2.7 The Exponential Function

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The real exponential function is de_ned by

exp(x) = 1+x+x2=2! + x3=3! + x4=4! + _ _ _

70 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

or more formally as the in_nite sum:

exp(x) =

1 X

n=0

xn=n!

We write

exp(x) as ex

for reasons which will become apparent shortly.

2.7.1 Digression: In_nite Series

Expressing functions by in_nite series is something you must get used to; the

thing you need to realise is that almost all the functions that you use other

than polynomials and ratios of polynomials are given by these in_nite series

(called power series in the above case, because they have di_erent powers of

x in them). When you calculate sin(12o) or p768:3 or ln(35:4) by pressing

the buttons on your calculator, it produces a number on the display. It gets

it from taking the _rst k terms in a power series for the function. A better

calculator will take more terms; you get the k you pay for.

It is very convenient to have a formula for sin, cos, ex and all the other

functions as an in_nite series, because it is easy to add on some number

of terms, and to stop when the increment is so small it doesn't make any

di_erence to the answer. There is, however, a fundamental problem with this

approach.

If you add the terms:

1 + 1=2 + 1=4 + 1=8 + 1=16 + _ _ _

or more formally if you calculate

1 X

n=0

1=2n

you rapidly get something pretty close to 2. Ten terms gets you to within

one tenth of a percent of the answer, which is, of course, 2.

Suppose you add up the _rst few terms of the series

1 + 1=2 + 1=3 + 1=4 + 1=5 + _ _ _

2.7. THE EXPONENTIAL FUNCTION 71

or more explicitly you try to compute a _nite number of terms of

1 X

n=1

1=n

You _nd that you seem to be getting to the answer rather slowly, but it

is easy enough to put it on a computer and _nd a few thousand terms very

quickly. If you do this, you discover that after about ten thousand terms, you

are only getting increments in the fourth decimal place (of course!) and after

a million terms, you get increments only in the sixth place. If your calculator

is working to single precision and it does this sort of thing, it will conclude

that the series sums to 16.695311. This is what I get on my computer if I

sum ten million terms, which takes less than ten seconds. The question is:

how far out is the result? Could it get up to 20 if we kept going on a higher

precision machine?

The answer is that the result is about as far out after ten million terms as it

is after two. The series actually diverges and goes o_ to in_nity.

If you didn't know whether a series converged or diverged, it would be possible

for you to calculate a number to six places of decimals in a few seconds,

and to get a result which is absolutely and totally wrong, by assuming it

converges because the increments have fallen below the precision of your machine.

For this reason, it is of very considerable practical importance to be

able to decide if a series converges. It is also useful to work out how fast

it converges by getting a bound on the error as a function of the number of

terms used in the sum.

If you have a power series expansion for a (real) function, then it will, of

course have an x in it, and when you plug in a value for the x and add up

the series, you get the value of f(x). It may happen that the series converges

nicely for some values of x and goes o_ its head for others. To see the kind

of thing that could happen, take a look at the function

f(x) =

1

1 + x

You would have to be crazy to evaluate this by a power series, but there

might be other functions which behave the way this one does, so bear with

me.

It is not too hard to persuade yourself that the equation

1

1 + x

= 1􀀀x+x2

􀀀x3 +x4

􀀀x5 +_ _ _

72 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

holds for at least some x. If you do the 'multiplication':

(1 + x)(1 􀀀 x + x2

􀀀 x3 + x4

􀀀 x5 + _ _ _)

you get

(1 􀀀 x + x2

􀀀 x3 + x4

􀀀 x5 + _ _ _) + (x􀀀x2 +x3

􀀀x4 +_ _ _)

and it certainly looks plausible that all terms cancel except for the initial 1.

So cross multiplication seems to work. What more could you want?

If you put x = 1=2 you _nd, if you investigate the matter carefully, that the

series does converge. If you put x = 􀀀1 the sum goes o_ to in_nity, but it

should anyway. If you put x = 1 you get

1 􀀀 1 + 1􀀀1 + 1􀀀1 + _ _ _

which is supposed to sum to 1=2. There would seem to be some reasonable

doubt about this.

Exercise 2.7.1 Put x = 2. How do you feel about the resulting series converging

to 1=3?

I am, I hope, reminding you of _rst and second year material, and I hope

even more that you have some recollections of how to test for convergence of

in_nite series. If not, look it up in a good book.

The best sort of result we can hope for is that a power series converges for

every value of x, and that we can get a handle on estimating a bound for the

error after n terms. This bound will usually depend on the x.

The situation for exp(x) is fairly good: the Taylor-MacLaurin theorem tells

us that the error at the stage n is not bigger than the (n + 1)th term, for

negative x. This can, indeed, be made as small as desired by making n big

enough. You can satisfy yourself by some heavy thought that the situation

for positive x is also under control. So the formula

exp(x) = 1+x+x2=2! + x3=3! + x4=4! + _ _ _

is one we can feel relatively secure about.

Exercise 2.7.2 How many terms would you need to calculate exp(100) to

four places of decimals? How about exp(-100)?

2.7. THE EXPONENTIAL FUNCTION 73

2.7.2 Back to Real exp

The following exercise is important and will explain why we write exp(x) as

ex.

Exercise 2.7.3 Write down the _rst four terms of exp(x) and the _rst four

terms of exp(y). Multiply them together and collect up to show you have

rather more than the _rst four terms of exp(x + y).

Produce an argument to convince a sceptical friend that you can say with

con_dence that

8x; y 2 R exp(x) exp(y) = exp(x + y)

Another thing we can do is to show that if we di_erentiate the function

exp(x), we recover the function. This assumes that if we have an in_nite

series and we di_erentiate it term by term, the resulting series will converge

to the derivative of the function. You might like to brood on this to decide

whether you think this is going to happen (a) always (b) sometimes (c)

never. The answer cannot be (c) because the exponential function actually

IS di_erentiable, and is indeed its own derivative, just as you would hope

from di_erentiating the series termwise.

Exercise 2.7.4 You recall, I hope, computing the Fourier series for a square

wave. The series consists of di_erentiable functions, so you can di_erentiate

the series expansion termwise. But you clearly can't di_erentiate the square

wave at the discontinuities.

What do you get if you di_erentiate the series termwise and take limits?

I said earlier that everything you could do for R you could also do for C . You

can certainly add and multiply complex numbers, and you can divide them

except by zero. So the terms in the series

1 + z +z2=2! + z3=3! + z4=4! + _ _ _

are all respectable complex numbers. We can ask if the series converges to

some complex number when we stick a particular value of z in.

We know that it works if the value of z is a real number. What if it isn't?

74 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

The answer is that all the arguments go through, and the series converges

for every value of z. I am not going to give a formal proof of this as it is a

fair amount of work and anyway, you are probably not much of a mind to

do formal proofs. But it is instructive to work it out in a few cases. Let us

therefore calculate exp(i).

We get the series:

exp(i) = 1+i + (􀀀1)=2! 􀀀 i=3! + 1=4! 􀀀 i=5! 􀀀 1=6! + _ _ _

Separating the real and imaginary parts we get:

exp(i) = 1􀀀1=2! + 1=4! 􀀀 1=6! + _ _ _+i􀀀i=3! + i=5! 􀀀 i=7! + _ _ _

or

exp(i) = (1􀀀1=2! + 1=4! 􀀀 1=6! + _ _ _) + i(1 􀀀 1=3! + 1=5! 􀀀 1=7! + _ _ _)

You may or may not recognise the separate series as representing terms you

know. If you calculate the Taylor-MacLaurin expansions by

f(x) = f(0) + xf0(0)=1! + x2f00(0)=2! + x3f000(0)=3! + _ _ _

for the functions cos(x); sin(x) you will immediately recognise

exp(i) = cos 1 + i sin 1

By putting ix in place of i you get:

exp(ix) = cos(x) + i sin(x)

This gives us, when x = _, Euler's Formula:

ei_ + 1 = 0

This links up the _ve most interesting numbers in Mathematics, 0; 1; e; i; _,

in the most remarkable formula there is. Since e seems to be all about what

you get if you want a function f satisfying f0 = f, and _ is all about circles,

it is decidedly mysterious.

Thinking about this gives you a creepy feeling up the back of the spine: it is

as though you went exploring the Mandelbrot set and found a picture of an

old bloke with a stern look and long white whiskers looking out at you. It

might incline you to be better behaved henceforth. I have, therefore, some

reservations about the next exercise:

2.7. THE EXPONENTIAL FUNCTION 75

Exercise 2.7.5 (Don't do this if you watch the X-Files)

Euler's formula might either (a) be in no need of an explanation, just a

proof, or (b) be explained by God having a silly sense of humour, just like

most intelligent people or (c) have a more prosaic explanation.

The exponential function is a procedure for turning vector _elds into ows;

if you take the vector _eld which is given by

V _ x

y _ = _ 0 􀀀1

1 0 __ x

y _

you call the matrix A and then the ow is given as

etA

This is basic to the theory of systems of ODE's. You can verify this particular

case by exponentiating the matrix tA using the standard power series formula

for the exponential of a real number x and replace x by tA. Since all you

have to do with x is to multiply it by itself, divide by a non-zero real number,

add a _nite set of these things and take limits, and since all of these can be

done with matrices, this all makes sense.

Draw a picture of the vector _eld and the resulting ow.

Identify the matrix as a complex number.

Deduce that eit = cos t + i sin t is little more than the observation that a

tangent to a circle is always orthogonal to the radius, together with the observation

that exponentiation is about solving ODE's by Euler's method taken

to the limit.

If you watch the X-Files (The Truth is out there, the Lies are in the programme),

you might prefer to have the mystery preserved. Actually, there is

still heaps of mystery left, indeed it's the charm of Mathematics5.

Since the argument that exp(x + y) = exp(x) exp(y) (the 'index law') goes

through for the complex numbers just as it does for the reals, we can write

exp(x + iy) = exp(x)(cos(y) + i sin(y))

5Engineers who want to preserve mystery are a bit of a worry.

76 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

And the index law justi_es our writing

ex+iy = ex(cos y + i sin y)

If we write our complex number out in Polar form, z = (r; _) we have that

z = x + iy = r cos _ + ir sin _ = rei_

This is a quite common notation; it needs a bit of explanation and I have

just given you one.

2.7.3 Back to Complex exp and Complex ln

We are now ready to look at the exponential map

exp : C 􀀀! C

It is certainly much more complicated that the real exponential, and the

extra complications will turn out to be very useful.

To see what it does, notice that exp takes the real axis to the positive real

numbers. 0 goes to 1, and all the negative real numbers get squashed into

the space between 0 and 1. It takes the imaginary axis and wraps it around

the unit circle. The map eiy is a periodic function: think of the imaginary

axis as a long line made out of chewing gum, and note that the chewing gum

line is picked up by exp and wrapped (without stretching) in the positive

(anticlockwise) direction around the unit circle. The negative imaginary

numbers are wrapped around in the opposite direction.

I have indicated the start of this process on the axes, as if we have almost

got the exponential function but not quite, restricted to the axes. This is

_gure 2.22.

What happens to the rest of the plane? The image by exp of the axes will

cover only the positive real axis and the unit circle, but the unit circle gets

covered in_nitely often. The number 1 + 0i also has an in_nite number of

points sent to it. Does anything at all get sent to the origin? To -2? These

are all good questions to ask.

I start o_ to answer some of them in the following example of how to compute

the e_ect of the complex exponential.

2.7. THE EXPONENTIAL FUNCTION 77

Figure 2.22: The start of the exponential function, restricted to the axes

Example 2.7.1 What is the image by the exponential map of the unit square?

Solution The lower edge, the point x + 0i for 0 _ x _ 1 gets sent to the

line segment ex + 0i since cos 0 + i sin 0 = 1. The top edge gets sent to

ex(cos 1 + i sin 1) Since 1 is 1 radian, this goes to a line at about 570 and of

radii from 1 to e. The left hand edge, the part of the imaginary axis between

0 and 1, goes to the corresponding arc of the unit circle, and the right hand

edge of the unit square, the points 1+iy, for 0 _ y _ 1 goes to e(cos y+i sin y)

which is an arc of a circle of radius e. The result is shown in _gure 2.23

The _gure following shows the results of applying the exponential map to

the bigger square centred on the origin of side 2 units:

Exercise 2.7.6 Mark in the image of the axes on _gure 2.24

The image by exp of the unit disk is shown in _gure 2.25

I have marked on the X and Y axes to make it clearer where it is.

Exercise 2.7.7 What is the inverse of exp of a point in the spotty region of

_gure 2.25 which is closest to the origin?

78 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

Figure 2.23: The image by the exponential function of the unit square

Figure 2.24: The image by the exponential function of the 4 times unit square

2.7. THE EXPONENTIAL FUNCTION 79

Figure 2.25: The image by the exponential function of the unit disk

Note that this is taking the complex logarithm of the point!

A little experimenting is called for, and is quite fun; strips which are vertical

and big enough, get mapped onto disks about the origin with a hole at the

centre. If the height of the strips is too small, they get mapped into sectors

of disks with a hole at the centre. You can easily see that there is no way to

actually get any point to cover the origin, although you can get as close as

you like to it. If the strip is very high, you go around several times.

Suppose you wanted a logarithm for 􀀀1 + i which is sitting in the second

quadrant. That is to say, you want something, anything which is mapped to

it by exp. Then we have that

ex(cos y + i sin y) = (􀀀1 + i)

It is easier to express (-1+i) in polars as (p2; 3_=4). Then we have

ex(cos y + i sin y) = p2(cos(3_=4) + i sin(3_=4))

which tells us that

x = ln(p2); y = 3_=4

does the job.

80 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

So does

x = ln(p2); y = (2n_ + 3_=4)

for every integer n. There is a vertical line of points in C , at x value ln(p2)

and y values separated by 2_ which all get sent to the same point, 􀀀1 + i.

If we want the logarithm of 􀀀2, we write it as 2(cos _ + i sin _) and see that

ln 2 + i_ does it nicely. So does ln 2 + i(2n + 1)_ for every integer n.

Exercise 2.7.8 Check this claim by exponentiating ln 2 + i(2n + 1)_

The conclusion that we come to is that every point in C except 0 has an

in_nite number of logarithms, so we have the same problem as for z2, only

much worse, if we insist on having a logarithm function. Our Riemann surface

for the exponential and logarithm functions has not just two but an in_nite

number of leaves, joined together like an in_nite ascending spiral staircase.

The leaves all have their centres joined together: this one is a little di_cult

to draw. Think of a set of cones, one for each integer, all with a common

vertex, nested inside each other, with cuts as in the diagram for z2 giving a

path from each cone to the one lower down- for ever.

Exercise 2.7.9 Draw a bit of the Riemann surface for the exponential and

logarithm functions. Show how you can make some branch cuts to get a piece

of it which maps to C with the negative real axis removed. Show that there

are in_nitely many such pieces.

It is common to de_ne a Principal Branch of the logarithm, often called Log,

by insisting that we restrict attention to answers which lie in the horizontal

strip with 􀀀_ < y < _. Alternatively, think of what exp does to such a strip.

The word 'branch' suggests to me either trees or banks, and neither seems

to have much to do with a piece of the plane which is mapped to a piece of a

thing like an in_nite ascending spiral staircase, the central column of which is

non-existent. It is, as explained earlier for the squaring function, rather old

fashioned terminology. The exponential function onto the Riemann surface

is a good test of your ability to visualise things. You know you are getting

close when you feel dizzy just thinking about it.

The log function is a proper inverse to exp providing we regard exp as going

from C to this Riemann surface. And if we don't, we get the usual mess, as

seen in the case of the square function.

2.8. OTHER POWERS 81

Exercise 2.7.10 Show that the function Log(1+z

1􀀀z ) takes the interior of the

unit disk to the horizontal strip 􀀀_=2 < y < _=2.

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