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Type another comma, 2, another comma, and 0—like this:,2,0
Kirjoita toinen pilkku, 2, toinen pilkku ja 0 seuraavasti:,2,0
How is "of course" separated by commas? How is comma separated by the way?
Miten "tietysti" erotetaan pilkulla? Miten pilkku erotetaan toisella tavalla?
The vocative is normally separated from the rest of the sentence by a comma or commas.
Vokatiivi on normaalisti erotettu muusta lauseesta pilkulla tai pilkuilla.
Part of the sentence before the comma is a condition, and after the comma is an effect.
Osa lauseesta ennen pilkkua on ehto, ja pilkulla on vaikutus.
As the MCS crossed northwestern Pennsylvania, it formed into a distinctive comma shape.
Ylittäessään luoteista Pennsylvaniaa se otti omalaatuisen pilkun muodon.
For example a numeric field may only allow the digits 0–9, the decimal point and perhaps a minus sign or commas.
Tässä järjestelmässä luvut muodostetaan numeroista 0–9 sekä desimaalierottimesta, joka on maasta riippuen yleensä pilkku tai piste.
It is identical to the plain quote, except that symbols prefixed with a comma are replaced with those symbols' values as variables.
Se merkitsee muutoin samaa kuin tavalliset lainausmerkit, paitsi että symbolit, joiden edessä on pilkku, korvataan kyseisten symbolien arvoilla muuttujina.
His system, intended to be compatible with typography, is based on a single line, displaying numbers representing intervals between notes and dots and commas indicating rhythmic values.
Se perustui yhteen riviin, jossa nuottien välisiä intervalleja edustavat numerot, sekä pisteisiin ja pilkkuihin, jotka osoittivat rytmin.
Here I put a comma, there, when it's necessary to cut the view, I put a parenthesis; there I end it with a period and start on another theme.
Tähän laitan pilkun, tuonne, jos se on näkymän kannalta tarpeellista, laitan lainausmerkit; sitten päätän sen pisteellä ja aloitan uudella teemalla.
In English, a comma is used to separate a dependent clause from the independent clause if the dependent clause comes first: After I fed the cat, I brushed my clothes.
Jos yhteinen subjekti sen sijaan ilmenee pelkästään predikaatin persoonamuodosta, se tulkitaan tavallisesti yhteiseksi lauseenjäseneksi eikä lauseiden väliin merkitä pilkkua: ”Ensin haukkaan pikaisesti jotain ja menen sitten kylpyyn.”
The most common shintai are man-made objects like mirrors, swords, jewels (for example comma-shaped stones called magatama), gohei (wands used during religious rites), and sculptures of kami called shinzō (神像), but they can be also natural objects such as stones, mountains, trees and waterfalls.
Yleisimmät shintait ovat ihmisen valmistamia esineitä, kuten peilejä, miekkoja, jalokiviä (esimerkiksi pilkun muotoisia magatama-kiviä), uskonnollisissa rituaaleissa käytettyjä gohei-sauvoja tai kameja esittäviä veistoksia, joita kutsutaan nimellä shinzō (jap.
Expressed mathematically, this assumption may be written as ∫ Ω ′ L ( α A , α A , ν , ξ μ ) d 4 ξ − ∫ Ω L ( φ A , φ A , ν , x μ ) d 4 x = 0 {\displaystyle \int _{\Omega ^{\prime }}L\left(\alpha ^{A},{\alpha ^{A}}_{,\nu },\xi ^{\mu }\right)d^{4}\xi -\int _{\Omega }L\left(\varphi ^{A},{\varphi ^{A}}_{,\nu },x^{\mu }\right)d^{4}x=0} where the comma subscript indicates a partial derivative with respect to the coordinate(s) that follows the comma, e.g. φ A , σ = ∂ φ A ∂ x σ . {\displaystyle {\varphi ^{A}}_{,\sigma }={\frac {\partial \varphi ^{A}}{\partial x^{\sigma }}}\,.} Since ξ is a dummy variable of integration, and since the change in the boundary Ω is infinitesimal by assumption, the two integrals may be combined using the four-dimensional version of the divergence theorem into the following form ∫ Ω { + ∂ ∂ x σ } d 4 x = 0 . {\displaystyle \int _{\Omega }\left\{\left+{\frac {\partial }{\partial x^{\sigma }}}\left\right\}d^{4}x=0\,.} The difference in Lagrangians can be written to first-order in the infinitesimal variations as = ∂ L ∂ φ A δ ¯ φ A + ∂ L ∂ φ A , σ δ ¯ φ A , σ . {\displaystyle \left={\frac {\partial L}{\partial \varphi ^{A}}}{\bar {\delta }}\varphi ^{A}+{\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}{\bar {\delta }}{\varphi ^{A}}_{,\sigma }\,.} However, because the variations are defined at the same point as described above, the variation and the derivative can be done in reverse order; they commute δ ¯ φ A , σ = δ ¯ ∂ φ A ∂ x σ = ∂ ∂ x σ ( δ ¯ φ A ) . {\displaystyle {\bar {\delta }}{\varphi ^{A}}_{,\sigma }={\bar {\delta }}{\frac {\partial \varphi ^{A}}{\partial x^{\sigma }}}={\frac {\partial }{\partial x^{\sigma }}}({\bar {\delta }}\varphi ^{A})\,.} Using the Euler–Lagrange field equations ∂ ∂ x σ ( ∂ L ∂ φ A , σ ) = ∂ L ∂ φ A {\displaystyle {\frac {\partial }{\partial x^{\sigma }}}\left({\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}\right)={\frac {\partial L}{\partial \varphi ^{A}}}} the difference in Lagrangians can be written neatly as = ∂ ∂ x σ ( ∂ L ∂ φ A , σ ) δ ¯ φ A + ∂ L ∂ φ A , σ δ ¯ φ A , σ = ∂ ∂ x σ ( ∂ L ∂ φ A , σ δ ¯ φ A ) . {\displaystyle {\begin{aligned}&\left\\={}&{\frac {\partial }{\partial x^{\sigma }}}\left({\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}\right){\bar {\delta }}\varphi ^{A}+{\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}{\bar {\delta }}{\varphi ^{A}}_{,\sigma }={\frac {\partial }{\partial x^{\sigma }}}\left({\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}{\bar {\delta }}\varphi ^{A}\right).\end{aligned}}} Thus, the change in the action can be written as ∫ Ω ∂ ∂ x σ { ∂ L ∂ φ A , σ δ ¯ φ A + L ( φ A , φ A , ν , x μ ) δ x σ } d 4 x = 0 . {\displaystyle \int _{\Omega }{\frac {\partial }{\partial x^{\sigma }}}\left\{{\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}{\bar {\delta }}\varphi ^{A}+L\left(\varphi ^{A},{\varphi ^{A}}_{,\nu },x^{\mu }\right)\delta x^{\sigma }\right\}d^{4}x=0\,.} Since this holds for any region Ω, the integrand must be zero ∂ ∂ x σ { ∂ L ∂ φ A , σ δ ¯ φ A + L ( φ A , φ A , ν , x μ ) δ x σ } = 0 . {\displaystyle {\frac {\partial }{\partial x^{\sigma }}}\left\{{\frac {\partial L}{\partial {\varphi ^{A}}_{,\sigma }}}{\bar {\delta }}\varphi ^{A}+L\left(\varphi ^{A},{\varphi ^{A}}_{,\nu },x^{\mu }\right)\delta x^{\sigma }\right\}=0\,.} For any combination of the various symmetry transformations, the perturbation can be written δ x μ = ε X μ {\displaystyle \delta x^{\mu }=\varepsilon X^{\mu }} δ φ A = ε Ψ A = δ ¯ φ A + ε L X φ A {\displaystyle \delta \varphi ^{A}=\varepsilon \Psi ^{A}={\bar {\delta }}\varphi ^{A}+\varepsilon {\mathcal {L}}_{X}\varphi ^{A}} where L X φ A {\displaystyle {\mathcal {L}}_{X}\varphi ^{A}} is the Lie derivative of φA in the Xμ direction.
Matemaattisesti tämä oletus voidaan kirjoittaa seuraavasti: ∫ Ω ′ L ( α A , α A , ν , ξ μ ) d 4 ξ − ∫ Ω L ( ϕ A , ϕ A , ν , x μ ) d 4 x = 0 {\displaystyle \int _{\Omega ^{\prime }}L\left(\alpha ^{A},{\alpha ^{A}}_{,\nu },\xi ^{\mu }\right)d^{4}\xi -\int _{\Omega }L\left(\phi ^{A},{\phi ^{A}}_{,\nu },x^{\mu }\right)d^{4}x=0} missä muuttujien jälkeen yläpuolelle kirjoitetut pilkut tarkoittavat osittaisderivaattoja niiden koordinaattien suhteen, jotka seuraavat pilkun jälkeen, toisin sanoen ϕ A , σ = ∂ ϕ A ∂ x σ . {\displaystyle {\phi ^{A}}_{,\sigma }={\frac {\partial \phi ^{A}}{\partial x^{\sigma }}}\,.} Koska ξ on pelkkä integroimisvakio ja koska rajan Ω muutos oletettiin infinitesimaaliseksi, nämä kaksi integraalia voidaan yhdistää divergenssilauseen neliulotteisen version mukaisesti seuraavaan muotoon: ∫ Ω { + ∂ ∂ x σ } d 4 x = 0 . {\displaystyle \int _{\Omega }\left\{\left+{\frac {\partial }{\partial x^{\sigma }}}\left\right\}d^{4}x=0\,.} Lagrangen funktioiden erotus voidaan kirjoittaa ensimmäisessä kertaluvuissa infinitesimaalisilla muutoksilla: = ∂ L ∂ ϕ A δ ¯ ϕ A + ∂ L ∂ ϕ A , σ δ ¯ ϕ A , σ . {\displaystyle \left={\frac {\partial L}{\partial \phi ^{A}}}{\bar {\delta }}\phi ^{A}+{\frac {\partial L}{\partial {\phi ^{A}}_{,\sigma }}}{\bar {\delta }}{\phi ^{A}}_{,\sigma }\,.} Koska nämä muutokset kuitenkin on määritelty samassa edellä selityssä pisteessä, muutokset ja derivoinnit voidaan suorittaa myös päinvastaisessa järjestyksessä; ne kommutoivat: δ ¯ ϕ A , σ = δ ¯ ∂ ϕ A ∂ x σ = ∂ ∂ x σ ( δ ¯ ϕ A ) . {\displaystyle {\bar {\delta }}{\phi ^{A}}_{,\sigma }={\bar {\delta }}{\frac {\partial \phi ^{A}}{\partial x^{\sigma }}}={\frac {\partial }{\partial x^{\sigma }}}\left({\bar {\delta }}\phi ^{A}\right)\,.} Käyttämällä Eulerin-Lagrangen kenttäyhtälöä ∂ ∂ x σ ( ∂ L ∂ ϕ A , σ ) = ∂ L ∂ ϕ A {\displaystyle {\frac {\partial }{\partial x^{\sigma }}}\left({\frac {\partial L}{\partial {\phi ^{A}}_{,\sigma }}}\right)={\frac {\partial L}{\partial \phi ^{A}}}} Lagrangen funktioiden erotus voidaan kirjoittaa yksinkertaisesti muotoon = ∂ ∂ x σ ( ∂ L ∂ ϕ A , σ ) δ ¯ ϕ A + ∂ L ∂ ϕ A , σ δ ¯ ϕ A , σ = ∂ ∂ x σ ( ∂ L ∂ ϕ A , σ δ ¯ ϕ A ) . {\displaystyle \left={\frac {\partial }{\partial x^{\sigma }}}\left({\frac {\partial L}{\partial {\phi ^{A}}_{,\sigma }}}\right){\bar {\delta }}\phi ^{A}+{\frac {\partial L}{\partial {\phi ^{A}}_{,\sigma }}}{\bar {\delta }}{\phi ^{A}}_{,\sigma }={\frac {\partial }{\partial x^{\sigma }}}\left({\frac {\partial L}{\partial {\phi ^{A}}_{,\sigma }}}{\bar {\delta }}\phi ^{A}\right)\,.} Näin ollen aktion muutokseksi saadaan ∫ Ω ∂ ∂ x σ { ∂ L ∂ ϕ A , σ δ ¯ ϕ A + L ( ϕ A , ϕ A , ν , x μ ) δ x σ } d 4 x = 0 . {\displaystyle \int _{\Omega }{\frac {\partial }{\partial x^{\sigma }}}\left\{{\frac {\partial L}{\partial {\phi ^{A}}_{,\sigma }}}{\bar {\delta }}\phi ^{A}+L\left(\phi ^{A},{\phi ^{A}}_{,\nu },x^{\mu }\right)\delta x^{\sigma }\right\}d^{4}x=0\,.} Koska tämä pätee missä tahansa alueessa Ω, integrandin on oltava nolla ∂ ∂ x σ { ∂ L ∂ ϕ A , σ δ ¯ ϕ A + L ( ϕ A , ϕ A , ν , x μ ) δ x σ } = 0 . {\displaystyle {\frac {\partial }{\partial x^{\sigma }}}\left\{{\frac {\partial L}{\partial {\phi ^{A}}_{,\sigma }}}{\bar {\delta }}\phi ^{A}+L\left(\phi ^{A},{\phi ^{A}}_{,\nu },x^{\mu }\right)\delta x^{\sigma }\right\}=0\,.} .
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