A problem involving power series

18 October, 2016 in 246A - complex analysis, math.CV, question | Tags: power series | by Terence Tao

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

$\displaystyle \sum_{n=0}^\infty (-1)^n\frac{x^n}{n!}$

has very rapidly decaying coefficients (of order ${O(1/n!)}$ ), leading to an infinite radius of convergence; also, as the series converges to ${e^{-x}}$ , the series decays very rapidly as ${x}$ approaches ${+\infty}$ . The problem is whether this is essentially the only example of this type. More precisely:

Problem 1 Let ${a_0, a_1, \dots}$ be a bounded sequence of real numbers, and suppose that the power series

$\displaystyle f(x) := \sum_{n=0}^\infty a_n\frac{x^n}{n!}$

(which has an infinite radius of convergence) decays like ${O(e^{-x})}$ as ${x \rightarrow +\infty}$ , in the sense that the function ${e^x f(x)}$ remains bounded as ${x \rightarrow +\infty}$ . Must the sequence ${a_n}$ be of the form ${a_n = C (-1)^n}$ for some constant ${C}$ ?

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.)

67 comments

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18 October, 2016 at 1:38 pm

Anonymous

I 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

Anonymous

e^x *sin x is not bounded..

18 October, 2016 at 1:42 pm

Anonymous

Sorry, misunderstood.

18 October, 2016 at 2:27 pm

Steve Severin

Could 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 Severin

Of coursem 0 < alpha <=1.

18 October, 2016 at 2:32 pm

Anonymous

Then 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 Severin

Thanks 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 Severin

I 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 Severin

Wish I could retract this. I will stop posting. Sigh.

18 October, 2016 at 3:27 pm

Chenyang Yu

I 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

anonymous

what if we take $f(x)=e^{-2x}$ i.e. $a_n= 2^{n} (-1)^{n}$ ?

18 October, 2016 at 7:35 pm

Lior Silberman

Then the $a_n$ aren’t bounded.

18 October, 2016 at 7:57 pm

anonymous

Oh, 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

Alexander

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

19 October, 2016 at 1:27 am

maxbaroi

Take $a_n = \frac{(-1)^n}{n!}$ . It satisfies $f = O(e^{-x})$ (compare two consecutive terms of the powers series $e^{-x]}$ , with the consecutive terms in the power series of $f(x)$ ). 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.

[ $\sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n$ 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

maxbaroi

I’m sorry, I was referring to $\sum_{n=0}^\infty \frac{(-1)^n}{n!n!}x^n$ .

19 October, 2016 at 10:35 pm

maxbaroi

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

20 October, 2016 at 8:29 am

Terence Tao

The series $\sum_{n=0}^\infty \frac{(-1)^n}{n! n!} x^n = J_0(2 \sqrt{x})$ is not dominated by $\sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n = e^{-x}$ , 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 $x^{-1/4}$ rather than $e^{-x}$ .)

20 October, 2016 at 12:36 pm

maxbaroi

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

19 October, 2016 at 4:28 am

Iosif Pinelis

Terry 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

Anonymous

It seems that your proof still applies for the more general case of polynomial growth of the coefficients $a_n$ with $n$ (i.e. $a_n = O(n^p)$ for any fixed $p > 0$ .)

22 October, 2016 at 4:41 pm

Iosif Pinelis

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

25 October, 2016 at 7:06 pm

Anonymous

In addition to the polynomial growth of the $a_n$ ‘s, it seems that the boundedness restriction of $e^x f(x)$ can be relaxed to $o(x)$ as $x \to +\infty$ (because in your proof it implies that $(u+1)g(u) = o(1)$ as $u+1 \to 0+$ – which is sufficient to rule out the possibility of a pole singularity of $g$ at $-1$ .)

24 April, 2018 at 8:38 am

marccof

Hi, 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 Pinelis

The 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

marccof

Thanks you !

19 October, 2016 at 7:25 am

Iosif Pinelis

I 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

Anonymous

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

19 October, 2016 at 8:05 am

Anonymous

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

19 October, 2016 at 8:12 am

Lior Silberman

$\exp(-x/2)$ does not decay as fast as $\exp(-x)$ .

19 October, 2016 at 8:26 am

Iosif Pinelis

Thank you. I have removed that conjecture.

19 October, 2016 at 8:25 am

Lior Silberman

Iosif: How do you show that the singularity at $-1$ is a pole? It’s true that the integral $\int_0^\infty e^{-sx}f(x)dx$ is bounded above by $\frac{1}{1+\Re(s)}$ in the domain $\Re(s)>-1$ — but why is that enough to show that the singularity is a pole? Perhaps for $s$ tending to $-1$ from the other side the growth is faster?

For example, the function $\exp(-1/z)$ has an essential singularity at zero but is nevertheless bounded on the half-plane $\Re(z)>0$ .

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

19 October, 2016 at 2:07 pm

Iosif Pinelis

Thank 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

Anonymous

It seems that your lemma shows that $-1$ is not an essential singularity of $g$ . The fact that $-1$ is a removable singularity then follows from your earlier result that $g(u) = O(1)$ on the half plane $\Re (u) > -1$ (thereby avoiding the possibility of being a pole.)

19 October, 2016 at 5:03 pm

Iosif Pinelis

Anonymous: This is mainly so, except that I only showed that $g(u)=O(1)$ for $u>-1$ , but that is still enough.

19 October, 2016 at 6:31 pm

Anonymous

In fact, to exclude the possibility of a pole of $g$ at $-1$ , it is sufficient to show only that $g$ is bounded on a sequence converging to $-1$ .

19 October, 2016 at 8:14 am

Lior Silberman

It’s not hard to show the Laplace transform continues to a function on the puncture place $\mathbb{C}\setminus\{-1\}$ , with good enough decay to shift contours and invert the transform. If the singularity were a pole, the function would be of the form $P(x)\exp(-x)$ where P is a polynomial, and the boundedness shows that $P$ is constant. How do you show the singularity is a pole?

19 October, 2016 at 9:46 am

Anonymous

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

19 October, 2016 at 9:39 am

Anonymous

Is 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 $f(z) = \sum_n a_n z^n$ be a nontrivial entire function, and denote $m_f(r) := min_{|z| = r} |f(z)|$ for each $r \geq 0$ . Then if $|a_n| \leq b_n$ where $b_n$ is a given sequence decaying to zero “sufficiently fast”, then there is a function $g(r)$ defined for $r \geq 0$ and depending only on the sequence $b_n$ such that $m_f(r)$ is not decaying to zero faster than $g(r)$ .
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 $L^1(\mathbb R)$ 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 Manugal

Let $\mathbf{a}$ be a random variable and $\mathbf{x}$ be Poisson distributed with mean $x$. The expectation is $\mathbb{E}[\mathbf{a}] = e^{-x} \sum a_n \frac{x^n}{n!}$ . The Poisson distribution as with mean $x$ as $x \to \infty$ has that lump moving off to infinity. And $\frac{x^n}{n!}$ should concentrate around $n = x$ . That’s of course pretty bogus. But that’s what I’ve been thinking.

19 October, 2016 at 9:57 pm

Isn’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

Anonymous

Unfortunately, it is only given that $g(x)$ remains bounded as $x \to +\infty$ (but no information given on its possible behavior as $x\to -\infty$ or for complex valued $x$ .)

20 October, 2016 at 9:37 am

Anonymous

Since |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

Anonymous

The example $g(z) : = \exp(-z^{2 n})$ shows that a non-constant entire function can be bounded (with faster than exponential decay) on arbitrarily many straight lines (on which $z^n$ is real) in the complex plane.

21 October, 2016 at 7:03 am

Anonymous

That’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 Percival

It’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

Anonymous

In order to show uniqueness of such linear combination, it seems necessary to find an orthogonal system of polynomials of $e^{-x}$ on $[0, \infty)$ with respect to some weight function.

22 October, 2016 at 9:29 am

Lior Silberman

Colin: 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

rsj

It’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

Anonymous

See the above comment of Av and its reply.

20 October, 2016 at 4:01 am

Anonymous

This 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

elianto84

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

23 October, 2016 at 9:54 am

sergei

Dear 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 Lee

Here is my rough idea: Let $V$ be the space entire functions such that the sequence $(f^{(n)}(0) : n \geq 0)$ satisfies the given condition. Also we define

$\displaystyle \| f \| = \max\{ \sup_{n\geq1} |f^{(n)}(0)|, \sup_{x \geq 0} |e^x f (x)| \}.$

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

$\displaystyle I(f) = x \mapsto - \int_{x}^{\infty} f(t) \, dt.$

It is easy to check that $I$ is an injective linear operator on $V$ . Thus $I$ is an isomorphism. The most important implication of this observation is that if $f \in V$ then $f' \in V$ , since $f = I(g)$ for some and hence $f' = g \in V$ . Now since $V$ is finite dimensional, there exist $n$ and $c_0, \cdots, c_n \in \Bbb{R}$ such that

$\displaystyle \sum_{k=0}^{n} c_k f^{(k)} = 0.$

But we know how to solve this differential equation, and the only possible choice is $f(x) = c e^{-x}$ for some $c \in \Bbb{R}$ .

25 October, 2016 at 5:03 am

Sangchul Lee

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

25 October, 2016 at 5:19 am

Iosif Pinelis

Your idea seems interesting. (You don’t need to check that $V$ is a Banach space; that is, you don’t need to check that $V$ 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 $\ell^1$ is compact (which is of course false.)

25 October, 2016 at 5:30 am

Sangchul Lee

Thank 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 Pinelis

It seems that in the proof of the compactness you would have to engage the boundedness of $e^x|f(x)|$ . 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 Lee

I agree with you. Now I am now more prone to think that complex analysis is inevitable, thinking that proving $\dim V < \infty$ is essentially equivalent to proving that the Laplace transform $\mathcal{L}f(s)$ has pole at $s = -1$ .

27 October, 2016 at 2:45 am

Paolo Bonzini

Functions 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?

13 March, 2019 at 8:02 pm

Iker Martinez

I thought a purely real variable solution.
We have that $e^{x}$ and $f(x)$ are both analytic, so their product $\phi(x)=e^{x}\cdot f(x)$ is, therefore $\phi(x)=\overset{\infty}{\underset{n=0}{\sum}}c_n x^n$ , where $c_n= a_0\cdot\frac{1}{(n-1)!}+...+\frac{a_i}{i!}\cdot\frac{1}{(n-i)!}+...+\frac{a_n}{n!}$ .

Since $\underset{x\to \infty}{\lim}\phi(x)=\underset{x\to \infty}{\lim}\dfrac{f(x)}{e^{-x}}=C$ , for every $\epsilon>0$ there are $R>0$ and $N\in\mathbb{N}$ such that:

$\displaystyle \Bigg|\overset{k}{\underset{n=0}{\sum}}c_n x^n - C \Bigg|\leq \Bigg|\overset{k}{\underset{n=0}{\sum}}c_n x^n - \overset{\infty}{\underset{n=0}{\sum}}c_n x^n \Bigg|+ \Bigg|\overset{\infty}{\underset{n=0}{\sum}}c_n x^n - C \Bigg|\leq \dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon.$

for all $k> N$ and all $x> R$ . Thus,

$\displaystyle C- \epsilon < \overset{k}{\underset{n=0}{\sum}}c_n x^n N, x > R$

this implies that,

$\displaystyle 0=\underset{x\to \infty}{\lim}\dfrac{C- \epsilon}{x^k} < c_k N.$

From this we deduce that $\phi(x)=\overset{N}{\underset{n=0}{\sum}}c_n x^n$ and because $\underset{x\to \infty}{\lim}\phi(x)=C$ , $\phi(x)$ must be constant $C.$ Hence, $f(x)=Ce^{-x}$ , where the equations $a_n=C(-1)^n$ follows.

13 March, 2019 at 8:46 pm

Iker Martinez

I dont know how to write math here, so I posted my solution here, https://math.stackexchange.com/questions/3147563/a-real-variable-solution-of-a-problem-posted-on-terry-taos-blog/3147565#3147565.

2 May, 2019 at 12:20 am

amsmath

There are three flaws in this “proof”. First, your coefficients seem wrong to me. Second, the limit of $\phi$ at $\infty$ might not exist. Third, you assume that the series converges uniformly on $(R,\infty)$ , which you cannot do.

26 January, 2021 at 7:46 pm

Terence Tao

Eugenia Malinnikova has just pointed out to me that the statement of this problem is similar to the Hardy uncertainty principle, and the solution to the problem that had been worked out in the comments is in fact very similar to Hardy’s second proof of that principle in his original paper.

14 June, 2024 at 1:48 pm

Terence Tao

As an experiment, I have explained the solution to this problem to ChatGPT at this link. Through an interactive discussion, GPT was able to expand out my sketch of a proof into a full LaTeX document, which one can find here. This experiment was meant to illustrate a recent comment I made in this interview, envisioning how in the future a mathematician might produce a proof through conversation with a GPT, though ideally the GPT should also be able to provide a running formalization of the proof in a language such as Lean in order to guarantee 100% correctness of the argument (this is the main reason why I think the full workflow that I envision is still a few years away).

15 June, 2024 at 11:44 am

Anonymous

If you use double dollar signs instead of single dollar signs when messaging ChatGPT, your latex will render in your own message.

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A problem involving power series

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A problem involving power series

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