Talk:Transcendental number

Question about definition.

If a number is not the root of any polynomial with integer and rational coefficients, but is the root of a number with irrational coefficients (say the root of 2.) Is it transcendental?

The current definition only specifies that the coefficients should be integer or rational, while the citation I added defines it as:

"A transcendental number is not a root of any polynomial with algebraic coefficients"

Replacing Integer and rational with algebraic would have the added advantage of succinctness.

I await the input from a mathematician, thank you.--TZubiri (talk) 04:07, 20 May 2020 (UTC)

Furthermore, another source[1] defines it as:

Definition 2.2(Transcendental numbers).A number is transcendental if it is not a root of any nonzero polyno-mial with integer coefficients.

Thus excluding non-integer rational numbers.

[1] http://www.math.toronto.edu/vohuan/Notes/mas216_report.pdf

--TZubiri (talk) 04:58, 20 May 2020 (UTC)

If a polynomial has rational coefficients you can multiply through by the least common multiple of the denominators to obtain a polynomial with integer coefficients that has the same roots.

If you allow a polynomial P to have algebraic coefficients there will be a polynomial, possibly of higher degree, with integer coefficients that includes all the roots of P among its roots. For example, if P(x) = √2 x²−3x+1, then the polynomial (√2 x²−3x+1)(√2 x²+3x−1) has integer coefficients (after expanding) and includes the roots of P among its roots. Generalizing this example is not completely straightforward, but nonetheless can be done, as implied by Property 1 listed at the article on algebraic extensions. Hence roots of polynomials with algebraic coefficients are still algebraic.

Although algebraic coefficients still give algebraic roots, you get a circularity if you try to mention algebraic coefficients in the definition of algebraic numbers. Either integer or rational coefficients are fine—it doesn't matter which one you use in the definition. Will Orrick (talk) 17:20, 7 November 2022 (UTC)

Move of Mahler's classification material

I've opened a discussion of this move at Talk:Transcendental_number_theory, but I know that far more people follow this article than Transcendental number theory, so I wanted to mention it here too. Will Orrick (talk) 21:58, 5 November 2022 (UTC)

I should add that I'm not trying to get other people to do all the work. I'm happy to do the grunt work once it is decided what should be done. Will Orrick (talk) 17:22, 7 November 2022 (UTC)

Liouville constant infinite sum renders poorly in Chrome

Not sure if this is something that falls under the purview of this community to do anything about, but here we go. The result of the LaTeX (or whatever equivalent is actually used) markup for the infinite sum representation of the Liouville constant renders poorly in Google Chrome when the page is viewed at 100% zoom. Here it is:

${\begin{aligned}L_{b}&=\sum _{n=1}^{\infty }10^{-n!}\\&=10^{-1}+10^{-2}+10^{-6}+10^{-24}+10^{-120}+10^{-720}+10^{-5040}+10^{-40320}+\ldots \\&=0.{\textbf {1}}{\textbf {1}}000{\textbf {1}}00000000000000000{\textbf {1}}00000000000000000000000000000000000000000000000000000\ \ldots \\\end{aligned}}$

On my machine, the equal sign in the $n=1$ under the $\sum$ symbol is missing its top bar, so it appears to be a minus sign. It doesn't seem to be the case here in the talk page, so I dunno if it has something to do with the way that the rendered image is being formatted or laid out in the page or what. I don't know if this is just my computer, but I thought I'd mention it. Does anyone else have this problem with the same browser? Modus Ponens (talk) 16:37, 18 October 2023 (UTC)

@Modus Ponens: I have tried this on four browsers: on my computer, Firefox and Brave, and on my phone Privacy Wall and Samsung Internet. In both of the browsers on my phone it is absolutely fine. In both of the browsers on my computer at first glance it looks fine, but looking more closely reveals that the top bar of the equals sign is there, but somewhat fainter than the rest of the text, including the bottom bar of the same equals sign. Unfortunately I have no idea how to correct the problem, but that certainly shows that there is some kind of problem which goes beyond your computer, and indeed beyond Chrome. You may be able to find someone who knows enough to be able to help if you post a query at Wikipedia:Village pump (technical). JBW (talk) 21:35, 18 October 2023 (UTC)

@JBW: You know what, I just tried it on the same browser on my work computer and it's rendering correctly here. Not sure what's causing it to happen on my home computer. Modus Ponens (talk) 02:41, 27 October 2023 (UTC)

I can't verify this in Chrome on my macOS Monterey system. –LaundryPizza03 (dc̄) 00:56, 20 October 2023 (UTC)

Question about the proof

Why is there no $c_{0}$ term in $Q$? Why it starts with $c_{1}$ for this reason? $${\begin{aligned}P&=c_{0}\left(\int _{0}^{\infty }f_{k}(x)e^{-x}\,\mathrm {d} x\right)+c_{1}e\left(\int _{1}^{\infty }f_{k}(x)e^{-x}\,\mathrm {d} x\right)+c_{2}e^{2}\left(\int _{2}^{\infty }f_{k}(x)e^{-x}\,\mathrm {d} x\right)+\cdots +c_{n}e^{n}\left(\int _{n}^{\infty }f_{k}(x)e^{-x}\,\mathrm {d} x\right)\\Q&=c_{1}e\left(\int _{0}^{1}f_{k}(x)e^{-x}\,\mathrm {d} x\right)+c_{2}e^{2}\left(\int _{0}^{2}f_{k}(x)e^{-x}\,\mathrm {d} x\right)+\cdots +c_{n}e^{n}\left(\int _{0}^{n}f_{k}(x)e^{-x}\,\mathrm {d} x\right)\end{aligned}}$$ Euglenos sandara (talk) 20:08, 21 March 2024 (UTC)

In the expression for $Q$, $c_{1}$ is the coefficient of a term involving an integral from 0 to 1, $c_{2}$ is the coefficient of a term involving an integral from 0 to 2, and so on. Presumably $c_{0}$ would be the coefficient of a term involving an integral from 0 to 0, which would simply be 0. JBW (talk) 00:06, 22 March 2024 (UTC)

Adding universal parabolic constant to the examples section

Can we add Universal parabolic constant to the examples section? Zeyn1 (talk) 15:13, 5 September 2024 (UTC)

The list is already quite long and its transcendence follows from that of $\ln a$ with $a\in {\overline {\mathbb {Q} }}\setminus \{0,1\}$. I don't see why a further entry or an explicit mention would be necessary. MuCephei (talk) 19:49, 5 September 2024 (UTC)