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Talk:Exponentiation

Article milestones Date Process Result April 7, 2007 Good article nominee Listed September 8, 2007 Good article reassessment Kept August 5, 2021 Good article reassessment Delisted

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The result was: rejectedbyTheleekycauldron (talk) 10:57, 25 December 2022 (UTC)[reply]

• ... that exponentiation is basically repeated multiplication? Source: https://mathinsight.org/exponentiation_basic_rules
  • Reviewed:
  • Comment: My first nomination submission.

Created by Slaythe (talk). Self-nominated at 01:05, 24 December 2022 (UTC).[reply]

• Sorry, Slaythe, but it's not eligible, because it hasn't been created, 5x expanded, or designated a Good Article within the last week. Please see WP:DYKCRIT. It you get it promoted to Good Article status, you may resubmit a DYK nomination. MANdARAX • XAЯAbИAM 01:34, 24 December 2022 (UTC)[reply]

This article uses 3 different notations for multiplication. IMO, either ${\displaystyle \times }$ must be replaced with ${\displaystyle \cdot ,}$or${\displaystyle \cdot }$ must be replaced by ${\displaystyle \times ;}$ in any case, some occurrences of ${\displaystyle \cdot }$ should be removed, especially in exponents. As I have no clear opinion on the best choice, I wait for a consensus here. For clarification (see the preceding thread), I have added an explanatory footnote. D.Lazard (talk) 10:28, 9 January 2023 (UTC)[reply]

As you wrote in the first thread, the primary issue with using ${\displaystyle \cdot }$ is that it becomes ambiguous whether this represents ellipses or multiplication (to be fair, it's not really ambiguous, because one could figure out from context that one dot means multiplication and three means ellipses). So, I think that defaulting to ${\displaystyle \times }$ in this article is probably best, even though this notation feels quite elementary-school-y. Duckmather (talk) 15:20, 9 January 2023 (UTC)[reply]

Now that I know what is meant, I would suspect that most uninitiated readers (in the UK at least) would recognize a ${\displaystyle \times }$ sign as a multiplication sign, but that the ${\displaystyle \cdot }$ sign for multiplication would be much less familiar, and might well be a source of confusion. Additionally, I have come to this article in following-up work on the biographical WP article on William Oughtred, and I find that the introduction of the ${\displaystyle \times }$ sign, in W.O.'s Clavis Arithmeticae (1631), followed on very soon after, and in the context of, the description of logarithms (in the English edition of John Napier's Description of the Admirable Table of Logarithmes (S. Waterson, London 1618), Appendix, at p. 4 (Google)). (An explanation of the sign is given in William Forster's Forster's Arithmetick (1673), at pp. 43-44 and pp. 113-14 (Google).) Hence there is an historical association between Exponentiation and this ${\displaystyle \times }$ usage which some readers may want to understand. I find some explanations in F. Cajori's History of Mathematics (Macmillan, New York/London 1919), at pp. 157-58 (Internet Archive). Perhaps the pedagogic example is the better for being elementary? Eebahgum (talk) 18:38, 9 January 2023 (UTC)[reply]

I adjusted the intro to be consistent with /times. I left the rest of the article as is. Emschorsch (talk) 04:49, 24 July 2023 (UTC)[reply]

In the section:

[...]

and the imaginary partofz satisfies

-π < Im (z) < π [this does not make sense to me: Isn't it a condition on the Arg(z) or equivalently on the Im(log(z))? Since log(z)=log(|z|)+i(Arg(z)+2nπ), n in Z and the principal value of log(z) can be defined as Log(z) when chosing -π < Im(log((z))=Arg((z)) < π, i.e. n=0]

[...] 217.10.52.10 (talk) 09:52, 20 April 2023 (UTC)[reply]

 Fixed. D.Lazard (talk) 16:12, 21 April 2023 (UTC)[reply]

In the top of the page, the article demonstrates how exponents of rational numbers correspond to nth-roots by proving that b^(1/2) == sqrt(b). Part of this proof relies upon the property that (b^M)*(b^N) == b^(M+N) (see excerpt pasted below). However, the article only proved this property based on the definition that natural-number exponents are equivalent to repeated multiplication. This proof does not apply when M or N are rational because rational exponents are not defined as a repeated multiplication.

Therefore, the article needs to have a separate proof that (b^M)*(b^N) == b^(M+N) when M and N are rational numbers.

Proving this property holds true for rational numbers is a fair bit more complicated than proving it holds true for natural numbers, but it's not so complicated as to be out of the scope of a wiki article. One such proof can be found on this stack overflow page[1]. unfortunately i do not know of any proofs that meet wikipedia's credibility requirements.

this is the specific excerpt from the wiki article that i take issue with: "Using the fact that multiplying makes exponents add gives b^(r+r) == b". It is located at the top of the page. Snickerbockers (talk) 16:14, 24 November 2023 (UTC)[reply]

The general case is an easy generalization of the given case: if M and N are rational numbers, one may reduce them to the same denominator, that is ${\displaystyle M={\frac {m}{q}}}$ and ${\displaystyle N={\frac {n}{q}},}$ with ${\displaystyle m,n,q}$ integers. Setting ${\displaystyle y=x^{1/q},}$ one has

${\displaystyle x^{M}x^{N}=y^{m}y^{n}=y^{m+n}=x^{\frac {m+n}{q}}=x^{M+N}.}$

D.Lazard (talk) 17:08, 24 November 2023 (UTC)[reply]

${\displaystyle \left(a\cdot b\right)^{r}=a^{r}\cdot b^{r}\cdot e^{-i2\pi nr}}$
${\displaystyle \left(a^{r}\right)^{s}=a^{r\cdot s}\cdot e^{-i2\pi ns}}$
取らぬタヌキ (talk) 21:41, 4 January 2024 (UTC)[reply]

${\displaystyle x^{-a}={\frac {1}{x^{a}}}}$

There are no branches.

${\displaystyle x^{a+b}=x^{a}\cdot x^{b}}$

There are no branches.

${\displaystyle \left(a\cdot b\right)^{x}=a^{x}\cdot b^{x}\cdot e^{-i2\pi nx}}$

There are branches.

${\displaystyle \left(x^{a}\right)^{b}=x^{a\cdot b}\cdot e^{-i2\pi nb}}$

There are branches.

These will help you understand the following:

${\displaystyle {\sqrt {x^{2}}}=x\cdot e^{-i\pi n}=\left\{x,\quad -x\right\}}$

If it is a real number domain, the square root cannot take a negative value, so ${\displaystyle {\sqrt {x^{2}}}=|x|}$.

Also,

${\displaystyle (e^{i2\pi n})^{i2\pi m}=e^{-4\pi ^{2}nm}\cdot e^{4\pi ^{2}nm}=1}$

取らぬタヌキ (talk) 19:37, 5 January 2024 (UTC)[reply]

We must understand correctly that a exponentiation is a multivalued function. The following formula transformation is important.

${\displaystyle x=x\cdot e^{-i2\pi n}}$

${\displaystyle x^{a}=x^{a}\cdot e^{-i2\pi na}}$

${\displaystyle \left(x^{a}\right)^{b}=x^{a\cdot b}\cdot e^{-i2\pi \left(n_{1}a+n_{2}\right)b}}$

Furthermore,

${\displaystyle x=x\cdot e^{-i2\pi n}}$

${\displaystyle \ln x=\ln x-i2\pi n}$

${\displaystyle \ln x^{a}=a\ln x-i2\pi \left(n_{1}a+n_{2}\right)}$

取らぬタヌキ (talk) 23:36, 10 January 2024 (UTC)[reply]