and just randomly specify additional values. That's no real challenge. The difficulty is to do this in a way that is some sense "natural" or "obvious." We saw that in the previous article - Fill In The Gaps - where we added a value for a single point, and it was "natural" for that value to be 1. Anything else felt like it would be contrived.

In a later we make this concept more precise.


<<<< Prev <<<<
Fill In The Gaps
:
>>>> Next >>>>
Square Root By Long Division ...


You should follow me on twitter @ColinTheMathmo

Comments

I've decided no longer to include comments directly via the Disqus (or any other) system. Instead, I'd be more than delighted to get emails from people who wish to make comments or engage in discussion. Comments will then be integrated into the page as and when they are appropriate.

If the number of emails/comments gets too large to handle then I might return to a semi-automated system. We'll see.


Contents

 

Links on this page

 
Site hosted by Colin and Rachel Wright:
  • Maths, Design, Juggling, Computing,
  • Embroidery, Proof-reading,
  • and other clever stuff.

Suggest a change ( <-- What does this mean?) / Send me email
Front Page / All pages by date / Site overview / Top of page

Universally Browser Friendly     Quotation from
Tim Berners-Lee
    Valid HTML 3.2!