Radius of convergence.
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Let's think about a not-too mysterious
function.
Well let's think about the function f of x
equals 1 over 1 plus x squared.
I can write down a power series for this
function around zero.
So, I'll begin with a power series for 1
over 1 minus x.
That's the sum and goes from 0 to infinity
of
x to the n.
Now we'll replace x by negative x.
And then I'll find that 1 over 1 plus x is
the
sum, n goes from 0 to infinity of negative
x to the n.
Which I could write as the sum and goes
from 0 to
infinity of negative 1 to the n, times x
to the n.
And now to get from 1 over 1 plus x to 1
over 1 plus x squared, I'll just replace x
by x squared.
So 1 over 1 plus x squared is the sum n
goes from 0
to infinity of negative 1 to the n times x
squared to the n.
Which I could also write as the sum, n
goes from 0 to
infinity of negative 1 to the n times x to
the 2n power.
What's the radius of convergence of that
power series?
I was being super sloppy when I wrote this
out.
I wrote
one over one minus x equals this power
series,
but that's not true unless I include this
statement.
I should have written this down.
If the absolute value of x is less than 1.
And if that's true than these are equal.
The radius of convergence of this series
for 1 over 1 plus
x squared is also 1, and we can see that
in the graph.
Here I've graphed the function one over
one plus x squared, and
take a look at it's Taylor Series
expansion around the point zero.
Here is just the first term, which is a
constant term.
Here's through the quadratic term.
Here is through the x to the fourth term,
through x to the sixth term, and so on.
And you can see that it's doing an
increasingly good
job of approximating the function between
minus 1 and 1.
But over here it's actually doing an
increasingly bad job.
1 over 1 plus x squared looks like a
really nice function.
So if it's such a nice function, why isn't
the Taylor Series
centered around zero doing a better job of
approximating the function far away?
Well here I've graphed y equals sine of x.
And honestly, I mean, qualitatively at
least this isn't so different
looking from the graph of 1 over 1 plus x
squared.
The graph of 1 over 1 plus X squared had
kind of a
big bump in the middle, and this thing's
just got lots of bumps.
And yet the Taylor Series is just totally
different experience.
Let's start writing down the Taylor Series
around the origin.
So here is the point around which I'm expanding.
Here's just the constant term in red.
Here's through the linear term.
Here's through the cubic term.
And I'm going to keep on going.
And as I add more and more terms, in my
Taylor
Series expansion, I'm getting increasingly
good
approximations to sine of x everywhere.
Sine of x is equal to its Taylor Series
centered around
zero for all values of x.
Whereas 1 over 1 plus x squared, its
Taylor Series around zero
is only equal to 1 over 1 plus x squared
if x is less than 1.
And we know other examples like that.
Like this, this is the graph y equals 1
over 1 minus x.
And I can start right around the Taylor
Series around this point.
Here's the constant term, the linear term.
Take more and more terms
and, and yeah, I mean the radiance of
convergence here is one.
But maybe that isn't so surprising.
I mean, 1 over 1 minus x has a bad point
at 1, alright,
I don't want to plug in x equals 1 because
then I'll be dividing by 0.
This function's not defined at one.
So if I start a Taylor Series around zero
and I imagine
how big do I really expect that radius of
convergence to be.
Well, you know an interval of radius
one centered around zero bumps into the
problem point.
So maybe I don't really expect the the
radius
of convergence to be more than 1, at 0.
So maybe 1 over 1 minus x has a problem
when x equals
1, but 1 over 1 plus x squared doesn't
have any problem at 1.
It doesn't have any problem anywhere.
This function is defined for all x.
So if 1 over 1 plus x squared looks just
as nice as sine of x,
why is the Taylor Series for this function
centered around zero, so
much worse than the Taylor Series for sine
of x centered around zero?
Taylor series for sine of x converges to
sine of x everywhere, and the Taylor
Series
for 1 over 1 plus x squared is only good
in an interval of radius one.
Well let's see what happens if we write
down a Taylor Series expansion for 1
over 1 plus X squared, but not centered
around zero, but centered around some
other point.
Let us write down the Taylor Series
expansion centered around x equals one.
See what happens.
So here we go, there's the constant term.
There's the linear terms.
And we keep on going.
We add more and more terms to the Taylor
Series expansion centered around one.
And I mean, we're doing an increasingly
good job,
but not over the whole real line again,
right?
But at least the radius of convergence is
bigger now.
Well how
much bigger?
The Taylor Series for 1 over 1 plus x
squared centered at zero has radius one.
The Taylor Series centered at one turns
out to have radius the square root of 2.
And we had this idea that maybe places
where the
function's undefined, these bad points,
where the function doesn't exist.
Maybe those are somehow to blame for the
radius of convergence.
And that's what happened in the case of 1
over 1 minus x, alright.
The radius of convergence around x equals
0 was 1,
because this function has this bad point
at x equals 1.
Function's not defined if I plug in x
equals 1.
Maybe I'm bumping into a bad point here
too.
Well let's diagram the situation this way.
Here's the real line.
And around zero I've got this well it's an
interval but I've drawn it as a circle.
It's a circle
of radius one.
So if that's representing the radius of
convergence of my Taylor Series
around the point zero for the function 1
over 1 plus x squared.
Now around the point one, alright, I've
got
a different radius and convergence it's
bigger as
the square root of 2, so here is a circle
of radius the square root of 2.
And I'm placing it so that its center is
at one.
And the idea
here is try to get a sense of whether or
not maybe I'm bumping into a bad point.
Is there's some point on the real line,
which is distance
1, from 0, and distance the square root of
2 from 1.
But there isn't such a point on the real
line.
Okay, but when I draw them as circles, the
circles are touching at these two points.
So where is that point?
Well those points
are in the complex plane.
That point there is i and that point there
is minus i.
Alright, so those are square roots of
minus 1.
Those are imaginary numbers.
Well what happens if I evaluate 1 over 1
plus x squared, when x
equals i or x equals minus i?
So, i squared is minus 1.
So what's f of i?
Well, f is 1 over 1 plus its input,
squared, so that would be 1 over 1 plus
what's i squared, it's negative one.
1 over 1 plus minus 1, that is not
defined, that's dividing by 0.
So there is a bad point.
There is a point where the function is
undefined.
It's just that the bad point for this
function isn't a real point.
It's an imaginary input.
All right, if I evaluate
this function at i or minus i, it's
undefined.
And yet the bad point in that complex
plane is
messing up my radius of convergence, even
along the real line.
We're beginning to get a glimpse of the
important role that complex
numbers play even in the theory of just
real value Taylor Series.
Additional evidence comes for example by
this equation, e to
the i x equals cos x plus i sin x.
And again i is a squared of minus 1.
And you can interpret this really as a
statement of power series.
So here I've written down the Taylor
Series for e
to the x, but with x replaced with i x.
Here I've just written down cos x and here
I've written down sin x, but I've
multiplied by i.
And you can expand this out, using the
fact that
i squared is to minus one to get this
equality.
And you can do kind of fun things like
substitute in say
pi in for x and conclude that e to the i
pi is cos pi plus i sin pi.
So co sin pie is minus 1, sin pie is 0.
So e to the i pie is negative 1.
Taylor Series aren't just a jumping off
point for calculus, and for numerical
approximations.
Taylor Series are also our, our first step
into the theory of complex analysis.
And that theory is more natural than it
might seem at first.
I mean, if the complex numbers are
affecting the radius of
convergence of my real power series, then
they must be really there.
I mean, they're not as imaginary as people
might think.
So Taylor Series aren't just a jumping
off point for calculus and for numerical
approximations.
Taylor Series, are also our, our first
step into the theory
of complex analysis.
And that theory is more natural than it
might seem at first.
I mean, if the complex numbers are
affecting the radius of
convergence of my real power series then
they must be really there.
I mean, they're not as imaginary as people
might think.
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