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This is a whimsical question, motivated purely by curiosity rather than for any application.

We are all familiar with countless mathematical results which use Archimedes' constant $\pi$ either in their statement, or in their proof. However, in almost all cases that I am aware of, the precise numerical value of $\pi = 3.14159\dots$ does not play any significant role in the arguments; in some hypothetical alternate mathematical universe in which $\pi$ somehow managed to take on some different positive real numerical value, most proofs involving $\pi$ would continue to be locally valid despite the global inconsistency created.

So my question is whether there are any examples of mathematical theorems where a numerical bound on $\pi$, e.g., $\pi \geq 3$ or $\pi \leq 22/7$, or even much cruder bounds such as $\pi \geq 1$, actually played an important role in the proof, in that the proof would fail or require significant modification if such a bound was not known. (Here I exclude the really trivial bounds $0 < \pi < +\infty$, which certainly are implicitly used all over the place.) To put it another way: how many digits of $\pi$ do we need to know in order to do modern mathematics?

This can be contrasted with the analogous question for Euler's number $e=2.71828\dots$, where one can certainly think of places where it was important to know that, for instance, $e \geq 2$ (and certainly $e>1$ is extremely important!). But I can't think of many examples involving $\pi$; this constant seems to be a lot less "willing" to behave like a genuinely dimensionless quantity than say $e$, $\sqrt{2}$, or the golden ratio $\phi$, in that it rarely interacts in an additive fashion with such quantities, and its presence often cancels itself out at the end of a calculation.

To avoid some degenerate answers, let me exclude the following categories of theorems from this question:

  1. Theorems about $\pi$ itself, such as Lindemann's theorem that $\pi$ is transcendental, or mathematical coincidences involving $\pi$ such as the Feynman point.
  2. Theorems involving a lengthy numerical calculation involving expressions that contain $\pi$. An example might be a PDE result which relied on some numerical estimation of an integral expression involving $\pi$, or an eigenvalue of an operator which also involved $\pi$ somehow. [I would waive this exclusion though if one could make a clear case that the numerical value of $\pi$ played a particularly decisive role in the calculation, as opposed to the appearance of $\pi$ merely being an artefact of one's normalization conventions (e.g., for the Fourier transform) that played no actual role in the final theorem.]
  3. Theorems arbitrarily comparing numerical quantities, at least one of which involves $\pi$ implicitly or explicitly. An example of this might be a result comparing the volume of a cube and a ball in some dimension, but without any motivation for why it would be interesting to know which one is larger.
  4. Theorems in which a numerical bound involving $\pi$ was replaced with a slightly weaker numerical bound not involving $\pi$, purely to make the final conclusion look nicer. (Suggested by JoshuaZ in the comments.)
  5. Theorems that gratiutiously use $\pi$ in their proof, but for which the proof can be easily modified to avoid mention of $\pi$ (or related quantities). (Suggested by Christian Remling in the comments; for instance, the statement that a square inscribes a circle of diameter equal to the sidelength can be proven directly without explicit mention of $\pi$.)

I'll start with one near-example to this question: in some sieve-theoretic applications in analytic number theory, it is useful to know that the average value of $|\sin x|$, namely $\frac{2}{\pi}$, is strictly less than one (see, e.g., the end of Section 4 of this blog post of mine on a Banach algebra proof of the prime number theorem). Ostensibly this is using the lower bound $\pi > 2$. However, one doesn't actually need to know about $\pi$ to see this fact, since it is immediate just from the inspection of the graph of $|\sin x|$ that its average value is going to be less than one! The fact that this average value also happens to be evaluated explicitly as $\frac{2}{\pi}$ is, as far as I know, inessential to the applications that I am aware of, and so the appearance of $\pi$ here is (almost literally) tangential. (Instead, one can view this argument as providing a proof of the lower bound $\pi>2$, and thus is basically a result of the first category of exclusions; one can also view this particular example as also belonging to the fifth category.)

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    $\begingroup$ I was going to answer Apéry's proof of irrationality of $\zeta(3)$. But I looked at it and in fact it does not fit here. Some asymptotics in there involve $\pi$, but the ones needed for the irrationality proof involve only $e$ and $\sqrt2$. (The $\pi$s all cancel.) $\endgroup$ Commented May 12 at 16:58
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    $\begingroup$ Does the Basel problem count? I have read, that Euler knew the numerical value of $\pi^2/6$ so that he could guess and prove it: eulerarchive.maa.org/hedi/HEDI-2003-12.pdf $\endgroup$ Commented May 12 at 16:58
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    $\begingroup$ Somewhat along the lines of your near-example, Bourgain's Sum-free subset theorem goes through a painful case analysis where the inner product of various trigonometric series is taken against test functions and one needs a few digits of pi to conclude that the results are sufficient to conclude the argument. See 3.15 through 3.23 at link.springer.com/article/10.1007/BF02774027. $\endgroup$
    – Mark Lewko
    Commented May 12 at 16:58
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    $\begingroup$ @MarkLewko Ah, yes, this is exactly the kind of thing I had in mind! It might be fun to see exactly what bounds on $\pi$ are needed here in order to obtain Bourgain's final conclusion (which I already had a fondness for, as it is a good candidate for minimizing the ratio (degree of improvement over trivial bound)/(mathematical strength of author)). $\endgroup$
    – Terry Tao
    Commented May 12 at 17:25
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    $\begingroup$ Searching for uses of the bounds in Lean, I can see that $\pi>3$ is used for proving that the ring of integers of $\mathbb{Q}[\zeta_3]$ and $\mathbb{Q}[\zeta_5]$ is a PID. Not sure if it fits the requirements. The upper bound I found only in here where it seems to be used for upper bounds on diagonal Ramsey numbers. We could search for such uses in other proof assistants. $\endgroup$ Commented May 12 at 17:31

9 Answers 9

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I'm not sure if this qualifies, but there are cases where it is important to know bounds for $\zeta(2k)$. In particular, if you are willing to equate $\zeta(2)$ with $\pi^2/6$, then here is an example where one needs a numerical bound for $\pi$.

Definition: Let ${\mathcal D}=(D_n)_{n\ge1}$ be a sequence of positive integers. We say that a prime $p$ is a primitive divisor of $D_n$ if $p\mid D_n$ and $p\nmid D_m$ for all $m<n$. The Zsigmondy set of $\mathcal D$ is $$Z(\mathcal D)=\{n\ge1:\text{$D_n$ has no primitive divisors}\}.$$

Theorem: Let $\mathcal D$ be an elliptic divisibility sequence (definition below). Then $Z(\mathcal D)$ is finite.

One step in proof (ZBL0654.10019 Lemma 9) uses the fact that $2-\zeta(2)>0$, i.e., $\pi<3.464$. More precisely, the quantity $$ n^2 - \sum_{\substack{k\mid n\\k<n\\}} k^2 \ge \left(2-\frac{\pi^2}{6}\right)n^2 \tag{$*$}$$ needs to grow as fast as some positive multiple of $n^2$.

Definition: Let $E/\mathbb Q$ be an elliptic curve given by a Weierstrass equation, let $P\in E(\mathbb Q)$ be a non-torsion point, and write the $x$-coordinate of the multiples of $P$ as $x(nP)=A_n/D_n^2$. Then the sequence $(D_n)_{n\ge1}$ is called an elliptic divisibility sequence.

Addendum: The estimate $(*)$ uses the fact that $2>\zeta(2)=\pi^2/6$, but it would have sufficed to know that \begin{aligned} 0 &< \limsup_{n\ge1} \left\{2-\sum_{k\mid n}\frac{1}{k^2}\right\} = \limsup_{n\ge1} \left\{2-\prod_{p\mid n}\left(1+\frac{1}{p^2}\right)\right\}\\ &= \limsup_{n\ge1} \left\{2-\prod_{p\mid n}\left(1-\frac{1}{p^4}\right)\left(1-\frac{1}{p^2}\right)^{-1}\right\}\\ &= 2-\frac{\zeta(2)}{\zeta(4)} = 2 - \frac{\pi^2}{15}. \end{aligned} Thus the required inequality is $\pi<\sqrt{30}\approx5.477$, which means that Terry's $2<\pi<4$ would suffice.

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    $\begingroup$ Yes, this qualifies! I had thought to add a calculation in one of my own recent papers (page 21 of arxiv.org/pdf/2309.02325) where I needed the fact that $\zeta(2) \zeta(3)/\zeta(6) = 1.94359\dots$ was less than $2$, though this expression is not purely expressible in terms of $\pi$ so this is less compelling of an example. $\endgroup$
    – Terry Tao
    Commented May 12 at 18:48
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    $\begingroup$ One could argue this falls under exception 5, since just the trivial estimate $\sum \frac{1}{n^2} < 2 \sum \frac{1}{n(n+1)}=2$ suffices, with no need to mention $\pi$. I suppose it's a matter of taste though, since 'numerical bounds for $\pi$' is equivalent to 'numerical bounds for $\sum \frac{1}{n^2}$', but at least this way one never needs to actually calculate $\pi$ or invoke the Basel problem. $\endgroup$ Commented May 13 at 15:56
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Viazovska's two Annals papers on sphere packings require verification of several inequalities involving $\pi$. For example from Section 5 of the first paper:

(Someone might test how sensitive these bounds are to the value of $\pi$. Maybe there are more examples concerning modular forms. How about $e^{\pi\sqrt{163}}$?) These are auxiliary inequalities to prove one of the two main inequalities, which Romik (2023) gave proofs that do not rely on computer calculations but still requires proving some bounds (see Equation 20 and Lemma 2, 3) involving $\pi$, $e$ and $\Gamma(1/4)$. It seems the second paper still rely on computer calculations though.

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    $\begingroup$ Very nice example! In retrospect it makes a lot of sense that questions about optimal sphere packing would require bounds on $\pi$, although the way $\pi$ comes into Viazovska's work seems to be highly nontrivial in this case, and not just through the obvious relation between $\pi$ and the volume of a ball. $\endgroup$
    – Terry Tao
    Commented May 13 at 14:32
  • $\begingroup$ Yeah, I'd say the appearance of $\pi$ here is from evaluation of Fourier series of modular forms on vertical lines in the upper half plane. $\endgroup$
    – Junyan Xu
    Commented May 13 at 16:27
  • $\begingroup$ Seewoo Lee just announced an algebraic proof in the 24 dimensional case as well! x.com/antimath3/status/1805046798765174904 $\endgroup$
    – Junyan Xu
    Commented Jun 24 at 3:13
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    $\begingroup$ @TerryTao I tend to view the appearance of 𝜋 in the formulas for the optimal density of sphere packing as an artifact of the way that human mathematicians have decided to formulate the question. As you said, the 𝜋 just comes from the formula for the volume of a unit ball. It may be more natural to measure the "density" of a sphere packing by the number of spheres per unit volume instead of by the ratio of the volume of space the spheres occupy. When the question is formulated in that way, 𝜋 doesn't appear in the answer (in dimensions 1, 2, 3, 8 and 24 where the answer is known). $\endgroup$
    – Dan Romik
    Commented 16 hours ago
  • $\begingroup$ ... From that perspective, the fact that numerical bounds for 𝜋 end up being relevant to the workings of the proof of Viazovska’s theorem does actually seem pretty mysterious, at least to me. $\endgroup$
    – Dan Romik
    Commented 16 hours ago
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A lot of Minkowski related examples might work here. For example:

Let $D$ be the Discriminant of an algebraic number field $K$ with degree $n$ over $ \mathbb{Q}$, where $r_1$ and $r_2$ are the numbers of real and complex embeddings. Then every class in the ideal class group of $K$ contains an integral ideal of norm at most $$\sqrt{|D|} \left(\frac{4}{\pi}\right)^{r_2} \frac{n!}{n^n} \ . $$

Understanding how this behaves for various fields sometimes uses that $\pi <4$. Other estimates connect this bound to Stirling's formula for $n!$.

Another related direction, also coming from Minkowski type arguments is the Minkowski-style proof of Lagrange's four square theorem, where $\pi >2$ is used. More accurate estimates might be used for results about other similar quadratic forms.

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    $\begingroup$ Nice! In retrospect I guess I can see that the bounds $2 < \pi < 4$ (which geometrically correspond to the optimal ways to inscribe or circumscribe a disk by a square) would play some non-trivial role in mathematics. Perhaps the bound $\pi>3$ (which geometrically corresponds to the optimal way to inscribe a disk by a regular hexagon) would also show up for similar reasons. But I wonder how much of modern mathematics would actually be lost if for some reason the best numerical accuracy we had for $\pi$ was $2 < \pi < 4$... $\endgroup$
    – Terry Tao
    Commented May 12 at 17:27
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    $\begingroup$ This was the example that came to my mind when I first read Terry's question. When I took a first course in algebraic number theory many years ago, applying this inequality was the only time during the whole course that I had to pull out my calculator. For example, sometimes we used the Minkowski formula to prove that the class number of a particular number field was 1. But I just took a quick look at the assigned homework questions, and none of them needed much numerical accuracy for $\pi$. $\endgroup$ Commented May 13 at 3:42
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    $\begingroup$ This answer mathoverflow.net/a/26504/3332 apparently requires tighter bounds to prove certain number fields have no unramified extensions. (found through math.stackexchange.com/q/699128/12932) $\endgroup$
    – Junyan Xu
    Commented May 13 at 7:29
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    $\begingroup$ In some cases tighter bounds on $\pi$ enter the reckoning. For example, $\pi^2>480/49$ enters into rendering the class group of $\mathbb{Q}[\sqrt{30}]$. $\endgroup$ Commented May 13 at 19:01
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    $\begingroup$ Make that $\mathbb{Q}[\sqrt{-30}]$. $\endgroup$ Commented May 13 at 20:01
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How about an example of where the numerical value of $\pi$ is what makes a problem not solved? The moving sofa problem. See Dan Romik's paper.

An original upper bound on the moving sofa constant given by Hammersley is $2\sqrt{2}\approx 2.828$ which has been improved to $2.37$ by Kallus and Romik.

There is an example given below of a sofa with area $\pi/2+2/\pi\approx 2.2074$. Note that this is not a solution because $\pi/2+2/\pi<2.37.$

Actually, $x/2+2/x=2.37$ has two solutions, $x\approx 1.09843,3.64157.$ In particular for $x\in (1.09843,3.64157)$ we have that $x/2+2/x<2.37$.

The example given below is not optimal because $\pi\in(1.09843,3.64157)$.

Sofa

I believe this qualifies because this shape is a natural attempt and one of the first things one would try. The problem is interesting and nontrivial precisely because one of the first things one would try doesn't work. The $\pi$ shows up due to circular shape and hence isn't an arbitrary choice.

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  • $\begingroup$ The numerology in this answer is amusing, but the numerical value of 𝜋 is not why the problem is unsolved, and the fact that $\pi/2+2/\pi<2.37$ is not why Hammersley's sofa, which has area $\pi/2+2/\pi$, is not the solution to the moving sofa problem. The number 2.37 is only an upper bound, so its existence does not preclude the possibility that Hammersley's sofa might be the solution. On the other hand, Hammersley's sofa is indeed not the solution, but that's for a different reason than what you said - namely, because another shape, known as Gerver's sofa, has a larger area of around 2.2195. $\endgroup$
    – Dan Romik
    Commented yesterday
  • $\begingroup$ So, one can make the correct statement that the numerical value of 𝜋 (together with the numerical value 2.2195 of Gerver's sofa) is why Hammersley's sofa is not the one with maximal area. But it is not why the problem is unsolved. The problem is unsolved simply because no one has been able to prove the conjectured maximality of Gerver's sofa, or to disprove it. This has nothing to do with my and Yoav Kallus's upper bound of 2.37, which is simply the best known bound at present. $\endgroup$
    – Dan Romik
    Commented yesterday
  • $\begingroup$ (By the way, the figure in your answer shows Gerver's sofa, not Hammersley's sofa. They look pretty similar to each other though.) $\endgroup$
    – Dan Romik
    Commented yesterday
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In the paper, Space vectors forming rational angles, by Kiran S. Kedlaya, Alexander Kolpakov, Bjorn Poonen, and Michael Rubinstein, the authors classify all sets of nonzero vectors in $\mathbb{R}^3$ such that the angle formed by each pair is a rational multiple of $\pi$. Surprisingly, the proof relies heavily on high-precision floating-point calculations, and $\pi$ is heavily implicated in these calculations. Exactly how many digits of $\pi$ are needed, though, I'm not sure; I'll see if I can contact the authors for more detailed information.

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    $\begingroup$ Thanks for alerting me to this paper. Conway had me work on the rational tetrahedra problem in 1975. I got nowhere on it. Nice to know Kedlaya et al have laid it to rest. $\endgroup$ Commented May 13 at 6:47
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    $\begingroup$ If you're referring to page 13-14 I think they don't actually use the value of $\pi$. The cosine values of rational multiples of $\pi$ are cyclotomic integers and in particular solvable by radicals, so you don't need $\pi$ to express them. Also, the numerical computations are to filter out non-examples and identify candidates, and eventually the candidates are verified to be actual examples by proving strict identities. It should also be possible to prove the non-examples symbolically, because it's just verifying non-identities in a lattice of rank at most $\phi(420)=\phi(280)=96$. $\endgroup$
    – Junyan Xu
    Commented May 13 at 7:51
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    $\begingroup$ @JunyanXu I just checked with Kiran Kedlaya. He said that they did in fact calculate rational multiples of $\pi$ numerically and then take the cosine. But he also emphasized that this wasn't the only way to do the calculation; they just took the most expedient route. As for how many digits of precision, his off-the-cuff guess was 50 digits or so. $\endgroup$ Commented May 13 at 12:08
  • $\begingroup$ Yeah I have no doubt they performed the computation, as the code is provided. I just want to remark the problem tackled is essentially discrete, and it happens in this case that numerical computations involving $\pi$ offer a more effective way to disprove such discrete identities than a symbolic approach. To me, the results in the paper depend on the value of $\pi$ as much as $\cos(\pi/3)=1/2$ depends on it. $\endgroup$
    – Junyan Xu
    Commented May 13 at 16:15
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    $\begingroup$ @JunyanXu I agree with your first and second sentences but I don't think your third sentence is fair. The problem is at the edge of our computational abilities, so if we care about producing an actual proof (as opposed to just claiming that we could produce a proof one way or the other if we had more powerful computers), then computational efficiency becomes relevant. Or to put it another way, if we didn't know $\pi$ to enough decimal places, then the problem might still be open today. $\endgroup$ Commented May 14 at 0:28
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We have a deterministic recursive relation to determine all (natural) prime numbers in ascending order:

$p_1=\lim\limits_{k\to\infty}[\zeta(k)-1]^{-1/k}(=2)$

$p_{n\ge2}=\lim\limits_{k\to\infty}\left[\zeta(k)\prod\limits_{m=1}^{n-1}\left(1-\dfrac{1}{p_m^k}\right)-1\right]^{-1/k}$

This is, essentially, just an implementation of the Sieve of Erastothenes (see the addendum below.).

Given the infinite sequence $\zeta(2)=\pi^2/6,\zeta(4)=\pi^4/90,\zeta(6)=\pi^6/945,...$ and a sufficiently accurate rendering of $\pi$ (or, if you prefer, of $\pi^2$), we could in principle use this formula to churn out all primes in succession.

When I tried this decades ago with a calculator having eight decimal place precision, I got as far as

$2,3,5,7,11. \tag{Oh well.}$

(Using exact values of $\zeta(2k)$ up to $k=6$ and Microsoft Excel for the calculations I extended the list above to include $13$ and $17$.)


Addendum:

Here is how the formuka is derived (I will make no discovery claim, someone(s) must gave exolored this before I was born!)

For $k>1$ we have

$\zeta(k)=1+(1/2^k)+(1/3^k)+(1/4^k)+(1/5^k)+...$

If we subtract the $1$ we see that the leading term that remains is $1/2^k$ which makes $[\zeta(k)-1]^{-1/k}$ converge to $2$. As this is the next natural number after $1$ it cannot gave any factors between and so is a prime.

Now multiply the $\zeta$ function by $1-(1/2^k)$ to get rid of the term with $2$. Then all terms with multiples of $2$ go away just as they would with the Sieve of Erastothenes, leaving

$\zeta(k)[1-1/(2^k)]=1+(1/3^k)+(1/5^k)+(1/7^k)+(1/9^k)+...$

Now subtracting $1$ and raising to the power of $-1/k$ gives convergence to $3$, and all possible prime factors below $3$ are removed by the multiplication/sieving process so $3$ must be the next prime.

We keep going. To explore numbers not divisible by either of thevprimes $2$ and $3$, we tack on a factor of $1-(1/3^k)$ which removes not only $3$ but all its remaining multiples:

$\zeta(k)[1-1/(2^k)][1-1/(3^k)]=1+(1/5^k)+(1/7^k)+(1/11^k)+(1/13^k)+...$

where the $1/5^k$ term renders $5$ as the first number greater tgan $1$ but not divisible by the lower primes $2,3$,therefore the third prime. In principle we can iterate this process indefinitely, but in practice the formula is ill-conditioned and we require extremely precise values for the $\zeta$ functions at various $k$ to keep on track in a numerical computation.

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  • $\begingroup$ How do you derive this formula? $\endgroup$ Commented May 13 at 12:41
  • $\begingroup$ Yes I first upvoted the after numerical check I was not so sure and downvoted. If you like to expand on your comment or give a reference for the formula, that would be nice and I would upvote again. $\endgroup$ Commented May 13 at 12:55
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    $\begingroup$ sagecell.sagemath.org/?q=xrmruh $\endgroup$ Commented May 13 at 12:57
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    $\begingroup$ @mathoverflowUser for $k<\infty$ your expression should be below the prime. You miss that for $13$ because you truncate the $\zeta$ function sum a little early. But using the truncated sum together with the full infinite series sum allows you to bracket the limit and prove (to your accuracy) the prime $13$. Higher than $13$, however, would require including more terms in the $\zeta$ function sum. $\endgroup$ Commented May 13 at 13:51
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    $\begingroup$ This is an interesting example: if one works numerically by evaluating only finitely many values of $k$ (and either rigorously estimating the error, or just assuming that it is sufficiently small), then these formulae do require numerical bounds on $\pi$; but if one is genuinely evaluating the limit, then I don't think numerical bounds on $\pi$ actually makes an appearance, because the asymptotics for Bernoulli numbers also involve $\pi$, and all the factors of $\pi$ should cancel out (indeed, the arguments become circular at this point). $\endgroup$
    – Terry Tao
    Commented May 13 at 14:41
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In another answer I explored the relationship between precise values of $\pi$, the zeta function and prime numbers. In this one I consider the role of bounding $\pi$ on proving the class group of an integer domain.

The integers of the imaginary quadratic field $\mathbb{Q}[\sqrt{-30}]$ have three ramified primes $2,3,5$, which factor into ideals as:

$(2)=(2,\sqrt{-30})^2$

$(3)=(3,\sqrt{-30})^2$

$(5)=(5,\sqrt{-30})^2$

The ideals on the right sides of these equations are not principal, but by construction their squares are, and the trilinear product of these three ideals is also principal. These conditions guarantee that the class group of $\mathbb{Q}[\sqrt{-30}]$ will contain the Klein four-group.

Now to prove that the class group of this field is the Klein four-group only, we would show that the Minkowski bound

$M=(2/\pi)\sqrt{120}<7\iff\pi^2>480/49.$

There is a neat way to prove this bound. In this answer a Fourier series is used to render the series

$\pi/2\sqrt2=+(1/1)+(1/3)-(1/5)-(1/7)...$

where the denominators are consecutive odd numbers and the indicated sign pattern from the first four terms repeats. By evaluating groups of four terms following the initial $1$ we can show that any truncation after $4n+1$ terms underestimates $\pi/2\sqrt2$ and the error falls roughly in proportion to $1/n^2$. Well, for $n=2$ this lower bound leads to

$\pi/2\sqrt2>1.10773...\implies\pi^2>9.8165...$

versus $480/49=9.7959...$, proving the claim.

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André Voros defined a sequence of real numbers $\Lambda_1$, $\Lambda_2$, ..., which unlike the classical Keiper-Li coefficients can be expressed in closed form using a power of $\pi$.

The main claim of 2204.01036 and other earlier work is that the Riemann Hypothesis is true if and only if some asymptotic tail of this sequence is positive.

The positivity of the first three numbers translates into the following inequalities: $$ \pi > 3 $$ $$ \pi^4 5^7 < 2^7 3^{11} $$ $$ \pi^8 5^{25} > 2^3 3^{22} 7^{11} $$

Below is a screenshot of the actual numbers, taken from page 12 of 1703.02844. They are all defined in terms of Bernoulli numbers by a convoluted but completely explicit formula.

Screenshot from Voros

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I post this answer because it gives a the necessary lower bound for $\pi$ coming from harmonic analysis. In particular H. Helson in his article "Hankel forms" asked if there exists an analogue of Nehari's theorem for the infinite dimensional polydisc.

This is equivalent to ask that there exists a dimension independent constant $C>0$ such that if the Hankel form $H_\psi(f,g) = \langle \psi f , g \rangle_{H^2(\mathbb{T}^d)} $ is bounded, (here $H^2(\mathbb{T}^d$) is the Hardy space in the $d$-dimensional polydisc) then there exists a bounded function $\varphi \in L^\infty(\mathbb{T}^d)$ such that $ \psi = P_+\varphi $, where $P_+$ is the Riesz projection, and $$ \Vert \varphi \Vert_\infty \leq C \Vert H_\psi \Vert = C \sup_{\Vert f \Vert \leq 1, \Vert g \Vert\leq 1} |H_\psi(f,g)|. $$

In the article ``A lower bound in Nehari's theorem'' Seip and Ortega-Cerda gave a negative answer to this question showing that the optimal such $C$ in dimension $d$ is bigger than $(\pi^2/8)^{d/4}$. This blows up as $d\to \infty$ since $\pi > 2\sqrt{2}$.

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