In a metric space, a Cauchy sequence is a sequence U whose far terms are withina vanishing distance of each other... This is to say that,given any small positive quantity ,there's an integer N() such that,for any p and q both larger than N(),the distance from U(p) to U(q) is always less than .
The concept was first introduced byAugustin Cauchy (1789-1857) as a nice way tocharacterize 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 asa metric space in which every Cauchy sequence converges.
FollowingGeorg Cantor (1845-1918)one usuallydefines real numbers, as equivalence classes ofrational Cauchy sequences. Two sequencesU and V are considered equivalent if the limit ofU(n)-V(n) is zero.
For example, the constant sequence U(n) = 1 is a Cauchy sequenceequivalent to the Cauchy sequence V(n) = 1-(0.1) , whose first terms are:
So, both sequences define the same real number (the number 1). Any real number which has a finite decimal expansion also has an infiniteone, 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.
In the realm of real numbers, proving that a sequence converges and proving it's aCauchy sequence are just two aspects of the same thing. Therefore, we'll choose an exampleof a sequence in the the field of rationals (a notoriously incomplete space,as was first glimpsed by a disciple ofPythagoras, probablyHippasus of Metapontum, about 2500 years ago).
Consider the rational sequence u, recursively defined via:
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 thatu(2n+1)-u(2n) tends toward zero as n tends toinfinity, 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 rationalCauchy sequence.
On the irrationality of the constant of Pythagoras
The above should come as no surprise to anyone who knows about theirrational limitof u (namely 2),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 promptedthe sacrifice to the gods of100 oxen (a so-calledhecatomb)...
(2016-05-24) Exploring the boundary between convergent and divergent series.
Convergence tests for seriesamount to a minor art form. They're an unavoidable part of an undergraduate education in mathematics.
As discussedelsewhere 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 =
an
The series is said to be convergent, of sum A, iff the sequenceof its partial sums is a convergent sequence of limit A. In which case, we write:
A =
an
In the particular case of a real series whose terms are alternativelypositive and negative, this happens if and only if the terms of the seriestend to zero. Otherwise, this necessary condition is not sufficient,as demonstrated by the case of theharmonic 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 arecomplex,hypercomplex or, more generally,belong to anormed vector space (the norm of a vector being corresponds to the absolute value of a number).
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 settledby comparing the series with an integral, because of the following indefinite integrals.
x
dt
=
Lk+1(x)
L0(t) L1(t) ... Lk(t)
x
dt
=
Lk(x) 1-m
if m 1
L0(t) L1(t) ... Lk(t)m
1-m
Other cases are settled by termwise comparisons with series of this type.
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).
(2021-07-13) If the series of term an > 0 converges, so does the series an / n.
Because the two positive series an and 1/n2 are convergent, all their partial sums are boundedby their respective sums A and B. We may then use the Cauchy-Schwartzinequality to obtain, for any m :
an / n
2
≤
an
(1/n) 2
≤ A B = A
2 6
So, all partial sums of the targeted series are bounded by A½ 6.
Generalization :
The same argument proves the convergence of the series an/nk if k > ½.
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 serieswhich is convergent but not absolutely convergent (that is, the series of absolute valuesdoes not converge). That's known as theRiemann series theorem or also Riemann's rearrangement theorem.
This applies here, because the series involved converges toa well-known constant while the series of absolute valuesis the harmonic series, which has beenknown 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 thepartial sum fall below S, then use positive terms again to go above S, switch to negativeterms to go below S again, etc.
Note that, as advertised above, it is always possible to add enough terms of the series tomake 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 (soboth the seriesof negative terms and the series of positive terms must be divergent, or else the wholeseries would not be convergent).
In this process (at least after the first step) the difference between S and any partial sumnever exceeds the magnitude of the term added at the latest "switch" from negative topositive (or vice-versa).Since the magnitudes of such terms tend to zero, partial sums tend toward S.S is therefore the sum of therebuilt series.
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 latterterm is stronger: some unconditionally convergent series may exist which arenot absolutely convergent. ( Consider the series whose n-th term is (1/n) un if the Euclidean norm is used withan 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 thefollowing so-called harmonic series diverges:
Around 1350, Nicole Oresme (1323-1382) published one elementary way to do so, based on the remark thatthe 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:
By grouping equal terms in that series, we see that the partial sumup to the term of rank q=2p 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 isinfinity.
(2012-10-25)
Such a counterintuitive monstrosity can be constructed very simplyfrom any convergent series with decreasing positiveterms 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.
Strictly speaking, the two series described are merely nonincreasing but minor details could be adjusted to make them decreasing ones.
(Brent Watts of Hickory, NC.2001-04-13) How do you show that the sequence f n : x xnconverges for each x in the closed interval [0,1]but that the convergence isn't uniform?
Thesimple 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 isthe 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: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 ofgn(x) is always zero (x=0 included,sincegn(0)=0 for any n).Yet, the integral of gn is always equal to 1, for any n>1.
This is why the notion ofuniform convergence was introduced: We say that a sequenceof functions fn defined on somedomain of definition D convergesuniformly to its limit f when it's always possible for any positive quantity to exhibit a number N() suchthat whenever n is more than N(), the quantity|fn(x)-f(x)| is less than ,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 specifywhich 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) limitof 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 beshown directly that it's not possible for a given(small enough) quantity >0to find an N such that f n(x) would be within of its limit for any x whenever n>N. Indeed, for 0<<1,any x in[1/n,1[is such that fn(x) exceeds .
The convergence is uniform within any intervalstrictly smaller than [0,1[. It's not uniform over [0,1[ itself, although the limit is continuous.
brentw (Brent Watts of Hickory, NC.2001-04-14) [...] Explain the concept of Darboux integrals.
Before Lebesgue took aradically different (and better) approach, several definitions of integration wereproposed 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 ≤
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 tends to zero, regardless of the chosen subdivisions (the limit being the integral of f on [a,b]). Viz:
Cauchy: (xk+1-xk) f (xk) [This definition is now obsolete.]
Riemann: (xk+1-xk) f (sk) where sk may be anywhere between xk and xk+1
Darboux (lower): (xk+1-xk)Lk where Lk is thegreatest lower bound of f (x) for x in [xk+1-xk]
Darboux (upper): (xk+1-xk)Uk where Uk is theleast 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 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 intohorizontal slices rather thanvertical oneswould 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 of such horizontal slices,which may be quite intricate...
õ
f (x) dx
= sign (b-a)
({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 measurabledomain of integration I which need not be an interval (such as I = [a,b] ). That entails a simpler general relation:
I
f (x) dx
=
({ x I : f (x) = y }) y dy
Unlike Riemann integration, Lebesgue integration doesn't depend on the topology or ordering of the realnumbers 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 isthe Dirichlet function, f = 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 () has zero Lebesgue measure.
Therefore, 1Q has a zero Lebesgue integral. ( : The integrand in the above right-hand-side is always zero.) It's not Riemann-integrable becauseits two Darboux integrals are different (one is zero, the other is b-a).
brentw (Brent Watts of Hickory, NC.2000-11-25) How do I evaluate the Fourier series of the function f (x) = x(2-x) in the interval 0 < x < 2 ?
If a function f (x) = [f (x-) +f (x+)] has period 2, then its Fourier expansion is defined via:
f (x)
ao
[ancos(nx) + bnsin(nx) ]
2
The coefficients an and bn are twice theaverage values of cos(nx)f (x) and sin(nx)f (x). They're given byEuler's formulas :
an
1
f (x) cos(nx) dx
bn
1
f (x) sin(nx) dx
For aneven function, like the one at hand, theb-coefficients are allzero and we are only concerned with the first formula, giving thea-coefficients.(Conversely, thea-coefficients would all be zero for anodd function.)
In the case at hand, we integrate by parts twice over the interval [0,2] when n is nonzero (for n = 0 we just integrate a quadratic function). Thus, an is -4/n if n 0, whereas ao is 4/3. All told, we obtain:
x (2-x)
22
4
cos(nx)
[ For x between 0 and 2 ]
3
n2
The Basel Problem :
For x = 0, the above may serve asa proof of the famous result at right, obtained byEuler in 1735: The sumof the reciprocals of all nonzero perfect squares is equal to /6
2
1
6
n2
/6 before he could justify that.
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 2which equals 2x/ when x is in the interval [-/2,/2]. Euler's formulas give its Fourier expansion:
f (x)
8
(-1)n
sin(2n+1)x
2
(2n+1)2
Integrate that to obtain the expansion of a primitive g(x) of f (x), namely:
g(x)
C
8
(-1)n
cos(2n+1)x
2
(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). Withg(0)=0, we haveg(x)=x/for x between 0 and /2.Because of the symmetry about x=/2,the average C is:
C = g(/2) = /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/32for the sum we were after.
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 (calledDirichlet's Beta Function) may be defined by analytic continuation over the entire complex plane. It has no singularities.
(z)
(-1)n
(2n+1)z
The above shows that (3) =/32. Differentiating f (x), instead of integrating it, would have given(1) = /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 withrepeated integrations. Every other time, such an integration gives anexact expression for the alternating sum of some new power of thereciprocals of odd integers. In other words, we obtain the value of (k) for any odd k, and it happens to be a rationalmultiple of :
(1) = /4
The general expression is
(2n+1)
2n+1
| E2n |
2(2n)!
(3) = 3/32
(5) = 55/1536
(7) = 617/184320
(9) = 2779/8257536
(11) = 5052111/14863564800
In this, | E2n | is a nonnegative integer. TheEuler number En is defined as the coefficient of zn/n! in the Taylor expansion of 1/ch(z) [wherech is thehyperbolic cosine function;ch(z) = (ez +e-z )/2]. Starting at the index n = 0, the sequence of (signed) Euler numbers is:
We may also consider thesecant function 1/cos(z) which has the same expansion as 1/ch(z) except that all the coefficients are positive, so that:
¼ x / cos(½ x) = n (2n+1) x 2n+1
There does not seem to be any similar expression for even powers. In fact, (2) is currently defined as an independentfundamental mathematical constant, the so-calledCatalan Constant:G = 0.915965594177219015...
This is the exact opposite of the situation for nonalternating sums, whereeven powers correspond to an exact expression in terms of a rational multiple of thematching power of , whereas odd powers do not [that's anopen conjecture].
brentw (Brent Watts of Hickory, NC.2000-11-21) How do you prove the following relation?
The relation mn u(m) v(n) = [mu(m) ] [nv(n) ] holds whenever the series involved areabsolutely convergent(which is clearly the case here).Therefore, we only have to establish the following simpler equality:
(/a) coth(a)
1
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 obtainedfor the continuous even function of period 2which equals cosh(ax) if x is in the interval []. When x is in that interval, the Fourier expansion has two equivalentforms (using a-m = am ):
cosh(ax)
ao
amcos(mx) = ½
amcos(mx)
2
Euler's formulas give am =[2a(-1)m/]sinh(a) / (m2+a2). At the point x = ,we have cos(mx)= (-1)mand the above relation thus translates into the desired equality [just divide both sides by (a/) sinh(a)].
brentw (Brent Watts of Hickory, NC.2001-04-14) How do you use the Fourier series for the functionf(x) = ex for x in]0,2[to find the sum [S] of the series 1/(k2 + 1) ?[ k=1 to ]
UseEuler's formulas to compute the Fourier coefficients of f(x).Note that if you consider f as a function of period2 equal to exp(x) in]0,2[,it has a jump discontinuity at any point x=2n(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 limitat such points of discontinuity,in particular the value at point 0 is[exp(2)+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 2 is thus (exp(2)-1)(1-ki)/(1+k2). The Fourier coefficients are obtained by multiplying this by 1/ and using the real and imaginary part separately. All told:
f(x) =[exp(2)-1]/(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(2)+1] / 2 = [exp(2)-1] / ( ½ + S )
where S is the sum we were after. Therefore:
S = /2 - ½ + /[ exp(2)-1 ] = 1.076674047...
That's also a special case (a = 1) of the relation obtainedabove in the form:
coth() = 1 + 2 S
brentw (Brent Watts of Hickory, NC.2000-11-28) [...] Please explain the Gibbs phenomenon of Fourier series.
At a point x where a function f has a jump discontinuity, any partial sum ofits 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 jumpJ=|f(x+)-f(x-)|.
This amplitude isnot reduced by adding more terms of the Fourier series.It's not difficult to prove that, with n terms, the maximum value of the overshootoccurs at/near a distance of /2n on either side of x.(You may do the computation with any convenient function having a jump J;I suggestf(x)=sign(x)J/2 between- and .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 overshootoscillation is about 8.948987% of the jump J.This value is precisely(2G/-1)/2,where G is known as theWilbraham-Gibbs Constant:
G
=
sin d
=
1.8519370519824661703610533701579913633458...
This is sometimes called "the 9% overshoot",as it is about 9% of thetotal 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 seriesabout a discontinuity of f.For a small h, you can always increase the number of Fourier terms so that Gibbsoscillations 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...
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.
If the function f has a single singularity q betweena andb(a<b),the Cauchy principal value of its integral froma tob is the limit(whenever it exists),as tends to 0+, of the sum of the integral froma to q-and the integral from q+ tob. Also, if the intervalof integration is with asingularity 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 betweena andb(a andb excluded, unless both are infinite),the PV of its integral may be obtained bysplitting the interval [a,b] into a sequence of intervals each containinga single singularity. The above applies to each of these, and the PV of the integralover the entire interval is obtained by adding the principal values over all suchsubintervals.
The fact that the principal value is used may be indicated by the letters PVbefore 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 principalvalue is used whenever needed, and some authors don't bother to insist on thiswith special typography.
brentw (Brent Watts of Hickory, NC.2000-11-21) How do I solve the differential equation2(1-x)y" + (1+x)y' + [x - 3 - (x-1)exp(x)]y = 0about the pole x=1?
The singularity at x = 1 is a regularFuchsian one (which means that, if the coefficientof y" is normalized to 1, then the coefficient of y' has at most a single poleat 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 = h [ a(0) + a(1) h + a(2) h + a(3) h + ... ]
In the above, m is not necessarily an integer, so that aFrobenius seriesis more general than either aTaylor series (for which m is a natural integer)or aLaurent series (for which m is any integer).In the DE we're asked to study, we have:
-2h y" + (2+h) y' + [h-2-hexp(1+h)] y = 0
Themethod of Frobenius is simply to expand the above LHS in termsof powers of h to obtain a sequence of equations that will successively givethe values of a=a(0), b=a(1), c=a(2), d=a(3), etc.Let's do it.The above LHS is h multiplied by:
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 theindex m (this is a general feature of the method and this firstcritical equation is called theindicial 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 towork 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 theusual problem you encounter after N stepswhen 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 cand 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 solutionas a Frobenius seriesF(h),then a=0 and c=1 to get a linearly independent solutionG(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 thisparticular case.)
brentw (Brent Watts of Hickory, NC.2000-11-21) How to determine the Laurent series of a function about a singular point.
For each singular point (orpole) z,you want to expandf (z+h). If a pole has multiplicity n, thenh f (z+h)is normally an analytic function. Compute itsTaylor expansion about h=0 and divide that seriesby hto 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 h2 to obtain an analytic function about h = 0, namely:
g(h) = h2f (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 expansionof 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 + ...
i times the sum of the residues for those poles.This can be a practical way to computeeasily manydefinite integrals that would otherwise be difficult to obtain. Examplesfollow.
Many textbook insist on the following format (which I don't recommend):
It's sometimes desirable to present the function globallyas the sum of so-called simple elements about every poleand an entire function (which is just a polynomial whenthe function is rational,like this one).
f (z) = 1/(z-1)2 1/(z-1) + 1/z + 0
That type of reduction allows the immediateintegrationof rational functions:
f (z) dz = -1/(z-1) ln(z-1) + ln(z)
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 applyCauchy's residue theorem[the fundamental theorem of complex analysis] to multivalued functions(like the square root function involved here), it is important to specifya so-called "cut" in the complex plane were the function is allowed to be discontinuous,so that it iseverywhere else continuous and single-valued.
In the case of the square-root function, it is not possible to give a continuous definitionvalid aroundany path encircling the origin.Therefore, a so-called "branch-cut" linemust 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 is in[0,2[, the "square root" of the complex numberz = r exp(i) is simplyz r exp(i/2) (the notation r being unambiguous because r is a positivereal number and its square root is thusdefined as the onlypositivereal 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 ofRiemann surfaces,which are beyond the scope of our current discussion).
With the above definition of the square root of a complex argument,we may thus apply theResidue Theorem to the functionf(z)=1/(1+z)z on any contour which does not cross thepositive 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 smallerthan the reciprocal of thelength of the path of integration.Therefore, the contribution of the outer semicircle to thecontour 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 infinitelysmall 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 integralwe seek.
Cauchy's Theorem states that the contour integral equals2i 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 tozero ofh f(i+h), namely1/2ii,so the value of the contour integral isi = 2(1+i)/2.As stated above, this is (1+i) times the integral we want.Therefore, the value of that integral is exactly2,or about 2.22144146907918312350794049503...
For what values of does this integral converge? What's the value of the integral when it converges?
Theprevious article deals with the special case = -1/2. In general, we see that the integral makes sense in the neighborhood of zeroif-1 and it converges in the neighborhood of when 1.All told, the integral converges when is in the open interval ]-1,1[.
We apply the above method tof(z) = z/(1+z2)[defining z with the positive x-axis asbranch cut] on the same contour (pictured at right).The smaller semicircle is useless when is positive andit has a vanishing contribution otherwise(when-1).The contribution of the outer semicircle is vanishingly small also(when 1) because|f(z)| multiplied by thelength 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 integralwe are after, whereas the negative part of the axis contributesexp(i) as much.All told therefore, Cauchy's theorem tells us that our integral is2i/(1+exp(i))times the residue of f at the pole z = i.
The residue at z = i is the limit, as h tends to zero,ofh f(i+h), which is simplyexp(i)/2i.This makes the integral equal to
exp(i) / (1+exp(i))
That boils down to /(2 cos /2) so that we obtain:
x dx
=
[ -1 1 ]
0
1 + x2
2 cos (/2)
We may notice that this final result is an even function of ,which we could have predicted with the simple change of variable y = 1/x...
(2007-05-08) Holomorphic functions, entire functions, meromorphic functions, etc.
Holomorphic Functions (analytic functionsof a complex variable)
A complex function f of a complex variablewhich is differentiable about every point z of its domainis called ananalytic function,anholomorphic function,or a complex differentiable function :
f (z+h) f (z)
= f ' (z)
h 0
h
The existence of such a derivative function ( f ' ) is amuch more restrictive condition in the complex realm thanamong real numbers, since it implies essentially that f is differentiable infinitely many times andpossesses a convergent expansion as a Taylor series about every point x inside its domain of definition.
In this, both u and v are real functions of the two real variables x and y. The above differentiability of f as afunction 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.
if / 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 thatboth u and v are necessarily differentiable infinitely many timeswith respect to x and y as soon as they are known to be differentiableonce).
An holomorphic function defined over the whole complex plane (without any singularities) is called an entire function. Liouville's Boundedness Theorem asserts that such a function can onlybe bounded if it's constant.
Meromorphic Functions
A meromorphic functionf over some domain Dis an holomorphic function on the domain obtained by removingfrom 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 existswhere the function f (z) multiplied by some power of (z-x) is simply an holomorphicfunction (i.e., without singularities).
Etymologically,meros( = part) is opposed toholos( = whole).
(2021-06-22) 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 preservesangles and planar orientation). This turns out to be a necessry condition for the analyticity of the function f.
As advertised above, the existence of a (first) derivative with respect toa 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 thecomplex 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 thatthe following limit is well-defined as the (complex) quantity h tends to zero.
f (z+h) f (z)
= f ' (z)
h 0
h
This implies, in particular, that the limit exists when h=x is real and when h=iyis imaginary, which says that both u and v have partial derivatives with respectto either x or y. Furthermore, the limits for h=x and h=iy as eitherx or y tends to zero must be identical (or else, there would not be awell-defined limit in the neighborhood of h=0) which implies the following:
Cauchy-Riemann Equations
u
=
v
and
v
=
u
x
y
x
y
Contrary to Riemann's belief, the converse isn't quite true as the existence of partial derivativessatisfying the Cauchy-Riemann equations doesn't implythat 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 0 (and f = 0 when y = 0)
y
When the partial derivatives are continuous (which isn't the case in the above example) a theorem due toClairault (1740) applies, which states that the values ofpartial 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. (: The continuous function z5/|z|4 isn't analytic anywhere.)
Generalizations of the Looman-Menchoff theorem have been devised, culminating in the detailed results published in Russian by GrigoriiKhaimovich Sindalovskii (1928-2020) in 1985:
(2021-06-22) 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. Usingpartial 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
=
i
and
=
+ i
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) + iy(u+iv) = (xuyv) + i(xv+yu)
Thus, loosely speaking, an analytic function depends only on z, not z*.
The first Wirtinger derivative coincides with the ordinary derivativein 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 theLog 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
=
i
and
=
+ i
zk
xk
yk
zk*
xk
yk
(2021-08-09) Fundamental Theorem of Complex Analysis.
If f is holomorphic, in a simply-connected neighborhood of point a, then itsvalue at point a is given by an integral around any contour C encircling a within that neighborhood:
f (a) =
1 2i
C
f (z) z-a
dz
Differentiate both sides n times with respect to a to obtain a corollary :
f(n)(a) =
n! 2i
C
f (z) (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 contourcontributes to the integral exactly the half-sum of what it would if it was on either side!
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 toinfer 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.
(2021-08-09) 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.
(2019-08-23) Existence of a biholomorphic mapping from U to the open unit disk.
A biholomorphic mapping is a bijectiveholomorphic mappingwhose inverse is also holomophic. Such a mapping betweenthe 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 propersubset U of . That statement is now known as the Riemann mapping theorem.
(2021-08-05) 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) 0 :
f (z) =
g (k z) g (0) 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) =
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 (1 q z) 2
=
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 notpossible when u and v are inside the unit disk, since the modulus of q2 u v is then strictly less than unity.
Injectivity does fail everywhere on the boundary. [: 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).
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 sufficientto establish Bieberbach's conjecture, a fact which would be put togood 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 p such that:
p(z) p = f (z p)
Proof : Let's definep using the binomial theorem for exponent 1/p in a smallenough neighborhood of zero, in the following way:
p(z) =
b pk+1 z pk+1
= z
1/p k
y k
Where y =
f (z p) z p
1 = z p
a n z p(n-2)
The other possible solutions would be deduced from that one by multiplying it into a nontrivialp-th root of unity, , but this would make the derivative atthe origin be 1, which is ruled outfor 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 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 p so defined is injective: p(u) = p(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 p to get p (u) / p (v) = v/u, so that the former ratio can only be unity if the latter is.
Apply this result, for any positive integer p, to our previous family parametrized by q = r exp(i ), with 0 < r ≤ 1, to obtain a 3-parameter family of schlicht functions (2 continuous parameters and a discrete one):
f(z) =
z (1 r e i z p) 2
=
n r n-1 e i (n-1) z p (n-1)+1
The basic Koebe function is retrieved for r = 1, = 0 and p = 1.
Bieberbach's conjecture = De Branges's theorem (1985)
(2021-08-08). (Nevanlinna, 1921) Bieberbach's conjecture was first proved in this special case.
A 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].
(2021-07-27) The modulus in a compact region is largest on the boundary.
Thus, for any holomorphic function f ofa complex variable z, | f (z) | can't have a strict local maximumanywhere 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:
Consider, for example, the sequence of functions gn on [0,1]for which gn(x) is n2x when x is in [0,1/n],
With the above definition of the square root of a complex argument,we may thus apply theResidue Theorem to the function
. That statement is now known as the Riemann mapping theorem.