# Tag Info

73

Philosophically, there is essentially only one way to prove that a number is irrational/transcendental, which is to use the fact that there is no integer between 0 and 1. That is, one assumes that the number in question is rational/algebraic, and constructs some quantity that can be shown to be bounded away from 0, less than 1, and also an integer. To get ...

44

There are number theorists who understand this subject much better than I do. However, I feel obliged to post an incomplete answer quickly before people have a chance to close this question. There are a lot more connections known between $\pi$ and $e$ and other numbers than between $\gamma$ and other numbers. We can get proofs of their irrationality by ...

17

At the present time, we do not even know how to prove that the Euler-Mascheroni constant $\gamma=\lim_{n\to\infty} \sum_{k=1}^n\frac{1}{k} - \log n$ is irrational, much less transcendental; although it is conjectured to be transcendental. The reason you won't find a lot about this topic online (or in the research literature) is because so little is known ...

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Barry, this post is coming a "little bit" after your class ended, but I want to direct you to a paper on the Lindemann-Weierstrass theorem by Beukers, Bezivin, and Robba in the Amer. Math. Monthly in 1990: https://www.jstor.org/stable/pdf/2324683.pdf. They give a proof of LW theorem that does not explicitly involve a choice of an auxiliary prime number, and ...

14

Irrationality measure is a question about approximation by rationals. The continued fraction expansion gives the best approximations and controls their quality. Irrationality measure is a kind of asymptotic growth of the continued fraction expansion. Asking about the irrationality measure of a particular number is asking properties of its continued fraction ...

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This is most likely open, since alredy $e^{\pi^2}$ is not known to be transcendental. As an added difficulty, I don't think that $\frac{\pi^2}{12 \log 2}$ is known to be transcendental either. There are very few, very limited, tricks to prove this kind of result: things like taking $(-1)^{-i}$ and $i^i$ and applying Gelfond–Schneider, or building the ...

9

Let $\{t(i)\}_0^\infty$ be the Thue-Morse sequence. (It starts $0,1,1,0,1,0,0,1,\ldots$.). I claim that your sequence is described by $a(n)=1-t(n-1)$ where $n$ is a positive integer. (It is similar to sequence A010059 in the OEIS, but its index starts at $1$ instead of at $0$.) The proof is as follows: For $n=1$, $a(1)=1-t(1-1)=1$. We now consider $n\... 9 No. Choose a field$k$and$D=k[u^2,u^3,v^2,v^3,uv]\subset k[u,v]$, so$D$is a noetherian domain. In$D[x,y]$, choose$f=(ux+vy)^2$and$g=(ux+vy)^3$; they are clearly algebraically dependent (but$ux+vy\notin D[x,y]$). Write$K=k(u,v)=\mathrm{Frac}(D)$. Claim: there is no$P\in D[x,y]$such that$f,g\in D[P]$. By contradiction, let$P\in D[x,y]$such ... 8 The answer is no. For instance, let$a=3$and$b\neq 3$be the real number satisfying$3^b=b^3$. Clearly$b$is not an integer. It follows that$b$is irrational -- indeed, if$b$was a non-integer rational,$3^b$would be irrational, while$b^3$would be rational. Finally,$b$is transcendental, since otherwise$b$would be algebraic irrational,$b^3$would ... 7 It seems that the transcendence degree of$k(G)$over$k$should be the dimension of the$\mathbb Q$-vector space$G \otimes_\mathbb Z \mathbb Q$. Indeed, passing from$G$to$G \otimes \mathbb Q$corresponds to adding roots of existing elements, so it does not alter the transcendence degree. Thus we are reduced to$\mathbb Q$-vector spaces. Choosing a basis ... 7 The reverse implication is true in a considerably more general setting (Burchnall-Chaundy theory). Namely, for any pair$(U,V)$of commuting meromorphic coefficient differential operators in one variable of order at least one, there is a two-variable polynomial$P(z,w)$such that$P(U,V)=0$(the polynomial evaluation is unambiguous because$U$and$V$... 7 Answering Q1, I believe there is a transcendental tiling. Let us begin with this tiling with congruent convex pentagons: Notice that one can rearrange any of the triples of pentagons that form the regular hexagon, by rotating any triple we want by 180 degrees. Thus we can have two kinds of triples: pointing up or down. ... 6 I understand your question as a historical one: how did Lambert solve the trinomial equation? Let me try to walk you through his derivation in his 1758 paper "Observationes variae in mathesin puram". In paragraph 35 (see here) Lambert first explains his method of successive inequalities for the simplest case$m=1$, so to solve$x+px=q$for$x$. He ... 5 First notice that $$f'(Q)=\frac{1}{2}\sum_{d=1}^\infty \binom{2d}{d}Q^d = \frac{(1-4Q)^{-1/2}-1}{2Q}.$$ It follows that differentiation of$q=Q\exp(2f(Q))$with respect to$Q$gives $$q'=\frac{\exp(2f(Q))}{\sqrt{1-4Q}}.$$ Then $$\frac{q'}{q}=\frac{1}{Q\sqrt{1-4Q}}.$$ Solution to this differential equation is $$q = C\frac{\sqrt{1-4Q}- 1}{\sqrt{1-4Q}+ 1}.$$ ... 5 The sequence of maps$\phi_{i_1}^{e_1},\dots,\phi_{i_{\ell}}^{e_{\ell}}$is not uniquely determined by the image set$\phi_{i_{\ell}}^{e_{\ell}}\circ\dots\circ\phi_{i_1}^{e_1}(\mathbb{Z})$. One reason for this is that all commutators$\phi_i\phi_j\phi_i^{-1}\phi_j^{-1}$are translations$x\mapsto x+b$, so any two commutators commute with one another. Thus, ... 5 Schanuel's conjecture implies that if$x$is an algebraic irrational,$n^x$for natural numbers$n$are linearly independent over the rationals: see Will Sawin's answer to this recent question for a proof. It's easy to extend this to positive rationals. Thus Schanuel implies that if the$a_i$and$b_i$are rational,$x$will either be rational or ... 5 The answer is no. Example. Let$f=\Gamma(z)$, Euler's Gamma function, and$g(z)=\Gamma^2(z)$. Evidently they are algebraicaly dependent. Then$g'=2ff'$. Suppose that$f'$and$g'$are algebraically dependent, that is there is an equation$F(f',g')=0$where$F$is a polynomial with constant coefficients. Then$F(f',2ff')=0$but this is an algebraic ... 5 No.$M\mathrel{:=}k(x,y)$has a$k$-automorphism$\sigma:x\mapsto 1/x,\,y\mapsto 1/y$, of order 2. Let$G\mathrel{:=}\langle\sigma\rangle$, and put$L\mathrel{:=}M^{G}$, the fixed field. The elements$x+1/x$and$y+1/y$of$L$are algebraically independent over$k$, hence$L$has transcendency degree 2. However, no non-constant polynomial$g\in k[x,y]$can ... 4 If you were buying a random sequence from a specialized firm for you poker website, and if you were handed the first 1,000,000 digits of$\pi$, you would be entitled to go ask for a refund. Indeed, gifted players on your website could figure out the pattern and use it to win games. This intuition is formalized by the fact that arbitrarily long sequences of ... 4 The numbers$a_j$do not have to be algebraic. Theorem. If$a_j,\; 1\leq j\leq n$are$Q$-linearly independent then$e^{a_jz}$are algebraically independent over$C(z)$Proof. Let $$F(x_1,\ldots,x_n)=\sum_j c_j(z)x_1^{m_{j,1}}\ldots x_n^{m_{j,n}},$$ where$c_j\in C(z)$, and$(m_{j,1}\ldots,m_{j,n})\neq(m_{k,1}\ldots,m_{k,n})$for every pair$j\neq k$. ... 3 Unless I'm being stupid, it seems to me the result is false since it would imply (together with ) that all height-$s$primes of$\mathbb{Q}[X_{1},\dotsc,X_{n}]$can be generated by$s$elements .  Let$A$be a ring,$\mathfrak{a}$an ideal of$A$, and$f$a nonzerodivisor of$A$. Then$\mathfrak{a}$and$f\mathfrak{a}$have the same number of ... 3 The group is not free, because it is solvable. The subgroup of maps that are just translations, without scaling, is normal and abelian, with abelian quotient. 3 I'm not sure this really constitutes an answer, but I think that it may not be fruitful to study the type of generalized question you discussed at the start of your post; you should be able to prove rather easily that for any real number$r>1$you can find a sequence of primes$p_i$with exponents$e_i=p_i$such that the infinite product$\prod_{i=1}^\...

3

$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}$No. As explained in this question, in $\mathbb{Q}[x,y]$, the condition $\operatorname{Jac}(f,g)=0$ implies that there exists an $h \in \mathbb{Q}[x,y]$ such that $f$ and $g$ are in $\mathbb{Q}[h]$. We now need some lemmas that are basically variants of Gauss's lemma, with multiplication replaced by composition. Recall ...

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This recent result Root separation for trinomials by Koiran also uses Baker's Theorem. In my PhD thesis, we use this result to demonstrate a polynomial time algorithm to isolate real roots of integer trinomials.

2

The irrationality measure of the Champernowne constant $C_b$ in base $b>2$ is exactly $b$.

2

A quick thought for your Question 2. If the limit for $\xi_f$ exists, then as you noted we'll get $\xi_f ^2 - 1 = f(\xi_f +1)$. If we define $g(x) = f(x+1) - x^2 +1 = f(x+1) - (x+1)^2 + 2(x+1)$, then we have $x^2 - 1 = f(x+1)$ iff $g(x) = 0$. So your second question more or less reduces to saying we have a function $g$, and we would like to know when the ...

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The first Weyl algebra over $C$ is isomorphic to the algebra of polynomials $C[x_1,x_2]$ equipped with a new multiplication, as follows: define a linear operator $L$ on $C[x_1,x_2,y_1,y_2]$ to be the composite of partial differentiation w.r.t. $x_2$ and of partial differentiation w.r.t. $y_1$. Then $exp(L)$ defines an associative binary operation on $C[x_1,... 1 This can be checked by computing the Wronskian which results in the determinantal valuation $$e^{(a_1+\cdots+a_n)z}V(a_1,\dots,a_n)$$ where$V:=V(a_1,\dots,a_n)$is the determinant of the Vandermonde matrix $$V=\prod_{1\leq i<j\leq n}(a_j-a_i).$$ So, it can only vanish when there are duplicates$a_i=a_j$for some$i,j\$. There is no need to have algebraic ...

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