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Final Answers
© 2000-2023   Gérard P. Michon, Ph.D.

Analysis,  Complex Analysis

 Michon
 

See also, on this site:

Related Links (Outside this Site)

Complex Variables, Complex Analysis   by  John H. Mathews  (2000).
Complex Variables, Contour Integration  by  Joceline Lega  (1998).

Bibliography :

Complex Analysis (1953-1979)  Lars Ahlfors (1907-1996).  Fields medal (1936). Complex Analysis (2017)  by  Valery S. Serov  (University of Oulou, Finland).
 
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Convergence,  Series,  Complex Analysis


brentw (Brent Watts, NC. 2000-12-07)   Defining  Real  Numbers...
What's the definition of a Cauchy sequence?

In a  metric space,  a  Cauchy sequence  is a sequence  U  whose  far terms  are within a vanishing distance of each other...  This is to say that, given any small positive quantity e, there's an integer N(e) such that, for any p and q both larger than N(e), the distance from U(p) to U(q)  is always less than e.

In this, the distance from x to y is usually  |x-y|  using the ordinary norm,  but more exotic possibilities exist, including the p-adic metric.

 Augustin Cauchy 
 1789-1857 The concept was first introduced by Augustin Cauchy (1789-1857)  as a nice way to characterize convergent sequences of real numbers  without  referring explicitely to the  limits  they converge to.  ( Cours d'analyse de l'Ecole Polytechnique,  1821.)

A convergent sequence is always a  Cauchy sequence.  The converse is only true in a  complete  space  (like the real numbers); it's not true for the rationals.  In fact a  complete metric space  can be defined as a metric space in which  every Cauchy sequence converges.

Following Georg Cantor (1845-1918) one usually defines real numbers, as equivalence classes of rational Cauchy sequences.  Two sequences U and V are considered equivalent if the limit of U(n)-V(n) is zero.

For example, the constant sequence  U(n) = 1  is a Cauchy sequence equivalent to the Cauchy sequence  V(n) = 1-(0.1)n ,  whose first terms are:

0     0.9     0.99     0.999     0.9999     0.99999     0.999999     0.9999999

So, both sequences define the  same  real number  (the number 1).  Any real number which has a finite decimal expansion also has an infinite one, ending with infinitely many nines.  Other  numbers have just a  single  decimal expansion.  This confuses many beginners, as they wrestle with the  definition  of real numbers.

There's another (equally valid) way to define real numbers, which predates the above.  It's based on  Dedekind cuts  which turn out to be more difficult to manipulate than Cauchy sequences...

On 2001-03-11, Brent asked for:
[A specific] example of how to prove whether a given sequence is a Cauchy sequence or not.

In the realm of real numbers, proving that a sequence converges and proving it's a Cauchy sequence are just two aspects of the same thing.  Therefore, we'll choose an example of a sequence in the the field of rationals (a notoriously incomplete space, as was first glimpsed by a disciple of Pythagoras, probably Hippasus of Metapontum, about 2500 years ago).

Consider the rational sequence u, recursively defined via:

u(0)  =  1       and       u(n+1)   =   u(n) / 2  +  1 / u(n)
 
u(1) = 3/2 ,  u(2) = 17/12 ,  u(3) = 577/408 ,  u(4) = 665857/470832   etc.

First you may want to prove that u(2n) is an increasing sequence and that u(2n+1) is a decreasing one, whereas u(2m+1) is greater than any u(2n) for any pair n,m. With the additional remark that u(2n+1)-u(2n) tends toward zero as n tends to infinity, you've got all the ingredients to prove that, for p and q greater than n, |u(p)-u(q)|  is less than  |u(n)-u(n+1)|  and thus tends to zero when n tends to infinity.  In other words, the sequence u is a rational Cauchy sequence

On the irrationality of the constant of Pythagoras

The above should come as no surprise to anyone who knows about the irrational limit of u (namely Ö2), a "special" number which was not at all taken for granted 2500 years ago: The irrationality of what is still sometimes referred to as the  constant of Pythagoras  is said to have prompted the sacrifice to the gods of 100 oxen (a so-called hecatomb)...


(2016-05-24)   Determining the convergence of a series.
Exploring the boundary between convergent and divergent series.

Convergence tests  for series amount to a minor art form.  They're an unavoidable part of an undergraduate education in mathematics.

As discussed elsewhere on this site,  there's much more to a series of terms  an  than the sequence  Am  of its partial sums.  However, that's what  classical analysis  focuses on  (a viewpoint formalized by  Cauchy in 1821):

Am   =    m    an
å
n=0

The series is said to be convergent, of sum  A,  iff  the sequence of its partial sums is a convergent sequence of limit  A.  In which case, we write:

A    =    ¥    an
å
n=0

In the particular case of a real series whose terms are alternatively positive and negative, this happens if and only if the terms of the series tend to zero.  Otherwise, this necessary condition is not sufficient, as demonstrated by the case of the harmonic series.

A series is said to be  absolutely convergent  when the series formed by the absolute values of its terms converges.  The same terminology can also be used for series whose terms are complex, hypercomplex or, more generally, belong to a normed vector space  (the norm of a vector being corresponds to the absolute value of a number).

 Come back later, we're
 still working on this one...

Convergence Tests for Positive Series :

The ratio test.

The root test.

Comparison with an integral.

In asymptotic analysis, critical cases often involve cascades of logarithms:

L0 (x)   =   x         and         Lk+1 (x)   =   Log | Lk (x) |
 
an   ~   a  /  ( L0m0 (n) L1m1 (n) ... Lkmk (n) )

If all exponents are equal to 1, then the series diverges.  Otherwise, the series converges  iff  the first exponent which differs from 1 is greater than 1.

Proof :   The only critical case is when all exponents are equal to  1,  with the possible exception of the last one.  This can be settled by comparing the series with an integral,  because of the following  indefinite integrals.

ó
õ
 x dt    =    Lk+1 (x)
Vinculum
L0(t) L1(t) ... Lk(t)
ó
õ
 x dt =Lk (x) 1-m       if  m ¹ 1
Vinculum Vinculum
L0(t) L1(t) ... Lk(t)m 1-m

Other cases are settled by termwise comparisons with series of this type.  QED

The idea for this convergence test originates with  Joseph Bertrand (1822-1900)  who introduced it for the case  k = 2  in 1842  (cf. Série de Bertrand).

Convergent series   |   Cauchy condensation test
 
Generalized Bertrand series  (MathOverflow, 2010-01-05).


(2021-07-13)   A piece of mathematical folklore:
If the series of term  an > 0  converges,  so does the series  Öan / n.

Because the two positive series  an  and  1/n2  are convergent,  all their  partial sums  are bounded by their respective sums  A  and  B.  We may then use the  Cauchy-Schwartz inequality  to obtain,  for any  m :

(   m
å

n = 0
 Öan / n   )2
 
    ≤     m
å

n = 0
  an    m
å

n = 0
  (1/n) 2     ≤     A B   =   A  p2
Vinculum
6

So,  all partial sums of the targeted series are bounded by   A½ p/Ö6.   QED

Generalization :

The same argument proves the convergence of the series  Öan/nk  if  k >  ½.

Höder's inequality (Rogers 1888, Hölder 1889)   |   Minkowski's inequality


johnrp (John P. of Middletown, NJ. 2000-10-14)
Can you rearrange the following infinite series so that its sum equals 43?
1 -1/2 +1/3 -1/4 +1/5 -1/6 +1/7 -1/8 +1/9 -1/10 ...

Yes.  In fact such a thing can be done for any "target sum" S (here S=43) with any series which is convergent but not absolutely convergent (that is, the series of absolute values does not converge).  That's known as the Riemann series theorem  or also  Riemann's rearrangement theorem.

This applies here,  because the series involved converges to a well-known constant while the series of absolute values is the  harmonic series, which has been known to diverge since the 14th century  (at least).  Let's discuss the construction in general terms:

Take just as many positive terms of the series as necessary to exceed S (that's always possible, as explained below), then take as many negative terms to have the partial sum fall below S, then use positive terms again to go above S, switch to negative terms to go below S again, etc.

Note that, as advertised above, it is always possible to add enough terms of the series to make up for any (positive or negative) difference between the current sum and the target S. That's because the series of the absolute values is divergent (so both the series of negative terms and the series of positive terms must be divergent, or else the whole series would not be convergent).

In this process (at least after the first step) the difference between S and any partial sum never exceeds the magnitude of the term added at the latest "switch" from negative to positive (or vice-versa). Since the magnitudes of such terms tend to zero, partial sums tend toward S. S is therefore the sum of the rebuilt series.

The above reasoning is due to  Bernhard Rieman (1826-1866).

In 1910,  Waclaw Sierpinski (1882-1969)  further showed that any target sum could be achieved by rearranging terms of only one sign  (e.g.,  just the negative terms)  in any convergent series which isn't absolutely convergent.

When a convergent series remains convergent  (with unchanged sum)  regardless of the order of its terms,  it's said to be  unconditionally convergent.  This is the case when the series is  absolutely convergent  (i.e.,  when the series formed by the norms of its terms converges).  In a vector-space of finite dimension, there are no others.  However, in a space of  infinite dimension,  the latter term is stronger:  some  unconditionally convergent  series may exist which are not  absolutely convergent.  (HINT:  Consider the series whose n-th term is  (1/n) un  if the Euclidean norm is used with an infinite set of unit vectors  u  which are pairwise orthogonal).

Oresme's proof of the divergence of the harmonic series :

To apply the above to the question at hand, we have to show that the following so-called  harmonic series  diverges:

1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + ...

   Nicole Oresme 
 1323-1382
Around 1350,  Nicole Oresme (1323-1382)  published one elementary way to do so,  based on the remark that the series is bounded from below by the series obtained by replacing 1/n with 1/q,  where  q  is the lowest power of  2  greater than or equal to  n:

1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 + 1/16 + ...

By grouping equal terms in that series, we see that the partial sum up to the term of rank  q=2 is simply equal to 1+p/2. The partial sum of the harmonic series up to that rank is therefore no less than 1+p/2.  This means that such partial sums will eventually exceed any preset bound,  no matter how high.  The series (slowly)  diverges ;  its limit is infinityQED

The Harmonic Series Diverges Again and Again   by  Steven J. Kifowit  &  Terra A. Stamps   (2006)
Riemann series theorem   |   The Mercator series (1668)
 
Riemann Rearrangement Theorem  by  Robin Whitty  (Theorem of the Day #218).


(2012-10-25)   Decreasing divergent series whose minimum converges!

Such a counterintuitive monstrosity can be constructed very simply from any convergent series with decreasing positive terms  un   (e.g.,  un = 1/n).

We'll build two divergent series of terms  an  and  bn  such that:

un   =   min ( an , bn )

Let's choose a sequence of indices  p(0) = 0 < p(1) < p(2) < p(3) < ...
such that  p(i+1) - p(i) > 1/ui

  • If   p(2i) ≤ n < p(2i+1) ,  then   an = up(2i)   and   bn = un
  • If   p(2i+1) ≤ n < p(2i+2) ,  then   an = un   and   bn = up(2i+1)

The sum of all terms  an  when  n  goes from  p(2i)  to  p(2i+1)-1  is greater than  1, by construction, so the  an  series diverges!  So does the  bn  series, for similar reasons.  QED

Strictly speaking, the two series described are merely  nonincreasing  but minor details could be adjusted to make them  decreasing  ones.

Two Divergent Series Whose Minimum Converges  (March 2010)

 Nonuniform convergence
 to a function which is 
 zero except when x=1...
(Brent Watts of Hickory, NC. 2001-04-13)
How do you show that the sequence  f n : x ® xn converges for each x in the closed interval [0,1] but that the convergence isn't uniform?

The simple convergence of a sequence of functions is just pointwise convergence. In this case, the limit of xn is clearly 0 when x is in [0,1[ and 1 when x=1. The sequence  f n  thus converges and its limit is the function  f  defined over [0,1] which is zero everywhere except at point 1, where  f (1) = 1.

Now, simple convergence does not tell you much about the limit. The limit of continuous functions may not be continuous (this is what happens here). Worse, the integral of the limit may not be equal to the limit of the integrals:  A sequence of functions with 
 a unit integral, whose 
 nonuniform limit is zero! Consider, for example, the sequence of functions gn on [0,1] for which gn(x) is n2x when x is in [0,1/n], n(2-nx) when x is in [1/n,2/n] and zero elsewhere. The pointwise limit of gn(x) is always zero (x=0 included, since gn(0)=0 for any n). Yet, the integral of gn is always equal to 1, for any n>1.

This is why the notion of uniform convergence was introduced: We say that a sequence of functions fn defined on some domain of definition D converges uniformly to its limit f when it's always possible for any positive quantity e to exhibit a number N(e) such that whenever n is more than N(e), the quantity |fn(x)-f(x)| is less than e, for any x in D.  (Note that a "domain of definition" is not necessarily a "domain" in the sense of an open region, ita est.  Whenever it's critical, make sure to specify which meaning of "domain" you have in mind.)

Uniform convergence does imply that the integral of the (uniform) limit is the limit of the integrals.  It also implies that the (uniform) limit of continuous functions is continuous. Since you have a discontinuous limit here, the convergence can't possibly be uniform...

The above settles the question, but it can also be shown directly that it's not possible for a given (small enough) quantity e>0 to find an N such that  f n(x)  would be within e of its limit for any x whenever n>N.  Indeed, for 0<e<1, any x in [e1/n,1[ is such that  fn(x) exceeds eQED

The convergence is uniform within any interval strictly smaller than [0,1[.  It's not uniform over [0,1[ itself,  although the limit  is  continuous.

 Gaston Darboux
Gaston Darboux

brentw (Brent Watts of Hickory, NC. 2001-04-14)
[...]  Explain the concept of  Darboux integrals.

Before Lebesgue took a radically different  (and better)  approach,  several definitions of integration were proposed which involved dividing the interval of integration  [a,b]  into a finite number of arbitrarily small segments :
a = x0 < x1 <  ...  < xn = b     where  xk+1 - xk  ≤  e

Each author then defined a certain finite sum  (see below)  depending on a given function  f  which was said to be  integrable  (in the sense of Cauchy, Riemann, etc.) if that sum has a limit  as e tends to zero,  regardless of the chosen subdivisions  (the limit being the  integral  of  f  on  [a,b]).  Viz:

  • Cauchy: å (xk+1-xkf (xk)   [This definition is now obsolete.]
  • Riemann: å (xk+1-xkf (sk)
    where sk may be anywhere between xk and xk+1
  • Darboux (lower): å (xk+1-xk) Lk
    where Lk is the greatest lower bound of  f (x)  for x in [xk+1-xk]
  • Darboux (upper): å (xk+1-xk) Uk
    where Uk is the least upper bound of  f (x)  for x in [xk+1-xk]

The last two sums correspond to the lower and upper Darboux integrals.  The nice thing is that a function  f  is Riemann-integrable  if and only if  its lower Darboux integral equals its upper Darboux integral.

The Riemann Integral was introduced by  Bernhard Riemann (1826-1866) in his  Habilitationsschrift  (1854)

 Henri Lebesgue
Henri Lebesgue

(2014-08-09)   Theory of Integration   (Lebesgue, 1902)
Measure Theory.  Lebesgue measure.  Lebesgue Integrals.

The foundations of the modern theory of integration were laid down in the  1902  doctoral dissertation of  Henri Lebesgue (1875-1941)  under  Emile Borel (1871-1956).

Lebesgue realized that slicing the "area" delimited by a function into horizontal slices rather than vertical ones would lead to a notion of integral that is far more satisfying than all  previous attempts.

The caveat is that the  Lebesgue Integral  requires a careful definition of the  measure  l  of such horizontal slices, which may be quite intricate...

ó b
õa
  f (x)  dx    =   sign (b-a)   ó +¥
õ
l ({x :  (x-a)(x-b) < 0  &  f (x) = y})  y dy
 

Both sides are equal if they are well-defined as Riemann integrals  (possibly  improper  ones,  understood as  limits).  However,  the right-hand-side  can make sense even when the left-hand-side doesn't.  In that case,  the left-hand-side becomes a notation for the new concept of  Lebesgue integral.

The above remains valid, with the usual sign convention, when  a > b.  That important practical point fully justifies the added complication of the above definition,  but Lebesgue integrals are best viewed as integrals over a non-oriented  measurable  domain of integration  I  which need not be an  interval  (such as  I = [a,b] ).  That entails a simpler general relation:

ó 
õ
I
  f (x)  dx    =    ó +¥
õ
l ({ x Î I  :  f (x) = y })  y dy

Unlike Riemann integration,  Lebesgue integration doesn't depend on the topology or ordering of the real numbers and is thus easily generalized to other realms...  This is fundamentally different from the equally-important generalization of  oriented  integration over smooth manifolds,  intimately related to  differential forms.

The classic example of a function which is Lebesgue-integrable but not Riemann-integrable is the  Dirichlet functionf = 1Q  (the  indicator  of the rationals)  whereby  f (x)  is  1  when  x  is rational and  0  otherwise.  Being  countable  the set of the rationals  (Q)  has zero  Lebesgue measure.

Therefore,  1Q  has a zero Lebesgue integral.  (HINT :  The integrand in the above right-hand-side is always zero.)  It's not Riemann-integrable because its two  Darboux integrals  are different  (one is zero,  the other is  b-a).

Lebesgue integration   |   Lebesgue measure   |   Lebesgue-Stieltjes integration
 
Lebesgue Integral Overview (26:22)  by  Peyam Tabrizian (Dr. Peyam, 2018-02-09).


 Joseph Fourier 
 (1768-1830)  Leonhard Euler 
 (1707-1783) brentw (Brent Watts of Hickory, NC. 2000-11-25)
How do I evaluate the Fourier series of the function
 f (x) = x(2p-x)   in the interval  0 < x < 2p ?

If a function   f (x)  =  ½ [ f (x-) + f (x+)]   has period  2p,  then its  Fourier expansion  is defined via:
 
 f (x)
 

=
  ao    +  
¥
å
n=1
    
[ an cos(nx)  +  bn sin(nx) ]
 
vinculum +

=
2 +
The coefficients  an  and  bn  are  twice  the average values of  cos(nx) f (x)  and  sin(nx) f (x).  They're given by Euler's formulas :

an
 

=
1  
ò  2p 
 
0
 
 f (x) cos(nx) dx
 
vinculum

=
p

bn
 

=
1  
ò  2p 
 
0
 
 f (x) sin(nx) dx
 
vinculum

=
p

For an even function, like the one at hand, the b-coefficients are all zero and we are only concerned with the first formula, giving the a-coefficients. (Conversely, the a-coefficients would all be zero for an odd function.)

In the case at hand, we  integrate by parts  twice over the interval  [0,2p]  when  n  is nonzero  (for  n = 0  we just integrate a quadratic function).  Thus,  an  is  -4/n2  if  n ¹ 0,  whereas  ao  is  4p2/3.  All told, we obtain:

 
x (2p-x)
 

=
2p2  
 4 
 
 
¥
å
n=1
   cos(nx)            [ For x between 0 and 2p ]
vinculum - vinculum

=
3 n2

The Basel Problem :

For  x = 0,  the above may serve as a proof of the famous result at right,  obtained by Euler in 1735:  The sum of the reciprocals of all nonzero perfect squares is equal to  p2/6   p2
=
¥
å
n=1
   1
vinculum vinculum
6
=
n2

 Coat-of-Arms of the 
 BERNOULLI Family 
 (Basel, Switzerland) The problem of finding the  exact  value of that sum was posed by Pietro Mengoli in 1644.  It was once known as the Basel Problem, after the hometown of Jacob Bernoulli, who was first in a long list of notorious mathematicians  (including Leibniz)  who failed to discover the above solution.  Euler first worked it out numerically to 20 decimal places  (earlier in 1735)  using the  Euler-Maclaurin formula.   He identified the value as  p2/6  before he could justify that.

Six Ways to Sum the Basel Series  by Dan Kalman.
The College Mathematics Journal, 24, 5 (Nov. 1993) pp. 402-421.
 
Cauchy's Proof of the Basel Problem (9:58)  Rise to the Equation  (2021-08-21).

p3 =
¥
å

n = 0
   (-1)n
vinculum vinculum
32
=
(2n+1)3

brentw (Brent Watts of Hickory, NC. 2001-03-05)
How does one prove the relation at right?

Consider the odd function  f (x)  of period 2p which equals  2x/p  when  x  is in the  interval  [-p/2,p/2].  Euler's formulas  give its Fourier expansion:

 The function f(x) is +1 or -1 when 
 x is an odd multiple of pi/2.
 
 f (x) 
 
=
8   ¥
å

n = 0
   (-1)n  
  sin(2n+1)x
 
vinculum vinculum

=
p2 (2n+1)2

Integrate that to obtain the expansion of a primitive g(x) of  f (x), namely:
 
g(x) 
 

=
 
 C 
 
8   ¥
å

n = 0
   (-1)n  
  cos(2n+1)x
 
- vinculum vinculum

=
p2 (2n+1)3

The constant  C  is the average of g(x) over one full period. It depends on which value we choose for g(0).  With g(0)=0, we have g(x)=x2/p for x between 0 and p/2. Because of the symmetry about x=p/2, the average C is:

C   =   g(p/2)   =   p/4.

Plug this value of C in the above relation at point x=0 (where g(x)=0 and cos((2n+1)x)=1), to obtain the value p3/32 for the sum we were after.  QED

Dirichlet's Beta Function  (and Euler Numbers) :

Consider the following expression, which generalizes the above.  Here,  we're primarily concerned with integral (positive) values of  z,  but this function  b  (called Dirichlet's Beta Function)  may be defined  by analytic continuation  over the entire complex plane.  It has no singularities.

 
b(z)
 
=
¥
å

n = 0
   (-1)n
vinculum

=
(2n+1)z

The above shows that  b(3) = p3/32.  Differentiating  f (x),  instead of integrating it,  would have given b(1) = p/4,  a result which is commonly obtained by computing the value of the  arctangent function  at  x=1,  using its  Taylor expansion  about 0.

It's worth noting that the above method may be carried further with repeated integrations.  Every other time, such an integration gives an exact expression for the alternating sum of some new power of the reciprocals of odd integers.  In other words, we obtain the value of  b(k)  for any  odd  k,  and it happens to be a rational multiple of  p k :

b(1)  =  p/4 The general expression is
 
b(2n+1)
 
=
   (p/2) 2n+1  
  | E2n |
 
vinculum

=
2(2n)!
b(3)  =  p3/32
b(5)  =  5p5/1536
b(7)  =  61p7/184320
b(9)  =  277p9/8257536
b(11)  =  50521p11/14863564800

In this, | E2n | is a nonnegative integer.  The Euler number  En  is defined as the coefficient of  zn/n!  in the Taylor expansion of  1/ch(z)  [where ch is the hyperbolic cosine function; ch(z) = (e+e-z )/2].  Starting at the index  n = 0,  the sequence of  (signed)  Euler numbers is:

1, 0, -1, 0, 5, 0, -61, 0, 1385, 0, -50521, 0, 2702765, 0, -199360981, 0, 19391512145, 0, -2404879675441, 0, 370371188237525, 0, ... (A122045)

We may also consider the secant function  1/cos(z)  which has the same expansion as  1/ch(z)  except that all the coefficients are positive, so that:

¼ px  /  cos(½ px)   =     å n   b(2n+1)  x 2n+1

There does not seem to be any similar expression for even powers.  In fact,  b(2)  is currently defined as an independent fundamental mathematical constant,  the so-called Catalan Constant: G = 0.915965594177219015...

This is the exact opposite of the situation for nonalternating sums, where even powers correspond to an exact expression in terms of a rational multiple of the matching power of p, whereas odd powers do not  [that's an open conjecture].


brentw (Brent Watts of Hickory, NC. 2000-11-21)
How do you prove the following relation?
 
(p2/ab) coth(pa) coth(pb)
 

=
¥
å
m =
   
¥
å
n =
   1
vinculum

=
(m2 + a2 ) (n2 + b2 )

The relation   åm ån u(m) v(n)  =  [ åmu(m) ]   [ ånv(n) ]   holds whenever the series involved are absolutely convergent (which is clearly the case here). Therefore, we only have to establish the following simpler equality:

 
(p/a) coth(pa)
 

=
¥
å
m =
       1
vinculum

=
m2 + a2

The sum on the right-hand side looks like a series of Fourier coefficients.  For what periodic function?  Well,  it's not difficult to see that the correct denominator is obtained for the continuous even function of period 2p which equals  cosh(ax)  if x is in the interval [-p,p].  When x is in that interval, the Fourier expansion has two equivalent forms  (using  a-m  =  a):

 
cosh(ax)
 

=
  ao    +  
¥
å
m = 1
    
am cos(mx)   =   ½
 
¥
å
m =
 
am cos(mx)
 
vinculum +

=
2 +

Euler's formulas  give   am = [2a(-1)m/p] sinh(pa) / (m2 + a2 ).  At the point x = p, we have  cos(mx)= (-1)m   and the above relation thus translates into the desired equality  [just divide both sides by  (a/p) sinh(pa)].


brentw (Brent Watts of Hickory, NC. 2001-04-14)
How do you use the Fourier series for the function f(x) = ex for x in ]0,2p[ to find the sum [S] of the series 1/(k2 + 1) ? [ k=1 to ¥ ]

Use Euler's formulas to compute the Fourier coefficients of f(x). Note that if you consider f as a function of period 2p equal to exp(x) in ]0,2p[, it has a jump discontinuity at any point x=2np (where n is any integer). This means (and it's important for the rest) that the Fourier series converge to the half-sum of the left limit and the right limit at such points of discontinuity, in particular the value at point 0 is [exp(2p)+1]/2.

Now,  the computation of the Fourier coefficients is easy if you remark that  exp(x)cos(kx)  and  exp(x)sin(kx)  are the real and imaginary parts of  exp((1+ki)x)   (it's clear we'll only need the real part, but I'll pretend I didn't notice).  The indefinite integral of that is simply  exp((1+ki)x)/(1+ki),  which we may also express as  exp((1+ki)x) (1-ki)/(1+k2).  The definite integral from  0  to  2p  is thus  (exp(2p)-1)(1-ki)/(1+k2).  The Fourier coefficients are obtained by multiplying this by  1/p and using the real and imaginary part separately.  All told:

f(x) = [exp(2p)-1]/p (½ + SUM[ k=1 to ¥, (cos(kx) - k sin(kx))/(1+k2) ] )

All you have to do is apply this to x=0 (this is why we did not really need the coefficients of sin(kx)).  With the above remark to the effect that the LHS really is  [f(x-)+f(x+)]/2  at any jump discontinuity like x=0,  we obtain:

[exp(2p)+1] / 2   =   [exp(2p)-1] / p ( ½ + S )

where  S  is the sum we were after.  Therefore:

S   =   p/2 - ½ + p / [ exp(2p)-1 ]   =   1.076674047...

That's also a special case  (a = 1)  of the relation obtained above in the form:

p coth(p)   =   1 + 2 S


brentw (Brent Watts of Hickory, NC. 2000-11-28)   Overshoot
[...] Please explain the Gibbs phenomenon of Fourier series.

At a point x where a function f has a jump discontinuity, any partial sum of its Fourier series adds up to a function that has an "overshoot" (i.e., a dampened oscillation) whose initial amplitude is about 9% of the value of the jump J=|f(x+)-f(x-)|.

This amplitude is not reduced by adding more terms of the Fourier series. It's not difficult to prove that, with n terms, the maximum value of the overshoot occurs at/near a distance of p/2n on either side of x. (You may do the computation with any convenient function having a jump J; I suggest f(x)=sign(x)J/2 between -p and p. Adding a continuous function to that would put you back to the "general" case without changing the nature or amplitude of the Gibbs oscillations.)

When n tends to infinity, the maximum reached by the first overshoot oscillation is about 8.948987% of the jump J. This value is precisely (2G/p-1)/2, where G is known as the Wilbraham-Gibbs Constant:

G  =   ó p
õ
0  
sin(q)/q dq  
=   1.8519370519824661703610533701579913633458...

This is sometimes called "the 9% overshoot", as it is about 9% of the total jump J.  [It's 18% (17.89797...%) when half the jump (J/2) is used as a unit.]

This tells you exactly what kind of convergence is expected from a Fourier series about a discontinuity of  f. For a small h, you can always increase the number of Fourier terms so that Gibbs oscillations are mostly confined to the very beginning of the interval [x,x+h].

This resembles the convergence to zero of the sequence of functions  f (n,x)  defined as being equal to  4nx(1-nx)  for x between 0 and 1/n, and zero elsewhere.  f (n,x) always reaches a maximum value of 1 for x=1/2n.  That sequence does converge to zero, but it's  not  uniform convergence!  Same thing with partial Fourier sums in a neighborhood of a jump discontinuity...

Josiah Willard Gibbs, Jr. (1839-1903)


ó  b
 f (x) dx
PV
õ
a
brentw (Brent Watts of Hickory, NC. 2000-12-08)
What is the Cauchy principal value (PV) of an integral?

If f has no singularities, the principal value (PV) is just the ordinary integral.

 Augustin Cauchy 
 (1789-1857)

If the function f has a single singularity q between a and b (a<b), the Cauchy principal value of its integral from a to b is the limit (whenever it exists), as e tends to 0+, of the sum of the integral from a to q-e and the integral from q+e to b.  Also, if the interval of integration is ]-¥,+¥[ with a singularity at ¥ the principal value is the limit, whenever its exists, of the integral in the interval ]-A,+A[ as A tends to infinity.

When f has a discrete number of singularities between a and b (a and b excluded, unless both are infinite), the PV of its integral may be obtained by splitting the interval [a,b] into a sequence of intervals each containing a single singularity.  The above applies to each of these, and the PV of the integral over the entire interval is obtained by adding the principal values over all such subintervals.

The fact that the principal value is used may be indicated by the letters PV before the integral sign, or by crossing with a small horizontal dash the integral sign (see illustration above). However, it is more or less universally understood that the Cauchy principal value is used whenever needed, and some authors don't bother to insist on this with special typography.

Soldner constant   |   Cauchy integration formula


brentw (Brent Watts of Hickory, NC. 2000-11-21)
How do I solve the differential equation 2(1-x)y" + (1+x)y' + [x - 3 - (x-1)2exp(x)]y = 0 about the pole x=1?

The singularity at  x = 1  is a  regular  Fuchsian  one  (which means that,  if the coefficient of y" is normalized to 1,  then the coefficient of y' has at most a single pole at x=1 and the coefficient of y has at most a double pole at x=1).

Therefore,  the  method of Frobenius  is applicable.  It consists in finding a solution in the form of a so-called  Frobenius series  of the following form  (where h=x-x0 in general, and h=x-1 here)  with a(0) nonzero:

y = hm [ a(0) + a(1) h + a(2) h2 + a(3) h3 + ... ]

In the above, m is not necessarily an integer, so that a Frobenius series is more general than either a Taylor series (for which m is a natural integer) or a Laurent series (for which m is any integer). In the DE we're asked to study, we have:

-2h y" + (2+h) y' + [h-2-h2exp(1+h)] y = 0

The method of Frobenius is simply to expand the above LHS in terms of powers of h to obtain a sequence of equations that will successively give the values of a=a(0), b=a(1), c=a(2), d=a(3), etc. Let's do it. The above LHS is hm-1 multiplied by:

[-2am(m-2)] +[a(m-2)-2b(m2-1)]h +[a+b(m-1)-2cm(m+2)]h2 +O(h3)

We have to successively equate to zero all the square brackets.  Since a is nonzero, the first square bracket gives us the acceptable value(s) of the index m (this is a general feature of the method and this first critical equation is called the indicial equation).  Generally, the indicial equation has two roots (for a second-degree DE) and this gives you a pair of independent solutions.  Usually, when the roots differ by an integral value (like here) you've got (somewhat) bad news, since the Frobenius method is only guaranteed to work for the "larger" of the two values of m.  However, "by accident" you're in luck here:

The case m=0 gives b=a (second bracket). Then, the third bracket gives zero for the coefficient of c  (that's the usual problem you encounter after N steps when starting with the smaller root, if the two roots differ by N)  but it so happens that the rest of the bracket is zero too!  (That's exceptional!) So you can continue with an arbitrary value of c and obtain d as a linear combination of a and c using the next bracket (which I was too lazy to work out, since I knew  tough  problems could not occur past that point).

The way to proceed from here is to first use a=1 and c=0 to get the first solution as a Frobenius series F(h), then a=0 and c=1 to get a linearly independent solution G(h). The general solution about the singularity x=1 is then  a F(x-1) + c G(x-1). (You don't have to bother with the index m=2 in this particular case.)


brentw (Brent Watts of Hickory, NC. 2000-11-21)   Laurent Series
How to determine the Laurent series of a function about a singular point.

For each singular point  (or pole)  zo, you want to expand  f (zo+h).  If a pole has multiplicity  n,  then hn  f (zo+h) is normally an analytic function.  Compute its Taylor expansion about h=0 and divide that series by hn to have the  Laurent series  about that particular pole.

Let's present the computation step-by-step for the following simple example:

f (z)   =   1 / [z (z-1)2 ]

Let's examine the double pole at  z = 1.

First we compute   f (1+h)   [in the neighborhood of the pole, h is  small ].  It's just a matter of replacing  z  by  (1+h).  Nothing to it:

f (1+h)   =   1 / [ (1+h)((1+h)-1)2 ]   =   1 / [ (1+h)h2 ]

Multiply this by  h to obtain an analytic function about  h = 0,  namely:

g(h)   =   h2  f (1+h)   =   1 / (1+h)

The Taylor expansion of  g  is well known:

g(h)   =   1 - h + h2 - h3 + h4 - h5 + h6 - h7 +   ...

Since  f (1+h) = g(h)/h2, we divide the above by  h2  to obtain the Laurent expansion of  f (1+h)  about h=0 or, equivalently, of  f (z)  about z=1 :

f (1+h) = 1/h2 - 1/h + 1 - h + h2 - h3 + h4 - h5 +   ...

Usually, we're only concerned with the coefficient of 1/h, which is called the residue for that pole (here it's equal to -1).  The integral of a function along any closed contour encircling a certain number of poles is equal to  2pi  times the sum of the residues for those poles. This can be a practical way to compute easily many definite integrals that would otherwise be difficult to obtain.  Examples follow.

Many textbook insist on the following format  (which I don't recommend):

f (z)   =   1/(z-1)2 - 1/(z-1) + 1 - (z-1) + (z-1)2 - (z-1)3 +   ...

It's sometimes desirable to present the function globally as the sum of so-called  simple elements  about every pole and an  entire  function  (which is just a polynomial when the function is  rational, like this one).

f (z)   =   1/(z-1)2  -  1/(z-1)  +  1/z  +  0

That type of reduction allows the immediate integration of rational functions:

ò   f (z) dz   =   -1/(z-1)  -  ln(z-1)  +  ln(z)

The field of formal Laurent series   |   Pierre Laurent (1813-1854; X1830)


ò  ¥   dx
vinculum
 
0
(1+x2 )Öx
yourm0mz ( 2001-12-15)
How do you find this definite integral?
(I am using the positive x-axis as a branch cut.)

When attempting to apply Cauchy's residue theorem  Augustin Cauchy 
 (1789-1857) [the fundamental theorem of complex analysis] to multivalued functions (like the square root function involved here), it is important to specify a so-called "cut" in the complex plane were the function is allowed to be discontinuous, so that it is everywhere else continuous and single-valued.

In the case of the square-root function, it is not possible to give a continuous definition valid around any path encircling the origin. Therefore, a so-called "branch-cut" line must be specified which goes from the origin to infinity. The usual choice is [indeed] to use the positive x-axis for that purpose. This choice means that, when the angle q is in [0,2p[, the "square root" of the complex number z = r exp(iq) is simply Öz º Ör exp(iq/2)  (the notation Ör being unambiguous because r is a positive real number and its square root is thus defined as the only positive real number of which it is the square). This definition does present a discontinuity when crossing the positive real axis (a difficulty avoided only with the introduction of Riemann surfaces, which are beyond the scope of our current discussion).

Path With the above definition of the square root of a complex argument, we may thus apply the Residue Theorem to the function f(z)=1/(1+z2)Öz on any contour which does not cross the positive real axis. We may choose the contour pictured at left, which does not encircle the origin [this would be a no-no, regardless of the chosen "branch cut"] but encloses the pole at +i when the outer circle is big enough.

On the outer semicircle, the quantity |f(z)| eventually becomes much smaller than the reciprocal of the length of the path of integration. Therefore, the contribution of the outer semicircle to the contour integral tends to zero as the radius tends to infinity. The smaller semicircle is introduced to avoid the singularity at the origin, but its contribution to the contour integral is infinitely small when its radius is infinitely small. What remains, therefore, is the contribution of the two straight parts of the contour. The integral along the right part is exactly the integral we are asked to compute, whereas the left part contributes i times that quantity. All told, the limit of the contour integral is (1+i) times the integral we seek.

Cauchy's Theorem states that the contour integral equals 2pi times the sum of the residues it encircles.

In this case, there's only one such residue, at the pole i. The value of the residue at pole i  is the limit as h tends to zero of f(i+h), namely 1/2iÖi, so the value of the contour integral is i = 2(1+i)/2. As stated above, this is (1+i) times the integral we want. Therefore, the value of that integral is exactly p/Ö2, or about 2.22144146907918312350794049503...


ò  ¥   xa dx
vinculum
 
0
1 + x2
For what values of a does this integral converge?
What's the value of the integral when it converges?

The previous article deals with the special case a = -1/2.
In general, we see that the integral makes sense in the neighborhood of zero if a>-1 and it converges in the neighborhood of when a<1. All told, the integral converges when a is in the open interval ]-1,1[.

Path

We apply the above method to f(z) = za/(1+z2) [defining za with the positive x-axis as branch cut] on the same contour (pictured at right). The smaller semicircle is useless when a is positive and it has a vanishing contribution otherwise (when a>-1). The contribution of the outer semicircle is vanishingly small also (when a<1) because |f(z)| multiplied by the length of the semicircle becomes vanishingly small when the radius becomes large enough. On the other hand, the contribution of the entire positive x-axis is the integral we are after, whereas the negative part of the axis contributes exp(ipa) as much. All told therefore, Cauchy's theorem tells us that our integral is 2pi/(1+exp(iap)) times the residue of  f at the pole z = i.

The residue at z = i  is the limit, as h tends to zero, of f(i+h), which is simply exp(iap/2)/2i. This makes the integral equal to

p exp(iap/2) / (1+exp(iap))

That boils down to   p/(2 cos ap/2)   so that we obtain:

ó
õ
 ¥   xa dx     =     p [ -1 < a < 1 ]
vinculum vinculum
0 1 + x2 2 cos (ap/2)

We may notice that this final result is an even function of a, which we could have predicted with the simple change of variable  y = 1/x...


(2007-05-08)   The Vocabulary of Complex Analysis
Holomorphic functions, entire functions, meromorphic functions, etc.

Holomorphic Functions
(analytic functions of a complex variable)

A complex function  f  of a complex variable which is differentiable about every point  z  of its domain is called an analytic function, an holomorphic function, or a  complex differentiable function :

lim    f (z+h)  -  f (z)     =   f ' (z)
Vinculum
h ® 0 h

The existence of such a  derivative function  ( f ' )  is a much more restrictive condition in the complex realm than among real numbers, since it implies essentially that  f  is differentiable  infinitely many times  and possesses a convergent expansion as a  Taylor series  about every point  x  inside  its domain of definition.

 f (z)    =  
¥
å
n = 0
   
 f (n) (x)
  (z-x) n   where   f (0) = f
vinculum
n!
 
 f (n+1)  =  [ f (n) ]'

We may break down complex quantities into their real and imaginary parts:

f (z)   =   f (x+iy)   =   u + i v   =   u(x,y) + i v(x,y)

In this, both  u  and  v  are real functions of the  two  real variables  x  and  y.  The above differentiability of  f  as a function of the complex variable  x+iy  implies the following equation in terms of the  partial derivatives  of  f  with respect to the two real variables  x  and  y.

i f / x   =   f / y

In terms of  u  and  v,  this translates into the following differential equations, known as the  Cauchy-Riemann equations,  which are a necessary condition for the related function  f  to be holomorphic  (note how these equations imply that both u and v are necessarily differentiable infinitely many times with respect to x and y as soon as they are known to be differentiable once).

u / x   =   v / y         and         v / x   =   - ¶u / y

Entire Functions and Liouville's Theorem

An  holomorphic function  defined over the whole complex plane  (without any singularities)  is called an  entire functionLiouville's Boundedness Theorem  asserts that such a function can only be bounded if it's  constant.

Meromorphic Functions

meromorphic function  f  over some domain D is an  holomorphic function  on the domain obtained by removing from D a discrete set of isolated ordinary  poles  (as opposed to  essential singularities).  That's to say that a neighborhood of any such singularity  x  exists where the function  f (z)  multiplied by some power of  (z-x)  is simply an holomorphic function (i.e., without singularities).

Etymologically, meros (meroV = part) is opposed to holos (holoV = whole).

Analytic expression   |   Numericana :   Analytic contination   |   Factorizations and pole expansions
 
Casorati-Sokhotski-Weierstrass theorem
 
Complex Analysis Overview (36:22)  by  Peyam Tabrizian (Dr. Peyam, 2018-07-22).


(2021-06-22)   Cauchy-Riemann equations  (d'Alembert-Euler conditions)
On the partial derivatives in an analytic function of a complex variable.

These first-order differential relations between the real and imaginary parts  of a complex-valued function  f = u+iv   of the complex variable  z = x+iy  indicate that the mapping which sends  (x,y)  to  (u,v)  is conformal  (which is to say that it preserves angles and planar orientation).  This turns out to be a necessry condition for the  analyticity  of the function  f.

The equations were first published in 1752 by  Jean-le-Rond d'Alembert (1717-1783)  in an essay on fluid mechanics.  They were first mentioned as a criterion for analyticity by Leonhard Euler (1707-1783)  in 1777.  Augustin Cauchy (1789-1857)  started using them as a cornerstone for his theory of functions in 1814.  They take center stage in the celebrated 1851 dissertation of  Bernhard Riemann (1826-1866).

As advertised above, the existence of a (first) derivative with respect to a complex variable is a much stronger condition than its counterpart for real variables.  Let's examine more precisely what this entails...

Let  f  be a differentiable complex function of the complex variable  z.  Let's call u and v the  the real and imaginary parts  of  f  and let's introduces as x and y the  real variables  which z reduces to:

  •   z      =   x  +  i y
  • f (z)   =   u(x,y)  +  i v(x.y)

The differentiability of  f  about point  z  says that the following limit is well-defined as the  (complex)  quantity  h  tends to zero.

lim    f (z+h)  -  f (z)     =   f ' (z)
Vinculum
h ® 0 h

This implies, in particular, that the limit exists when h=x is real and when h=iy is imaginary, which says that both u and v have partial derivatives with respect to either x or y.  Furthermore,  the limits for h=x and h=iy as either x or y tends to zero must be identical  (or else, there would not be a well-defined limit in the neighborhood of h=0)  which implies the following:

Cauchy-Riemann Equations
u    =    v           and           v    =   -  u
Vinculum Vinculum Vinculum Vinculum
x y x y

Contrary to Riemann's belief,  the converse isn't quite true  as the existence of partial derivatives satisfying the  Cauchy-Riemann equations  doesn't imply that  u+iv  is differentiable.  A related fact had also escaped Cauchy in 1821  (see  Cauchy's mistake).  One counterexample due to  Carl Johannes Thomae (1870) is the following function, which meets all conditions but doesn't have a derivative at point zero  (where it's not even continuous):

 f (x+iy)   =   sin ( 4 Arctg  x  )     when  y ¹ 0   (and  f = 0 when y = 0)
Vinculum
y

When the partial derivatives are continuous  (which isn't the case in the above example)  a theorem due to Clairault (1740) applies, which states that the values of partial derivatives with respect to several variables do not depend on the order of the differentiations.

Continuity of the partial derivatives isn't required to apply the  Looman-Menchoff theorem  which states that a function  known  to be continuous with partial derivatives in a  neighborhood  of  z  is holomorphic in that neighborhood if and only if it satisfies the  Cauchy-Riemann equations.

Satisfying the Cauchy-Riemann equations only at point  z  isn't enough.  (HINT:  The continuous function  z5/|z|4  isn't analytic anywhere.)

 G. Kh. Sindalovskii

Generalizations of the  Looman-Menchoff theorem  have been devised,  culminating in the detailed results published in Russian by  Grigorii Khaimovich Sindalovskii (1928-2020)  in 1985:

On the Cauchy-Riemann conditions in the class of functions with summable modulus and some boundary properties of holomorphic functions
Mat. Sb. 128 (170) No. 3 (1985), pp. 364-382

Cauchy-Riemann equations (1851)   |   Augustin Cauchy (1789-1857)   |   Bernhard Riemann (1826-1866)
Looman-Menchoff theorem (1923, 1936)   |   Herman Looman (1896-1983)   |   Dmitrii Menchoff (1892-1988)
 
Cauchy-Riemann equations (17:08)  by  Dr. Peyam (2017-10-27).   |   Cauchy's mistake (1821)


(2021-06-22)   Wirtinger Derivatives  =  Wirtinger Operators   (1926)
Applying the methods of calculus to non-analytic complex functions.

Consider any function  w  of the two variables x and y.  Let's introduce, as a new pair of variables, z=x+iy and z*=x-iy.  Using partial derivatives:

dw   =   wx dx  +  wy dy   =   wz dz  +  wz* dz*

Since   dx = ½ (dz + dz*)   and   dy = ½ (i dz* - i dz)   we obtain:

½ wx (dz + dz*)  +  ½ wy (i dz* - i dz)   =   uz dz  +  uz* dz*

Equating separately the coefficients of  dz  and  dz*  yields :

  • wz   =   ½ (wx - i wy )
  • wz*  =   ½ (wx + i wy )

Let's translate this into the standard  del notations  for partial derivatives:

Wirtinger Derivatives for a Single Complex Variable
  =   ½ (    -  i   )     and       =   ½ (    + i     )
Vinculum Vinculum Vinculum Vinculum Vinculum Vinculum
z x y z* x y

Because the Wirtinger derivatives are just partial derivatives,  they obey the same  standard rules of calculus  as straight derivatives,  including linearity, product rule and chain rule.  Interestingly,  the condition  z* f = 0  is equivalent to the  Cauchy-Riemann equations  because:

z* (u+iv)   =   x (u+iv)  +  i y (u+iv)   =   (x u - y v) + i (x v + y u)

Thus,  loosely speaking,  an analytic function depends only on  z,  not z*.

The first Wirtinger derivative coincides with the ordinary derivative in the case of analytic functions but it's also defined for non-analytic functions  (whenever partial derivatives exist).

Wirtinger derivatives of some functions
f (z)  =  f (x+iy) f / z
zn n zn-1
Analytic f (z) f ' (z) = df / dz
(z*)n 0
Analytic f (z*) 0
x  =  Re (z)  =  ½ (z*+z) 1 / 2
y  =  Im (z)  =  ½ (iz*-iz) -i / 2
|z| = (z z*)½ ½ z* / |z|
|z|n ½ n z* |z|n-2
Analytic  f (|z|) ½ z* f ' (|z|) / |z|
Log |z| 1 / z

This last example is of particular interest since the Log function is not well-defined in the complex plane, but Log |z|  is.  That real-value function is not analytic anywhere,  but it's  harmonic  everywhere  (except at the origin).

Complex functions of several complex variables :

The generalization to  n  complex variables   z1 ... zn   is straightforward:

Wirtinger Derivatives for Several Complex Variables
     =    ½ (      -  i     )     and          =    ½ (      + i       )
Vinculum Vinculum Vinculum Vinculum Vinculum Vinculum
zk xk yk zk* xk yk

Wirtinger derivatives (1926)   |   Wilhelm Wirtinger (1865-1945)
 
Intuition behind Wirtinger derivatives  (MathStackExchange, 2013-02-26).


(2021-08-09)   Cauchy Integral Formula
Fundamental Theorem of Complex Analysis.

If  f  is  holomorphic,  in a  simply-connected neighborhood of point  a,  then its value at point  a  is given by an integral around any contour  C  encircling  a  within that neighborhood:

 f (a)   =    1
Vinculum
2pi
  òC    f (z)
Vinculum
z-a
  dz

Differentiate both sides  n  times with respect to  a  to obtain a  corollary :

 f (n) (a)   =    n!
Vinculum
2pi
  òC    f (z)
Vinculum
(z-a) n+1
  dz

In the plain version stated above,  the contour avoids all zeros of  f.  In some cases,  that requirement is inconvenient  (as we may not know  a priori  where those zeros are).  A zero on the contour contributes to the integral exactly the  half-sum  of what it would if it was on either side!

Cauchy principal value


(2021-08-10)   [ Cauchy's ]  Argument Principle   (1831 & 1855)
Contours of the  logarithmic derivative  yield   2pi  [ #(zeros) - #(poles) ]

#{zeros of  f  in K} - #{poles of  f  in K}   =    1
Vinculum
2pi
  òK    f ' (z)
Vinculum
f (z)
  dz

In 1831, Augustin Cauchy (1789-1857)  stated this theorem for  holomorphic  functions only  (no poles).  He extended it to  meromorphic  functions  (poles and zeros)  in 1855.

The  argument principle  is used in the proof of  Rouché's theorem  (1862).

Even when the computation of the contour integral is  marred  with substantial  rounding errors  it's easy to infer the result with absolute accuracy knowing only integers can be involved.

The method has been used to locate  nontrivial zeros of the zeta function:  First a rectangular contour is used which spans the full width of the  critical strip  (0 < x < 1)  The result is then compared to what's obtained with an infinitesimally narrow central section thereof,  counting all the zeros arbitrarily close to the critical line at  x =  ½.

Logarithmic derivative (Numericana)   |   Argument principle (Cauchy 1831 & 1855)
 
Strecker-Nyquist stability criterion (1930, 1932)   |   Felix Strecker (1892-1951)   |   Harry Nyquist (1889-1976)

 Eugene Rouche (1832-1910)
Eugène Rouché

X (2021-08-09)   Rouché's theorem   (1862)
For  f  and  g  holomorphic  in  K :   If | g | < | f |  on  K,  then  f + g  has as many zeros as  f  in  K.

We assume the boundary  K  of the region  K  is a simple loop.  Multiple zeros are counted with their multiplicities.

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Winding number (Numericana)   |   Rouché's theorem (1862)   |   Eugène Rouché (1832-1910; X1852)


(2019-08-23)   Riemann Mapping Theorem   (1851, Carathéodory 1912)
Existence of a  biholomorphic mapping  from  U  to the open unit disk.

biholomorphic mapping  is a  bijective  holomorphic  mapping whose inverse is also holomophic.  Such a mapping between the open unit disk and a  proper part  U  of the complex plane is called a  Riemmann mapping.  It's necessarily a  conformal map  (i.e., it preserves locally the angles of lines and the orientation of surfaces,  but not necessarily their areas).

In his thesis (1851)  Bernhard Riemann (1826-1866)  stated that such a mapping exists for  any  nonempty simply-connected open proper subset  U  of  C.  That statement is now known as the  Riemann mapping theorem.

Using the  Schwarz lemmaHenri Poincaré  proved that a Riemann mapping is esentially unique.

Riemann's original flawed proof was based on  Dirichlet's principle,  which  Weierstrass  would later find not to be universally valid.  The first correct proof was obtained by  Constantin Carathéodory (1873-1950)  in 1912.

Riemann mapping theorem
 
Schwarz lemma   |   Hermann Schwarz (1843-1921)


(2021-08-08)   Univalent functions  are  holomorphic  injections.
Their  derivatives  never vanish.

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Schlicht function   |   De Branges's theorem (1984)   |   Louis De Branges de Bourcia (1932-)
 
Bieberbachiana   |   Talk (2016-09-01).


(2021-08-05)   Schlicht functions  are  univalent functions  from the open unit disk to the complex plane,  normalized  to   f (0) = 0  and  f ' (0) = 1.

Any  univalent function  g  reduces to a  schlicht function  f  using an  affine transformation  made legitimate by  the fact  that  g' (0) ¹ 0 :

f (z)   =    g (k z) - g (0)
Vinculum
k  g' (0)

In this,  k > 0  is at most equal to the  radius of convergence  R  of  g.  Thus, the radius of convergence of  f  is at least equal to 1,  as required by law.

General properties of univalent functions are often just stated in terms of  schlicht functions,  whose radius of convergence is always at least 1:

With   a0 = 0   and   a1 = 1 ,     f (z)   =    ¥
å

n = 0
  an zn       when  | z | < 1

A major fact about  schlicht functions  was called  Bieberbach's conjecture  for  68 years  (1916-1984).  It's now known as  De Branges's theorem :

| an |   ≤   n

This result is  sharp,  as equality is achieved in the following example...

Key example :   An important family of  schlicht functions  consists of the  rotated Koebe functions  which depend on a complex parameter  q  of unit norm  (i.e.,  | q | = 1).  The basic  Koebe function  is the case  q = 1.

 fq (z)   =    z
 Vinculum
(1 - q z) 2
   =     ¥
å

n = 0
  n qn-1 zn

To see that  f  is injective,  consider that  f (u) = f (v)  implies:

0    =    u (1 - q v) 2  -  v (1 - q u) 2    =    (u - v) [1 - q2 u v]

If  u  and  v  are different,  this requires the square bracket to vanish,  which is not possible when u  and  v  are inside the unit disk,  since the modulus of  qu v  is then  strictly  less than unity.  QED

Injectivity does fail  everywhere  on the boundary.  [HINT:  If  u  is on the unit circle,  so is  v = 1/(q2 u).]  This doesn't prevent  f  from being  schlicht,  but makes those functions  borderline  cases  ( | q | < 1  would be less tight).

 Gerard Michon

Robertson's lemma  (1936)

The special case  p = 2  of the lemma below appears  (without proof)  in the 1936 paper where  M.I.S. Robertson  proposed a new conjecture sufficient to establish  Bieberbach's conjecture,  a fact which would be put to good use by  Louis de Branges  in his celebrated proof thereof  (1984-85).

Lemma :   For a given schlicht function  f  and any positive integer  p,  there is a  unique  schlicht function  Fp  such that:

Fp (z) p   =   f ( z p )

Proof :   Let's  define  Fp   using the  binomial theorem  for exponent  1/p  in a small enough neighborhood of zero,  in the following way:

Fp (z)   =    ¥
å

k = 0
  b pk+1  z pk+1    =   z  ¥
å

k = 0
  ( 1/p
k
 )    y k

Where  y   =     f ( z p )
Vinculum
z p
 - 1   =   z p   ¥
å

n = 2
  a n  z p(n-2)

The other possible solutions would be deduced from that one by multiplying it into a nontrivial p-th root of unity,  w,  but this would make the derivative at the origin be  w ¹ 1,  which is ruled out for a schlicht function.

With the issue of ambiguity so resolved in a neighborhood of zero where the intermediate  (binomial)  series converges  (because  | y | < 1)  the defining algebraic relation between  f  and  F  persists for the latter by  analytic continuation  to the  b  series for as long as the former is defined without singularities,  which is the case whenever  | z | < 1.  Thus the  radius of convergence  of the  b  series is always at least 1,  as required.

It remains only to show that the  Fp  so defined is injective:  Fp (u) = Fp (v)  implies that the p-th powers of the two sides are equal.  which means:
f ( u p )  =  f ( v p )   and,  since  f  is injective,   u p = v p.

Plug this equality into the  b  series for  Fp  to get    F(u) / F(v)  =  v/u,  so that the former ratio can only be unity if the latter is.  QED

Apply this result,  for any positive integer  p,  to our previous family parametrized by  q = r exp(i q),  with  0 < r ≤ 1,  to obtain a 3-parameter family of  schlicht functions  (2 continuous parameters and a discrete one):

 f (z)   =    z
 Vinculum
(1 - r e iq z p ) 2
   =     ¥
å

n = 0
  n r n-1 e i (n-1) q z p (n-1)+1

The basic  Koebe function  is retrieved for  r = 1,  q = 0  and  p = 1.

One-quarter theorem  (Koebe 1907, Bieberbach 1916)

Milin  (1964)

Isaak Milin (1919-1992)...

Robertson  (1936)

Malcolm Irving Slingsby Robertson (1906-1998, PhD 1934)

Bieberbach's conjecture  =  De Branges's theorem (1985)

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Schlicht function   |   De Branges's theorem (1984)
 
A proof of the Bieberbach conjecture (1985)  by  Louis De Branges de Bourcia (1932-).
 
Bieberbachiana  and  Talk  by  Tomas Schonbek  (FAU, 2016-09-01)


(2021-08-08)   Starlike  schlicht functions.   (Nevanlinna, 1921)
Bieberbach's conjecture  was first proved in this special case.

holomorphic function  is  called  starlike  when its domain is  radially convex  (or  star convex)  which neans that:  If it contains  z,  then it also contains  t z  for any  t  in the  interval  [0,1].

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Carathéodory's lemma (1907)   |   Constantin Carathéodory (1873-1950).
 
Nevanlinna's criterion (1920)   |   Rolf Nevanlinna (1895-1980).


(2021-07-27)   Maximum modulus principle,  for holomorphic functions:
The modulus in a compact region is largest on the boundary.

Thus,  for any  holomorphic  function  f  of a complex variable  z,  | f (z) |  can't have a strict local maximum anywhere in the complex plane.

Various extensions of the principle have been proposed which apply to boundaries of non-compact regions.  The most popular is probably this:

The Phragmén-Lindelöf principle   (1908)

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Complex modulus   |   Maximum modulus principle
 
Phragmén-Lindelöf principle (1908)   |   Edvard Phragmén (1863-1937)   |   Ernst Lindelöf (1870-1946)

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