My colleague Tom Liggett recently posed to me the following problem about power series in one real variable . Observe that the power series

has very rapidly decaying coefficients (of order ), leading to an infinite radius of convergence; also, as the series converges to , the series decays very rapidly as approaches . The problem is whether this is essentially the only example of this type. More precisely:

**Problem 1** Let be a bounded sequence of real numbers, and suppose that the power series

(which has an infinite radius of convergence) decays like as , in the sense that the function remains bounded as . Must the sequence be of the form for some constant ?

As it turns out, the problem has a very nice solution using complex analysis methods, which by coincidence I happen to be teaching right now. I am therefore posing as a challenge to my complex analysis students and to other readers of this blog to answer the above problem by complex methods; feel free to post solutions in the comments below (and in particular, if you don’t want to be spoiled, you should probably refrain from reading the comments). In fact, the *only* way I know how to solve this problem currently is by complex methods; I would be interested in seeing a purely real-variable solution that is not simply a thinly disguised version of a complex-variable argument.

(To be fair to my students, the complex variable argument does require one additional tool that is not directly covered in my notes. That tool can be found here.)

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18 October, 2016 at 1:38 pm

AnonymousI think the answer is of course not, just consider the sine or cosine, so some coefficients are 0s.

19 October, 2016 at 12:48 pm

Anonymouse^x *sin x is not bounded..

18 October, 2016 at 1:42 pm

AnonymousSorry, misunderstood.

18 October, 2016 at 2:27 pm

Steve SeverinCould not a_n be of the form C*(-alpha)^n and still satisfy the conditions described in the problem?

18 October, 2016 at 2:28 pm

Steve SeverinOf coursem 0 < alpha <=1.

18 October, 2016 at 2:32 pm

AnonymousThen the decay as x tends to infinity won’t be fast enough, since it would give f(x)=exp(-alpha*x) with alpha in (0,1)

18 October, 2016 at 2:51 pm

Steve SeverinThanks that makes sense. So we need a function that balances out (or nulls out) exp(x) but whose Taylor coefficients multplied by n! are bounded. I am not a mathematician, just like looking at Terry’s output, sort of like a visitor to an art gallery.

18 October, 2016 at 4:31 pm

Steve SeverinI suppose we could add any polynomial (finite degree) to the first series in the post and it would still be kosher. This is probably what Terry means by *essentially* the only example.

18 October, 2016 at 4:35 pm

Steve SeverinWish I could retract this. I will stop posting. Sigh.

18 October, 2016 at 3:27 pm

Chenyang YuI will give an outline of my thoughts. First observe that this series expansion is very similar to that of the definition of Bernoulli numbers, and in that definition the series equals x/(e^x-1). In fact, we can assume they are equal, and try to prove that for n is odd, an = 0. Then the terms left are the even terms, which coincide with Bernoulli numbers. I have some ideas about it and I am doing detailed proof.

18 October, 2016 at 4:07 pm

anonymouswhat if we take i.e. ?

18 October, 2016 at 7:35 pm

Lior SilbermanThen the aren’t bounded.

18 October, 2016 at 7:57 pm

anonymousOh, yes, I missed the boundedness hypothesis. Sorry, my bad. Thank you Lior.

18 October, 2016 at 10:52 pm

Can't Sleep... Why am I doing math to sleep!?!2^-x? E^-xlog2 should result in a_n = 1*(log2)^n.

Or does that violate the big O bound? Nope, Les mental hospital rules hold.

19 October, 2016 at 1:03 am

AlexanderIf we use complex analysis methods, it is then natural to consider complex coefficients.

19 October, 2016 at 1:27 am

maxbaroiTake . It satisfies (compare two consecutive terms of the powers series , with the consecutive terms in the power series of ). This should be fine since both series are easy to show to be absolutely. The radius of converge is all the reals by the ratio test.

I don’t think this argument uses anything trickier that undergrad calc, so I’, probably wrong.

[ is alternating in sign, but the magnitudes are not monotone decreasing, so one cannot easily perform a comparison test. -T.]19 October, 2016 at 9:11 am

maxbaroiI’m sorry, I was referring to .

19 October, 2016 at 10:35 pm

maxbaroiWould some please explain to me why my reasoning is spurious?

20 October, 2016 at 8:29 am

Terence TaoThe series is not dominated by , due to the non-monotone nature of the magnitudes of the terms in the alternating series (so their consecutive differences can fluctuate in sign). See for instance this Wolfram Alpha plot. (From the asymptotics of Bessel functions, your power series in fact decays like rather than .)

20 October, 2016 at 12:36 pm

maxbaroiThank you. The fools we (I) make ourselves when not considering asymptotic analysis.

19 October, 2016 at 4:28 am

Iosif PinelisTerry Tao’s hint about the Laplace transform was indeed all that was needed. :-) A “complex” solution to this nice problem can be found at https://www.dropbox.com/s/vsclr67uwx64oiy/power-series.pdf?dl=0 . (Sorry, it was too hard for me to use the online latex tool.)

21 October, 2016 at 6:24 pm

AnonymousIt seems that your proof still applies for the more general case of polynomial growth of the coefficients with (i.e. for any fixed .)

22 October, 2016 at 4:41 pm

Iosif PinelisThis is a good point. Indeed, it seems easy to extend the proof to cover the more general case of polynomial growth of the ‘s.

25 October, 2016 at 7:06 pm

AnonymousIn addition to the polynomial growth of the ‘s, it seems that the boundedness restriction of can be relaxed to as (because in your proof it implies that as – which is sufficient to rule out the possibility of a pole singularity of at .)

24 April, 2018 at 8:38 am

marccofHi, do you still have your proof somewhere ? The dropbox link is dead and I would be interested in seeing the solution to this question.

Thanks !

24 April, 2018 at 9:02 am

Iosif PinelisThe Dropbox link is no longer valid. The SelectedWorks link https://works.bepress.com/iosif-pinelis/14/ should work and be of more permanent nature.

26 April, 2018 at 1:38 am

marccofThanks you !

19 October, 2016 at 7:25 am

Iosif PinelisI have added a remark and a conjecture in the file at that link, https://www.dropbox.com/s/vsclr67uwx64oiy/power-series.pdf?dl=0 .

19 October, 2016 at 7:58 am

Anonymousisn’t exp(-1/2*x) a counterexample to your conjecture?

19 October, 2016 at 8:05 am

AnonymousI mean its coefficients, a_n = (-1)^n*(1/2)^n

19 October, 2016 at 8:12 am

Lior Silbermandoes not decay as fast as .

19 October, 2016 at 8:26 am

Iosif PinelisThank you. I have removed that conjecture.

19 October, 2016 at 8:25 am

Lior SilbermanIosif: How do you show that the singularity at is a pole? It’s true that the integral is bounded above by in the domain — but why is that enough to show that the singularity is a pole? Perhaps for tending to from the other side the growth is faster?

For example, the function has an essential singularity at zero but is nevertheless bounded on the half-plane .

I’ve been stuck on this point since Terry published the problem.

19 October, 2016 at 2:07 pm

Iosif PinelisThank you Lior for this comment. I had indeed overlooked this point. Now I have added a lemma in the file at https://www.dropbox.com/s/vsclr67uwx64oiy/power-series.pdf?dl=0 that shows that $-1$ is indeed a pole of $g$. The proof of the lemma is indeed not quite trivial, and I hope I didn’t make a major error there.

19 October, 2016 at 3:51 pm

AnonymousIt seems that your lemma shows that is not an essential singularity of . The fact that is a removable singularity then follows from your earlier result that on the half plane (thereby avoiding the possibility of being a pole.)

19 October, 2016 at 5:03 pm

Iosif PinelisAnonymous: This is mainly so, except that I only showed that for , but that is still enough.

19 October, 2016 at 6:31 pm

AnonymousIn fact, to exclude the possibility of a pole of at , it is sufficient to show only that is bounded on a sequence converging to .

19 October, 2016 at 8:14 am

Lior SilbermanIt’s not hard to show the Laplace transform continues to a function on the puncture place , with good enough decay to shift contours and invert the transform. If the singularity were a pole, the function would be of the form where P is a polynomial, and the boundedness shows that is constant. How do you show the singularity is a pole?

19 October, 2016 at 9:46 am

AnonymousIt is sufficient to show that this singularity is isolated and bounded in some neighborhood to be removable.

19 October, 2016 at 9:39 am

AnonymousIs it possible that this problem is a special case of a more general principle:

“A nontrivial entire function and its Taylor coefficients can’t both(!) decay to zero arbitrarily fast”? More precisely, let be a nontrivial entire function, and denote for each . Then if where is a given sequence decaying to zero “sufficiently fast”, then there is a function defined for and depending only on the sequence such that is not decaying to zero faster than .

Since the Taylor coefficients of an entire function are its Fourier coefficients on the unit circle, they can be viewed (in some sense) as its “spectral coefficients” – so this principle for the case of entire functions is similar to the case of functions in which can’t decay both(!) with their Fourier transform (playing the role of “spectral coefficients”) faster then Gaussian decay. So in a sense, this problem may be interpreted as a special case of some “uncertainty principle” for entire functions.

19 October, 2016 at 7:41 pm

Jhon ManugalLet be a random variable and be Poisson distributed with mean $x$. The expectation is . The Poisson distribution as with mean as has that lump moving off to infinity. And should concentrate around . That’s of course pretty bogus. But that’s what I’ve been thinking.

19 October, 2016 at 9:57 pm

AvIsn’t it just implied from Liouville’s theorem for g(x) = exp(x) * f(x) which is limited and analytical => constant

=> 0 = g'(x) = (exp(x) * f(x))’ = exp(x) * (f(x) + f'(x))

=> a_n + a_{n+1} = 0

=> QED

19 October, 2016 at 11:55 pm

AnonymousUnfortunately, it is only given that remains bounded as (but no information given on its possible behavior as or for complex valued .)

20 October, 2016 at 9:37 am

AnonymousSince |f(z)| <= M e^(|z|), where M is the guaranteed bound for the sequence a_n, it follows that |g(z)| <= |e^z M e^(-z)|= M when z is real and negative. That shows g is bounded on the real line at least.

20 October, 2016 at 11:25 pm

AnonymousThe example shows that a non-constant entire function can be bounded (with faster than exponential decay) on arbitrarily many straight lines (on which is real) in the complex plane.

21 October, 2016 at 7:03 am

AnonymousThat’s true, but the sequence a_n corresponding to f(z)=exp^(-z)exp(-z^(2m)) is not bounded as we are given here except for m=0.

20 October, 2016 at 1:01 am

Colin PercivalIt’s almost 2AM so I’m not able to flesh out the details right now, but I have a feeling that there’s an argument based on writing f(x) as a linear combination of exp(-n x). Providing such a linear combination exists, the answer falls right out, of course.

20 October, 2016 at 1:43 am

AnonymousIn order to show uniqueness of such linear combination, it seems necessary to find an orthogonal system of polynomials of on with respect to some weight function.

22 October, 2016 at 9:29 am

Lior SilbermanColin: that’s exactly what the Laplace transform does, except that the decomposition has a continuous rather than discrete parametrization because the domain is non-compact.

20 October, 2016 at 3:19 am

rsjIt’s been 20 years since I took a complex analysis class, but IIRC if e^x*f(x) is an entire bounded analytic function on the complex plane it must be constant. e^x f(x) = C. Then f(x) = Ce^(-x)

To show that an entire bounded analytic function must be constant, look at the expression of the n’th derivative via the cauchy integral formula and show that they are all zero by taking large circular paths in the integral. The integral will have a radius term to a power in the denominator which is unbounded and the numerator will be bounded by the function bound.

20 October, 2016 at 3:41 am

AnonymousSee the above comment of Av and its reply.

20 October, 2016 at 4:01 am

AnonymousThis problem seems to be closely related to the theory of Quasi-analytic classes and the Denjoy-Carleman theorem (see e.g. Chapter 19 in Rudin’s book “Real and complex analysis”, 1987).

20 October, 2016 at 12:00 pm

elianto84Is Ramanujan master theorem (https://en.wikipedia.org/wiki/Ramanujan%27s_master_theorem) considered a real technique? ;)

23 October, 2016 at 9:54 am

sergeiDear Terence Tao! Very much glad,what you draw attention on me! Big part of time expensed on translation from English into Russian and inside out,and retrieval of key-words for reflection(train of thought). In ours positions is many similarity points(fulcrums,points of intersection). Now I assured in,what not barely so,not vain be stay to have nothing to do on first floor of Erdesh tower twenty years.(1) Through Cauchy’s theorem-that through channel,connect three partitions of science: mathematic,theoretical physics,physics of elementarys particles. On another ending of through channel is such turn of Vica-place of return at zero’s of Riman dzeta- function.Confirmation-limit of Shvinger.(2) Necessary to prepare section[a,b] at four-impulse integration over widening:[a,b]=>max[xj-x(j-1)]=>[t,x,y,z] I wait reply. Sergei. Thanks!

25 October, 2016 at 4:37 am

Sangchul LeeHere is my rough idea: Let be the space entire functions such that the sequence satisfies the given condition. Also we define

It is not hard to check that this is indeed a norm with respect to which is a Banach space. Moreover, using the diagonalization argument, it is not hard to check that the unit ball of is compact. So is finite dimensional. Next, we consider the linear map on defined by

It is easy to check that is an injective linear operator on . Thus is an isomorphism. The most important implication of this observation is that if then , since for some and hence . Now since is finite dimensional, there exist and such that

But we know how to solve this differential equation, and the only possible choice is for some .

25 October, 2016 at 5:03 am

Sangchul LeeOops, ignore this. My claim on compactness of the unit ball was too bold.

25 October, 2016 at 5:19 am

Iosif PinelisYour idea seems interesting. (You don’t need to check that is a Banach space; that is, you don’t need to check that is complete.) However, I don’t see how you can prove the compactness by the diagonalization argument alone — this would look similar to trying to prove that the unit ball is compact (which is of course false.)

25 October, 2016 at 5:30 am

Sangchul LeeThank you for pointing out. I definitely agree with you; in fact, I realized my fault right after I posted my idea. I have a faint impression that this may be salvaged by invoking some heavy machinery in functional analysis, though I am not sure at this point. I should take some caffeine and see if there is a way this approach can be fixed.

25 October, 2016 at 6:17 am

Iosif PinelisIt seems that in the proof of the compactness you would have to engage the boundedness of . But how to do that without complex analysis? The complex analysis argument (the only one I know) is pretty involved and varied. So, I think it would be very surprising and educational if one can do here without complex analysis.

25 October, 2016 at 9:36 am

Sangchul LeeI agree with you. Now I am now more prone to think that complex analysis is inevitable, thinking that proving is essentially equivalent to proving that the Laplace transform has pole at .

27 October, 2016 at 2:45 am

Paolo BonziniFunctions like tanh x or 1/cosh x would satisfy the property for real x. Does that rule out a solution that doesn’t use complex analysis?