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Radius of convergence.

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[SOUND]

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Let's think about a not-too mysterious
function.

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Well let's think about the function f of x
equals 1 over 1 plus x squared.

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I can write down a power series for this
function around zero.

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So, I'll begin with a power series for 1
over 1 minus x.

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That's the sum and goes from 0 to infinity
of

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x to the n.
Now we'll replace x by negative x.

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And then I'll find that 1 over 1 plus x is
the

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sum, n goes from 0 to infinity of negative
x to the n.

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00:00:40,890 --> 00:00:43,420
Which I could write as the sum and goes
from 0 to

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infinity of negative 1 to the n, times x
to the n.

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And now to get from 1 over 1 plus x to 1

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over 1 plus x squared, I'll just replace x
by x squared.

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So 1 over 1 plus x squared is the sum n
goes from 0

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to infinity of negative 1 to the n times x
squared to the n.

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Which I could also write as the sum, n
goes from 0 to

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infinity of negative 1 to the n times x to
the 2n power.

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What's the radius of convergence of that
power series?

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I was being super sloppy when I wrote this
out.

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I wrote

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one over one minus x equals this power
series,

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but that's not true unless I include this
statement.

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I should have written this down.

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If the absolute value of x is less than 1.

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And if that's true than these are equal.

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The radius of convergence of this series
for 1 over 1 plus

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x squared is also 1, and we can see that
in the graph.

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Here I've graphed the function one over
one plus x squared, and

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take a look at it's Taylor Series
expansion around the point zero.

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Here is just the first term, which is a
constant term.

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Here's through the quadratic term.

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Here is through the x to the fourth term,
through x to the sixth term, and so on.

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And you can see that it's doing an
increasingly good

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job of approximating the function between
minus 1 and 1.

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But over here it's actually doing an
increasingly bad job.

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1 over 1 plus x squared looks like a
really nice function.

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So if it's such a nice function, why isn't
the Taylor Series

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centered around zero doing a better job of
approximating the function far away?

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Well here I've graphed y equals sine of x.

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And honestly, I mean, qualitatively at
least this isn't so different

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looking from the graph of 1 over 1 plus x
squared.

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The graph of 1 over 1 plus X squared had
kind of a

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big bump in the middle, and this thing's
just got lots of bumps.

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And yet the Taylor Series is just totally
different experience.

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Let's start writing down the Taylor Series
around the origin.

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So here is the point around which I'm expanding.

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Here's just the constant term in red.
Here's through the linear term.

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Here's through the cubic term.
And I'm going to keep on going.

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And as I add more and more terms, in my
Taylor

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Series expansion, I'm getting increasingly
good

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approximations to sine of x everywhere.

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Sine of x is equal to its Taylor Series
centered around

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zero for all values of x.

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Whereas 1 over 1 plus x squared, its
Taylor Series around zero

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is only equal to 1 over 1 plus x squared
if x is less than 1.

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And we know other examples like that.

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Like this, this is the graph y equals 1
over 1 minus x.

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And I can start right around the Taylor
Series around this point.

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Here's the constant term, the linear term.

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Take more and more terms

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and, and yeah, I mean the radiance of
convergence here is one.

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But maybe that isn't so surprising.

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I mean, 1 over 1 minus x has a bad point
at 1, alright,

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I don't want to plug in x equals 1 because
then I'll be dividing by 0.

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This function's not defined at one.

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So if I start a Taylor Series around zero
and I imagine

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how big do I really expect that radius of
convergence to be.

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Well, you know an interval of radius

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one centered around zero bumps into the
problem point.

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So maybe I don't really expect the the
radius

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of convergence to be more than 1, at 0.

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So maybe 1 over 1 minus x has a problem
when x equals

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1, but 1 over 1 plus x squared doesn't
have any problem at 1.

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It doesn't have any problem anywhere.
This function is defined for all x.

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So if 1 over 1 plus x squared looks just
as nice as sine of x,

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why is the Taylor Series for this function
centered around zero, so

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much worse than the Taylor Series for sine
of x centered around zero?

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Taylor series for sine of x converges to

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sine of x everywhere, and the Taylor
Series

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for 1 over 1 plus x squared is only good
in an interval of radius one.

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Well let's see what happens if we write
down a Taylor Series expansion for 1

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over 1 plus X squared, but not centered

84
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around zero, but centered around some
other point.

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Let us write down the Taylor Series

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expansion centered around x equals one.
See what happens.

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So here we go, there's the constant term.

88
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There's the linear terms.
And we keep on going.

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We add more and more terms to the Taylor
Series expansion centered around one.

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And I mean, we're doing an increasingly
good job,

91
00:05:01,410 --> 00:05:04,380
but not over the whole real line again,
right?

92
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But at least the radius of convergence is
bigger now.

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Well how

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much bigger?

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The Taylor Series for 1 over 1 plus x
squared centered at zero has radius one.

96
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The Taylor Series centered at one turns
out to have radius the square root of 2.

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And we had this idea that maybe places
where the

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function's undefined, these bad points,
where the function doesn't exist.

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Maybe those are somehow to blame for the
radius of convergence.

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And that's what happened in the case of 1

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over 1 minus x, alright.

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The radius of convergence around x equals
0 was 1,

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because this function has this bad point
at x equals 1.

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Function's not defined if I plug in x
equals 1.

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Maybe I'm bumping into a bad point here
too.

106
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Well let's diagram the situation this way.
Here's the real line.

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And around zero I've got this well it's an
interval but I've drawn it as a circle.

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It's a circle

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of radius one.

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So if that's representing the radius of
convergence of my Taylor Series

111
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around the point zero for the function 1
over 1 plus x squared.

112
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Now around the point one, alright, I've
got

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a different radius and convergence it's
bigger as

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the square root of 2, so here is a circle
of radius the square root of 2.

115
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And I'm placing it so that its center is
at one.

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And the idea

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here is try to get a sense of whether or
not maybe I'm bumping into a bad point.

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Is there's some point on the real line,
which is distance

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1, from 0, and distance the square root of
2 from 1.

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But there isn't such a point on the real
line.

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Okay, but when I draw them as circles, the
circles are touching at these two points.

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So where is that point?

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Well those points

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are in the complex plane.

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That point there is i and that point there
is minus i.

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Alright, so those are square roots of
minus 1.

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Those are imaginary numbers.

128
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Well what happens if I evaluate 1 over 1
plus x squared, when x

129
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equals i or x equals minus i?
So, i squared is minus 1.

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So what's f of i?
Well, f is 1 over 1 plus its input,

131
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squared, so that would be 1 over 1 plus
what's i squared, it's negative one.

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1 over 1 plus minus 1, that is not
defined, that's dividing by 0.

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So there is a bad point.

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There is a point where the function is
undefined.

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It's just that the bad point for this
function isn't a real point.

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It's an imaginary input.

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All right, if I evaluate

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this function at i or minus i, it's
undefined.

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And yet the bad point in that complex
plane is

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messing up my radius of convergence, even
along the real line.

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We're beginning to get a glimpse of the
important role that complex

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numbers play even in the theory of just
real value Taylor Series.

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Additional evidence comes for example by
this equation, e to

144
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the i x equals cos x plus i sin x.

145
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And again i is a squared of minus 1.

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And you can interpret this really as a
statement of power series.

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So here I've written down the Taylor
Series for e

148
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to the x, but with x replaced with i x.

149
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Here I've just written down cos x and here

150
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I've written down sin x, but I've
multiplied by i.

151
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And you can expand this out, using the
fact that

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i squared is to minus one to get this
equality.

153
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And you can do kind of fun things like
substitute in say

154
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pi in for x and conclude that e to the i
pi is cos pi plus i sin pi.

155
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So co sin pie is minus 1, sin pie is 0.
So e to the i pie is negative 1.

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Taylor Series aren't just a jumping off

157
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point for calculus, and for numerical
approximations.

158
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Taylor Series are also our, our first step
into the theory of complex analysis.

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And that theory is more natural than it
might seem at first.

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I mean, if the complex numbers are
affecting the radius of

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convergence of my real power series, then
they must be really there.

162
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I mean, they're not as imaginary as people
might think.

163
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So Taylor Series aren't just a jumping

164
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off point for calculus and for numerical
approximations.

165
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Taylor Series, are also our, our first
step into the theory

166
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of complex analysis.

167
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And that theory is more natural than it
might seem at first.

168
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I mean, if the complex numbers are
affecting the radius of

169
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convergence of my real power series then
they must be really there.

170
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I mean, they're not as imaginary as people
might think.

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[SOUND]