Background: It is currently unknown whether $e$ is normal. A natural way to approach this question is to find a class to which $e$ belongs, and prove all members of that class are normal. For example, if we want to know whether $\sqrt{2}$ is normal, it makes sense to consider the class of irrational algebraic numbers, but it is still an open problem whether every irrational algebraic number is normal. Finding a counter-example to this conjecture, or a similar conjecture for an appropriate class containing $e$, would be quite useful to understanding the problem in general.

Differential Rings with Composition: One natural class containing $e$ seems to be numbers definable without parameters in the language of differential rings with composition, say in the space of analytic functions on $\mathbb{C}$. The language of differential rings with composition is $(0,1,+,*,\partial,\circ)$, where $0$, $1$ are the additive and multiplicative identities (constant functions), $+$ and $*$ are addition and multiplication, $\partial$ is a derivation (in this case differentiation), and $\circ$ is composition. We can define the constant function $e$ in this language by the formula $$\psi(x) : \exists f [(\partial f = f) \wedge (f \circ 0 = 1) \wedge (f \circ 1 = x)]$$

Question: Is there a non-normal irrational number definable without parameters in the ring of analytic functions on $\mathbb{C}$ in the language of differential rings with composition?

I expect this is quite a hard question, so answers slightly modifying the question would also be welcome. For example, perhaps it helps to generalize to algebraic elements in this differential ring with composition, rather than just definable elements, or to work in a different differential ring into which $\mathbb{R}$ embeds. This is a larger class of numbers than algebraic numbers, so in principle it should be an easier question to answer in the positive than the corresponding question of whether there is a non-normal irrational algbraic number.

  • $\begingroup$ If anyone has suggestions for additional relevant tags, that would be much appreciated. $\endgroup$
    – James
    Mar 15 '19 at 23:50
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    $\begingroup$ I'm fairly certain that you can define the integers, which allows you to define the rationals. If you do put in the absolute value function then the reals as a subset of the complex numbers would be definable (as the closure of the rationals, essentially). Now at this point every computable real number would be definable, including many non-normal numbers, since the cut of any computable real number is definable in the rational. $\endgroup$ Mar 16 '19 at 3:41
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    $\begingroup$ I don’t know enough complex analysis to be sure, but I suspect $\mathbb R$ is indeed definable in the structure. Since we can define $\mathbb Z$ with its usual arithmetic structure and $\mathbb Q$, we can also define $X=\{f(0):\forall n\in\mathbb N\,f(1/n)\in\mathbb Q\}$. This is a subring of $\mathbb R$ containing $\mathbb Q$. Could it be that $X=\mathbb R$? Or if not, what about $\{f(q):q\in\mathbb Q,\forall n\in\mathbb N\,f(1/n)\in\mathbb Q\}$? If it helps, we may also use, say, arbitrary $\Sigma^0_k$-definable sequences of rationals converging to $0$ in place of $1/n$. $\endgroup$ Mar 16 '19 at 16:08
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    $\begingroup$ So far as I can tell, $\partial$ is superfluous here. We have $$g=\partial f \iff \forall a\, \exists h\, \forall z\, [\forall x[a \circ x=a] \implies$$ $$f \circ z = f \circ a + (z-a)(h \circ z)\ \wedge\ g \circ a = h \circ a]$$ $\endgroup$
    – Matt F.
    Mar 19 '19 at 19:49
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    $\begingroup$ @MichaelBächtold That’s just a name for the formula defined after the colon. $\endgroup$ Mar 20 '19 at 13:47

We can define Greg Martin's absolutely abnormal number in this language, based on Martin's paper as in the arxiv or the American Mathematical Monthly.

Our number which is abnormal in all bases is $\alpha=M(1)$, where we will define the function $$M(z) = \prod_{j=2}^\infty \left( 1-\frac{z^{2^j}}{d_j}\right)$$ using a sequence of $d$'s defined by $d_2=4$ and $$d_{j+1} = (j+1)^{d_j/j}$$ for $j\ge 2$. The $d$'s increase so super-exponentially that $M$ is an analytic function.

Before we can define $M$ we need some preliminaries. They are slightly tricky because we can not define a global function $x^k$, but can define it when $k$ is a natural number. Similarly, we can not define $j<k$ globally, but can define it when $j$ and $k$ are both natural numbers. \begin{align} z=e^y &:= \exists f [\partial f = f \wedge f \circ 0 = 1 \wedge f \circ y = z]\\ n\in Z &:= \forall y [e^y=1 \implies e^{ny}=1]\\ n\in N &:= \exists w,x,y,z [n=w^2+x^2+y^2+z^2 \wedge w,x,y,z\in Z]\\ j < k &:= \exists i[i,j,k\in N \wedge i+j+1=k]\\ Power(p,k) &:= \forall u[\partial u = 1 \land u\circ0=0 \implies u\partial p = kp \wedge p \circ 1 = 1]\\ y=x^k &:= \exists p[Power(p,k) \wedge y=p\circ x]\\ Legendre(q,n) &:= \forall u[\partial u = 1 \land u\circ0=0 \implies \partial((1-u^2)\partial q)+n(n+1)q=0]\\ Poly(p,k) &:= \forall n,q,v[k < n \wedge Legendre(q,n) \wedge \partial v = pq \implies v \circ (-1) = v \circ 1]\\ Coef(f,k,c) &:= \exists p,q,r [f=pq+r \wedge Power(p,k) \wedge q \circ 0 = c \wedge Poly(r,k-1)]\\ \end{align}

Now we will abbreviate $M_k$ for the $k$th coefficient in the power series expansion of $M$ at 0, i.e. the number such that $Coef(M,k,M_k)$. We will abbreviate $d_k$ for $-1/M_{2^k}$.

Then, finally, $\alpha$ is defined by \begin{align} \exists M[& M \circ 1=\alpha\\ &\wedge M_0=1 \wedge M_1=0 \wedge M_2=0 \wedge M_3=0 \wedge d_2=4\\ &\wedge \forall j[1<j\implies d_{j+1}^j=(j+1)^{d_j}]\\ &\wedge \forall j,k[j<2^k \implies M_{j+2^k} = M_j M_{2^k}]] \end{align}

The validity of these definitions uses several facts:

  • for $Power(p,k)$ and $Legendre(q,n)$, the fact that they solve standard differential equations;
  • for $Poly(p,k)$, the completeness of the Legendre polynomials;
  • for $M$, the fact that analytic functions are uniquely determined by their power series coefficients.
  • $\begingroup$ Great answer! Thank you! $\endgroup$
    – James
    Mar 19 '19 at 22:16
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    $\begingroup$ This is a clever use of Legendre polynomials. Anyway, once you have Coef, you can easily define $\mathbb R$, for example $x\in\mathbb R$ iff $\exists f\,(f\circ1=x\land\forall k\in\mathbb N\,\exists q\in\mathbb Q\,\mathit{Coef}(f,k,q))$. This implies that the structure is biinterpretable with the standard model of second-order arithmetic. In particular, one can define every real number whose (say) decimal expansion is definable in second-order arithmetic. $\endgroup$ Mar 20 '19 at 13:44

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