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This question is possibly a bit more philosophical $-$ compatible with the Christmas season, which is an appropriate moment to look at the world from a more universal angle... My last question with a similar flavor was a long time ago.

Some mathematical constants come in pairs with their reciprocal. One of the first examples that comes to mind is the golden ratio, sometimes defined as $\phi=\frac{\sqrt{5}+1}{2}$, sometimes as $\phi=\frac{\sqrt{5}-1}{2}$, with the role of both being perfectly ambivalent. This happens of course because, up to sign, $\phi$ and $1/\phi$ are Galois conjugates. Similarly for $i$ and $-i$ (and in fact any pair of conjugate complex numbers of norm $1$, with a special mention of the 3rd roots of unity), and one could argue that $\phi $ and $i$ and $e^{2\pi i/3}$ are essentially just the simplest non trivial cases of solutions to $x^2+px\pm1=0$.

But before you say "this is trivial, and after all, how should one even define a constant?", let me state that for $e$ and $e^{-1}$ the situation is very similar. While exponential growth is of course quite different from exponential decay, the same principle is at work in both. We may reduce this to a simple sign change in the defining formulae $\lim\limits_{n\to\infty}\left(1\pm\frac1n\right)^n$, or take just the sign change between $e^x$ and $e^{-x}$.

Is that already all there is to say about "ambivalent" constants?

On the other side, a striking example of a non ambivalent one is $\pi$, which is a period while $1/\pi$ probably isn't, likewise e.g. for $\ln2$.

Wolfgang
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    The cousin of $\pi$ is $2\pi$ not $1/\pi$... – Dirk Dec 25 '17 at 17:50
  • @Dirk In which sense do you mean "cousin"? To me, a factor 2 seems too trivial (or rather say hazardous) for distinguishing. – Wolfgang Dec 25 '17 at 18:21
  • It's just that it would have been equally possible that $2\pi$ would have the role $\pi$ has today... – Dirk Dec 25 '17 at 18:24
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    $\pi$ was defined as the ratio of circumference and diameter of a circle. When doing geometry, the diameter is much more important than the radius, so this definition makes sense. When doing calculus, the diameter plays no role at all, while the radius is important, most importantly in Cauchy's integral formula. So today one would define $\pi$ as 6.28... . – Jan-Christoph Schlage-Puchta Dec 30 '17 at 09:20
  • @Jan-ChristophSchlage-Puchta I agree. Likewise normalizing Gaussian functions, many inner products, most formulae involving the Gamma function (even though $\Gamma(1/2)=\sqrt{\pi}$) etc. But the distinction ${\pi}$ vs. ${2\pi}$ is not the gist of my question, which is limited to reciprocals. – Wolfgang Dec 30 '17 at 10:11
  • You may want to add $e^{\gamma}$ and $e^{-\gamma}$. – Anixx Oct 15 '19 at 21:43
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    Ambivalent constants are those which are likely to be arguments of logarithm. – Anixx Oct 15 '19 at 21:47
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    For Europeans around 1600-1650, $\pi$ vs $1/\pi$ looked more ambivalent: the first two infinite series for it were actually infinite products. https://en.wikipedia.org/w/index.php?title=Pi&mobileaction=toggle_view_desktop#Infinite_series –  Aug 28 '21 at 21:53

3 Answers3

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Ambivalent constants are those which are likely to be arguments of logarithm. This is because inversion of logarithm's argument merely changes the sign of the logarithm.

All constants usually have negative or reciprocal pair, depending on their typical role as being an argument of logarithm or exponential:

  • $\gamma$ has negative pair $-\gamma$ because it is usually an argument of exponential function rather than logarithm.

  • $\pi$ has negative pair $-\pi$ because it is also an argument of exponential function (Euler's formula).

  • $i$ has both reciprocal and negative pairs, which coincide, because it can be seen on the both sides of Euler's formula: $i=e^{i\pi/2}$.

In general, periods are typically arguments of exponential function, so have a negative pair and their exponentials usually have a reciprocal pair.

Anixx
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    This doesn't seem to address any of the Galois conjugate examples mentioned in the question. – JoshuaZ Aug 28 '21 at 21:18
  • @JoshuaZ well, the fact that the golden ratio has a reciprocal conjugate may indicate that its logarithm may be interesting. https://www.fq.math.ca/Papers1/55-5/Brown.pdf – Anixx Aug 28 '21 at 21:21
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I think we get insight here by asking a more specific question:

  • Geometrically, why is $e$ vs $1/e$ ambivalent when $\pi$ vs $1/\pi$ is not?

It’s easier to talk about $\pi$ first. Consider a square $S$ inscribed in a circle $C$, and their areas.

enter image description here

Then are these ratios ambivalent, or is one more distinguished: $$\frac{area(C)}{area(S)} \ \text{ or } \ \frac{area(S)}{area(C)}\ ?$$

Maybe surprisingly, the answer is clear: the first ratio is more distinguished, because we can approximate it by approximating the area of the circle with the area of smaller squares. More generally, it’s easier to approximate curved areas with polygonal areas than vice versa. (Archimedes would have been ridiculous to calculate $\pi$ by filling a square with small circles.) And therefore $\pi/2$ is more distinguished than $2/\pi$.

We get the same distinction with the circle inscribed in the square, or using perimeters instead of areas.

Talking about $e$ geometrically requires a little more work. Let $h$ be a hyperbola with asymptotes $x$ and $y$. Let $HX$ and $H'X'$ be line segments from $h$ to $x$ which are parallel to $y$.

enter image description here

Suppose the area of the curved region $HH'X'X$ equals the area of the parallelogram between $H$ and the asymptotes. Then are these ratios ambivalent, or is one more distinguished: $$\frac{HX}{H'X'} \ \text{ or }\ \ \frac{H'X'}{HX}\ ?$$

Asked this way, the answer is again clear: these two ratios are ambivalent, because they are both ratios of simple straight line segments. And therefore $1/e$ and $e$ are ambivalent.

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I'd say that every constant which is proved to exist but whose exact value is unknown, qualifies. The best example that comes to my mind is the de Bruijn-Newman constant $\Lambda$, which equals $0$ if and only if the Riemann Hypothesis is true. So far, it has only been proven that $0\leq\Lambda\leq 0.2$.

Sylvain JULIEN
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