5.2 Taylor Series

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De_nition 5.2.1 If f is an in_nitely di_erentiable function at some point

z0, then the Taylor Series for f about z0 is

f(z0) + f0(z0)(z 􀀀 z0) +

f2(z0)

2!

(z 􀀀 z0)2 +

f3(z0)

3!

(z 􀀀 z0)3 + _ _ _

or in more condensed notation:

1 X

0

fk(z0)

k!

(z 􀀀 z0)k

where f0 = f and 0! = 1! = 1.

Note what I didn't say: I didn't claim that the series was equal to f(z). In

general it isn't. For example, the function might be zero on [􀀀1; 1] and do

anything you liked outside that interval. Then if we took z0 = 0 we would

get zero for every coe_cient in the series, which would not tell us anything

about f(2). Why should it? Things are even worse than this however:

Exercise 5.2.1 The real function f is de_ned as follows: for x _ 0, f(x) =

0. For all other x, f(x) = e101e􀀀

1

x2 .

Verify that f is in_nitely di_erentiable everywhere. Verify that all derivatives

are zero at the origin. Deduce that the series about 0, evaluated at 1 is zero

and the value of f(1), e100, di_ers from it by rather a lot.

Draw the graph.

5.2. TAYLOR SERIES 135

I also didn't claim that the Taylor series for f converges. We have from

contemplating the above exercise the depressing conclusion that even when it

converges, it doesn't necessarily converge to anything of the slightest interest.

Exercise 5.2.2 There is a perfectly respectable function

eex

Compute its Taylor series about the origin. Likewise, investigate the Taylor

Series for

eee

x

Does it converge?

The question of whether Taylor Series have to converge could keep you busy

for quite a while, but I shall pass over this issue rather quickly. The situation

for analytic functions of a complex variable is so cheering by comparison that

it needs to be stated quickly as an antidote to the depression brought on by

thinking about the real case.

Theorem 5.1 (Taylor's Theorem)

If f is a function of a complex variable which is analytic in a disk of radius

R centred on w, then the Taylor series for f about w converges to f:

f(z) = f(w) + f0(w)(z 􀀀 w) +

f2(w)

2!

(z 􀀀 w)2 +

f3(w)

3!

(z 􀀀 w)3 + _ _ _

=

1 X

0

fk(w)

k!

(z 􀀀w)k

provided jz 􀀀 wj < R.

No Proof 2

It is usual to tell you that the convergence of the series is uniform on subdisks

of the given disk, which means that the N you _nd for some accuracy

" depends only on " and not on z. Unfortunately, this merely means that

on each subdisk there is for every " a 'worst case z' and we can pick the

N for that case and it will work for all. Of course, the worst case may be

136 CHAPTER 5. TAYLOR AND LAURENT SERIES

terrible, and the case we actually care about have much smaller N, so this is

of limited practical value sometimes.

Although Taylor's Theorem brings us a ray of cheer, note that it gives us no

practical information about how fast the series converges, although this may

be available in particular cases. I shall skip telling you how to _nd this out;

it is treated in almost all books on Complex Function Theory.

It is worth pointing out that the power series expansion of a function is

unique:

Theorem 5.2 (Uniqueness of Power Series)

1 X

0

akzk =

1 X

0

bkzk

) 8k; ak = bk

No Proof. 2

This doesn't mean that the Taylor series for a function about di_erent points

can't look di_erent.

We know that sin z has derivative cos z which in turn has derivative 􀀀sin z.

We also know the values of sin 0(0) and cos 0(1). This is enough:

sin(z) = sin(0) + sin0(0)z +

sin2(0)

2!

z2 + _ _ _

So

sin(z) = z 􀀀

z3

3!

+

z5

5! 􀀀_ _ _

as advertised.

The Taylor Series about 0 is often ( but not always) a good choice, and is

called the MacLaurin Series.

It is normally the case that if a Power series converges in a disk but not at

some point on the boundary, it will diverge for every point outside the disk.

This doesn't mean that there isn't a perfectly good power series about some

other point.

5.2. TAYLOR SERIES 137

Example 5.2.1 Take

f(z) =

1

1􀀀z

Then it is easy to see that

fk(z) =

k!

(1 􀀀 z)k+1

and hence that fk(0) = k!. It follows that the Taylor series is

f(z) = 1+z +z2 +z3 +_ _ _ =

1 X

0

zk

Now it is clear that this converges in a disk centred at 0 of radius 1, and

rather obvious that it doesn't converge at z = 1. At z = 2i we get

1 + 2i􀀀4􀀀8i + 16+_ _ _

which also diverges. If however we expand about i and evaluate at 2i we get

1

1 􀀀 z

=

1

1 􀀀 i

+ i

1

(1 􀀀 i)2

+ i2 1

(1 􀀀 i)3

+ _ _ _

which is

ei_=4

p2

+ i

ei2_=4

(p2)2

+ i2 ei3_=4

(p2)3

+ _ _ _

Now if we look at the modulus of each term we get:

1

p2

+

1

(p2)2

+

1

(p2)3

+ _ _ _

which is a geometric series with ratio less than 1 and hence converges.

If you were to 'straighten out' the series of complex numbers being added up

so that they all lay along the positive reals, then we would have a convergent

series. Now if you rotate them back into position a term at a time, you

would have to still have the series converge in the plane. (This is an intuitive

argument to show that absolute convergence implies convergence for series.

It is not hard to make it rigorous.) So the series converges.

Exercise 5.2.3 What is the radius of convergence (i.e. the radius of the

largest disk such that the series converges in the interior of the disk) for the

function sin z expanded about the point i? Draw a picture and take a ying

guess _rst, then prove your guess is correct.

138 CHAPTER 5. TAYLOR AND LAURENT SERIES

It is true that, on any common domain of convergence, power series can be

added, subtracted, multiplied and divided. The last operation may introduce

poles at the zeros of the divisor, just as for polynomials. All the others result

in new (convergent) power series. In fact thinking about power series as

'in_nitely long polynomials where the higher terms matter less and less' is

not a bad start. It clearly goes a bit wrong with division however.

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