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\chapter{The Fundamental Properties of Analytic Functions;
Taylor's, Laurent's and Liouville's Theorems}

\Section{5}{1}{Property of the elementary functions.}

The reader will be already familiar with the term elementary function,
as used (in text-books on Algebra, Trigonometry, and the Differential
Calculus) to denote certain analytical expressions* depending on a
variable z, the symbols involved therein being those of elementary
algebra together with exponentials, logarithms and the trigonometrical
functions; examples of such

expressions are

1  §

Z-, e~, iogz, arcsm '-.

Such combinations of the elementary functions of analysis have in
common a remarkable property, which will now be investigated.

Take as an example the function e .

Write e'=f z).

Then, if 2 be a fixed point and if z' be any other point, we have

f z')-f z) \ e~ -e' \, e' '- ' - 1 z -z z - z z - z

f, z' - z (z -z)-

and since the last series in brackets is uniformly convergent for all
values of it follows \hardsectionref{3}{7}) that, as z'->z, the quotient

z - z tends to the limit e, uniformly for all values of arg z - z).
This shews that tlie limit of

f z:)-f z)

z - z

is in this case independent of the path by which the point z tends
towards coincidence witJt z.

It wall be found that this property is shared by many of the
well-known elementary functions; namely, that iif(z) be one of these
functions and h. be

* The reader will observe that this is uot the sense in which the term
function is defined \hardsectionref{3}{1}) in this work. Thus e.g. .r - hj and | z \
are functions of z (-x + iy) in the sense of \hardsectionref{3}{1}, but are not
elementary functions of the type under consideration.

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any complex
number, the limiting value of

exists and is independent of the mode in luhich h fends to zero.

The reader will, however, easily prove that, ii f(z)=x -iy, where z =
x- iy, then lim' - - - - IXJ- is not independent of the mode in which
A- >0.

\Subsection{5}{1}{1}{Occasional failure of the jwoperty.}

For each of the elementary functions, however, there will be certain
points z at which this property will cease to hold good. Thus it does
not hold for the function l/( - a) at the point z = a, since

 i Qh\ z - a+h z-a

does not exist when z = a. Similarly it does not hold for the
functions log z

and z at the point z = Q.

These exceptional points are called singular points or singulaHties of
the function f(z) under consideration; at other points f(z) is said
to be analytic.

The property does not hold good at any point for the function \ z.

\Subsection{5}{1}{2}{Cauchy's* definition of an analytic function of a complex variable.}

The property considered in § b\ \ will be taken as the basis of the
definition of an analytic function, which may be stated as follows.

Let a two-dimensional region in the -plane be given; and let w be a
function of z defined uniquely at all points of the region. Let z, z-
hz be values of the variable z at two points, and u, u + Bu the
corresponding values

of u. Then, if, at any point z within the area, - tends to a limit
when 8x->0,

By-*0, independently (where 8z = 8x + iBy), u is said to be a function
of z which is monogenic or analytic j" at the point. If the function
is analytic and one-valued at all points of the region, we say that
the function is analytic throughout the region.

We shall frequently use the word ' function ' alone to denote an
analytic function, as the functions studied in this work will be
almost exclusively analytic functions.

* See the memoir cited in § 5 "2.

t The words ' regular ' and ' holomorphic ' are sometimes used. A
distinctiou has been made by Borel between ' monogenic ' and '
analytic ' functions in the case of functions with an infinite number
of singularities. See \hardsubsectionref{5}{5}{1}.

X See \hardsectionref{5}{2} cor. 2, footnote.

6-2

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In the foregoing definition, the function u has been defined only
within a certain region in the -plane. As will be seen subsequently,
however, the function u can generally be defined for other values of 2
not included in this region; and (as in the case of the elementary
functions already discussed) may have singularities, for which the
fundamental property no longer holds, at certain points outside the
limits of the region.

We shall now state the definition of analytic functionality in a more
arithmetical form.

Let f(z) be analytic at z, and let e be an arbitrary positive number;
then we can find numbers I and h, h depending on e) such that

z -z I

whenever \ z' - z\ < h.

Vi f(z) is analytic at all points of a region, I obviously depends on
z; we consequently write 1 = f z).

Hence /(/) = f(z) + - z) f(z) + v z'- z\

where v is a function of z and z such that ! v < e when \ z -z\ < t>.

Example 1. Find the points at which the following functions are not
analytic :

z - \ (i) 2 . (ii) cosec2 (2 = /itt, w any integer). (iii) - - - - -
(3 = 2,3).

1

(iv) ez (j = 0). (V) z- ) zf (2 = 0,1).

Example 2. If z = x - iy, f(z) = u ii\ where u, v, x, y are real and /
is an analytic function, shew that

8m "bv cu cv /T - s

 :r =, : =-; . \addexamplecitation{Kiemann.}

ex oy oy ex

\Subsection{5}{1}{3}{An application of the modified Heine-Borel theorem.}

Let f(z) be analytic at all points of a continuum; and on any point z
of

the boundar ' of the continuum let numbers f (z), S (S depending on z)
exist

such that

\ f(z')-f(z)- z'-z)Mz)<e,z'-z\

whenever \ z - z < 8 and z is a point of the continuum or its
boundary.

[We write /i iz) instead of /' z) as the differential coefficient
might not exist when c approaches z from outside the boundary so
that/j (2) is not necessarily a unique derivate.]

The above inequality is obviously satisfied for all points z of the
continuum as well as boundary points.

Applying the two-dimensional form of the theorem of \hardsectionref{3}{6}, we see that
the region formed by the continuum and its boundary can be divided
into a jinite number of parts (squares with sides parallel to the axes
and their

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interiors, or portions of such squares) such that inside or on the
boundary of any j art there is one point z such that the inequality

fiz') - f z,) - (/ - z,)f, z,) .<e\ z'-z, \ is satisfied by all points
z inside or on the boundary of that part.

\Section{5}{2}{Cauchy's theorem* on the integral of a function round a contour.}

A simple closed curve C in the plane of the variable z is often called
a contour; ii A, B, D he points taken in order in the
counter-clockwise sense along the arc of the contour, and if f(z) be a
one-valued continuousf function of z (not necessarily analytic) at all
points on the arc, then the integral

f f z)dz or f f z)dz

taken round the contour, starting from the point A and returning to A
again, is called the integral of f(z) taken along the contour. Clearly
the value of the integral taken along the contour is unaltered if some
point in the contour other than A is taken as the starting-point.

We shall now prove a result due to Cauchy, which may be stated as
follows. If fiz) is a function of z, analytic at all points on and
inside a contour G, then

I f z)dz = 0.

For divide up the interior of C by lines parallel to the real and
imaginary

axes in the manner of \hardsubsectionref{5}{1}{3}; then the interior
of $C$ is divided into a number

of regions whose boundaries are squares Cj, C, ... Cm and other
regions

whose boundaries D,, D,, ... Dy are portions of sides of squares and
parts

of G; consider

Mr N r

X f(z)dz+ S f z)dz,

n = lJ(,C ) n=lJ(.D )

each of the paths of integration being taken counter-clockwise; in
the complete sum each side of each square appears twice as a path of
integration, and the integrals along it are taken in opposite
directions and consequently cancel §; the only parts of the sum which
survive are the integrals oi f z)

* Memoire sur les integrates definies prises entre des limites
imarjinaires (1825). The proof here given is that due to Goursat,
Trans. American Math. Soc. i. (1900), p. 14.

t It is sufficient for f(z) to be continuous when variations of z
along the arc only are considered.

:J: It is not necessary that f(z) should be analytic on C (it is
sufficient that it be continuous on and inside C), but if /( ) is not
analytic on C, the theorem is much harder to prove. This proof merely
assumes that /' [z) exists at all points on and inside C. Earlier
proofs made more extended assumptions; thus Cauchy's proof assumed the
continuity of f' z). Eiemann's proof made an equivalent assumption.
Goursat's first proof assumed that f(z) was uniformly differentiable
throughout C.

§ See § 4G, example.

/

J (On)

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taken along a number of arcs which together make up G, each arc being

taken in the same sense as in I f\ z)dz; these integrals therefore
just make

hc)'

up f z)dz.

Now consider 1 f z)dz. With the notation of \hardsubsectionref{5}{1}{2},

J (Cn)

f z) dz = [ f z,) + ( - z,)f' (.-,) + z- z,) v] dz

-' (C )

= /( i) - V ( 01 f dz +/' (z,) I zdz+l (z- z,) vdz.

\ f? = Mc = 0, f zdz

by the examples of \hardsectionref{4}{6}, since the end points of C coincide. Now let
In be the side of Cn and An the area of C . Then, using \hardsubsectionref{4}{6}{2},

TODO

But

l\ \ l n

= 0,

I

! (C )

In like manner

< e V2 . f \ dz\= eln \/2 . 4,, = 4e sj'2.

J Cn

I f z)dz\ \ i \ \ {z-z )vdz\

J (Dn) I J (Dn)

(Dn)

  4>e (An + In Xi)\ \ '2,

Avhere An is the area of the complete square of which Dn is part, In
is the side of this square and \ n is the length of the part of G
which lies inside this square. Hence, if \ be the whole length of G,
while I is the side of a square which encloses all the squares Gn and
Dn,

f(z)dz k S f z)dz + t \ f z)dz

|J(0 I n = \'J(Cn) I n = \ \ J Dn) |

( M N iV )

<4eV2 An+ S An' + l tXn\ (m = 1 m = 1 n=l )

< 4e V2 . I' + X).

Now e is arbitrarily small, and I, \ and 1 f z)dz are independent of
e.

I iC)'

It therefore follows from this inequality that the only value which I
f(z) dz can have is zero; and this is Cauchy's result.

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Corollary . If there are two jjaths z AZ and Zq,BZ from 2o to Z, and
if /(z) is a function of z analytic at all points on these curves and
throughout the domain enclosed by

these two paths, then / f(z) dz has the same value whether the path of
integration is

Z(iAZ (jv ZqBZ. This follows from the fact that ZqAZBzq is a contour,
and so the integral taken round it (which is the difference of the
integrals along zqAZ and ZoBZ) is zero.

Thus, if /(2) be an analytic function of z, the value of / f z)dz is
to a certain extent

J AB

independent of the choice of the arc AB, and depends only on the
terminal points A and B. It must be borne in mind that (his is only
the case whenf z) is an analytic function in the sense of \hardsubsectionref{5}{1}{2}.

Corollary 2. Suppose that two simple closed curves (7 and Cj are
given, such that Co completely encloses Cj, as e.g. would be the case
if Cq and Ci were confocal ellipses.

Suppose moreover that/ (2) is a function which is analytic* at all
points on Cq and Cj and throughout the ring-shaped region contained
between Cq and C . Then by drawing a network of intersecting lines in
this ring-shaped space, we can shew, exactly as in the theorem just
proved, that the integral

//(

dz

is zero, ivhere the integration is taken round the whole boundary of
the nng-shaped space; this boundary consisting of two curves Co and C
, the one described in the counter-clochvise direction aiid the other
described in the qlockwise direction.

Corollary 3. In general, if any connected region be given in the
3-plane, bounded by any number of simple closed curves Co, Ci, Co,
..., and if /(z) be any function of z which is analytic and one-valued
everywhere in this region, then

I

f z)dz

is zero, where the integral is taken round the whole boundary of the
region; this boundary consisting of the curves Co, Ci, ..., each
described in such a sense that the region is kept either uhvays on the
right or always on the left of a person walking in the sense in
question round the boundary.

An extension of Cauchy's theorem I f(z) dz = 0, to curves lying on a
cone whose vertex

is at the origin, has been made by Ravut (N'ouv. Annales de Math. (3)
xvi. (1897), pp. 365-7). Morera, Moid, del 1st. Lombardo, xxii.
(1889), p. 191, and Osgood, Bull.

Amer. Math. Soc. II. (1896), pp. 296-302, have shewn that the property
f z)dz =

may be taken as the property defining an analytic function, the other
properties being deducible from it. (See p. 110, example 16.)

Example. A ring-shaped region is bounded by the two circles | s | = 1
and | z | = 2 in the 2-plane. Verify that the value of I -, where the
integral is taken round the boundary of this region, is zero.

* The phrase 'analytic throughout a region' implies one-valuedness (§
5-12); that is to say that after z has described a closed path
surrounding Co, f(z) has returned to its initial value. A function
such as log z considered in the region 1 | | 2 will be said to be '
analytic at all points of the region. '

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For the boundary consists of the circumference ]2| = 1, described in
the clockwise direction, together with the circumference |s| = 2,
described in the counter-clockwise direction. Thus, if for points on
the iirst circumference we write 2 = e, and for points on the second
circumference we write z = '2e<'t>, then 6 and cp are real, and the
integral becomes

jo e' jo 2e'*

\Subsection{5}{2}{1}{The value of an analytic function at a 'point, expressed as an integral taken round a contour enclosing the point.}

Let C he a contour within and on which /( ) is an analytic function of
z. Then, if a be any point within the contour,

z - a is a function of z, which is analytic at all points within the
contour G except the point z = a.

Now, given e, we can find 3 such that

\ fi z)-f a)- z-a)f' a)\ \ \ z-a\

whenever | - a | < S; with the point a as centre describe a circle 7
of radius 7' < 8, r being so small that 7 lies wholly inside C

Then in the space between 7 and G f z)\ \ {z - a) is analytic, and so,
by \hardsectionref{5}{2} corollary 2, we have

f z)dz\ [f z)dz

c z - a Jy z - a where I and I denote integrals taken
counter-clockwise along the curves

G and 7 respectively.

But, since 1 2 - a | < S on 7, we have

r f(z) dz\ f f a)+iz- a)f (a) + v(z-a) J y z - a J y z - a

where \ v\ < e; and so

r m y j f dz f ( ) dz -\ vdz.

J C Z-a \ lyZ -a - Jy Jy

Now, if z be on 7, we may write

z - a = re, where r is the radius of the circle 7, and consequently

"2t ire dO

jyZ - a Jo

= I I de = 2771, y j, re'" Jo

and dz= ire' dd = 0;

also, by \hardsubsectionref{4}{6}{2},

vdz

< e . 27rr.

i

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r f(z)d2 \ 2 if( a) I = f €dz\ \ 2irre.

J c z - a \ J y 1

( r ( 5: 1 n.7: i /'

Thus

c z - a

But the left-hand side is independent of e, and so it must be zero,
since e is arbitrary; that is to say

    2771 ] c z - a

This remarkable result expresses the value of a function f z\ (which
is analytic on and inside (7) at any point a within a contour C, in
terms of an integral which depends only on the value of/( ) at points
on the contour itself.

Corollary. If f(z) is an analytic one-valued function of 2 in a
ring-shaped region bounded by two curves C and C", and a is a point in
the region, then

' 2ni J c -(i "tti c' Z - a

where C is the outer of the ciu-ves and the integi-als are taken
counter-clockwise.

\Subsection{5}{2}{2}{The derivates of an analytic function $f(z)$.}

The function/' (2'), which is the limit of

f z + h)-f z) h

as h tends to zero, is called the derivate of / z). We shall now shew
that f' z) is itself an analytic function of z, and consequently
itself possesses a derivate.

For if be a contour surrounding the point a, and situated entirely
within the region in which f(z) is analytic, we have

f(a+h)-f(a)

f (a) = hm --,

h o n

,, 0 2TTih [J c z-a-h (j z-a )

= Km -I- \ /( ) dz

h Q liri J c z-a) z- a~h)

Itti J c z - af h 27ri J c z - ay z-a- h)

Now, on C, f(z) is continuous and therefore bounded, and so is (z -
a)~; while we can take | h less than the upper bound of U - a |.

\

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Therefore

(z - a)- z - a - h) Then, if I be the length of C,

h f f z)dz

[chap. V

is bounded; let its upper bound be K.

TODO:missingpagenum?

lim

<lim !A|(27r)- 7 : = 0,

I h Q lirij c z -af z-a- h)

and consequently f (a) = - . |, .

   Ziri J c z - a)- '

a formula which expresses the value of the derivate of a function at a
point as an integral taken along a contour enclosing the point.

From this formula we have, if the points a and a + h are inside C,

f(a + h)-f'(a) J\ r f z)dz 1 1

h 27ri J c h

(z - a- hy z - ay

c (z - a - hy z - ay

f(z) dz

= 9:Z;f f )dz.

iTTl J c

... +hAh,

ZTTi J c \ Z - ay

and it is easily seen that J./ is a bounded function of z when \ h\ <
\ z - a.

Therefore, as h tends to zero, A~' /'( - + A) - /' (a) tends to a
limit, namely

J\ / f(z)dz 27ri J c z - ay '

Since /' (a) has a unique differential coefficient, it is an analytic
function of a; its derivate, which is represented by the expression
just given, is denoted by/" (a), and is called the second derivate of
/(a).

Similarly it can be shewn that /"(a) is an analytic function of a,
possessing a derivate equal to

2 f f(z)dz\ 27ri Jc z-ay'

this is denoted by f" (a), and is called the third derivate of /(a).
And in general an nth derivate/*"' (a) of /(ct) exists, expressible by
the integral

r. n r f(z)dz P 2'rTiJc z-aT+ '

and having itself a derivate of the form

(n + 1)! r f z)dz .

27ri Jciz- a)"+ ' the reader will see that this can be proved by
induction without difficulty.

 '

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A function which possesses a first derivate with respect to the
complex variable z at all points of a closed two-dimensional region in
the -plane therefore possesses derivates of all orders at all points
inside the region.

\Subsection{5}{2}{3}{Caiichys inequality for f " (a).}

Let f(z) be analytic on and inside a circle G with centre a and radius
?\ Let M be the upper bound oif z) on the circle. Then, by \hardsubsectionref{4}{6}{2},

!/""( )i;4/c; 'i'' '

M.n\

Example. l f(z) is analytic, z = x- iij and V- =;, + -2, .shew that

V2 log 1/(2) 1 = 0; andy2|/(e)|>0 unle.ss/(2) = or/' z) = Q.
\addexamplecitation{Trinity, 1910.}

\Section{5}{3}{Analytic functions represented by uniformly convergent series.}

 X)

Let X fn (z) be a series such that (i) it converges uniformly along a
=o

contour C, (ii) / (z) is analytic throughout C and its interior.

00 Then 2 fni ) converges, and the sum* of the series is an analytic

n =

function throughout C and its interior.

00

For let a be any point inside C; on C, let S fn z) = (z).

Then -. - dz = j- \ fn( )\

27ri J c z - a 27ri j c Im=o ] z -a

 .o \ 2mlc 2-a \ '

00

by* \hardsectionref{4}{7}. But this last series, by \hardsubsectionref{5}{2}{1}, is S fiid)', the series
under

n =

consideration therefore converges at all points inside C; let its sum
inside G (as well as on C) be called (z). Then the function is
analytic if it ha,s a unique differential coefficient at all points
inside G.

But if a and a + h be inside G,

 (a + h)- (a) \ J f ( ) (

A 27rt J c z - a) z - a - h)'

and hence, as in \hardsubsectionref{5}{2}{2}, lim W a- h) - a)] A~ ] exists and is equal to

7t-*0 * Since | 2 - a |~i is bounded when a is fixed and z is on C,
the uniformity of the convergence of S / z)l[z - a) follows from that
of 2 / [z).

n=0 n=0

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  - : 7 - dz; and therefore <I> (z) is analytic inside G. Further, by
27rt J c z- a)

transforming the last integral in the same way as we transformed the
first

00 ao

one, we see that <!>' (a) = S / ' (a), so that 2 fn (a) may be '
differentiated

n=0 n=0

term by term.'

If a series of analytic functions converges only at points of a curve
which is not closed nothing can be inferred as to the convergence of
the derived series*.

'COS 7hJ7

Thus 2 ( - )" - 2 - converges uniformly for real values of x \hardsubsectionref{3}{3}{4}).
But the derived

H = l 'i

  sin ij series 2 ( - )""* converges non-uniformly near A' = (2m+1)
tt, (m any integer); and

H = l 'ii

the derived series of this, viz. 2 ( - )"~ cos n.v, does not converge
at all.

(1=1

Corollary. By \hardsectionref{3}{7}, the sum of a power series is analytic inside its
circle of con- vergence.

\Subsection{5}{3}{1}{Analytic functions represented by integrals.}

Let f t, z) satisfy the following conditions when t lies on a certain
path of integration (a, h) and z is any point of a region >S' :

(i) f and ~ are continuous functions of t.

   dz

(ii) / is an analytic function of z.

df * (in) The continuity of ~- qua function of z is uniform with
respect to

the variable t.

rb .

Then I f(t, z)dt is an analytic function of z. For, by \hardsectionref{4}{2}, it has
the

, !* f* dt\ t, z) . unique derivate - - - - dt.

\Subsection{5}{3}{2}{Analytic functions represented by infinite integrals.}

From \hardsubsectionref{4}{4}{4} (II) corollary, it follows that I f (t, z) dt is an
analytic

J a

function of z at all points of a region >S' if (i) the integral
converges, (ii) f t, z) is an analytic function of z when t is on the
path of integration and z is on S,

(iii) - : ' is a continuous function of both variables, (iv) - - - dt

dz J a OZ

converges uniformly throughout 8.

For if these conditions are satisfied f t, z) dt has the unique
derivate

J a

J a

dz

* This might have been anticipated as the main theorem of this section
deals with uniformity of convergence over a two-dimensional region.

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A case of very great importance is afforded by the integral I e~' /(0
dt,

Jo where /(t) is continuous and \ f t)\ < Ke' where K, r are
independent of t; it is obvious from the conditions stated that the
integral is an analytic function of z when R z) r, > r. [Condition
(iv) is satisfied, by \hardsubsubsectionref{4}{4}{3}{1} (I),

r

since I fe"""''*' rf converges.]

Jo

\Section{5}{4}{Taylor's Theorem*TODO.}

Consider a function f z), which is analytic in the neighbourhood of a
point z = a. Let (7 be a circle with a as centre in the -plane, which
does not have any singular point of the function f(z) on or inside it
; so that f(z) is analytic at all points on and inside C. Let z = a +
h be any point inside the circle C. Then, by \hardsubsectionref{5}{2}{1}, we have

    Itti J c z-a-h

 -rriJc Xz-a iz-af ' ' ' z - ay- ' z - a)'' ' z - a - h)]

f z) But when z is on C, the modulus of - - -j is continuous, and so,

z - a - h

by \hardsubsectionref{3}{6}{1} cor. (ii), will not exceed some finite number M. Therefore,
by \hardsubsectionref{4}{6}{2},

1 f f(z)dz.h''+

27riJc z-ay'+' z-a-h) " 27r [rJ '

where R is the radius of the circle C, so that 'IttR is the length of
the path of integration in the last integral, and R = \ z - a\ for
points z on the cir- cumference of C.

The right-hand side of the last inequality tends to zero as ?? - > oo
. We have therefore

/(a + /0=/(a) + / /'(a)-h|,/"(a)+...-f |/-'(a) + ..., which we can
write

f(z)=f a) + (z - a)/' (a) + ~ ~ff" a) + ... + ri /< ) (a) + ....

This result is known as Taylors Theorem; and the proof given is due
to Cauchy. It follows that the radius of convergence of a poiuer
series is always

* The formal expansion was first published by Dr Brook Taylor (1715)
in his Methodus Incrementonim.

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at least so large as only just to exclude from the interior of the
circle of con- vergence the nearest singularity of the function
represented by the series. And by \hardsectionref{5}{3} corollary, it follows that the
radius of convergence is not larger than the number just specified.
Hence the radius of convergence is just such as to exclude from the
interior of the circle that singularity of the function which is
nearest to a.

At this stage we may introduce some terms which will be frequently
used.

If f(a) = 0, the function f(z) is said to have a zero at the point z =
a. If at such a point f (a) is different from zero, the zero of f(a)
is said to be simple; if, however,/' a),f" a), .../"*"'' (a) are all
zero, so that the Taylor's expansion of f(z) at z = a begins with a
term in (z - a)", then the function f(z) is said to have a zero of the
nth. order at the point z = a.

Example 1. Find the function / (s), which is analytic throughout the
circle C and its interior, whose centre is at the origin and whose
radius is unity, and has the value

a - cos 6 . sin 6

a -2acos6 + l a -2acos0 + l

(where a> 1 and 6 is the vectorial angle) at points on the
circumference of C.

[We have

f z) dz

/( )(0) = .f - - ' ' 2mjc z"

n ! /"St

27nJo

\ n\ n e~"i9d6 \ n \ f dz f d"" 1 "1

~2ffjo a-e<9 ~2iriJcz''(a-z)~~\ \ ds a-zJ

e- 'O.idd. -7- - ~;r ., (puttuig z = e*e) a- -2a cos + 1 ' ° '

Therefore by Maclaurin's Theorem*,

)l=0 "

or/(2) = (a-2)~' for all points within the circle.

This example raises the interesting question, Will it still be
convenient to define f(z) as (a-2)~ at points outside the circle ?
This will be discussed in \hardsubsectionref{5}{5}{1}.]

Example 2. Prove that the arithmetic mean of all values of 2"" 2 cv,
for points z on

the circumference of the circle \ z\ = l, is a, if Sc? " is analytic
throughout the circle and its interior.

/ (") (0) [Let 2 v2''=/(2), so that a, =;- . Then, writing z = e',
and calling C the circle

K=0 "

277 jo 2" ~ 2ni j c 2"* ~ n\ """ -'

z'i * The re8ult/ 2) =/(0) +2/' (0) + -/" (0) + ..., obtained By
putting a = in Taylor's Theorem,

is usually called Maclaurin's Theorem; it was discovered by Stirling
(1717) and published by Maclaurin (1742) in his Fluxions.

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Example 3. Let $f(z) = z^{r}$; then $f(z+h)$ is an analytic function
of $h$ when $\absval{h} < \absval{z}$ for all values of $r$; and so
$(z + h)^{r} = z^{r} + rz^{r-1} h + \frac{ r (r-1) }{2} z^{r-2} h^{2}
+ \cdots, $ this series converging when $\absval{h} < \absval{z}$.
This is the binomial theorem.\index{Binomial theorem}

Example 4. Prove that if h is a positive constant, and (1 - 2zh- h?) ~
in expanded in the form

\ + hP z) + h''P.2 z) + h P z) + (A),

(where P (2) is easily seen to be a polynomial of degree n in z), then
this series converges so long as z is in the interior of an ellipse
whose foci are the points z = \ and 2= -1, and whose semi-major axis
is (A + A"').

Let the series be first regarded as a function of A. It is a power
series in A, and therefore converges so long as the point A lies
within a circle in the /; -plane. The centre of this circle is the
point A = 0, and its circumference will be such as to pass through
that

singularity of (1 - 2zh- h' )~ which is nearest to A = 0.

But 1 - 22A -1- A2 = A - 2 + (22 - 1 )5>. /i \ 2 \ ( 2 \ 1 )ij,

so the singularities of (1 - 22A-|-A-)~2 are the points h=z - z' - ) '
and h=z + z - ) . [These singularities are branch points (see \hardsectionref{5}{7}).]

Thus the series (A) converges so long as | A | is less than both
|2-(22-l) | and |2-f(22\ l) |.

Draw an ellipse in the 2-plane passing through the point 2 and having
its foci at +L Let a be its semi-major axis, and 6 the eccentric angle
of 2 on it.

Then 2 = a cos -f i (a - 1 ) sin 9,

which gives 2 ± (22 - 1 )i = a + (a2 \ i) (cos + 1 sin 6),

so i2±(22-l)i | = a + (a2\ i)4.

Thus the series (A) converges so long as A is less than the smaller of
the numbers a-|-(a2- 1) and a- a - 1)2, i.e. so long as A is less than
a-(a2\ i)5. But A = a - (a2- 1) when a = |(A-f-A~i).

Therefore the series (A) converges so long as 2 is within an ellipse
whose foci are 1 and - 1, and whose semi-major axis is h h + h~ ).

\Subsection{5}{4}{1}{Forms of the remainder in Taylor's series.}

Let f(x) be a real function of a real variable; and let it have
continuous differential coefficients of the first n orders when a x a
+ h.

If O i l, we have

fl (71-1 hm ) /i n -f\ n-\

It li. 7! (1 - ') "/'"" ( + *> = ifr /'"' ( + "'> - ''/' ( + "')

Integrating this between the limits and 1, we have

n-l hm /! hn /I \ f\ n-i

f(a + h)=f a)+ -,f (a)+ ] ', f'Ha + th)dt.

ni = im: Jo n-L)l

Let Rn = J~iyi ] \ l - 0' - V'"* ( + ih) dt;

and let j-j be a positive integer such that p n.

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Then Rn = r - f ' ( " y~" ' ( " O"" /'"* ( + th) dt. Let U, L be the
upper and lower bounds of (1 - )' -p/("' (a + th). Then

f ' X (1 - 0 "' dt<\ \ t)P-' . (1 - O' -P/"'' (a + th) dt<[ U(l- t)P-'
dt. Jo Jo Jo

Since (1 - t)' ~P / '" (a + th) is a continuous function it passes
through all

values between U and L, and hence we can find 6 such that 1, and

[ I- ) -y <' ' (a + th) dt = -1(1- ey-Pf " (a + Oh). Jo

Therefore R = .J \ y;(1 - f- /*"' (a + 6h).

A" Writing p = n, we get Rn = - /"" (a + Oh), which is Lagrange s form
for

A" Me remainder; and writing j9 = 1, we get Rn = - \, y, (1 -
)''-'/''" (a + A),

which is Cauchys form for the remainder. Taking n = \ in this result,
we get

f a h)-f a) = hf a + 6h) \ i f(x) is continuous when a x a + h; this
result is usually known as the First Mean Value Theorem (see also §
4-14).

Darboux gave in 1876 Journal de Math. (3) ii. p. 291) a form for the
remainder in Taylor's Series, which is applicable to complex variables
and resembles the above form given by Lagrange for the case of real
variables.

\Section{5}{5}{The Process of Continuation.}

Near every point P, Zq, in the neighbourhood of which a function f z)
is analytic, we have seen that an expansion exists for the function as
a sei'ies of ascending positive integral powers of z - Zq), the
coefficients in which involve the successive derivates of the function
at z .

Now let A be the singularity of f(z) which is nearest to P. Then the
circle within which this expansion is valid has P for centre and PA
for radius.

Suppose that we are merely given the values of a function at all
points of the circumference of a circle slightly smaller than the
circle of convergence and concentric with it together with the
condition that the function is to be analytic throughout the interior
of the larger circle. Then the preceding theorems enable us to find
its value at all points within the smaller circle and to determine the
coefficients in the Taylor series proceeding in powers of z - Zq. The
question arises, Is it possible to define the function at points
outside the circle in such a way that the function is analytic
throughout a larger domain than the interior of the circle ?,

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In other words, given a potver series which converges and represents a
function only at poiiits within a circle, to define hy means of it the
values of the function at points outside the circle.

For this purpose choose any point TODO within the circle, not on the
line PA. We know the value of the function and all its derivates at
Pj, from the series, and so we can form the Taylor series (for the
same function) with Pi as origin, which will define a function
analytic throughout some circle of centre Pj. Now this circle will
extend as far as the singularity* which is nearest to Pi, which may or
may not be A; but in either case, this- new circle will iisuall '!
lie partly outside the old circle of convergence, and for jjoints in
the region which is included in the new circle but not in the old
circle, the new series may he used to define the values of the
function, although the old series failed to do so.

Similarly we can take any other point Po, in the region for which the
values of the function are now known, and form the Taylor series with
P as origin, which will in general enable us to define the function at
other points, at which its values were not previously known; and so
on.

This process is called continuation . By means of it, starting from a
representation of a function by any one power series we can find any
number of other power series, which between them define the value of
the function at all points of a domain, any point of which can be
reached from P without passing through a singularity of the function;
and the aggregate § of all the power series thus obtained constitutes
the analytical expression of the function.

It is important to know whether continuation by two different paths
fBQ, PB'Q will give the same final power series; it will be seen that
this is the case, if the function have no singularity inside the
closed curve PBQB'P, in the following way : Let P be any point on PBQ,
inside the circle C' with centre P; obtain the continuation of the
function with Pi as origin, and let it converge inside a circle Ci;
let P be any point inside both circles and also inside the curve
PBQB'P; let S, Si, Si be the power series with P, Pi, Pi as origins;
then|| *S'i = S'i' over a certain domain which will contain Pi, if Pi'
be taken sufficiently near Pi; and hence Si will be the continuation
of Si; for if Ti were the continuation of Si, we have Ti = Si over a
domain containing Pj, and so \hardsubsectionref{3}{7}{3}) corresponding coefficients in i
and Ti are the same. By carrying out such a process a sufficient
number of times, we deform the path PBQ into the path PB'Q if no
singular point is inside PBQB'P. The reader will convince himself by
drawing a figure that the process can be carried out in a finite
number of steps.

* Of the function defined by the new sei'ies.

+ The word ' usually ' must be taken as referring to the cases which
are likely to come under the reader's notice while studying the less
advanced parts of the subject.

X French, prolongement; German, Fortsetzung.

§ Such an aggregate of power series has been obtained for various
functions by M. J. M. Hill, by purely algebraical processes, Proc.
London Math. Soc. xxxv. (1903), pp. 388-416.

II Since each is equal to S.

W. M. A. 7

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Example. The series

1, Z 22 3

a a a" a* represents the function

/('-) = -- a - z

only for points z within the circle | I = | a | .

But any number of other power series exist, of the type

1 z-h z-hf z-hf

a-b' a-hf" a-bf' a-bf '-' '

if b/a is not real and positive these converge at points inside a
circle which is partly inside and partly outside | s | = | a j; these
series represent this same function at points outside this circle.

\Subsubsection{5}{5}{0}{1}{On functions to which the continuation-process cannot be applied.}

It is not always possible to carry out the process of continuation.
Take as an example the function /(2) defined by the power series

which clearly converges in the interior of a circle whose radius is
unity and w hose centre is at the origin.

Now it is obvious that, as -1-0, /(2) +qc; the point +1 is therefore
a singularity of/ (2).

But /(2)=22+/( 2)

and if z-- 0, f(z )- x and so /(s)- x, and hence the points for which
z- = l are singularities oi f z); the point 2= - 1 is therefore also
a singularity oif z). Similarly since

we see that if 2 is such that 2* = 1, then z is a singularity of/ (2)
; and, in general, any root of any of the equations

22=1, 2* = 1, 28 = 1, 2l =l, ...,

is a singularity of f z). But these points all lie on the circle | 2 |
= 1; and in any arc of this circle, however small, there are an
unlimited number of them. The attempt to carry out the process of
continuation will therefore be frustrated by the existence of this
imbroken front of singularities, beyond which it is impossible to
pass.

In such a case the function f(z) cannot be continued at all to points
2 situated outside the circle i 2 | = 1; such a function is called a
lacuvary f miction, and the circle is said to be a limiting circle for
the function.

\Subsection{5}{5}{1}{The identity of two functions. .,}

The two series

1 + 2 + ' + 2' + . . .

and - 1 + ( - 2) - ( - 2y- + ( - 2) - (2 - 2 + ...

do not both converge for any value, of z, and are distinct expansions.

Nevertheless, we generally say that they represent the same function,
on the

strength of the fact that they can both be represented by the same
rational

1 expression .

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This raises the question of the identity of two functions. When can
two different expansions be said to represent the same function ?

We might define a function (after Weierstrass), by means of the last
article, as consisting of one power series together with all the other
power series which can be derived from it by the process of
continuation. Two different analytical expressions will then define
the same function, if they represent power series derivable from each
other by continuation.

Since if a function is analytic (in the sense of Cauchy,\hardsubsectionref{5}{1}{2}) at
and near a point it can be expanded into a Taylor's series, and since
a convergent power series has a unique differential coefficient (§
5'3), it follows that the definition of Weierstrass is really
equivalent to that of Cauchy.

It is important to observe that the limit of a combination of analytic
functions can represent different analytic functions in different
parts of the plane. This can be seen by considering the series

5(' + j)\!,('"i)(TT7.-r;i )

The sum of the first n + 1 terms of this series is

1 / 1\ 1

z \ zJ' I + z' '

The series therefore converges for all values of z (zero excepted) not
on the circle j 2 | = 1. But, as w - > oo, | 2:'* | - > or j £" i -
oo according as | j is less or greater than unity; hence we see that
the sum to infinity of the series is

z when \ z\ < 1, and - when | j > 1. This series therefore represents
one

function at points in the interior of the circle | j = 1, and an
entirely different function at points outside the same circle. The
reader will see from \hardsectionref{5}{3} that this result is connected with the
non-uniformity of the convergence of the series near | j = 1.

It has been shewn by Borel* that if a region C is taken and a set of
points S such that points of the set S are arbitrarily near every
point of 6', it may be possible to define a function which has a
unique differential coefficient (i.e. is monogenic) at all points of C
which do not belong to *S'; but the function is not analytic in C in
the sense of Weierstrass.

Such a function is

.,, 00 w n exp(-expw*) f z)= 2 2 2; . . n=ip=oq=o z- p + qi)jn

* Proc. Math. Congress, Cambridge (1912), i. pp. 137-liJ8. Leqons sur
les fonctions mono- genes (1917). The functions are not monogenic
strictly in the sense of \hardsectionref{5}{1} because, in the example quoted, in
working out f z + h) -f(z)]jli, it must be su Dposed that R z + h) and
I z + ]i) are not botli rational fractions.

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\Section{5}{6}{Laurent's Theorem.}

A very important theorem was published in 1843 by Laurent*; it
relates to expansions of functions to which Taylor's Theorem cannot be
applied.

Let G and C be two concentric circles of centre a, of which C is the
inner; and let f(z) be a function which is analytic i* at all points
on G and G' and throughout the annulus between G and C. Let a + A be
any point in this ring-shaped space. Then we have \hardsubsectionref{5}{2}{1} corollary)

 ZTTt j cz - a - h liTi j c z - a - h

where the integrals are supposed taken in the positive or
counter-clockwise direction round the circles.

This can be written

We find, as in the proof of Taylor's Theorem, that

f(z)dz.h- r f(z)dz(z-ar+

c(z-a)"+' z-a-h) Jc z-a-h)h +'

tend to zero as n- cc; and thus we have

/(a + h) = cio + a h + ajr + ... + -~ + j + ...,

where + a = - / K, and 6 = 5- . I z - ay'-'f(z) dz.

This result is Laurent's Theorem; changing the notation, it can be
expressed in the following form: If f(z) be analytic on the concentric
circles G and G' of centre a, and throughout the annulus between them,
then at any point z of the annidus f(z) can he expanded in the form
$$
TODO
$$
where a = . j a.nd b = . [ (t - aT' f(t) dt.

An important case of Laurent's Theorem arises when there is only one
singularity within the inner circle G', namely at the centre a. In
this case the circle G' can be taken as small as we please, and so
Laurent's expansion is valid for all points in the interior of the
circle C, except the centre a.

* Comptes Rendus, xvii. (1843), pp. 348-349. t See \hardsectionref{5}{2} corollary 2,
footnote.

X We cannot write a- = fW (a)ln ! as in Taylor's Theorem since /(2) is
not necessarily analytic inside C.

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Example 1. Prove that

e ' = J x)+zJ x) + z' J x) + ...-irz-J,, x)- ...

1 /"St

where Jn (*) = s~ I ' ~ " )

 TT J

[For the function of z under consideration is analytic in any domain
which does not include the point z=Q; and so by Laurent's Theorem,

g2V z) = a,, + axZ + aoz'-+... + - ->r + ...,

where n = s - I - n and G = i, - . I e z dz,

and where 6' and C" are any circles with the origin as centre. Taking
C to be the circle of

radius unity, and writing = e', we have

1 Z"' "" 1 /' -'f

 -=--, / e:' sin .e- ' irf = jr- / COS (n6 - X H\ n 0) dB, 27ri. Jo
27r y

I sin (/i - i'sin ) tZ vanishes, as may be seen by writing in-cf) for
9. ThiLS

suice

a = J (A'), and \& = ( -)", since the function expanded is unaltered
if -z  l)e written for z, so that 6 = ( - )" Ai(.:i ), and the proof
is complete.]

Example 2. Shew that, in the annulus defined by|a|<|3|<i6|, the
function

r bz i

I'

\ \ {z-a) b-z) can be expanded in the form

*"+.?,* '(? + 6-.) c. * 1.3. ..(2 -1). 1.3... (2i + 2>i-l) /a\' " ' =
,!o 2 .ll l + n)l [b)

The function is one- valued and analytic in the annulus (see \hardsectionref{5}{7}),
for the branch-points 0, a neutralise each other, and so, by Laurent's
Theorem, if C denote the circle \ z\=r, where | o ' < / < | 6 |, the
coefficient of 2" in the required expansion is

1 f dz ( bz 1

27riJ c '- \ \ {z-a) b-z) ' Putting z = re', this becomes

1 [-' .,, 1.3... (2i-l) ?*<*" 1.3... (2i-l)a' -''

the series being absolutely convergent and uniformly convergent with
regard to 6.

The only terms which give integrals different from zero are those for
which k = l + n. So the coefficient of z" is

(2 -1) 1 . 3 ...(2 + 2/i-l) a \ Sn

1 r 'T * 1

2ir J 1=0

2Kl\ 2* + ™.(/+m) ! ¥*" b" '

Similarly it can be shewn that the coefficient of - is S a'\

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Example 3. Shew that

Z Z''

1 rsT

where = '" '' "> '''" cos ( - v) sin (9 - ?i(9 o?,

ZTT / 1 /"-"

and i = 5- / e'" + ")'=°' cos (i'-?Osi" - '<9 < <9-

ZTry

\Subsection{5}{6}{1}{T ie nature of the singularities of one-valued functions.}

Consider first a function f(z) which is analytic throughout a closed
region 8, except at a single point a inside the region.

Let it be possible to define a function z) such that (i) z) is analj
tic throughout S,

(ii) when a, /( ) = < ( ) + - 4-A . + ...+ "

z - a (z - a)- z - ay-

Then f(z) is said to have a 'pole of order n at a'; and the terms

h, -T + ... + -. are called the principal part of f(z) near a.

z - a. (z - a) (z - aY r r j \ /

By the definition of a singularity \hardsubsectionref{5}{1}{2}) a pole is a singularity.
If n = 1,

the singularity is called a simple pole.

Any singularity of a one-valued function other than a pole is called
an essential singularity.

If the essential singularity, a, is isolated (i.e. if a region, of
which a is an interior point, can be found containing no singularities
other than a), then a Laurent expansion can be found, in ascending and
descending powers of a valid when dk>\ z - a h, where A depends on the
other singularities of the function, and 8 is arbitrarily small. Hence
the ' principal part ' of a function near an isolated essential
singularity consists of an infinite series.

It should be noted that a pole is, by definition, an isolated
singularity, so that all singularities which are not isolated (e.g.
the limiting point of a sequence of poles) are essential
singularities.

There does not exist, in general, an expansion of a function valid
near a non-isolated singularity in the way that Laurent's expansion is
valid near an isolated singularity.

Corollary. If f(z) has a pole of order n at a, and z) = z - aYf z) z
a), i\ r a)= lim z-a)' f z), then y\ r z) is analytic at a.

Example 1. A function is not bounded near an isolated essential
singularity.

[Prove that if the function were bounded near z=a, the coefficients of
negative powers of 2 - a would all vanish.]

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c z\

Example 2. Find the singularities of the function e ~ 7 e - 1 .

At 2 = 0, the numerator is analytic, and the denominator has a simple
zero. Hence the function has a simple pole at 2 = 0.

Similarly there is a simple pole at each of the points mria ( = + 1,
+2, +3, ...); the denominator is analytic and does not vanish for
other values of z.

A.t z = a, the numerator has an isolated singularity, so Laurent's
Theorem is applicable, and the coefficients in the Laurent expansion
may be obtained from the quotient

c c

z- a 2 I (z - aV

Ml+'--" + ...)-l

which gives an expansion involving all positive and negative powers of
z - a). So there is an essential singularity at 2 = a.

Example 3. Shew that the function defined by the series

I wg"- (l +)t- )"-l

 =1 (2 -l) 2 -(l+/i-l)

has simple poles at the points 2 = (1 + ~i)e- ''' / ( =0, 1, 2, ... n
- \; ?i = l, 2, 3, ...). y \addexamplecitation{Math. Trip. 1899.}

\Subsection{5}{6}{2}{The 'point at infinity.'}

The behaviour of a function /C ') as | ' - oo can be treated in a
similar way to its behaviour as z tends to a finite limit.

If we write z = -,, so that large values of z are represented by small

values of z' in the '-plane, there is a one-one correspondence between
z and z, provided that neither is zero; and to make the
correspondence complete it is sometimes convenient to say that when z
is the origin, z is the * point at infinity.' But the reader must be
careful to observe that this is not a definite point, and any
proposition about it is really a proposition concerning the point / =
0.

Let/(2 ) = 4> z'). Then < z') is not defined at z = 0, but its
behaviour near z = is determined by its Taylor (or Laurent) expansion
in powers of z \ and we define < (0) as lim (/) if that limit exists.
For instance

the function (/) may have a zero of order m at the point 2' =; in
this case the Taylor expansion of ( ') will be of the form

and so the expansion of f(z) valid for sufficiently large values of |
.0 j will be of the form

In this case,/(0) is said to have a zero of order m at ' infinity.'

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Again, the function (f)(2') may have a pole of order m at the point z'
= 0; in this case

and so, for sufficiently large values of \ z\, f(z) can be expanded in
the form

N P

f z) = Az"' + Bz'''-' + Cz'"-- +...+Lz + M+- + - + ....

In this case,/(2 ) is said to have a pole of order m at ' infinity. '
Similarly f(z) is said to have an essential singularity at infinity,
if z) has an essential singularity at the point / = 0. Thus the
function e' has an

essential singularity at infinity, since the function e~' or

1 \ 1\ 1

has an essential singularity at z = 0.

Example. Discuss the function represented by the series

2 -, :; 5-5, ( >1).

Z 1

The function represented by this series has singularities at 2=- and
2= - i

 n=\, 2, 3, ...), since at each of these points the denominator of one
of the terms in the series is zero. These singularities are on the
imaginary axis, and have 3 = as a limiting point; so no Taylor or
Laurent expansion can be foi med for the function valid throughout any
region of which the origin is an interior point.

For values of z, other than these singularities, the series converges
absolutely, since the limit of the ratio of the (?i + l)th term to the
?ith is lim (?i+ l)~i a~ = 0. The function is

an even function of z (i.e. is unchanged if the sign of z be changed),
tends to zero as I 2 I - Qc, and is analytic on and outside a circle
C of radius greater than unity and centre at the origin. So, for
points outside this circle, it can be expanded in the form

h + j + b+

where, by Laurent's Theorem,

TODO

This double .series converges absolutely when | 2 | > 1, and if it be
rearranged in powers of 2 it converges uniformly.

Since the coefficient of 2 ~ Ms 2 -; and the only term which
furnishes a non-

H=o n !

zero integral is the term in z~, we have

(\ )fc-ia-2A- dz

b'k  1 - .

Itti J c =o, fo n\ a2*

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Therefore, when | 2 | > 1, the function can be expanded in the form

ill

e"' e * e

The function has a zero of the second order at infinity, since the
expansion begins with a term in z~' .

\Subsection{5}{6}{3}{Liouville's Theorem*.}

Let f(z) he analytic for all values of z and let \ f z)\ < K for all
values of z, where K is a constant (so that \ f(z) is bounded as | r |
- > x ). Then f(z) is a constant.

Let z, z' be any two points and let C be a contour such that z, z are
inside it. Then, by \hardsubsectionref{5}{2}{1},

take C to be a circle whose centre is z and whose radius is p 2 | / -
s |; on

C write \ \ = z pe' \ since jf - /| 2P when is on C it follows from §
4-62 that

= 2\ z -z\ Kp-K Make p- cc keeping z and z' fixed; then it is obvious
that/(/) -f(z) = 0; that is to say, f(z) is constant.

As will 1)0 seen in the next article, and again frequently in the
latter half of this volume (Chapters xx, xxi and xxii), Liouville's
theorem furnishes short and convenient proofs of some of the most
important results in Analysis.

\Subsection{5}{6}{4}{Functions with no essential singularities.}

We shall now shew that the only one-valued functions luhich have no
singidarities, except poles, at any jjoint (including oo ) are
rational functions.

For let f(z) be such a function; let its singularities in the finite
part of the plane be at the points Cj, c-j, ... Ck'. and let the
principal part \hardsubsectionref{5}{6}{1}) of its expansion at the pole Cr be

+ 1 ::; + . . . + - 7 -

Z - Cr (Z - Crf '" (Z - CyJ

Let the principal part of its expansion at the pole at infinity be

a- z + a-iZ + ... + anZ \

if there is not a pole at infinity, then all the coefficients in this
expansion

will be zero.

* This theorem, which is really due to Cauchy, Comptes Rendus, xix.
(1844), pp. 1377, 1378, was given this name by Borchardt, Journal fvr
Math, lxxxviii. (1880), pp. 277-310, who heard it in Liouville's
lectures in 1847.

%
% 106
%

Now the function

has clearly no singularities at the points c, c.j, ... Ck, or at
infinity; it is therefore analytic everywhere and is bounded as \ z -
>cc, and so, by Liou\ dlle's Theorem, is a constant; that is,

/(.) = + .. + a..= + . . . +,,.. + I l i + ( . +   + (i J

where C is constant; f(z) is therefore a rational function, and the
theorem is established.

It is evident from Liouville's theorem (combined with
\hardsubsectionref{3}{6}{1} corollary (ii)) that a function which is
analytic everywhere (including oc ) is merely a constant. Functions
which are analytic everywhere except at oc are of considerable
importance; they are known as integral functions*. Examples of such
functions are e, sin z, e . From \hardsectionref{5}{4} it is apparent
that there is no finite radius of convergence of a Taylor's series
which represents an integral function; and from the result of this
section it is evident that all integral functions (except mere
polynomials) have essential singularities at oo .

\Section{5}{7}{Many-valued functions.}

In all the previous work, the functions under consideration have had a
unique value (or limit) corresponding to each value (other than
singularities) of .

But functions may be defined which have more than one value for each
value of z; thus if 2- = r (cos 6 + i sin 6), the function 2- has the
two values

r* (cos 1(9 + t sin (9), r jcos \ 0 + lir) + i sin \ (6 + 27r)|;

and the function arc tan x (x real) has an unlimited number of values,
viz. Arc tan x + mr, where - tt < Arc tan x <- 'tt and n is any
integer; further

examples of many- valued functions are log z, z, sin z~).

Either of the two functions which z represents is. however, analytic
except at = 0, and we can apply to them the theorems of this chapter;
and the two functions are called ' branches of the many-valued
function z'-.' There will be certain points in general at which two or
more branches coincide or at which one branch has an infinite limit;
these points are called ' branch-points.' Thus z- has a branch-point
at; and, if we consider the change in z as z describes a circle
counter-clockwise round 0, we see that 6

* Vxench, fonction entilre; Germa,n, game Funktion.

%
% 107
%

increases by 27r, r remains unchanged, and either branch of the
function passes over into the other branch. This will be found to be a
general characteristic of branch-points. It is not the purpose of this
book to give a full discussion of the properties of many-valued
functions, as we shall always have to consider particular branches of
functions in regions not containing branch- points, so that there will
be comparatively little difficulty in seeing whether or not Cauchy's
Theorem may be applied.

Thus we cannot apply Cauchy's Theorem to such a function as z when the
path of iutegi-ation is a circle surrounding tlie origin; but it is
permissible to apply it to one of

the branches of z when the path of integration is like that shewn in §
6 '24, for through- out the contour and its interior the function has
a single definite value.

Example. Prove that if the different values of a, corresponding to a
given value of z, are represented on an Argand diagram, the
representative points will be the vertices of an equiangular polygon
inscribed in an equiangular spiral, the angle of the spiral being
independent of a.
\addexamplecitation{Math. Trip. 1899.}

The idea of the different braiickes of a function helps us to
understand such a paradox as the following.

Consider the function y = ' i

for which ~=x log x).

AVhen x is negative and real, is not real. But if x is negative and of
the form

P (where p and q are positive or negative integers), y is real.

Z'j -f- 1

If therefore we draw the real c\ irve

we have for negative values of ./ a set of conjugate points, one point
corresponding to each rational value of x with an odd denominator;
and then we might think of proceeding to form the tangent as the limit
of the chord, just as if the curve were continuous; and

thus --, when derived from the inclination of the tangent to the axis
of x, would appear

to be real. The question thus arises, Why does the ordinary process of
differentiation

give a non-real value for - ? The explanation is, that these conjugate
points do not all

arise from the same branch of the function y = x' . We have in fact

y - 5

where k is any integer. To each value of k corresponds one branch of
the function y. Now in order to get a real value of y when x is
negative, we have to choose a suitable value for k : and this value of
k varies as we go from one conjugate point to an adjacent one. So the
conjugate points do not represent values of y arising from the same
branch of the

function y=x, and consequently we cannot expect the value of - when
evaluated

for a definite branch to be given by the tangent of the inclination to
the axis of x of the line joining two arbitrarily close members of the
series of conjugate points.

%
% 108
%

REFERENCES.

E. Goursat, Cours dJ Analyse, ii. (Paris, 1911), Chs. xiv and xvi.

J. Hadamard, La Serie de Taylor et son prolongement analytique
(Scientia, 1901).

E. LiNDELOF, Le Calmd des Residus (Paris, 1905).

C. J. DE LA Vallee Poussin, Covrs d' Analyse Infinitesimale, i. (Paris
and Louvain, 1914), Ch. X.

E. BoREL, Lecons stir les Fonctions Entieres (Paris, 1900).

G. N. Watson, Complex Integration and Cauchy's Theorem (Camb. Math.
Tracts, no. 15, 1914).

Miscellaneous Examples.

1. Obtain the expansion

/(.)=/( ) + 2 1 -/ (- + -3-3-r/ ( -2-j + -25751- ' \ \ r-] ' and
determine the circumstances and range of its vaUdity.

2. Obtain, under suitable circumstances, the expansion

+ .... (Corey, Ann. of Math. (2), i. (1900), p. 77.)

3. Shew that for the series

  1

)i=o -r~

the region of convergence consists of two distinct areas, namely
outside and inside a circle of radius unity, and that in each of these
the series represents one function and represents it completely.

(Weierstrass, Berliner Monatsherichte., 1880, p. 731; Ges. Werke, 11.
(1895), p. 227.)

4. Shew that the function

2 s"'

tends to infinity as 2- -exp i-niplm !) along the radius through the
point; where m is any integer and p takes the values 0, 1, 2, ..'.
(ni I - 1).

Deduce that the function cannot be continued beyond the unit circle.

(Lerch, Sitz. BOhm. Acad., 1885-6, pp. 571-582.)

5. Shew that, if z-- 1 is not a positive real number, then

-". .s:)"<'-")- /:'=-('-'- '-**-

\addexamplecitation{Jacobi and Scheibner.}

%
% 109
%

6. Shew that, if s - 1 is not a positive real number, then

,,,, m, m(m+ ) to (m + 1) ... (to + ?i - 1)

(1-2) - =l+-2 + - 2 "" +-+ -1 .

n : J

\addexamplecitation{Jacobi and Scheibner.}

7. Shew that, if z and 1-2 are not negative real numbers, then

  Jo m+l [ m + 3 (m+3) ... (m + 27i- 1) J

+ "') (m+l)(w + 3)...(m + 27i-l)Jo ' -

\addexamplecitation{Jacobi and Scheibner.}

8. If, in the expansion of (a + i2 + a22 )'" by the multinomial
theorem, the remainder after n terms be denoted by R (2), so that

(a4- i2 + a2S-)'" = o + - i2 + -'-l2-"" + --- + n-iS"'"' + (s), shew
that

/4(.) = ( + a.. + ..r j, - (a + .M+a, r " '

9. If (ao + i2 + a22 )""'"' /'(ao + i< + 2 -)'"o?<

y

\addexamplecitation{Scheibner.}

be expanded in ascending powers of 2 in the form

Jl2 + .4222+...,

shew that the remainder after n-\ terms is

(ao + ai2 + a22 )~'"" I (ao + a, + a2 )'" o .i-(2?'i + ? + 1) 2 -i
""'c C

\addexamplecitation{Scheibner*.}

10. Shew that the series

where X (2)= - 1 +2- f, + I",- - + (- )", '

and where (2) is analytic near 2 = 0, is convergent near the point 2 =
; and shew that if the sum of the series be denoted by/(2), then/(2)
satisfies the differential equation

/'(--)=/( )-0(4

(Pincherle, Rend, dei Lincei (5), v. (1896), p. 27.)

11. Shew that the arithmetic mean of the squares of the moduli of all
the values of the series "2 0, on a circle |2| = r, situated within
its circle of convergence, is equal

 i=0

to the sum of the squares of the moduli of the separate terms.

(Gutzmer, Math. Ann. xxxii. (1888), pp. .596-600.)

* The results of examples 5, 6 and 7 are special cases of formulae
contained in Jacobi's dissertation (Berlin, 182.5) published in bis
Ges. Werke, in. (1884), pp. 1-44. Jacobi's formulae were generalised
by Scheibner, Lelpziger Berichte, xlv. (1893), pp. 432-443,

%
% 110
%

12. Shew that the series

2 e-2(am) -m-l m = l

converges when | 2 | < 1; and that, when a > 0, the function which it
represents can also be represented when | s ! < 1 by the integral

/ay f" e-"" dx n-J Jo ex\ 2 x '

and that it has no singularities except at the point z=l.

(Lerch, Monatshefte fiir Math, und Phys. viii.) 13. Shew that the
series

2

\ \,, 2 r z\ zj \

 z + z )-V- 2 ( i\ 2 \ 2y'zi)(2v + iv'zif' iv-2v'z- i) 2v + 2v'z-'
xf]'

in which the summation extends over all integral values of v, v',
except the combination (i' = 0, v' = 0), converges absolutely for all
values of z except purely imaginary values; and that its sum is + 1
or - 1, according as the real part of z is positive or negative.

(Weierstrass, Berliner Monatsherichte, 1880, p. 735.)

14. Shew that sin \ u ( + -)[ can be expanded in a series of the type

a, + aiZ + aoy-+...+- + j, + ...,

in which the coefficients, both of s" and of z~'\ are

1 / 2t

-- I sm (2m cos ) cos Ji o? . 2ir J )

n=i n-z' + a-

shew that/ (2) is finite and continuous for all real values of z, but
cannot be expanded as a Maclaurin's series in ascending powers of z;
and explain this apparent anomaly.

[For other cases of failure of Maclaurin's theorem, see a posthumous
memoir by Cellerier, Bidl. des Set. Math. (2), xiv. (1890), pp.
145-599; Lerch, Journal fiir Math. cm. (1888), pp. 126-138;
Pringsheim, Math. Ann. XLii. (1893), pp. 153-184; and Du Bois
Reymond, Miinehener Sitzungsberichte, vi. (1876), p. 235.]

16. If f(z) be a continuous one- valued function of z throughout a
two-dimensional region, and if

   f z)dz =

h

for all closed contours C lying inside the region, then f(z) is an
analytic function of z throughout the interior of the region.

[Let a be any point of the region and let

F z)=\'f z)dz

J a

It follows from the data that F(z) has the unique derivate f z). Hence
F z) is analytic \hardsectionref{5}{1}) and so \hardsubsectionref{5}{2}{2}) its derivate /(z) is also
analytic. This important converse of Cauchy's theorem is due to
Morera, Rendiconti del R. 1st. Lomhardo Milano), xxil. (1889), p.
191.]


\documentclass[12pt]{article}
\usepackage{pylatex}
\usepackage{mpllatex}
\usepackage{geometry}
\usepackage{amsmath}
\usepackage{pgf}
\usepackage{caption}
\usepackage{hyperref}
\usepackage{examples}

% portrait
\geometry{papersize={210mm,297mm},hmargin=2cm,tmargin=1.0cm,bmargin=1.5cm}
\parskip=8pt plus 4pt minus 2pt

% landscape
% \geometry{papersize={297mm,210mm},hmargin=2cm,tmargin=1.0cm,bmargin=1.5cm}
% \parskip=6pt plus 3pt minus 2pt

\begin{document}

\section*{A mixed Maple-Python example}

This example demonstrates a cooperative effort where Maple is used to do the analytic computations while Python is used to plot the data.

The example chosen here is to find and plot the solution to the boundary value problem defined by
\begin{align*}
   \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} +10 y = 0\quad\quad\text{with }y(0)=3,\> y'(0)=0
\end{align*}

This example requires two passes, once for Maple and once for Python (and in that order). This example can be run using

\vspace{5pt}

\begin{lstlisting}
   mpllatex.sh -x -i mixed
   pylatex.sh  -x -i mixed
   pdflatex          mixed
\end{lstlisting}

\vspace{5pt}

Note that the last pair of commands could also be combined as {\small\tt pylatex.sh -i mixed}.

\subsection*{The Maple code}

Here Maple is used to first find the general solution of th differential equation. The boundary conitions are then imposed and finally a uniform sampling of the solution is written to a file for later use by Python and Matplotlib.

\begin{maple}
   # a second order ode
   ode := diff(y(x), x, x) + 2*diff(y(x),x) + 10*y(x) = 0:  # mpl (ans.101,ode)

   # find the general solution
   ans := dsolve(ode):                     # mpl (ans.102,ans)

   # set initial conditions
   ics := y(0) = 3, (D(y))(0) = 0:
   tmp := {ics}:                           # mpl (ans.103,tmp)

   # find the particular solution
    f := rhs(dsolve([ics, ode])):          # mpl (ans.104,f)
   df := diff(f,x):                        # mpl (ans.105,df)

    y := x -> f:
   dy := x -> df:

   # now sample y and dy at selected points
   a,b,n := 0.0,2.0*Pi,300:        # domain and number of samples
   dx := (b-a)/n:                  # uniform step

   fd := fopen ("mixed.txt", WRITE):
   for i from 0 to n by 1 do
      x := a + dx*i:
      fprintf(fd,"% .10e % .10e % .10e\n",x,evalf(y(x)),evalf(dy(x))):
   end do:
   fclose(fd):
\end{maple}

The general solution of the differential equation is
\begin{equation*}
  \mpl{ans.102}
\end{equation*}
while the particular solution satifying the boundary conditions is given by
\vspace{5pt}
\begin{align*}
    y(x) &= \mpl{ans.104}
\end{align*}

\subsection*{The Python code}

This is a straighforward use of Matplotlib to plot two functions. The code reads the datafile created previously by Maple and then calls Matplotlib to plot that data.

\begin{python}
   import numpy as np
   import matplotlib.pyplot as plt

   plt.matplotlib.rc('text', usetex = True)
   plt.matplotlib.rc('grid', linestyle = 'dotted')
   plt.matplotlib.rc('figure', figsize = (5.5,4.1)) # (width,height) inches

   x, y, dy = np.loadtxt ('mixed.txt', unpack=True)

   plt.plot (x,y)
   plt.plot (x,dy)

   plt.xlim (0.0,4.0)

   plt.legend(('$y(x)$', '$dy(x)/dx$'), loc = 0)
   plt.xlabel('$x$')
   plt.ylabel('$y(x),\> dy/dx$')
   plt.grid(True)
   plt.tight_layout(0.5)

   plt.savefig('mixed-fig.pdf')
\end{python}

\vspace{10pt}

\begin{minipage}{\textwidth}
   \centering
   \IfFileExists{mixed-fig.pdf}%
   {\includegraphics[width=0.75\textwidth]{mixed-fig.pdf}}{Failed to create pdf plot.}
   \captionof{figure}{The function and its derivative.}
\end{minipage}

\end{document}


\documentclass{article}
\begin{document}
\section*{Quadratic equations}
The quadratic equation
\begin{equation}
  \label{quad}
  ax^2 + bx + c = 0,
\end{equation}
where \( a, b \) and \( c \) are constants and \( a \neq 0 \),
has two solutions for the variable \( x \):
\begin{equation}
  \label{root}
  x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}.
\end{equation}
If the \emph{discriminant} \( \Delta \) with
\[
  \Delta = b^2 - 4ac
\]
is zero, then the equation (\ref{quad}) has a double solution:
(\ref{root}) becomes
\[
  x = - \frac{b}{2a}.
\]
\end{document}


%!TEX root = index.tex
\section{Solving the Quartic}
\epigraph{The purpose of computation is insight, not numbers.}{Richard Hamming}

Moving on to the fourth degree, the idea is the same. The method gets much more complicated, and isn't quite useful in practice. You should aim at understanding the ideas and move on.

Consider a \emph{depressed quartic} with roots $ \gamma_1, \gamma_2, \gamma_3, \gamma_4$:
\begin{align*}
  P(x) = x^4 + a_2 x^2 + a_1 x + a_0
\end{align*}
As before, to simplify the calculations we're forcing $$ 0 =\gamma_1+ \gamma_2+ \gamma_3+ \gamma_4$$
If $ a_0 = 0$ we can factor out an $ x$ and reduce the quartic to a cubic. So we'll assume this is not the case.
\begin{questions}
  \item Express $ a_2, a_1, a_0$ in terms of $ \gamma_1, \gamma_2, \gamma_3, \gamma_4$.
\end{questions}
We need to find an intermediate third degree polynomial with roots $ \lambda_1, \lambda_2, \lambda_3$ which have \emph{fewer} symmetries than $ a_2, a_1, a_0$.

... and here they are:
  \begin{align*}
    \lambda_1 &= \gamma_1 \gamma_2 + \gamma_3 \gamma_4 \\
    \lambda_2 &= \gamma_1 \gamma_3 + \gamma_2 \gamma_4 \\
    \lambda_3 &= \gamma_1 \gamma_4 + \gamma_2 \gamma_3
  \end{align*}
\begin{questions}[resume]
  \item \begin{enumerate}
    \item How many ways are there to permute the 4 roots $ \gamma_1, \gamma_2, \gamma_3, \gamma_4$?
    \item Of these, which permutations leave \emph{all} the $\lambda_i$'s unchanged?
    \item What do the rest of the permutations do to the $\lambda_i$'s?
    \item Conclude that $\lambda_1 + \lambda_2 + \lambda_3$, $\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_3\lambda_1$ and $\lambda_1 \lambda_2 \lambda_3$ are symmetric in $ \gamma_1, \gamma_2, \gamma_3, \gamma_4$.
  \end{enumerate}
\end{questions}

\newpage
By the previous exercise the coefficients of the cubic polynomial whose roots are $ \lambda_1, \lambda_2, \lambda_3$ are polynomials in  $ a_2, a_1, a_0$.
\begin{questions}[resume]
  \item Verify the following identities (only for the intrepid)
    \begin{align*}
      a_2 &= \lambda_1 +\lambda_2 +\lambda_3\\
      -4 a_0 &= \lambda_1\lambda_2 +\lambda_1 \lambda_3 + \lambda_2 \lambda_3\\
      a_1^2 - 4a_0 a_2&=\lambda_1 \lambda_2 \lambda_3
    \end{align*}
    and hence $ \lambda_1, \lambda_2, \lambda_3$ are the roots of
    \begin{align*}
      R(x) = x^3 - a_2 x^2 - 4a_0x + (4a_0 a_2 - a_1^2)
    \end{align*}
\end{questions}

Thus we've reduced the problem from a quartic to a cubic. The last step is recovering $ \gamma_1, \gamma_2, \gamma_3, \gamma_4$ from $ \lambda_1, \lambda_2, \lambda_3$. For this notice that
  \begin{align*}
    \lambda_1 &= \gamma_1 \gamma_2 + \gamma_3 \gamma_4 \\
    a_0 &= \gamma_1 \gamma_2 \gamma_3 \gamma_4
  \end{align*}
and hence $ \gamma_1 \gamma_2$ and $ \gamma_3 \gamma_4$ are the roots of the quadratic $x^2 - \lambda_1 x + a_0$. Similarly for the others.  We have the identities
  \begin{align*}
    \lambda_1^2 = (\lambda_1\lambda_2).(\lambda_1\lambda_3).(\lambda_1\lambda_4)/a_0
  \end{align*}
which we can use to find $ \lambda_1^2$ and then test the two possible square roots to see which one works. The other $ \lambda_i$'s can be found by inspection.

\begin{questions}[resume]
  \item Find the roots of the following polynomials:
    \begin{enumerate}
      \item $x^4 - 1$
      \item $x^4 + 1$
    \end{enumerate}
\end{questions}


\documentclass[a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath, amssymb, amsthm}
\usepackage{gensymb}
\usepackage[nodayofweek]{datetime}

\usepackage{tikz}

% Set size of text area with total parameter
\usepackage[a4paper, total={135mm, 255mm}]{geometry}

\title{Complex Binomial Sequences}
\author{Dyson}
\date{\today}

\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\opi}{(1 + i)}
\newcommand{\mopi}{(-1 + i)}

\begin{document}

\maketitle

% Set paragraph spacing here to avoid messing with title
\setlength{\parindent}{0em}
\setlength{\parskip}{1em}

Throughout this paper, all $n \in \mathbb{N}$ and $0 \in \mathbb{N}$.

\section{Looking at $\opi^n$}

The sequence $\opi^n$ creates an interesting graph when plotted sequentially on an Argand diagram.

\begin{center}
\resizebox{135mm}{!}{
\begin{tikzpicture}
    \begin{scope}[thick,font=\scriptsize]
        % Axes
        \draw (-18,0) -- (18,0);
        \draw (0,-18) -- (0,18);

        % Axes labels
        \foreach \n in {-17,...,-1,1,2,...,17}{
            \draw (\n,-3pt) -- (\n,3pt);
            \draw (-3pt,\n) -- (3pt,\n);
        }

        % Draw the actual plot of the sequence
        \draw (1,0) -- (1,1) -- (0,2) -- (-2,2) -- (-4,0) -- (-4,-4) -- (0,-8) -- (8,-8) -- (16,0) -- (16,16);
    \end{scope}
\end{tikzpicture}
}
\end{center}

Initially, I confused this shape for a coarse approximation of a Golden Spiral. If this were the case, then the ratio between the moduli of successive elements of the sequence would tend to $\phi$ as $n$ grows to $\infty$. Here's an example:

\begin{align*}
\dfrac{\abs{\opi^3}}{\abs{\opi^2}} = \dfrac{\abs{-2 + 2i}}{\abs{2i}} = \dfrac{\sqrt{8}}{2} = \dfrac{2\sqrt{2}}{2} = \sqrt{2}
\end{align*}

The ratio is $\sqrt{2}$, which is not $\phi$. This ratio is true for all $\dfrac{\abs{\opi^{n + 1}}}{\abs{\opi^n}}$, meaning this shape is not a Golden Spiral.

\begin{proof}
Let $z_1$ and $z_2$ be two consecutive elements of the sequence $\opi^n$.

We want the ratio between the moduli of these elements, $\dfrac{\abs{z_2}}{\abs{z_1}}$.

Let $f(n) = \dfrac{\abs{\opi^{n + 1}}}{\abs{\opi^n}}$.

We can easily show that $f(0) = \dfrac{\abs{\opi^1}}{\abs{\opi^0}} = \dfrac{\abs{1 + i}}{\abs{1}} = \sqrt{2}$.

We can then show:
\begin{gather*}
f(n) = \dfrac{\abs{\opi^{n + 1}}}{\abs{\opi^n}}\\[0.5em]
= \dfrac{\abs{\opi\opi^n}}{\abs{\opi\opi^{n - 1}}} = \dfrac{\abs{\opi}}{\abs{\opi}} \times \dfrac{\abs{\opi^n}}{\abs{\opi^{n - 1}}}\\[0.5em]
= 1 \times \dfrac{\abs{\opi^n}}{\abs{\opi^{n - 1}}} = f(n - 1)
\end{gather*}

We can do this because for $z, w \in \mathbb{C}$, $\abs{zw} = \abs{z} \times \abs{w}$.

We've now shown that for $n > 0$, $f(n) = f(n - 1)$. This always recurs to the base case of $f(0)$ and means that all $f(n) = \sqrt{2}$.

This means that the ratio of the moduli between any two consecutive members of the sequence $\opi^n$ is always $\sqrt{2}$.
\end{proof}

\newpage

\section{Looking at $\mopi^n$}

If we look at the series $(1 - i)^n$ and plot it sequentially on an Argand diagram, then we get the same spiral as when plotting $\opi$ but reflected in the real axis.

However, if we look at $\mopi^n$, then we get something completely different.

\begin{center}
    \resizebox{135mm}{!}{
        \begin{tikzpicture}
        \begin{scope}[thick,font=\scriptsize]
        % Axes
        \draw (-18,0) -- (18,0);
        \draw (0,-18) -- (0,18);

        % Axes labels
        \foreach \n in {-17,...,-1,1,2,...,17}{
            \draw (\n,-3pt) -- (\n,3pt);
            \draw (-3pt,\n) -- (3pt,\n);
        }

        % Draw the actual plot of the sequence
        \draw (1,0) -- (-1,1) -- (0,-2) -- (2,2) -- (-4,0) -- (4,-4) -- (0,8) -- (-8,-8) -- (16,0) -- (-16,16);
        \end{scope}
        \end{tikzpicture}
    }
\end{center}

This is a very interesting plot. We can use a very similar proof to the one at the end of \S1 to prove that the ratio between the moduli of any two elements of the sequence is $\sqrt{2}$. However, the more interesting thing with this sequence is not the moduli of the elements, but the arguments. How does the angle of each point relative to the origin change?

Looking at the sequence $\arg(n)$ (in degrees) for $n \in \{0,1,...,10\}$, we get
% This text environment is to add spaces between each element. Yes, a text env in a math env is a bodge. Sue me.
$$\text{0, 135, 270, 405, 540, 675, 810, 945, 1080, 1215, 1350}$$

If we take each of these elements mod 360, we get
$$\text{0, 135, 270, 45, 180, 315, 90, 225, 0, 135, 270}$$

The arguments of the members of $\mopi^n$ rotate around the unit circle with a period of 8. This is the same period as $\opi^n$.

But, despite the two sequences having the same ratio of moduli, and the same rotation period, this second sequence creates a completely different graph. This is because the rotation of each term of $\opi^n$ relative to the last is 45. This means that it takes a whole rotation of 8 terms to get back to the initial angle.

However, the relative rotation of $\mopi^n$ is 135, meaning that we after 3 terms, we've rotated $45\degree$ from the initial value. This creates a very interesting pattern of triangles on the plot.

\end{document}


\chapter{Computer Engineering Theory}
This chapter should give you some idea about basic concepts of computer engineering, and the typical concepts it makes use of.

\section{Bits and Numbers}
There are different kinds of numbers - the most obvious example of different kinds of numbers are the \emph{natural} numbers (we use these numbers for counting objects - one apple, ten trees, etc), the \emph{real} numbers (these include fractions and all other non-integer numbers).

The natural numbers have limited use - it is hard to express fractional quantities by using natural numbers, and it becomes even worse if you attempt to perform arithmetics on that. This might not seem like a life problem - but some people have spent more than 10 years calculating just a few digits of $\pi$.

\emph{Computers} are the devices created by people which make arithmetic computations faster and easier to perform. But the initial problem that the computer engineers had to face was the complexity of implementing both the natural and the real numbers.

At first it seems that real and natural numbers could be well represented by gears and rotating mechanics - but this approach did not give mathematicans and physicists the precision they required in their calculations. Besides, the calculations could be much faster if they were performed electronically.

There is a way to represent real numbers in electronics - using so called analog computers. In these computers the voltage level would be somehow proportional to the numbers used in the calculation. But there was the precision problem again: it is not possible to reliably measure the voltage levels.

The next invention was the introduction of the non-decimal numeric systems - the kind of systems where number is represented by more or less than 10 symbols. For example the ternary system, which only uses symbols '0', '1', '2' to represent numbers, or the binary system that uses two symbols (typically '0' and '1', or 'true' and 'false'). This means that numbers could be represented with fewer mechanics and electronics.

A binary mechanical calculator would only require a single two-state switch. A binary electronic computer could only require two voltages: high voltage and low voltage, corresponding to '0' and '1'. The reduction in complexity gave major benefits, and also gave birth to the digital computers.

(missing text: binary system)

\section{Boolean Logic}
As part of its evolution mathematics has attempted to describe rules of basic logic, giving a formal description to all the common approaches people use for solving logic tasks.

This gave birth to boolean logic - a formal description that introduces a new kind of number, and new rules to work with it. The new kind of number is the boolean value - unlike the common numbers it only has two values. These values are commonly referred to as 'true' and 'false', or alternatively '1' and '0'.

The boolean logic is inseparatable from the computer engineering - all modern computers make use of the binary system, and it's vital to understand what kind of useful things can be done with the binary numbers.

There are few basic operations that can be performed on the boolean values:
\begin{enumerate}
	\item 'Or', which is somehow similar to '+' operation used on the usual numbers.
	\item 'And', which is similar to the multiplication ('*') operation.
	\item 'Not', the kind of a logic '-' (negation, not subtraction).
	\item 'Exclusive or', which is also similar to '+' operation, but with a quirk that will be explained below.
\end{enumerate}

It's very easy to illustrate properties of these operations by an example. The first line is the boolean logic example, and the lower line is the corresponding example from normal arithmetics:

$A or B$ is true if either A or B is true.
$A + B$ is non-zero if either A or B is not zero.

$A and B$ is true if both A and B are true.
$A * B$ is non-zero if both A and B are non-zero.

$not A$ is true when A is false, and vice-versa.
$-A$ is negative when A is positive, and vice-versa.

$A xor B$ is true if \emph{either} A \emph{or} B is true. It's false if they are both false, or both true.
$(A + B) mod 2$ (addition with modulus by 2).

\section{Electric Circuits}
All the modern computers make use of the electric circuits to perform the actual calculations. To gain a complete understanding of the computer the basic electric circuits theory must be known.

It's a common knowledge that a certain voltage causes a current to flow - this is not nearly enough to build a working computer though. As it was mentoined earlier all modern computers are digital, and use two voltage levels (that means they work in binary).

Two different voltage levels means that we must notice two things:
\begin{enumerate}
	\item Two different voltages will make two different currents run through our circuit (\emph{the current running through circuit must be controlled})
	\item A device that could control current or voltage must be created (\emph{it makes no difference whether you control current or you control voltage in the circuit - it's essentially the same thing}
\end{enumerate}

As we just noticed the current running through our circuit will change - that means we will have to deal with something like alternating circuits, but not exactly.

Alternating current is the kind of electric current that changes periodically, for example the common power outlets feed the current that changes its direction 50-60 times per second.

(insert more info about electronics).

The second thing we've noticed is that we need to create a kind of device that could control current running through it. Such device exists - in fact many of them exist, but the most commonly known one is the \emph{semiconducting transistor}. Its special electric properties allow to control current running through it by feeding another (much much smaller) current through it.

The transistor allows to implement all the basic boolean logic in an electric circuit. It is possible to construct electric circuits for all the basic boolean operations. These circuits can then be used as construction blocks for more complex devices - such as processors, controllers, etc.

\section{Processors}

\section{Digital Computer Architectures}

\section{Modern Computer}

\section{Spacecraft Computer Control}

\section{Operating Systems}

\section{Software Engineering}


\subsection{Integration by parts}

We have:

\(\dfrac{\delta y}{\delta x}=f(x)g(x)\)

We want that in terms of \(y\).

We know from the product rule of differentiation:

\(y=a(x)b(x)\)

Means that:

\(\dfrac{\delta y}{\delta x}=a'(x)b(x)+a(x)b'(x)\)

So let's relabel \(f(x)\) as \(h'(x)\)

\(\delta\)

\(\dfrac{\delta y}{\delta x}=h'(x)g(x)\)

\(\dfrac{\delta y}{\delta x}+h(x)g'(x)=h'(x)g(x)+h(x)g'(x)\)

\(y+\int h(x)g'(x)=\int h'(x)g(x)+h(x)g'(x)\)

\(y+\int h(x)g'(x)=h(x)g(x)\)

\(y=h(x)g(x)-\int h(x)g'(x)\)

For example:

\(\dfrac{\delta y}{\delta x}=x.\cos(x)\)

\(f(x)=\cos(x)\)

\(g(x)=x\)

\(h(x)=\sin(x)\)

\(g'(x)=1\)

So:

\(y=x\int \cos(x) dx-\int \sin(x)dx\)

\(y=x\sin(x)-\cos(x)+c\)


\subsection{Elliptic curves}

Of the form \(y^2=x^3+ax+b\).


%&LaTeX

\section{Let's Get Physical}

In this lab, you will use Matlab to explore the effects of summing
sinusoids. You will then investigate how a physical signal can be
considered to be composed of a sum of sinusoids --- its \emph{Fourier
  series}. You will be using the Matlab \texttt{AnalogSignal} class
and functions at
\url{http://faculty.washington.edu/stiber/pubs/Signal-Computing/}.

\subsection{Beating}

In this section, you will use the Matlab \texttt{AnalogSignal} class
(described previously in lab~1 and its
figure~\ref{fg:analogsignal-help}) to simulate an analog signal
generator. The constructor takes the following arguments:
\begin{lstlisting}[style=Matlab-editor,basicstyle=\mlttfamily\small]
% AnalogSignal(type, amplitude, frequency, dur)
%    Where:
%    type = 'sine', 'cosine', 'square', 'sawtooth', or 'triangle'
%    amplitude = signal amplitude
%    frequency = signal frequency
%    dur = signal duration
\end{lstlisting}

Remember that, though an \texttt{AnalogSignal} is simulating an analog
signal, in Matlab all functions are sampled at discrete points in
time. The crufty details of this are hidden away in the class's
implementation. Remember also that you can always plot an
\texttt{AnalogSignal} to see what you have; include such figures in
your report

\paragraph{Step 1.1} Verify that you can get a triangle wave. What is
the code to generate and plot a triangle wave ranging from -1 to 1
Volts with a frequency of 10Hz and a duration of 1 second?

\paragraph{Step 1.2} Next, verify that you can generate a sine wave of
1 second duration at a frequency of 440Hz, ranging from -1 to +1. What
is the Matlab code to do this? Use the \texttt{AnalogSignal}
\texttt{soundsc} method to play this as a sound. This pitch
corresponds to ``standard A'' on a musical scale --- A above middle
C. Use the Matlab \texttt{set(gca, 'XLim', [\itshape{Xmin},
  \itshape{Xmax}])} function call to set the X axis limits so that the
waveform is apparent (i.e., you're not just plotting a solid blob).

\paragraph{Step 1.3} Generate three sine waves of identical range and
duration, but with frequencies of 442, 444, and 448 Hz. Next, generate
the three sums of 440Hz and each of these new signals separately (so,
440Hz + 442Hz, 440Hz + 444Hz, and 440Hz + 448Hz) to generate beating
akin to tuning an instrument against the 440Hz standard. Play each sum
signal; can you hear the beating? Plot each sum signal for its full 1s
duration. The beating ``envelope'' should be obvious. What is the beat
frequency in each case? How does the beat frequency and amplitude
relate to the textbook discussion of beating?

\subsection{Fourier series representation of a physical signal}

\paragraph{Step 2.1} Recall that any periodic signal can be
represented as a sum of harmonic sinusoids.  The amplitudes of these
harmonics is collectively known as the Fourier Series. It may at first seem like
sums of sinusoids would be poor approximations of real periodic
signals, but this is not the case. We can illustrate this using a
triangle wave. The formula for synthesis of a triangle wave with
frequency $\omega_0$ is a sum of harmonically related sine waves (its
Fourier series):
  \[
  x(t) = \sum_{k=0}^{\infty}
  \left(
    \underbrace{ \frac{8}{\pi^2} \frac{(-1)^k}{(2k+ 1)^2} }_{ \text{amplitude} }
    \underbrace{ \sin((2k+1)\omega_0 t) }_{ (2k+1)^{th}\text{ harmonic} }
  \right)
  \]
  In this case, in the analog domain, we are dealing with frequencies
  in Hz, and so $\omega_0 = 2\pi f_0$. Notice that the Fourier Series
  of the triangle wave only uses odd harmonics (i.e., the only
  non-zero frequencies are $(2k+1)\omega_0=\omega_0, 3\omega_0,
  5\omega_0 \cdots$). Also notice that the resulting wave will have zero
  mean because there is no ``DC'' term (i.e., $2k+1 \neq 0$ for any
  integer k).

  Write a Matlab script that approximates a triangle wave by summing
  together the first 7 harmonics of its Fourier series; plot the
  resultant signal (i.e., use $f_0$, $2f_0$, $3f_0$, $\cdots 7f_0$,
  where $f_0=$10 Hz). How does this signal compare to the triangle
  wave computed directly in the previous step?

\paragraph{Step 2.2} Another way to view a signal is in the
\emph{frequency domain}. For a signal expressed in terms of its
Fourier series, the frequency representation is merely the
coefficients of the harmonics. Write a Matlab function or script to
compute and plot the spectrum of a triangle wave. You may find the
MATLAB function \texttt{stem} useful.  Note that you are \emph{not}
being asked to plot the triangle wave as a function of time; you
should plot the amplitudes of the component sinusoids as a function of
those sinusoids' frequencies (like the vertical lines in textbook
figure~1.12).  Use your code to plot the spectrum of the triangle wave
from the previous step (first 7 harmonics).


\section{Type classes}
\label{sec:typeclasses}

The first part of this section will give a short description of the concept of a type class. The second part will describe monoids in general and the type class \verb|Monoid|.

The concept of a type class was introduced as a construct that supports overloaded functions and \emph{\gls{adhoc_polymorphism}} \cite{Wadler}. Overloaded functions can be used with a variety of types, but with different definitions for the different types. For example, the function calls \verb|show 1| and \verb|show "hello"| use different \glspl{function-definition}. The function \verb|show 1| is of type \verb|Int -> String| and \verb|show "hello"| is of type \verb|String -> String|. The definition used depends on the type of the argument.  The next section will describe \gls{adhoc_polymorphism} and the relation to type classes in more detail.

\subsection{Polymorphism and type classes}
\label{sec:polymorphism}
%%In this section we will describe relation between type classes and polymorphism.
There are two types of polymorphism in Haskell \cite{Cardelli}: Parametric polymorphism and \gls{adhoc_polymorphism}. Type classes are used for \gls{adhoc_polymorphism}.
\begin{description}
\item[Parametric polymorphism] refers to functions which work over more than one type. For example, the library function \verb|length| returns the length of a list \verb|[]| that contains items of arbitrary type. It can be used to calculate the length of a list of integers, a list of strings, a list of booleans, etc. There are no constraints. The type of \verb|length| is
\begin{verbatim}
length :: [t] -> Int
\end{verbatim}
\verb|t| is a \emph{type variable}. That is, for any type \verb|t| the function \verb|length| has type \verb|[t] -> Int|. A type that contains a type variable is called \emph{polymorphic}. Hence, \verb|length| is a polymorphic type.
The \verb|length| function can be used with any type but it has a \emph{single} definition.  At compile time, the type variables are substituted with a concrete type. For example
\verb|[Int] -> Int|.
\item[Ad hoc polymorphism] is a synonym for function overloading or operator overloading. An overloaded function uses different function definitions depending on the types of the arguments. Suppose we want to define a function that converts a list containing items of arbitrary type (\verb|[t]|) to a string. We would write a function with the following \gls{typesignature}:
\begin{verbatim}
showlist :: [t] -> String
\end{verbatim}
It takes a list of arbitrary type \verb|t| and returns a string. The definition could look like this:
\begin{verbatim}
showlist [] = ""
showlist (x:xs) = show x ++ showlist xs
\end{verbatim}
We need a way to make sure that the function \verb|show| is defined for the type of the value \verb|x|. \verb|show| can't be a polymorphic type because the conversion depends on the type. There's no single definition that can convert an arbitrary type to a string.

There's a set of types. \verb|show| is defined over all members of this set. This set is called a type class. The type class \verb|Show| for example contains all types that can be converted to a string with the function \verb|show|. In order to prevent the application of the function \verb|showlist| with an argument that isn't a member of the type class \verb|Show|, we must constraint the type variable in the type signature declaration \verb|t|:
\begin{verbatim}
showlist :: Show t => [t] -> String
\end{verbatim}
\end{description}

Functions declared by the type class are defined over all members of the type class. And certain type classes exhibit properties that every definition of the corresponding functions must obey.

\subsection{Monoid}
\label{sec:monoid}

In mathematics, a \gls{monoid} is an algebraic structure with single associative binary operation and an identity element. Monoids are semigroups with identity \cite{wiki:monoid} \cite{renshaw}. Several elements of a monoid can always be reduced to a single element by applying the corresponding binary operator. It doesn't matter in which order we apply the operator, the result is always the same. This is called \emph{\gls{associativity}}. The set of elements has an identity element. For example, the set of natural numbers $\mathbb{N}$ form a monoid under multiplication.  The number 1 is the identity element. Multiplication of the identity and any other number $x$ results always in $x$.

In Haskell there is a type class for monoids. Types that form a monoid can become part of the \verb|Monoid| type class. For example, the type list \verb|[]| forms a monoid. Two values of type list can always concatenated to another list with the \verb|++| operator. The empty list \verb|[]| is the identity element.

The example in section \ref{sec:example} shows a plugin system that contains a monoid. Plugins can be composed with a binary operator. An arbitrary number of plugins can be composed to a single plugin. Because the composition operator is associative, plugins can be evaluated in arbitrary order.

\subsubsection{Functions of the type class monoid}

Members of the type class \verb|Monoid| have to implement the functions \verb|mempty| and \verb|mappend| amongst others (see appendix \ref{sec:monoiddefinition} for a complete declaration).
These functions have the following type signature declaration.
\begin{verbatim}
    mempty :: m
    mappend :: m -> m -> m
\end{verbatim}
The type variable \verb|m| is the type of the corresponding monoid.
\verb|mempty| returns the identity value. \verb|mappend| is the binary function that takes two values of the same type and returns another value of that type.

The type class \verb|Monoid| exhibits several laws. We will only describe the one that we will prove in the example of section \ref{sec:example}, the \emph{left identity law}.
When making monoid instances, we need to make sure that \verb|mempty| acts like the identity with respect to the \verb|mappend| function. This property can be expressed with the following equation:
\begin{equation}
  \label{eq:firstmonoidlaw}
  \text{mappend}(\text{mempty}, x) = x
\end{equation}
Equation \ref{eq:firstmonoidlaw} states that \verb|mempty| has to behave like the identity with respect to \verb|mappend|. When \verb|mappend| is applied with the identity and an other element \verb|x| of the monoid, it returns \verb|x|.

\subsubsection{Example monoid implementation}

There is a useful property of the \verb|Applicative| type class with respect to the \verb|Monoid| type class (the \verb|Applicative| type class is described in more detail in the appendix, section \ref{sec:applicatives}). The example in section \ref{sec:example} will use this property. If \verb|f| is an \verb|Applicative| and \verb|b| is a \verb|Monoid| then \verb|f b| is also a \verb|Monoid|. If a type is part of the \verb|Applicative| type class and the type contains a \verb|Monoid| we can create a \verb|Monoid| instance with the implementation in listing \ref{lst:monoidinstance1}. Figure \ref{fig:applicative_monoid} illustrates this property.

\begin{figure}
  \centering
     \includegraphics[width=0.7\textwidth]{monoid}
  \caption{An {\ttfamily Applicative} that encapsulates a {\ttfamily Monoid} is a {\ttfamily Monoid}}
  \label{fig:applicative_monoid}
\end{figure}

\lstset{
basicstyle=\ttfamily,
columns=fullflexible,
keepspaces=true,
captionpos=b
}
\begin{lstlisting}[caption={{\ttfamily Monoid} instance implementation of {\ttfamily IO}},label={lst:monoidinstance1}]

{-# LANGUAGE FlexibleInstances #-}
import Data.Monoid
import Control.Applicative

instance (Applicative f, Monoid a) => Monoid (f a) where
    mempty = pure mempty
    mappend = liftA2 mappend
\end{lstlisting}

\verb|mempty| is of type \verb|f a| . Hence \verb|pure mempty| has to be of type \verb|f a|.
As \verb|f| is an \verb|Applicative|, it implements \verb|pure|. The type of \verb|pure| is \verb|a -> f a| (see appendix \ref{sec:applicatives}). We call \verb|pure| with \verb|mempty| of type \verb|a|. We know that \verb|a| is part of \verb|Monoid| because of the type constraints. The compiler will use \verb|mempty| of \verb|a|. \verb|liftA2| is an utility function of \verb|Applicative|. It encapsulates the \verb|mappend| function in an applicative functor.

In section \ref{sec:example} we prove that the \gls{function-definition} of listing \ref{lst:monoidinstance1} obeys the first monoid law formed by equation \ref{eq:firstmonoidlaw} with  the verification technique equational reasoning.


\section{Polynomials}
\subsection{Definition and Basics}
A polynomial is defined as $P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0$
with names corresponding to their degree (constant, linear, quadratic, cubic, quartic).

The factored form is written as $P(x) = a(x-r)(x-p)\cdots(x-q)$.
The simplest and most useful polynomial is the quadratic. It can be written as $ax^2+bx+c$ and factored respectively.
The formula to solve for $x$ is $x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
The most important formula for polynomials is the Vieta Formulas.

\begin{formula}[Vieta Formulas]
  Sum of roots ($r_1+r_2+r_3+\cdots+r_n$): $-\frac{a_{n-1}}{a_{n}}$ \\
  Product of roots ($r_1  r_2  r_3 \cdots r_n$): $(-1)^n \cdot \frac{a_0}{a_n}$ \\
  Pairwise sums of $p$  ($p=2$: $r_1r_2+r_1r_3+r_1r_4+\cdots+r_{n-1}r_n$): $(-1)^p \cdot \frac{a_{n - p}}{a_{n}}$
\end{formula}

\begin{theorem}[Fundamental theorem of algebra]
  It states that a single variable polynomial with degree
  $n$ has exactly $n$ complex roots.
\end{theorem}

\begin{problem}
  Let $r,s,$ and $t$ be the roots of $3x^3-4x^2+5x+7=0$. (\ia, 8.20 pg.249)
  \begin{enumerate}
    \item Find $r+s+t$ ($\frac{4}{3}$).
    \item Find $r^2+s^2+t^2$ ($\frac{-14}{9}$).
    \item Find $\frac{1}{r} + \frac{1}{s} + \frac{1}{t}$ ($\frac{-5}{7}$).
  \end{enumerate}
\end{problem}

\subsection{Synthetic Division}
A simplification of traditional polynomial division. Note this only works when the coefficients of
the linear term in the divisor is $1$. It is also know as the \textbf{Ruffini's Rule}.

\textbf{Example:}
$\begin{array}{c|rrrrr} 3&\parbox[b]{0.2in}{\raggedleft 1}&\parbox[b]{0.2in}{\raggedleft -3}&\parbox[b]{0.2in}{\raggedleft 7}&\parbox[b]{0.2in}{\raggedleft -1}&\parbox[b]{0.2in}{\raggedleft 5}\\ &&{3}&{0}&21&60 \\\cline{2-6} \multicolumn{2}{r}{{1}}& {0}&7&20&\multicolumn{1}{|r}{65} \end{array}$

Which is the same as $(x^4 - 3x^3 + 7x^2 - x + 5) \div (x - 3) = x^3+7x+20+\frac{65}{x-3}$. Notice you work from left to right, and multiply to get the next number in the second row.
If your divisor doesn't have $1$ as its coefficient in the linear term, you can divide it by $1/n$ and in the end also multiply the quotient and remainder by $1/n$.

Usually, you write the result of polynomial division as $\frac{f(x)}{d(x)}=q(x)+\frac{r(x)}{d(x)}$.

\subsection{Rational Root Theorem}
\begin{theorem}[Rational Root Theorem]
  A rational root of a polynomial in the form $\pm\frac{p}{q}$ where $p$ and $q$ are relatively prime must follow the condition
  $p | a_0$ and $q | a_n$.
\end{theorem}

\subsection{Remainder Theorem}
\begin{theorem}[Remainder Theorem]
  When a polynomial $f(x)$ is divided by $x-a$, the remainder is determined by $f(a)$.
\end{theorem}

Theorem 1.3 can be proven using the form $\frac{f(x)}{d(x)}=q(x)+\frac{r(x)}{d(x)}$ and synthetic division.

\begin{proof}[Remainder Theorem]
  \begin{align*}
    f(x) &= (x-a)q(x) + r(x) \\
    &= (x-a)q(x)+c. \\
    f(a)&=(a-a)q(a)+c \\
    &=0 \cdot q(a) + c = c
  \end{align*}
  Thus, $f(a)$ always returns the remainder of $f(x) \div (x-a)$.
\end{proof}

\subsection{Factor Theorem}
\begin{theorem}[Factor Theorem]
given the expression $x-a$, it is a divisor of $p(x)$ if and only if $p(a)=0$. This can be proven with the remainder theorem.
\end{theorem}

\subsection{Miscellaneous}
To find the sum of the coefficients of a polynomial $P(x)$, plug $x=1$!
Example problem: Practice problem \#2

\subsection{Practice Problems}
Problems from \ia.
\begin{enumerate}
  \item \sout{6.11, pg. 181}
  \item 6.21 pg. 191
  \item 6.17, pg. 187
  \item 6.22, pg. 191
  \item 6.27, pg. 191
  \item 6.29, pg. 192
  \item $\star$ Challenge Problems, pg. 192
  \item \url{https://numbertheoryguydotcom.files.wordpress.com/2016/03/polynomials.pdf}
\end{enumerate}