Questions tagged [transcendence]
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65 questions
4
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On the irrationality of tetrations of algebraic numbers
$\sqrt{2}$ is irrational. By Gelfond-Schneider theorem, so is ${{\sqrt{2}}^\sqrt{2}}$. Is
${{\sqrt{2}}^\sqrt{2}}^\sqrt{2}$ irrational?
- 1,375
2
votes
0
answers
117
views
p-adic Lang-Rohrlich conjecture
A famous conjecture of Lang and Rohrlich describes all the algebraic relations between the values of the $\Gamma$ function at proper quotients in $\mathbb Q$ and $\pi$. My question is: does there ...
- 4,424
4
votes
1
answer
232
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What growth rate of $a(n)$ guarantees a transcendental $\prod 1+1/a(n)$?
Specific Question
Does there exist a growth rate for an increasing function $a(n)$, defined on the natural numbers, that guarantees the transcendence of the infinite product
$$
\prod_{n=1}^\infty \...
- 340
7
votes
1
answer
507
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Wikipedia claim: Baker's theorem is a generalization of Lindemann-Weierstrass theorem
Recall the Lindemann Weierstrass theorem: If $a_1,\cdots,a_n$ are $\mathbb Q$-linearly independent algebraic numbers then $e^{a_1},\cdots,e^{a_n}$ are algebraically independent.
Recall Baker's theorem:...
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5
votes
2
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774
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Were there any transcendental numbers other than pi, e, and logarithms that were known before the 18th century?
Is there any number that was known to humans before 18th century that would not be in the closure of algebraic numbers with pi and e and exp and ln operations, that turned out to be transcendental?
pi ...
- 5,856
1
vote
1
answer
161
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Transcendence degree and Krull dimension for homomorphic images of power series rings
Let $k$ be a field of characteristic zero and $I$ be a radical ideal of $k[[x_1,\ldots, x_n]]$. Let $P$ be a minimal prime ideal of the reduced ring $R:=k[[x_1,\ldots, x_n]]/I$. Then, $R_P$ is a field ...
- 470
3
votes
0
answers
145
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Exponential of Liouville Numbers
By Mahler classification of Transcendental real numbers (into the sets of $S$-, $T$- and $U$-numbers), we know that
Any Liouville number is a $U$-number.
$\log \alpha$ is either an $S$- or a $T$-...
- 545
1
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0
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93
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Siegel's method for transcendence measure quoted by Mahler
In "Zur Approximation der Exponentialfunktion und des Logarithmus. Teil II" Malher wrote (footnote 5 page 148) that the constant appearing in Satz 4 can be improved by a method communicated ...
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4
votes
1
answer
385
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Transcendence measure: of $\ln(a/b)$
In the book " Number Theory IV Transcendental Numbers" written by Parsin and Shafarevich (book, page 104) it is asserted that to explicit a transcendence measure of a complex number $w$, it ...
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1
vote
0
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137
views
Liouville numbers with some "special" convergents
Recall that a real number $\ell$ is called a Liouville number, if there exists an infinite sequence of rational numbers $(p_n/q_n)_n$ for which
$$
0<\left|\ell-\frac{p_n}{q_n}\right|<\dfrac{1}{...
- 545
4
votes
1
answer
345
views
Ground state energy of anharmonic oscillator: algebraic or transcendental?
Consider the quantum anharmonic oscillator, with Hamiltonian $H=p^2/2+q^2/2+gq^4$ for some real $g\geq 0$, with $p$ and $q$ obeying the usual Heisenberg commutation relations. For $g=0$, the ground ...
- 587
6
votes
1
answer
303
views
Reference request for results that involve the transcendence degree
Currently I'm reading "On the Decidability of the Real Exponential Field" by Macintyre and Wilkie and the Proof of Theorem 1.1 (page 462-464) uses two algebraic results that involve the ...
- 133
0
votes
2
answers
408
views
Transcendence on $ \alpha+f(\alpha), \alpha f(\alpha) $ and $ \alpha/f(\alpha) $ where $ \alpha$ is transcendental
Let $f(x)$ be a real transcendental function with algebraic coefficients. So $f(x)$ and $x$ are algebraically independent. Let $\alpha$ be a transcendental number, are the numbers $$\alpha+f(\alpha),\...
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15
votes
2
answers
2k
views
Why is it easier to prove $e$ is transcendental than $\pi$?
Why is it easier to prove $e$ is transcendental than $\pi$?
I noticed that the proofs of $\pi$'s transcendence are much longer and have more details to check than those of $e.$ My guess is that it's ...
- 765
6
votes
1
answer
199
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Transcendence of values of Fredholm series at algebraic arguments
Let $d$ be an integer greater than $1$ and let $f(z)=\sum_{n\ge0}z^{d^n}$ be the Fredholm series. It is well-known that the Roth-Ridout theorem implies that $f(r)$ is a transcendental number for all ...
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