Power series which diverges precisely at the roots of unity, converges elsewhere
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Is there a complex power series $sum a_nz^n$ with radius of convergence $1$ which diverges at the roots of unity (e.g., $z=e^{2pi itheta}$, $theta in mathbb{Q}$) and converges elsewhere on the unit circle ($z=e^{2pi itheta}$, $theta in mathbb{R} setminus mathbb{Q}$)?
I know $sum frac{z^n}{n}$ is a series with radius of convergence $1$ which converges everywhere on the unit circle except $1$. Perhaps we can play around with this to get the desired result.
I also know that $sum frac{z^{n!}}{n}$ diverges at the roots of unity, but I am not aware of a result that it converges at all other points on the unit circle.
Note similar questions have been asked here before, but they do not directly answer the question posed above.
complex-analysis convergence power-series conditional-convergence
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up vote
13
down vote
favorite
Is there a complex power series $sum a_nz^n$ with radius of convergence $1$ which diverges at the roots of unity (e.g., $z=e^{2pi itheta}$, $theta in mathbb{Q}$) and converges elsewhere on the unit circle ($z=e^{2pi itheta}$, $theta in mathbb{R} setminus mathbb{Q}$)?
I know $sum frac{z^n}{n}$ is a series with radius of convergence $1$ which converges everywhere on the unit circle except $1$. Perhaps we can play around with this to get the desired result.
I also know that $sum frac{z^{n!}}{n}$ diverges at the roots of unity, but I am not aware of a result that it converges at all other points on the unit circle.
Note similar questions have been asked here before, but they do not directly answer the question posed above.
complex-analysis convergence power-series conditional-convergence
For "unit disc" read "unit circle"?
– Lord Shark the Unknown
Nov 24 at 5:04
@LordSharktheUnknown Yes, thank you.
– MathematicsStudent1122
Nov 24 at 5:05
2
$sum frac{z^{n!}}{n}$ diverges not only for roots of unity, but at some other places too. Consider $z=exp(2pi i e)$ (not a root of unity, since $e$ is irrational). Then $z^{n!}=exp(2pi i e n!)$. We don't care about the integer part of $(e n!)$, only the fractional part. And ${e n!}=frac{1}{n+1}+frac{1}{(n+1)(n+2)}+cdots<frac{1}{n+1}+frac{1}{(n+1)^2}+cdots=frac{1}{n}$ converges to $0$. So the terms of $sum frac{z^{n!}}{n}$ will accumulate close to $1$, and the series will diverge.
– alex.jordan
Nov 26 at 6:32
A late addition to my comment above: a similar argument shows that series diverges for $z=exp(2pi i(e+r))$ for any rational $r$.
– alex.jordan
Nov 26 at 21:46
@MathematicsStudent1122 why did you already award the bounty? I gave an explicit example
– mathworker21
Nov 29 at 5:00
add a comment |
up vote
13
down vote
favorite
up vote
13
down vote
favorite
Is there a complex power series $sum a_nz^n$ with radius of convergence $1$ which diverges at the roots of unity (e.g., $z=e^{2pi itheta}$, $theta in mathbb{Q}$) and converges elsewhere on the unit circle ($z=e^{2pi itheta}$, $theta in mathbb{R} setminus mathbb{Q}$)?
I know $sum frac{z^n}{n}$ is a series with radius of convergence $1$ which converges everywhere on the unit circle except $1$. Perhaps we can play around with this to get the desired result.
I also know that $sum frac{z^{n!}}{n}$ diverges at the roots of unity, but I am not aware of a result that it converges at all other points on the unit circle.
Note similar questions have been asked here before, but they do not directly answer the question posed above.
complex-analysis convergence power-series conditional-convergence
Is there a complex power series $sum a_nz^n$ with radius of convergence $1$ which diverges at the roots of unity (e.g., $z=e^{2pi itheta}$, $theta in mathbb{Q}$) and converges elsewhere on the unit circle ($z=e^{2pi itheta}$, $theta in mathbb{R} setminus mathbb{Q}$)?
I know $sum frac{z^n}{n}$ is a series with radius of convergence $1$ which converges everywhere on the unit circle except $1$. Perhaps we can play around with this to get the desired result.
I also know that $sum frac{z^{n!}}{n}$ diverges at the roots of unity, but I am not aware of a result that it converges at all other points on the unit circle.
Note similar questions have been asked here before, but they do not directly answer the question posed above.
complex-analysis convergence power-series conditional-convergence
complex-analysis convergence power-series conditional-convergence
edited Nov 29 at 0:21
asked Nov 24 at 4:58
MathematicsStudent1122
7,71622263
7,71622263
For "unit disc" read "unit circle"?
– Lord Shark the Unknown
Nov 24 at 5:04
@LordSharktheUnknown Yes, thank you.
– MathematicsStudent1122
Nov 24 at 5:05
2
$sum frac{z^{n!}}{n}$ diverges not only for roots of unity, but at some other places too. Consider $z=exp(2pi i e)$ (not a root of unity, since $e$ is irrational). Then $z^{n!}=exp(2pi i e n!)$. We don't care about the integer part of $(e n!)$, only the fractional part. And ${e n!}=frac{1}{n+1}+frac{1}{(n+1)(n+2)}+cdots<frac{1}{n+1}+frac{1}{(n+1)^2}+cdots=frac{1}{n}$ converges to $0$. So the terms of $sum frac{z^{n!}}{n}$ will accumulate close to $1$, and the series will diverge.
– alex.jordan
Nov 26 at 6:32
A late addition to my comment above: a similar argument shows that series diverges for $z=exp(2pi i(e+r))$ for any rational $r$.
– alex.jordan
Nov 26 at 21:46
@MathematicsStudent1122 why did you already award the bounty? I gave an explicit example
– mathworker21
Nov 29 at 5:00
add a comment |
For "unit disc" read "unit circle"?
– Lord Shark the Unknown
Nov 24 at 5:04
@LordSharktheUnknown Yes, thank you.
– MathematicsStudent1122
Nov 24 at 5:05
2
$sum frac{z^{n!}}{n}$ diverges not only for roots of unity, but at some other places too. Consider $z=exp(2pi i e)$ (not a root of unity, since $e$ is irrational). Then $z^{n!}=exp(2pi i e n!)$. We don't care about the integer part of $(e n!)$, only the fractional part. And ${e n!}=frac{1}{n+1}+frac{1}{(n+1)(n+2)}+cdots<frac{1}{n+1}+frac{1}{(n+1)^2}+cdots=frac{1}{n}$ converges to $0$. So the terms of $sum frac{z^{n!}}{n}$ will accumulate close to $1$, and the series will diverge.
– alex.jordan
Nov 26 at 6:32
A late addition to my comment above: a similar argument shows that series diverges for $z=exp(2pi i(e+r))$ for any rational $r$.
– alex.jordan
Nov 26 at 21:46
@MathematicsStudent1122 why did you already award the bounty? I gave an explicit example
– mathworker21
Nov 29 at 5:00
For "unit disc" read "unit circle"?
– Lord Shark the Unknown
Nov 24 at 5:04
For "unit disc" read "unit circle"?
– Lord Shark the Unknown
Nov 24 at 5:04
@LordSharktheUnknown Yes, thank you.
– MathematicsStudent1122
Nov 24 at 5:05
@LordSharktheUnknown Yes, thank you.
– MathematicsStudent1122
Nov 24 at 5:05
2
2
$sum frac{z^{n!}}{n}$ diverges not only for roots of unity, but at some other places too. Consider $z=exp(2pi i e)$ (not a root of unity, since $e$ is irrational). Then $z^{n!}=exp(2pi i e n!)$. We don't care about the integer part of $(e n!)$, only the fractional part. And ${e n!}=frac{1}{n+1}+frac{1}{(n+1)(n+2)}+cdots<frac{1}{n+1}+frac{1}{(n+1)^2}+cdots=frac{1}{n}$ converges to $0$. So the terms of $sum frac{z^{n!}}{n}$ will accumulate close to $1$, and the series will diverge.
– alex.jordan
Nov 26 at 6:32
$sum frac{z^{n!}}{n}$ diverges not only for roots of unity, but at some other places too. Consider $z=exp(2pi i e)$ (not a root of unity, since $e$ is irrational). Then $z^{n!}=exp(2pi i e n!)$. We don't care about the integer part of $(e n!)$, only the fractional part. And ${e n!}=frac{1}{n+1}+frac{1}{(n+1)(n+2)}+cdots<frac{1}{n+1}+frac{1}{(n+1)^2}+cdots=frac{1}{n}$ converges to $0$. So the terms of $sum frac{z^{n!}}{n}$ will accumulate close to $1$, and the series will diverge.
– alex.jordan
Nov 26 at 6:32
A late addition to my comment above: a similar argument shows that series diverges for $z=exp(2pi i(e+r))$ for any rational $r$.
– alex.jordan
Nov 26 at 21:46
A late addition to my comment above: a similar argument shows that series diverges for $z=exp(2pi i(e+r))$ for any rational $r$.
– alex.jordan
Nov 26 at 21:46
@MathematicsStudent1122 why did you already award the bounty? I gave an explicit example
– mathworker21
Nov 29 at 5:00
@MathematicsStudent1122 why did you already award the bounty? I gave an explicit example
– mathworker21
Nov 29 at 5:00
add a comment |
2 Answers
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up vote
6
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accepted
Because your question asks whether this is possible when the set of divergence is a certain countable set, the answer is YES by Theorem 1 of [2] below.
Results including and related to what you’ve asked are discussed in the following Stack Exchange questions:
Behaviour of power series on their circle of convergence
Examples of Taylor series with interesting convergence along the boundary of convergence?
Complex power series divergent and convergent on dense subsets of the circumference of convergence?
Power series with funny behavior at the boundary
What are the subsets of the unit circle that can be the points in which a power series is convergent?
Because the Duke Mathematical Journal papers by Herzog/Piranian mentioned in some of the above are not freely available, I’ve included some relevant excerpts from them. Incidentally, I have not bothered to include excerpts from [3] because it is freely available.
In these excerpts, additional notes by me are enclosed in double square brackets [[ ... ]].
[1] Fritz Herzog and George Piranian, Sets of convergence of Taylor series I, Duke Mathematical Journal 16 #3 (September 1949), 529-534. MR 11,91f; Zbl 34.04806
Introduction (p. 529): Let $sum_n a_n z_n$ be a Taylor series of radius of convergence one, with $sum_n |a_n| = infty$ and $lim_n a_n = 0.$ We consider the point set $M$ on the unit circle $C,$ on which the series converges. As Landau [2; 13-14] [[= Landau (1929, Darstellung und Begründung einiger neuerer Ergebnisse der Funktionentheorie, 2nd edition)]] points out, the cardinal number of the set of such Taylor series is $mathfrak c$ [[at least $mathfrak c$ --- add any constant to a specific such Taylor series; at most $mathfrak c$ --- every such Taylor series represents a continuous function and there are $mathfrak c$ many continuous functions]], while the cardinal number of the set of subsets of $C$ is $mathfrak f$ [[= $2^{mathfrak c}$]]; hence, there exists a set $M$ on $C$ such that no Taylor series converges on $M$ and diverges on $C-M.$ It follows that if a set $M$ on $C$ is such that some Taylor series converges on $M$ and diverges on $C-M,$ the set must have certain special properties. Lusin [3] [[= Luzin (1911, Über eine potenzreihe)]] (see also Landau [2; 69-71]) has constructed a Taylor series whose coefficients tend to zero and which diverges on the entire unit circle $C.$ Sierpiński (see Landau [2; 71]) has modified Lusin's example to obtain divergence at all points of $C$ except one. For every closed arc $A$ on $C,$ Neder [5] [[= Neder (1919, Zur Konvergenz der Trigonometrischen Reihen, einschliesslich der Potenzreihen auf dem Konvergenzkreise, Ph.D. Dissertation)]] constructed a Taylor series which converges on $C-A$ and whose partial sums are unbounded at every point of $A.$ Mazurkiewicz [4] [[= Mazurkiewicz (1922, Sur les séries de puissances)]] used Neder's example to prove the following proposition: If $M$ is a closed set on $C,$ there exists a Taylor series which converges on $M$ and diverges on $C-M,$ and a Taylor series which diverges on $M$ and converges on $C-M.$ The present paper is devoted to the extension of these results. Its method is inspired by Lusin's example. Theorem 1 (p. 531): If $M$ is a set of type $F_{sigma}$ on the unit circle $C$ there exists a Taylor series which converges everywhere in $M$ and diverges everywhere in $C-M.$ Theorem 2 (p. 532): If $F$ is a closed set on the unit circle $C,$ there exists a Taylor series which converges uniformly in $F$ and diverges everywhere in $C-F.$ Theorem 3 (p. 533): If $M$ is a set on the unit circle $C,$ such that some Taylor series converges uniformly on $M$ and diverges on $C-M,$ then $M$ is a closed set. Theorem 4 (p. 533): If $M$ is a set on the unit circle $C,$ such that some Taylor series converges on $M$ and diverges on $C-M,$ then $M$ is of type $F_{sigma delta}.$ (last paragraph of the paper, at bottom of p. 533) [[References are on p. 534]] The "no-man's-land" between Theorems 1 and 4 is considerable; it consists of all sets of type $F_{sigma delta}$ on $C$ that are not of type $F_{sigma}.$ It is not known whether every set of type $F_{sigma delta}$ is the set of convergence of some Taylor series. However, if $M$ is the complement on the unit circle of an arbitrary denumerable set, there exists a Taylor series for which $M$ is the set of convergence (the construction of such a series will be described in a later paper); this implies that not every set of convergence of a Taylor series is of type $F_{sigma}.$ [[Note that if $D$ is a set that is dense in $C$ and we let $M = C-D,$ then $M$ is such a set. This follows from the Baire category theorem in the same way that one usually shows that the set of irrational numbers in $mathbb R$ is not an $F_{sigma}$ set.]]
[2] Fritz Herzog and George Piranian, Sets of convergence of Taylor series. II, Duke Mathematical Journal 20 #1 (March 1953), 41-54. MR 14,738b; Zbl 50.07802
(first two sentences of the paper, on p. 41) In an earlier paper [7] the authors have shown that if $M$ is a set of type $F_{sigma}$ on the unit circle $C,$ there exists a Taylor series which has $M$ as its set of convergence, i.e., which converges on $M$ and diverges on $C-M.$ The main purpose of the present paper is to exhibit Taylor series whose sets of convergence are not of type $F_{sigma}.$ Theorem 1 (p. 45): If $M$ is a denumerable set on the unit circle $C,$ there exists a function $f(z)$ with the following properties: (i) $f(z)$ is schlicht in the region $|z| < 1;$ (ii) the Taylor series of $f(z)$ diverges on $M$ and converges on $C-M;$ (iii) the partial sums of the Taylor series of $f(z)$ are uniformly bounded on $C;$ (iv) the set of vertices of the Mittag-Leffler star of $f(z)$ consists of the set $overline{M}.$ Theorem 2 (p. 48): If $M$ is a denumerable set on $C,$ and $N$ is a subset of $M,$ then there exists a function $f(z)$ with the following properties: (i) the Taylor series of $f(z)$ converges on $C-M,$ diverges on $M,$ and has uniformly bounded partial sums; (ii) $lim_{r rightarrow 1}f(re^{i theta})$ exists when $e^{i theta}$ lies in $C-N,$ does not exist when $e^{i theta}$ lies in $N.$ (from p. 50) In [7] it was proved that every set of type $F_{sigma}$ on $C$ is the set of convergence of some Taylor series. Together with the results established so far in this paper, this gives the result that every denumerable set on $C$ is both a set of convergence and a set of divergence. Since the construction in [7] can easily be modified to introduce arbitrarily large gaps, we also obtain the result that if $M$ is the union of a denumerable set and a set of type $G_{delta}$ on $C,$ then $M$ is the set of divergence of some Taylor series. As far as the general theory of sets of convergence is concerned, the following [[= Theorem 3]] is the most important conclusion that can be drawn from Theorems 1 and 2: Theorem 3 (p. 50): Not every set of convergence of Taylor series is of type $F_{sigma}.$ (first two sentences after Theorem 3 statement, on p. 50) This follows at once from the fact that if $M$ is a denumerable set, dense on $C,$ then $M$ is not of type $G_{delta}$ (see [6; 138]). It should be remarked that Theorem 3 can be deduced from Fejér’s work and should therefore have been included in [7]. Theorem 4 (p. 50): There exists a function $f(z)$ which is holomorphic, bounded and schlicht in $|z| < 1$ and whose Taylor series diverges on a set which is locally non-denumerable on $C.$ [["Locally non-denumerable on $C$" means that the set has uncountable intersection with every arc of $C.$ Since the sets in question are Borel sets, and uncountable Borel sets have cardinality $mathfrak c,$ it follows that "locally non-denumerable on $C$" can be replaced with "$mathfrak c$-dense in $C$".]] Theorem 5 (p. 51): There exists a function which is holomorphic in $|z| < 1$ and continuous in $|z| leq 1$ and whose Taylor series diverges on a set which is locally non-denumerable on $C.$ Theorem 6 (p. 51): There exists a function $f(z),$ bounded and schlicht in $|z| < 1,$ whose Taylor series converges everywhere on $C,$ but not uniformly on any arc of $C.$ Theorem 7 (p. 51): There exists a function $f(z)$ which is continuous in $|z| leq 1$ and whose Taylor series converges everywhere on $C,$ but not uniformly on any arc of $C.$ Definition (p. 52): If $f(z)$ is defined everywhere on $C,$ a point $z_0$ on $C$ lies in the set of boundedness of $f(z)$ provided $f(z)$ is bounded on some open arc of $C$ which contains the point $z_{0}.$ The set of unboundedness of $f(z)$ is the complement [[relative to $C$]] of the set of boundedness. Theorem 8 (p. 52): A necessary and sufficient condition for a set $M$ on $C$ to be the set of unboundedness of some Taylor series converging everywhere on $C$ is that $M$ be closed and nowhere dense on $C.$ [[I suspect that requiring the behavior to not just be unbounded at a point, but to actually approach infinity at the point (i.e. being locally unbounded and finite subsequent limits are not allowed) will be characterized by being a scattered set, as is the case at Is there a function whose all limits at rational points approach infinity? (see my comments there).]]
[3] Fritz Herzog and George Piranian, Some point sets associated with Taylor series, Michigan Mathematical Journal 3 #1 (1955-1956), 69-75. MR 17,834a; Zbl 70.29501
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Let $p_1 < p_2 < dots$ be the primes, and let $a_n = 0$ if $n$ is not prime and $a_n = frac{1}{k}$ for $n=p_k^2$.
Take any $frac{l}{q} in mathbb{Q}$. Note, by the prime number theorem for arithmetic progressions, that $$lim_{K to infty} frac{1}{K}sum_{k le K} e^{2pi i p_k^2frac{l}{q}} = frac{1}{phi(q)}sum_{substack{a le q \ (a,q) = 1}} e^{2pi i a^2 frac{l}{q}}.$$ By a standard summation by parts argument, it follows that $$lim_{K to infty} frac{1}{log K} sum_{k le K} frac{1}{k}e^{2pi i p_k^2frac{l}{q}} = frac{1}{phi(q)}sum_{substack{a le q \ (a,q) = 1}} e^{2pi i a^2 frac{l}{q}}.$$ Therefore, if we show that the sum $sum_{a le q \ (a,q) = 1} e^{2pi i a^2frac{l}{q}}$ is nonzero for any $q ge 1$ and $(l,q) = 1$, it follows that $sum_{k le K} frac{1}{k} e^{2pi i p_k^2frac{l}{q}}$ blows up in magnitude and in particular diverges. We show that the sum is nonzero below the fold.
Now take any $alpha in mathbb{R}setminusmathbb{Q}$. It's well known that $$lim_{K to infty} sum_{k le K} frac{1}{k}e^{2pi i p_k^2alpha}$$ exists, which clearly implies that $sum_{n le N} a_ne^{2pi i alpha n}$ converges as $N to infty$. This completes the argument.
--------------------------------------------------------------
Using the identity $sum_{d mid n} mu(d) = 1_{n=1}$, we see $$sum_{a le q \ (a,q) = 1} e^{2pi i a^2frac{l}{q}} = sum_{a le q} e^{2pi i a^2frac{l}{q}}sum_{d mid (a,q)} mu(d) = sum_{a le q} e^{2pi i a^2frac{l}{q}} sum_{substack{d mid a \ d mid q}} mu(d)$$ $$ = sum_{d mid q} mu(d) sum_{d mid a le q} e^{2pi i a^2frac{l}{q}} = sum_{d mid q} mu(d)sum_{j le q/d} e^{2pi i j^2frac{dl}{q/d}}$$ where the last equality was obtained by substituting $a = jd$. We have a linear combination of quadratic Gauss sums. Each $sum e^{2pi i j^2frac{dl}{q/d}}$ is either $0$ or $csqrt{q/d}$ for $c in {pm 1, pm i, pm (1+i), pm i(1+i)}$. If $q not equiv 2 pmod{4}$, then the term corresponding to $d=1$ is $csqrt{q}$ which cannot cancel with any of the other terms and we're done. If $q equiv 2 pmod{4}$, then the term $d=2$ will give a $-csqrt{q/2}$ term which cannot cancel with any of the other terms and we're done.
It's well known that $lim_{Ktoinfty}sum_{kle K}frac 1k e^{2pi ip^2kalpha}=0$. Really? Doesn't it bother you that irrationals have full measure and the series is a non-trivial Fourier series of an $L^2$-function?
– fedja
Nov 28 at 23:24
@fedja whoops. I meant exists. edited
– mathworker21
Nov 29 at 4:54
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2 Answers
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2 Answers
2
active
oldest
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active
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votes
active
oldest
votes
up vote
6
down vote
accepted
Because your question asks whether this is possible when the set of divergence is a certain countable set, the answer is YES by Theorem 1 of [2] below.
Results including and related to what you’ve asked are discussed in the following Stack Exchange questions:
Behaviour of power series on their circle of convergence
Examples of Taylor series with interesting convergence along the boundary of convergence?
Complex power series divergent and convergent on dense subsets of the circumference of convergence?
Power series with funny behavior at the boundary
What are the subsets of the unit circle that can be the points in which a power series is convergent?
Because the Duke Mathematical Journal papers by Herzog/Piranian mentioned in some of the above are not freely available, I’ve included some relevant excerpts from them. Incidentally, I have not bothered to include excerpts from [3] because it is freely available.
In these excerpts, additional notes by me are enclosed in double square brackets [[ ... ]].
[1] Fritz Herzog and George Piranian, Sets of convergence of Taylor series I, Duke Mathematical Journal 16 #3 (September 1949), 529-534. MR 11,91f; Zbl 34.04806
Introduction (p. 529): Let $sum_n a_n z_n$ be a Taylor series of radius of convergence one, with $sum_n |a_n| = infty$ and $lim_n a_n = 0.$ We consider the point set $M$ on the unit circle $C,$ on which the series converges. As Landau [2; 13-14] [[= Landau (1929, Darstellung und Begründung einiger neuerer Ergebnisse der Funktionentheorie, 2nd edition)]] points out, the cardinal number of the set of such Taylor series is $mathfrak c$ [[at least $mathfrak c$ --- add any constant to a specific such Taylor series; at most $mathfrak c$ --- every such Taylor series represents a continuous function and there are $mathfrak c$ many continuous functions]], while the cardinal number of the set of subsets of $C$ is $mathfrak f$ [[= $2^{mathfrak c}$]]; hence, there exists a set $M$ on $C$ such that no Taylor series converges on $M$ and diverges on $C-M.$ It follows that if a set $M$ on $C$ is such that some Taylor series converges on $M$ and diverges on $C-M,$ the set must have certain special properties. Lusin [3] [[= Luzin (1911, Über eine potenzreihe)]] (see also Landau [2; 69-71]) has constructed a Taylor series whose coefficients tend to zero and which diverges on the entire unit circle $C.$ Sierpiński (see Landau [2; 71]) has modified Lusin's example to obtain divergence at all points of $C$ except one. For every closed arc $A$ on $C,$ Neder [5] [[= Neder (1919, Zur Konvergenz der Trigonometrischen Reihen, einschliesslich der Potenzreihen auf dem Konvergenzkreise, Ph.D. Dissertation)]] constructed a Taylor series which converges on $C-A$ and whose partial sums are unbounded at every point of $A.$ Mazurkiewicz [4] [[= Mazurkiewicz (1922, Sur les séries de puissances)]] used Neder's example to prove the following proposition: If $M$ is a closed set on $C,$ there exists a Taylor series which converges on $M$ and diverges on $C-M,$ and a Taylor series which diverges on $M$ and converges on $C-M.$ The present paper is devoted to the extension of these results. Its method is inspired by Lusin's example. Theorem 1 (p. 531): If $M$ is a set of type $F_{sigma}$ on the unit circle $C$ there exists a Taylor series which converges everywhere in $M$ and diverges everywhere in $C-M.$ Theorem 2 (p. 532): If $F$ is a closed set on the unit circle $C,$ there exists a Taylor series which converges uniformly in $F$ and diverges everywhere in $C-F.$ Theorem 3 (p. 533): If $M$ is a set on the unit circle $C,$ such that some Taylor series converges uniformly on $M$ and diverges on $C-M,$ then $M$ is a closed set. Theorem 4 (p. 533): If $M$ is a set on the unit circle $C,$ such that some Taylor series converges on $M$ and diverges on $C-M,$ then $M$ is of type $F_{sigma delta}.$ (last paragraph of the paper, at bottom of p. 533) [[References are on p. 534]] The "no-man's-land" between Theorems 1 and 4 is considerable; it consists of all sets of type $F_{sigma delta}$ on $C$ that are not of type $F_{sigma}.$ It is not known whether every set of type $F_{sigma delta}$ is the set of convergence of some Taylor series. However, if $M$ is the complement on the unit circle of an arbitrary denumerable set, there exists a Taylor series for which $M$ is the set of convergence (the construction of such a series will be described in a later paper); this implies that not every set of convergence of a Taylor series is of type $F_{sigma}.$ [[Note that if $D$ is a set that is dense in $C$ and we let $M = C-D,$ then $M$ is such a set. This follows from the Baire category theorem in the same way that one usually shows that the set of irrational numbers in $mathbb R$ is not an $F_{sigma}$ set.]]
[2] Fritz Herzog and George Piranian, Sets of convergence of Taylor series. II, Duke Mathematical Journal 20 #1 (March 1953), 41-54. MR 14,738b; Zbl 50.07802
(first two sentences of the paper, on p. 41) In an earlier paper [7] the authors have shown that if $M$ is a set of type $F_{sigma}$ on the unit circle $C,$ there exists a Taylor series which has $M$ as its set of convergence, i.e., which converges on $M$ and diverges on $C-M.$ The main purpose of the present paper is to exhibit Taylor series whose sets of convergence are not of type $F_{sigma}.$ Theorem 1 (p. 45): If $M$ is a denumerable set on the unit circle $C,$ there exists a function $f(z)$ with the following properties: (i) $f(z)$ is schlicht in the region $|z| < 1;$ (ii) the Taylor series of $f(z)$ diverges on $M$ and converges on $C-M;$ (iii) the partial sums of the Taylor series of $f(z)$ are uniformly bounded on $C;$ (iv) the set of vertices of the Mittag-Leffler star of $f(z)$ consists of the set $overline{M}.$ Theorem 2 (p. 48): If $M$ is a denumerable set on $C,$ and $N$ is a subset of $M,$ then there exists a function $f(z)$ with the following properties: (i) the Taylor series of $f(z)$ converges on $C-M,$ diverges on $M,$ and has uniformly bounded partial sums; (ii) $lim_{r rightarrow 1}f(re^{i theta})$ exists when $e^{i theta}$ lies in $C-N,$ does not exist when $e^{i theta}$ lies in $N.$ (from p. 50) In [7] it was proved that every set of type $F_{sigma}$ on $C$ is the set of convergence of some Taylor series. Together with the results established so far in this paper, this gives the result that every denumerable set on $C$ is both a set of convergence and a set of divergence. Since the construction in [7] can easily be modified to introduce arbitrarily large gaps, we also obtain the result that if $M$ is the union of a denumerable set and a set of type $G_{delta}$ on $C,$ then $M$ is the set of divergence of some Taylor series. As far as the general theory of sets of convergence is concerned, the following [[= Theorem 3]] is the most important conclusion that can be drawn from Theorems 1 and 2: Theorem 3 (p. 50): Not every set of convergence of Taylor series is of type $F_{sigma}.$ (first two sentences after Theorem 3 statement, on p. 50) This follows at once from the fact that if $M$ is a denumerable set, dense on $C,$ then $M$ is not of type $G_{delta}$ (see [6; 138]). It should be remarked that Theorem 3 can be deduced from Fejér’s work and should therefore have been included in [7]. Theorem 4 (p. 50): There exists a function $f(z)$ which is holomorphic, bounded and schlicht in $|z| < 1$ and whose Taylor series diverges on a set which is locally non-denumerable on $C.$ [["Locally non-denumerable on $C$" means that the set has uncountable intersection with every arc of $C.$ Since the sets in question are Borel sets, and uncountable Borel sets have cardinality $mathfrak c,$ it follows that "locally non-denumerable on $C$" can be replaced with "$mathfrak c$-dense in $C$".]] Theorem 5 (p. 51): There exists a function which is holomorphic in $|z| < 1$ and continuous in $|z| leq 1$ and whose Taylor series diverges on a set which is locally non-denumerable on $C.$ Theorem 6 (p. 51): There exists a function $f(z),$ bounded and schlicht in $|z| < 1,$ whose Taylor series converges everywhere on $C,$ but not uniformly on any arc of $C.$ Theorem 7 (p. 51): There exists a function $f(z)$ which is continuous in $|z| leq 1$ and whose Taylor series converges everywhere on $C,$ but not uniformly on any arc of $C.$ Definition (p. 52): If $f(z)$ is defined everywhere on $C,$ a point $z_0$ on $C$ lies in the set of boundedness of $f(z)$ provided $f(z)$ is bounded on some open arc of $C$ which contains the point $z_{0}.$ The set of unboundedness of $f(z)$ is the complement [[relative to $C$]] of the set of boundedness. Theorem 8 (p. 52): A necessary and sufficient condition for a set $M$ on $C$ to be the set of unboundedness of some Taylor series converging everywhere on $C$ is that $M$ be closed and nowhere dense on $C.$ [[I suspect that requiring the behavior to not just be unbounded at a point, but to actually approach infinity at the point (i.e. being locally unbounded and finite subsequent limits are not allowed) will be characterized by being a scattered set, as is the case at Is there a function whose all limits at rational points approach infinity? (see my comments there).]]
[3] Fritz Herzog and George Piranian, Some point sets associated with Taylor series, Michigan Mathematical Journal 3 #1 (1955-1956), 69-75. MR 17,834a; Zbl 70.29501
add a comment |
up vote
6
down vote
accepted
Because your question asks whether this is possible when the set of divergence is a certain countable set, the answer is YES by Theorem 1 of [2] below.
Results including and related to what you’ve asked are discussed in the following Stack Exchange questions:
Behaviour of power series on their circle of convergence
Examples of Taylor series with interesting convergence along the boundary of convergence?
Complex power series divergent and convergent on dense subsets of the circumference of convergence?
Power series with funny behavior at the boundary
What are the subsets of the unit circle that can be the points in which a power series is convergent?
Because the Duke Mathematical Journal papers by Herzog/Piranian mentioned in some of the above are not freely available, I’ve included some relevant excerpts from them. Incidentally, I have not bothered to include excerpts from [3] because it is freely available.
In these excerpts, additional notes by me are enclosed in double square brackets [[ ... ]].
[1] Fritz Herzog and George Piranian, Sets of convergence of Taylor series I, Duke Mathematical Journal 16 #3 (September 1949), 529-534. MR 11,91f; Zbl 34.04806
Introduction (p. 529): Let $sum_n a_n z_n$ be a Taylor series of radius of convergence one, with $sum_n |a_n| = infty$ and $lim_n a_n = 0.$ We consider the point set $M$ on the unit circle $C,$ on which the series converges. As Landau [2; 13-14] [[= Landau (1929, Darstellung und Begründung einiger neuerer Ergebnisse der Funktionentheorie, 2nd edition)]] points out, the cardinal number of the set of such Taylor series is $mathfrak c$ [[at least $mathfrak c$ --- add any constant to a specific such Taylor series; at most $mathfrak c$ --- every such Taylor series represents a continuous function and there are $mathfrak c$ many continuous functions]], while the cardinal number of the set of subsets of $C$ is $mathfrak f$ [[= $2^{mathfrak c}$]]; hence, there exists a set $M$ on $C$ such that no Taylor series converges on $M$ and diverges on $C-M.$ It follows that if a set $M$ on $C$ is such that some Taylor series converges on $M$ and diverges on $C-M,$ the set must have certain special properties. Lusin [3] [[= Luzin (1911, Über eine potenzreihe)]] (see also Landau [2; 69-71]) has constructed a Taylor series whose coefficients tend to zero and which diverges on the entire unit circle $C.$ Sierpiński (see Landau [2; 71]) has modified Lusin's example to obtain divergence at all points of $C$ except one. For every closed arc $A$ on $C,$ Neder [5] [[= Neder (1919, Zur Konvergenz der Trigonometrischen Reihen, einschliesslich der Potenzreihen auf dem Konvergenzkreise, Ph.D. Dissertation)]] constructed a Taylor series which converges on $C-A$ and whose partial sums are unbounded at every point of $A.$ Mazurkiewicz [4] [[= Mazurkiewicz (1922, Sur les séries de puissances)]] used Neder's example to prove the following proposition: If $M$ is a closed set on $C,$ there exists a Taylor series which converges on $M$ and diverges on $C-M,$ and a Taylor series which diverges on $M$ and converges on $C-M.$ The present paper is devoted to the extension of these results. Its method is inspired by Lusin's example. Theorem 1 (p. 531): If $M$ is a set of type $F_{sigma}$ on the unit circle $C$ there exists a Taylor series which converges everywhere in $M$ and diverges everywhere in $C-M.$ Theorem 2 (p. 532): If $F$ is a closed set on the unit circle $C,$ there exists a Taylor series which converges uniformly in $F$ and diverges everywhere in $C-F.$ Theorem 3 (p. 533): If $M$ is a set on the unit circle $C,$ such that some Taylor series converges uniformly on $M$ and diverges on $C-M,$ then $M$ is a closed set. Theorem 4 (p. 533): If $M$ is a set on the unit circle $C,$ such that some Taylor series converges on $M$ and diverges on $C-M,$ then $M$ is of type $F_{sigma delta}.$ (last paragraph of the paper, at bottom of p. 533) [[References are on p. 534]] The "no-man's-land" between Theorems 1 and 4 is considerable; it consists of all sets of type $F_{sigma delta}$ on $C$ that are not of type $F_{sigma}.$ It is not known whether every set of type $F_{sigma delta}$ is the set of convergence of some Taylor series. However, if $M$ is the complement on the unit circle of an arbitrary denumerable set, there exists a Taylor series for which $M$ is the set of convergence (the construction of such a series will be described in a later paper); this implies that not every set of convergence of a Taylor series is of type $F_{sigma}.$ [[Note that if $D$ is a set that is dense in $C$ and we let $M = C-D,$ then $M$ is such a set. This follows from the Baire category theorem in the same way that one usually shows that the set of irrational numbers in $mathbb R$ is not an $F_{sigma}$ set.]]
[2] Fritz Herzog and George Piranian, Sets of convergence of Taylor series. II, Duke Mathematical Journal 20 #1 (March 1953), 41-54. MR 14,738b; Zbl 50.07802
(first two sentences of the paper, on p. 41) In an earlier paper [7] the authors have shown that if $M$ is a set of type $F_{sigma}$ on the unit circle $C,$ there exists a Taylor series which has $M$ as its set of convergence, i.e., which converges on $M$ and diverges on $C-M.$ The main purpose of the present paper is to exhibit Taylor series whose sets of convergence are not of type $F_{sigma}.$ Theorem 1 (p. 45): If $M$ is a denumerable set on the unit circle $C,$ there exists a function $f(z)$ with the following properties: (i) $f(z)$ is schlicht in the region $|z| < 1;$ (ii) the Taylor series of $f(z)$ diverges on $M$ and converges on $C-M;$ (iii) the partial sums of the Taylor series of $f(z)$ are uniformly bounded on $C;$ (iv) the set of vertices of the Mittag-Leffler star of $f(z)$ consists of the set $overline{M}.$ Theorem 2 (p. 48): If $M$ is a denumerable set on $C,$ and $N$ is a subset of $M,$ then there exists a function $f(z)$ with the following properties: (i) the Taylor series of $f(z)$ converges on $C-M,$ diverges on $M,$ and has uniformly bounded partial sums; (ii) $lim_{r rightarrow 1}f(re^{i theta})$ exists when $e^{i theta}$ lies in $C-N,$ does not exist when $e^{i theta}$ lies in $N.$ (from p. 50) In [7] it was proved that every set of type $F_{sigma}$ on $C$ is the set of convergence of some Taylor series. Together with the results established so far in this paper, this gives the result that every denumerable set on $C$ is both a set of convergence and a set of divergence. Since the construction in [7] can easily be modified to introduce arbitrarily large gaps, we also obtain the result that if $M$ is the union of a denumerable set and a set of type $G_{delta}$ on $C,$ then $M$ is the set of divergence of some Taylor series. As far as the general theory of sets of convergence is concerned, the following [[= Theorem 3]] is the most important conclusion that can be drawn from Theorems 1 and 2: Theorem 3 (p. 50): Not every set of convergence of Taylor series is of type $F_{sigma}.$ (first two sentences after Theorem 3 statement, on p. 50) This follows at once from the fact that if $M$ is a denumerable set, dense on $C,$ then $M$ is not of type $G_{delta}$ (see [6; 138]). It should be remarked that Theorem 3 can be deduced from Fejér’s work and should therefore have been included in [7]. Theorem 4 (p. 50): There exists a function $f(z)$ which is holomorphic, bounded and schlicht in $|z| < 1$ and whose Taylor series diverges on a set which is locally non-denumerable on $C.$ [["Locally non-denumerable on $C$" means that the set has uncountable intersection with every arc of $C.$ Since the sets in question are Borel sets, and uncountable Borel sets have cardinality $mathfrak c,$ it follows that "locally non-denumerable on $C$" can be replaced with "$mathfrak c$-dense in $C$".]] Theorem 5 (p. 51): There exists a function which is holomorphic in $|z| < 1$ and continuous in $|z| leq 1$ and whose Taylor series diverges on a set which is locally non-denumerable on $C.$ Theorem 6 (p. 51): There exists a function $f(z),$ bounded and schlicht in $|z| < 1,$ whose Taylor series converges everywhere on $C,$ but not uniformly on any arc of $C.$ Theorem 7 (p. 51): There exists a function $f(z)$ which is continuous in $|z| leq 1$ and whose Taylor series converges everywhere on $C,$ but not uniformly on any arc of $C.$ Definition (p. 52): If $f(z)$ is defined everywhere on $C,$ a point $z_0$ on $C$ lies in the set of boundedness of $f(z)$ provided $f(z)$ is bounded on some open arc of $C$ which contains the point $z_{0}.$ The set of unboundedness of $f(z)$ is the complement [[relative to $C$]] of the set of boundedness. Theorem 8 (p. 52): A necessary and sufficient condition for a set $M$ on $C$ to be the set of unboundedness of some Taylor series converging everywhere on $C$ is that $M$ be closed and nowhere dense on $C.$ [[I suspect that requiring the behavior to not just be unbounded at a point, but to actually approach infinity at the point (i.e. being locally unbounded and finite subsequent limits are not allowed) will be characterized by being a scattered set, as is the case at Is there a function whose all limits at rational points approach infinity? (see my comments there).]]
[3] Fritz Herzog and George Piranian, Some point sets associated with Taylor series, Michigan Mathematical Journal 3 #1 (1955-1956), 69-75. MR 17,834a; Zbl 70.29501
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Because your question asks whether this is possible when the set of divergence is a certain countable set, the answer is YES by Theorem 1 of [2] below.
Results including and related to what you’ve asked are discussed in the following Stack Exchange questions:
Behaviour of power series on their circle of convergence
Examples of Taylor series with interesting convergence along the boundary of convergence?
Complex power series divergent and convergent on dense subsets of the circumference of convergence?
Power series with funny behavior at the boundary
What are the subsets of the unit circle that can be the points in which a power series is convergent?
Because the Duke Mathematical Journal papers by Herzog/Piranian mentioned in some of the above are not freely available, I’ve included some relevant excerpts from them. Incidentally, I have not bothered to include excerpts from [3] because it is freely available.
In these excerpts, additional notes by me are enclosed in double square brackets [[ ... ]].
[1] Fritz Herzog and George Piranian, Sets of convergence of Taylor series I, Duke Mathematical Journal 16 #3 (September 1949), 529-534. MR 11,91f; Zbl 34.04806
Introduction (p. 529): Let $sum_n a_n z_n$ be a Taylor series of radius of convergence one, with $sum_n |a_n| = infty$ and $lim_n a_n = 0.$ We consider the point set $M$ on the unit circle $C,$ on which the series converges. As Landau [2; 13-14] [[= Landau (1929, Darstellung und Begründung einiger neuerer Ergebnisse der Funktionentheorie, 2nd edition)]] points out, the cardinal number of the set of such Taylor series is $mathfrak c$ [[at least $mathfrak c$ --- add any constant to a specific such Taylor series; at most $mathfrak c$ --- every such Taylor series represents a continuous function and there are $mathfrak c$ many continuous functions]], while the cardinal number of the set of subsets of $C$ is $mathfrak f$ [[= $2^{mathfrak c}$]]; hence, there exists a set $M$ on $C$ such that no Taylor series converges on $M$ and diverges on $C-M.$ It follows that if a set $M$ on $C$ is such that some Taylor series converges on $M$ and diverges on $C-M,$ the set must have certain special properties. Lusin [3] [[= Luzin (1911, Über eine potenzreihe)]] (see also Landau [2; 69-71]) has constructed a Taylor series whose coefficients tend to zero and which diverges on the entire unit circle $C.$ Sierpiński (see Landau [2; 71]) has modified Lusin's example to obtain divergence at all points of $C$ except one. For every closed arc $A$ on $C,$ Neder [5] [[= Neder (1919, Zur Konvergenz der Trigonometrischen Reihen, einschliesslich der Potenzreihen auf dem Konvergenzkreise, Ph.D. Dissertation)]] constructed a Taylor series which converges on $C-A$ and whose partial sums are unbounded at every point of $A.$ Mazurkiewicz [4] [[= Mazurkiewicz (1922, Sur les séries de puissances)]] used Neder's example to prove the following proposition: If $M$ is a closed set on $C,$ there exists a Taylor series which converges on $M$ and diverges on $C-M,$ and a Taylor series which diverges on $M$ and converges on $C-M.$ The present paper is devoted to the extension of these results. Its method is inspired by Lusin's example. Theorem 1 (p. 531): If $M$ is a set of type $F_{sigma}$ on the unit circle $C$ there exists a Taylor series which converges everywhere in $M$ and diverges everywhere in $C-M.$ Theorem 2 (p. 532): If $F$ is a closed set on the unit circle $C,$ there exists a Taylor series which converges uniformly in $F$ and diverges everywhere in $C-F.$ Theorem 3 (p. 533): If $M$ is a set on the unit circle $C,$ such that some Taylor series converges uniformly on $M$ and diverges on $C-M,$ then $M$ is a closed set. Theorem 4 (p. 533): If $M$ is a set on the unit circle $C,$ such that some Taylor series converges on $M$ and diverges on $C-M,$ then $M$ is of type $F_{sigma delta}.$ (last paragraph of the paper, at bottom of p. 533) [[References are on p. 534]] The "no-man's-land" between Theorems 1 and 4 is considerable; it consists of all sets of type $F_{sigma delta}$ on $C$ that are not of type $F_{sigma}.$ It is not known whether every set of type $F_{sigma delta}$ is the set of convergence of some Taylor series. However, if $M$ is the complement on the unit circle of an arbitrary denumerable set, there exists a Taylor series for which $M$ is the set of convergence (the construction of such a series will be described in a later paper); this implies that not every set of convergence of a Taylor series is of type $F_{sigma}.$ [[Note that if $D$ is a set that is dense in $C$ and we let $M = C-D,$ then $M$ is such a set. This follows from the Baire category theorem in the same way that one usually shows that the set of irrational numbers in $mathbb R$ is not an $F_{sigma}$ set.]]
[2] Fritz Herzog and George Piranian, Sets of convergence of Taylor series. II, Duke Mathematical Journal 20 #1 (March 1953), 41-54. MR 14,738b; Zbl 50.07802
(first two sentences of the paper, on p. 41) In an earlier paper [7] the authors have shown that if $M$ is a set of type $F_{sigma}$ on the unit circle $C,$ there exists a Taylor series which has $M$ as its set of convergence, i.e., which converges on $M$ and diverges on $C-M.$ The main purpose of the present paper is to exhibit Taylor series whose sets of convergence are not of type $F_{sigma}.$ Theorem 1 (p. 45): If $M$ is a denumerable set on the unit circle $C,$ there exists a function $f(z)$ with the following properties: (i) $f(z)$ is schlicht in the region $|z| < 1;$ (ii) the Taylor series of $f(z)$ diverges on $M$ and converges on $C-M;$ (iii) the partial sums of the Taylor series of $f(z)$ are uniformly bounded on $C;$ (iv) the set of vertices of the Mittag-Leffler star of $f(z)$ consists of the set $overline{M}.$ Theorem 2 (p. 48): If $M$ is a denumerable set on $C,$ and $N$ is a subset of $M,$ then there exists a function $f(z)$ with the following properties: (i) the Taylor series of $f(z)$ converges on $C-M,$ diverges on $M,$ and has uniformly bounded partial sums; (ii) $lim_{r rightarrow 1}f(re^{i theta})$ exists when $e^{i theta}$ lies in $C-N,$ does not exist when $e^{i theta}$ lies in $N.$ (from p. 50) In [7] it was proved that every set of type $F_{sigma}$ on $C$ is the set of convergence of some Taylor series. Together with the results established so far in this paper, this gives the result that every denumerable set on $C$ is both a set of convergence and a set of divergence. Since the construction in [7] can easily be modified to introduce arbitrarily large gaps, we also obtain the result that if $M$ is the union of a denumerable set and a set of type $G_{delta}$ on $C,$ then $M$ is the set of divergence of some Taylor series. As far as the general theory of sets of convergence is concerned, the following [[= Theorem 3]] is the most important conclusion that can be drawn from Theorems 1 and 2: Theorem 3 (p. 50): Not every set of convergence of Taylor series is of type $F_{sigma}.$ (first two sentences after Theorem 3 statement, on p. 50) This follows at once from the fact that if $M$ is a denumerable set, dense on $C,$ then $M$ is not of type $G_{delta}$ (see [6; 138]). It should be remarked that Theorem 3 can be deduced from Fejér’s work and should therefore have been included in [7]. Theorem 4 (p. 50): There exists a function $f(z)$ which is holomorphic, bounded and schlicht in $|z| < 1$ and whose Taylor series diverges on a set which is locally non-denumerable on $C.$ [["Locally non-denumerable on $C$" means that the set has uncountable intersection with every arc of $C.$ Since the sets in question are Borel sets, and uncountable Borel sets have cardinality $mathfrak c,$ it follows that "locally non-denumerable on $C$" can be replaced with "$mathfrak c$-dense in $C$".]] Theorem 5 (p. 51): There exists a function which is holomorphic in $|z| < 1$ and continuous in $|z| leq 1$ and whose Taylor series diverges on a set which is locally non-denumerable on $C.$ Theorem 6 (p. 51): There exists a function $f(z),$ bounded and schlicht in $|z| < 1,$ whose Taylor series converges everywhere on $C,$ but not uniformly on any arc of $C.$ Theorem 7 (p. 51): There exists a function $f(z)$ which is continuous in $|z| leq 1$ and whose Taylor series converges everywhere on $C,$ but not uniformly on any arc of $C.$ Definition (p. 52): If $f(z)$ is defined everywhere on $C,$ a point $z_0$ on $C$ lies in the set of boundedness of $f(z)$ provided $f(z)$ is bounded on some open arc of $C$ which contains the point $z_{0}.$ The set of unboundedness of $f(z)$ is the complement [[relative to $C$]] of the set of boundedness. Theorem 8 (p. 52): A necessary and sufficient condition for a set $M$ on $C$ to be the set of unboundedness of some Taylor series converging everywhere on $C$ is that $M$ be closed and nowhere dense on $C.$ [[I suspect that requiring the behavior to not just be unbounded at a point, but to actually approach infinity at the point (i.e. being locally unbounded and finite subsequent limits are not allowed) will be characterized by being a scattered set, as is the case at Is there a function whose all limits at rational points approach infinity? (see my comments there).]]
[3] Fritz Herzog and George Piranian, Some point sets associated with Taylor series, Michigan Mathematical Journal 3 #1 (1955-1956), 69-75. MR 17,834a; Zbl 70.29501
Because your question asks whether this is possible when the set of divergence is a certain countable set, the answer is YES by Theorem 1 of [2] below.
Results including and related to what you’ve asked are discussed in the following Stack Exchange questions:
Behaviour of power series on their circle of convergence
Examples of Taylor series with interesting convergence along the boundary of convergence?
Complex power series divergent and convergent on dense subsets of the circumference of convergence?
Power series with funny behavior at the boundary
What are the subsets of the unit circle that can be the points in which a power series is convergent?
Because the Duke Mathematical Journal papers by Herzog/Piranian mentioned in some of the above are not freely available, I’ve included some relevant excerpts from them. Incidentally, I have not bothered to include excerpts from [3] because it is freely available.
In these excerpts, additional notes by me are enclosed in double square brackets [[ ... ]].
[1] Fritz Herzog and George Piranian, Sets of convergence of Taylor series I, Duke Mathematical Journal 16 #3 (September 1949), 529-534. MR 11,91f; Zbl 34.04806
Introduction (p. 529): Let $sum_n a_n z_n$ be a Taylor series of radius of convergence one, with $sum_n |a_n| = infty$ and $lim_n a_n = 0.$ We consider the point set $M$ on the unit circle $C,$ on which the series converges. As Landau [2; 13-14] [[= Landau (1929, Darstellung und Begründung einiger neuerer Ergebnisse der Funktionentheorie, 2nd edition)]] points out, the cardinal number of the set of such Taylor series is $mathfrak c$ [[at least $mathfrak c$ --- add any constant to a specific such Taylor series; at most $mathfrak c$ --- every such Taylor series represents a continuous function and there are $mathfrak c$ many continuous functions]], while the cardinal number of the set of subsets of $C$ is $mathfrak f$ [[= $2^{mathfrak c}$]]; hence, there exists a set $M$ on $C$ such that no Taylor series converges on $M$ and diverges on $C-M.$ It follows that if a set $M$ on $C$ is such that some Taylor series converges on $M$ and diverges on $C-M,$ the set must have certain special properties. Lusin [3] [[= Luzin (1911, Über eine potenzreihe)]] (see also Landau [2; 69-71]) has constructed a Taylor series whose coefficients tend to zero and which diverges on the entire unit circle $C.$ Sierpiński (see Landau [2; 71]) has modified Lusin's example to obtain divergence at all points of $C$ except one. For every closed arc $A$ on $C,$ Neder [5] [[= Neder (1919, Zur Konvergenz der Trigonometrischen Reihen, einschliesslich der Potenzreihen auf dem Konvergenzkreise, Ph.D. Dissertation)]] constructed a Taylor series which converges on $C-A$ and whose partial sums are unbounded at every point of $A.$ Mazurkiewicz [4] [[= Mazurkiewicz (1922, Sur les séries de puissances)]] used Neder's example to prove the following proposition: If $M$ is a closed set on $C,$ there exists a Taylor series which converges on $M$ and diverges on $C-M,$ and a Taylor series which diverges on $M$ and converges on $C-M.$ The present paper is devoted to the extension of these results. Its method is inspired by Lusin's example. Theorem 1 (p. 531): If $M$ is a set of type $F_{sigma}$ on the unit circle $C$ there exists a Taylor series which converges everywhere in $M$ and diverges everywhere in $C-M.$ Theorem 2 (p. 532): If $F$ is a closed set on the unit circle $C,$ there exists a Taylor series which converges uniformly in $F$ and diverges everywhere in $C-F.$ Theorem 3 (p. 533): If $M$ is a set on the unit circle $C,$ such that some Taylor series converges uniformly on $M$ and diverges on $C-M,$ then $M$ is a closed set. Theorem 4 (p. 533): If $M$ is a set on the unit circle $C,$ such that some Taylor series converges on $M$ and diverges on $C-M,$ then $M$ is of type $F_{sigma delta}.$ (last paragraph of the paper, at bottom of p. 533) [[References are on p. 534]] The "no-man's-land" between Theorems 1 and 4 is considerable; it consists of all sets of type $F_{sigma delta}$ on $C$ that are not of type $F_{sigma}.$ It is not known whether every set of type $F_{sigma delta}$ is the set of convergence of some Taylor series. However, if $M$ is the complement on the unit circle of an arbitrary denumerable set, there exists a Taylor series for which $M$ is the set of convergence (the construction of such a series will be described in a later paper); this implies that not every set of convergence of a Taylor series is of type $F_{sigma}.$ [[Note that if $D$ is a set that is dense in $C$ and we let $M = C-D,$ then $M$ is such a set. This follows from the Baire category theorem in the same way that one usually shows that the set of irrational numbers in $mathbb R$ is not an $F_{sigma}$ set.]]
[2] Fritz Herzog and George Piranian, Sets of convergence of Taylor series. II, Duke Mathematical Journal 20 #1 (March 1953), 41-54. MR 14,738b; Zbl 50.07802
(first two sentences of the paper, on p. 41) In an earlier paper [7] the authors have shown that if $M$ is a set of type $F_{sigma}$ on the unit circle $C,$ there exists a Taylor series which has $M$ as its set of convergence, i.e., which converges on $M$ and diverges on $C-M.$ The main purpose of the present paper is to exhibit Taylor series whose sets of convergence are not of type $F_{sigma}.$ Theorem 1 (p. 45): If $M$ is a denumerable set on the unit circle $C,$ there exists a function $f(z)$ with the following properties: (i) $f(z)$ is schlicht in the region $|z| < 1;$ (ii) the Taylor series of $f(z)$ diverges on $M$ and converges on $C-M;$ (iii) the partial sums of the Taylor series of $f(z)$ are uniformly bounded on $C;$ (iv) the set of vertices of the Mittag-Leffler star of $f(z)$ consists of the set $overline{M}.$ Theorem 2 (p. 48): If $M$ is a denumerable set on $C,$ and $N$ is a subset of $M,$ then there exists a function $f(z)$ with the following properties: (i) the Taylor series of $f(z)$ converges on $C-M,$ diverges on $M,$ and has uniformly bounded partial sums; (ii) $lim_{r rightarrow 1}f(re^{i theta})$ exists when $e^{i theta}$ lies in $C-N,$ does not exist when $e^{i theta}$ lies in $N.$ (from p. 50) In [7] it was proved that every set of type $F_{sigma}$ on $C$ is the set of convergence of some Taylor series. Together with the results established so far in this paper, this gives the result that every denumerable set on $C$ is both a set of convergence and a set of divergence. Since the construction in [7] can easily be modified to introduce arbitrarily large gaps, we also obtain the result that if $M$ is the union of a denumerable set and a set of type $G_{delta}$ on $C,$ then $M$ is the set of divergence of some Taylor series. As far as the general theory of sets of convergence is concerned, the following [[= Theorem 3]] is the most important conclusion that can be drawn from Theorems 1 and 2: Theorem 3 (p. 50): Not every set of convergence of Taylor series is of type $F_{sigma}.$ (first two sentences after Theorem 3 statement, on p. 50) This follows at once from the fact that if $M$ is a denumerable set, dense on $C,$ then $M$ is not of type $G_{delta}$ (see [6; 138]). It should be remarked that Theorem 3 can be deduced from Fejér’s work and should therefore have been included in [7]. Theorem 4 (p. 50): There exists a function $f(z)$ which is holomorphic, bounded and schlicht in $|z| < 1$ and whose Taylor series diverges on a set which is locally non-denumerable on $C.$ [["Locally non-denumerable on $C$" means that the set has uncountable intersection with every arc of $C.$ Since the sets in question are Borel sets, and uncountable Borel sets have cardinality $mathfrak c,$ it follows that "locally non-denumerable on $C$" can be replaced with "$mathfrak c$-dense in $C$".]] Theorem 5 (p. 51): There exists a function which is holomorphic in $|z| < 1$ and continuous in $|z| leq 1$ and whose Taylor series diverges on a set which is locally non-denumerable on $C.$ Theorem 6 (p. 51): There exists a function $f(z),$ bounded and schlicht in $|z| < 1,$ whose Taylor series converges everywhere on $C,$ but not uniformly on any arc of $C.$ Theorem 7 (p. 51): There exists a function $f(z)$ which is continuous in $|z| leq 1$ and whose Taylor series converges everywhere on $C,$ but not uniformly on any arc of $C.$ Definition (p. 52): If $f(z)$ is defined everywhere on $C,$ a point $z_0$ on $C$ lies in the set of boundedness of $f(z)$ provided $f(z)$ is bounded on some open arc of $C$ which contains the point $z_{0}.$ The set of unboundedness of $f(z)$ is the complement [[relative to $C$]] of the set of boundedness. Theorem 8 (p. 52): A necessary and sufficient condition for a set $M$ on $C$ to be the set of unboundedness of some Taylor series converging everywhere on $C$ is that $M$ be closed and nowhere dense on $C.$ [[I suspect that requiring the behavior to not just be unbounded at a point, but to actually approach infinity at the point (i.e. being locally unbounded and finite subsequent limits are not allowed) will be characterized by being a scattered set, as is the case at Is there a function whose all limits at rational points approach infinity? (see my comments there).]]
[3] Fritz Herzog and George Piranian, Some point sets associated with Taylor series, Michigan Mathematical Journal 3 #1 (1955-1956), 69-75. MR 17,834a; Zbl 70.29501
answered Nov 26 at 11:20
Dave L. Renfro
24.3k33979
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Let $p_1 < p_2 < dots$ be the primes, and let $a_n = 0$ if $n$ is not prime and $a_n = frac{1}{k}$ for $n=p_k^2$.
Take any $frac{l}{q} in mathbb{Q}$. Note, by the prime number theorem for arithmetic progressions, that $$lim_{K to infty} frac{1}{K}sum_{k le K} e^{2pi i p_k^2frac{l}{q}} = frac{1}{phi(q)}sum_{substack{a le q \ (a,q) = 1}} e^{2pi i a^2 frac{l}{q}}.$$ By a standard summation by parts argument, it follows that $$lim_{K to infty} frac{1}{log K} sum_{k le K} frac{1}{k}e^{2pi i p_k^2frac{l}{q}} = frac{1}{phi(q)}sum_{substack{a le q \ (a,q) = 1}} e^{2pi i a^2 frac{l}{q}}.$$ Therefore, if we show that the sum $sum_{a le q \ (a,q) = 1} e^{2pi i a^2frac{l}{q}}$ is nonzero for any $q ge 1$ and $(l,q) = 1$, it follows that $sum_{k le K} frac{1}{k} e^{2pi i p_k^2frac{l}{q}}$ blows up in magnitude and in particular diverges. We show that the sum is nonzero below the fold.
Now take any $alpha in mathbb{R}setminusmathbb{Q}$. It's well known that $$lim_{K to infty} sum_{k le K} frac{1}{k}e^{2pi i p_k^2alpha}$$ exists, which clearly implies that $sum_{n le N} a_ne^{2pi i alpha n}$ converges as $N to infty$. This completes the argument.
--------------------------------------------------------------
Using the identity $sum_{d mid n} mu(d) = 1_{n=1}$, we see $$sum_{a le q \ (a,q) = 1} e^{2pi i a^2frac{l}{q}} = sum_{a le q} e^{2pi i a^2frac{l}{q}}sum_{d mid (a,q)} mu(d) = sum_{a le q} e^{2pi i a^2frac{l}{q}} sum_{substack{d mid a \ d mid q}} mu(d)$$ $$ = sum_{d mid q} mu(d) sum_{d mid a le q} e^{2pi i a^2frac{l}{q}} = sum_{d mid q} mu(d)sum_{j le q/d} e^{2pi i j^2frac{dl}{q/d}}$$ where the last equality was obtained by substituting $a = jd$. We have a linear combination of quadratic Gauss sums. Each $sum e^{2pi i j^2frac{dl}{q/d}}$ is either $0$ or $csqrt{q/d}$ for $c in {pm 1, pm i, pm (1+i), pm i(1+i)}$. If $q not equiv 2 pmod{4}$, then the term corresponding to $d=1$ is $csqrt{q}$ which cannot cancel with any of the other terms and we're done. If $q equiv 2 pmod{4}$, then the term $d=2$ will give a $-csqrt{q/2}$ term which cannot cancel with any of the other terms and we're done.
It's well known that $lim_{Ktoinfty}sum_{kle K}frac 1k e^{2pi ip^2kalpha}=0$. Really? Doesn't it bother you that irrationals have full measure and the series is a non-trivial Fourier series of an $L^2$-function?
– fedja
Nov 28 at 23:24
@fedja whoops. I meant exists. edited
– mathworker21
Nov 29 at 4:54
add a comment |
up vote
1
down vote
Let $p_1 < p_2 < dots$ be the primes, and let $a_n = 0$ if $n$ is not prime and $a_n = frac{1}{k}$ for $n=p_k^2$.
Take any $frac{l}{q} in mathbb{Q}$. Note, by the prime number theorem for arithmetic progressions, that $$lim_{K to infty} frac{1}{K}sum_{k le K} e^{2pi i p_k^2frac{l}{q}} = frac{1}{phi(q)}sum_{substack{a le q \ (a,q) = 1}} e^{2pi i a^2 frac{l}{q}}.$$ By a standard summation by parts argument, it follows that $$lim_{K to infty} frac{1}{log K} sum_{k le K} frac{1}{k}e^{2pi i p_k^2frac{l}{q}} = frac{1}{phi(q)}sum_{substack{a le q \ (a,q) = 1}} e^{2pi i a^2 frac{l}{q}}.$$ Therefore, if we show that the sum $sum_{a le q \ (a,q) = 1} e^{2pi i a^2frac{l}{q}}$ is nonzero for any $q ge 1$ and $(l,q) = 1$, it follows that $sum_{k le K} frac{1}{k} e^{2pi i p_k^2frac{l}{q}}$ blows up in magnitude and in particular diverges. We show that the sum is nonzero below the fold.
Now take any $alpha in mathbb{R}setminusmathbb{Q}$. It's well known that $$lim_{K to infty} sum_{k le K} frac{1}{k}e^{2pi i p_k^2alpha}$$ exists, which clearly implies that $sum_{n le N} a_ne^{2pi i alpha n}$ converges as $N to infty$. This completes the argument.
--------------------------------------------------------------
Using the identity $sum_{d mid n} mu(d) = 1_{n=1}$, we see $$sum_{a le q \ (a,q) = 1} e^{2pi i a^2frac{l}{q}} = sum_{a le q} e^{2pi i a^2frac{l}{q}}sum_{d mid (a,q)} mu(d) = sum_{a le q} e^{2pi i a^2frac{l}{q}} sum_{substack{d mid a \ d mid q}} mu(d)$$ $$ = sum_{d mid q} mu(d) sum_{d mid a le q} e^{2pi i a^2frac{l}{q}} = sum_{d mid q} mu(d)sum_{j le q/d} e^{2pi i j^2frac{dl}{q/d}}$$ where the last equality was obtained by substituting $a = jd$. We have a linear combination of quadratic Gauss sums. Each $sum e^{2pi i j^2frac{dl}{q/d}}$ is either $0$ or $csqrt{q/d}$ for $c in {pm 1, pm i, pm (1+i), pm i(1+i)}$. If $q not equiv 2 pmod{4}$, then the term corresponding to $d=1$ is $csqrt{q}$ which cannot cancel with any of the other terms and we're done. If $q equiv 2 pmod{4}$, then the term $d=2$ will give a $-csqrt{q/2}$ term which cannot cancel with any of the other terms and we're done.
It's well known that $lim_{Ktoinfty}sum_{kle K}frac 1k e^{2pi ip^2kalpha}=0$. Really? Doesn't it bother you that irrationals have full measure and the series is a non-trivial Fourier series of an $L^2$-function?
– fedja
Nov 28 at 23:24
@fedja whoops. I meant exists. edited
– mathworker21
Nov 29 at 4:54
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $p_1 < p_2 < dots$ be the primes, and let $a_n = 0$ if $n$ is not prime and $a_n = frac{1}{k}$ for $n=p_k^2$.
Take any $frac{l}{q} in mathbb{Q}$. Note, by the prime number theorem for arithmetic progressions, that $$lim_{K to infty} frac{1}{K}sum_{k le K} e^{2pi i p_k^2frac{l}{q}} = frac{1}{phi(q)}sum_{substack{a le q \ (a,q) = 1}} e^{2pi i a^2 frac{l}{q}}.$$ By a standard summation by parts argument, it follows that $$lim_{K to infty} frac{1}{log K} sum_{k le K} frac{1}{k}e^{2pi i p_k^2frac{l}{q}} = frac{1}{phi(q)}sum_{substack{a le q \ (a,q) = 1}} e^{2pi i a^2 frac{l}{q}}.$$ Therefore, if we show that the sum $sum_{a le q \ (a,q) = 1} e^{2pi i a^2frac{l}{q}}$ is nonzero for any $q ge 1$ and $(l,q) = 1$, it follows that $sum_{k le K} frac{1}{k} e^{2pi i p_k^2frac{l}{q}}$ blows up in magnitude and in particular diverges. We show that the sum is nonzero below the fold.
Now take any $alpha in mathbb{R}setminusmathbb{Q}$. It's well known that $$lim_{K to infty} sum_{k le K} frac{1}{k}e^{2pi i p_k^2alpha}$$ exists, which clearly implies that $sum_{n le N} a_ne^{2pi i alpha n}$ converges as $N to infty$. This completes the argument.
--------------------------------------------------------------
Using the identity $sum_{d mid n} mu(d) = 1_{n=1}$, we see $$sum_{a le q \ (a,q) = 1} e^{2pi i a^2frac{l}{q}} = sum_{a le q} e^{2pi i a^2frac{l}{q}}sum_{d mid (a,q)} mu(d) = sum_{a le q} e^{2pi i a^2frac{l}{q}} sum_{substack{d mid a \ d mid q}} mu(d)$$ $$ = sum_{d mid q} mu(d) sum_{d mid a le q} e^{2pi i a^2frac{l}{q}} = sum_{d mid q} mu(d)sum_{j le q/d} e^{2pi i j^2frac{dl}{q/d}}$$ where the last equality was obtained by substituting $a = jd$. We have a linear combination of quadratic Gauss sums. Each $sum e^{2pi i j^2frac{dl}{q/d}}$ is either $0$ or $csqrt{q/d}$ for $c in {pm 1, pm i, pm (1+i), pm i(1+i)}$. If $q not equiv 2 pmod{4}$, then the term corresponding to $d=1$ is $csqrt{q}$ which cannot cancel with any of the other terms and we're done. If $q equiv 2 pmod{4}$, then the term $d=2$ will give a $-csqrt{q/2}$ term which cannot cancel with any of the other terms and we're done.
Let $p_1 < p_2 < dots$ be the primes, and let $a_n = 0$ if $n$ is not prime and $a_n = frac{1}{k}$ for $n=p_k^2$.
Take any $frac{l}{q} in mathbb{Q}$. Note, by the prime number theorem for arithmetic progressions, that $$lim_{K to infty} frac{1}{K}sum_{k le K} e^{2pi i p_k^2frac{l}{q}} = frac{1}{phi(q)}sum_{substack{a le q \ (a,q) = 1}} e^{2pi i a^2 frac{l}{q}}.$$ By a standard summation by parts argument, it follows that $$lim_{K to infty} frac{1}{log K} sum_{k le K} frac{1}{k}e^{2pi i p_k^2frac{l}{q}} = frac{1}{phi(q)}sum_{substack{a le q \ (a,q) = 1}} e^{2pi i a^2 frac{l}{q}}.$$ Therefore, if we show that the sum $sum_{a le q \ (a,q) = 1} e^{2pi i a^2frac{l}{q}}$ is nonzero for any $q ge 1$ and $(l,q) = 1$, it follows that $sum_{k le K} frac{1}{k} e^{2pi i p_k^2frac{l}{q}}$ blows up in magnitude and in particular diverges. We show that the sum is nonzero below the fold.
Now take any $alpha in mathbb{R}setminusmathbb{Q}$. It's well known that $$lim_{K to infty} sum_{k le K} frac{1}{k}e^{2pi i p_k^2alpha}$$ exists, which clearly implies that $sum_{n le N} a_ne^{2pi i alpha n}$ converges as $N to infty$. This completes the argument.
--------------------------------------------------------------
Using the identity $sum_{d mid n} mu(d) = 1_{n=1}$, we see $$sum_{a le q \ (a,q) = 1} e^{2pi i a^2frac{l}{q}} = sum_{a le q} e^{2pi i a^2frac{l}{q}}sum_{d mid (a,q)} mu(d) = sum_{a le q} e^{2pi i a^2frac{l}{q}} sum_{substack{d mid a \ d mid q}} mu(d)$$ $$ = sum_{d mid q} mu(d) sum_{d mid a le q} e^{2pi i a^2frac{l}{q}} = sum_{d mid q} mu(d)sum_{j le q/d} e^{2pi i j^2frac{dl}{q/d}}$$ where the last equality was obtained by substituting $a = jd$. We have a linear combination of quadratic Gauss sums. Each $sum e^{2pi i j^2frac{dl}{q/d}}$ is either $0$ or $csqrt{q/d}$ for $c in {pm 1, pm i, pm (1+i), pm i(1+i)}$. If $q not equiv 2 pmod{4}$, then the term corresponding to $d=1$ is $csqrt{q}$ which cannot cancel with any of the other terms and we're done. If $q equiv 2 pmod{4}$, then the term $d=2$ will give a $-csqrt{q/2}$ term which cannot cancel with any of the other terms and we're done.
edited Nov 29 at 4:54
answered Nov 28 at 7:10
mathworker21
8,4711928
8,4711928
It's well known that $lim_{Ktoinfty}sum_{kle K}frac 1k e^{2pi ip^2kalpha}=0$. Really? Doesn't it bother you that irrationals have full measure and the series is a non-trivial Fourier series of an $L^2$-function?
– fedja
Nov 28 at 23:24
@fedja whoops. I meant exists. edited
– mathworker21
Nov 29 at 4:54
add a comment |
It's well known that $lim_{Ktoinfty}sum_{kle K}frac 1k e^{2pi ip^2kalpha}=0$. Really? Doesn't it bother you that irrationals have full measure and the series is a non-trivial Fourier series of an $L^2$-function?
– fedja
Nov 28 at 23:24
@fedja whoops. I meant exists. edited
– mathworker21
Nov 29 at 4:54
It's well known that $lim_{Ktoinfty}sum_{kle K}frac 1k e^{2pi ip^2kalpha}=0$. Really? Doesn't it bother you that irrationals have full measure and the series is a non-trivial Fourier series of an $L^2$-function?
– fedja
Nov 28 at 23:24
It's well known that $lim_{Ktoinfty}sum_{kle K}frac 1k e^{2pi ip^2kalpha}=0$. Really? Doesn't it bother you that irrationals have full measure and the series is a non-trivial Fourier series of an $L^2$-function?
– fedja
Nov 28 at 23:24
@fedja whoops. I meant exists. edited
– mathworker21
Nov 29 at 4:54
@fedja whoops. I meant exists. edited
– mathworker21
Nov 29 at 4:54
add a comment |
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For "unit disc" read "unit circle"?
– Lord Shark the Unknown
Nov 24 at 5:04
@LordSharktheUnknown Yes, thank you.
– MathematicsStudent1122
Nov 24 at 5:05
2
$sum frac{z^{n!}}{n}$ diverges not only for roots of unity, but at some other places too. Consider $z=exp(2pi i e)$ (not a root of unity, since $e$ is irrational). Then $z^{n!}=exp(2pi i e n!)$. We don't care about the integer part of $(e n!)$, only the fractional part. And ${e n!}=frac{1}{n+1}+frac{1}{(n+1)(n+2)}+cdots<frac{1}{n+1}+frac{1}{(n+1)^2}+cdots=frac{1}{n}$ converges to $0$. So the terms of $sum frac{z^{n!}}{n}$ will accumulate close to $1$, and the series will diverge.
– alex.jordan
Nov 26 at 6:32
A late addition to my comment above: a similar argument shows that series diverges for $z=exp(2pi i(e+r))$ for any rational $r$.
– alex.jordan
Nov 26 at 21:46
@MathematicsStudent1122 why did you already award the bounty? I gave an explicit example
– mathworker21
Nov 29 at 5:00