So when we were talking about Filling 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. |
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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.
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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."
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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." We saw that in the previous article -
"Filling 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 the next article we make this concept more precise.
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