Traduzione per "be combine" a finlandese
Frasi di contesto simile
Esempi di traduzione.
Multiple sources of potentially random data can be combined using XOR, and the unpredictability of the output is guaranteed to be at least as good as the best individual source.
Eri lähteistä saatuja potentiaalisesti satunnaisia syötteitä voidaan yhdistää XOR-operaatiolla, ja tuloksen ennakoimattomuus on tällöin varmasti vähintään yhtä hyvä kuin parhaan yksittäisen lähteen.
Notwithstanding its semi-legendary epic character, the LVG provides many important details, which can be combined with the sources closer to the period in question, such as Lazarus of Parpi and Procopius.
Puoliksi legendaarisesta ja eeppisestä luonteestaan huolimatta Vakhtang Gorgasalin elämä kertoo lukuisia yksityiskohtia, jotka voidaan yhdistää lähempänä kyseistä ajanjaksoa kirjoitettuihin lähteisiin, kuten Ghazar Parpetsiin ja Prokopiokseen.
In the Hamilton–Jacobi formulation of classical mechanics, velocity is given by v ( x , t ) = ∇ S ( x , t ) m {\displaystyle v(x,t)={\frac {\nabla S(x,t)}{m}}} where S ( x , t ) {\displaystyle S(x,t)} is a solution of the Hamilton-Jacobi equation − ∂ S ∂ t = ( ∇ S ) 2 2 m + V ~ ( 2 ) {\displaystyle -{\frac {\partial S}{\partial t}}={\frac {\left(\nabla S\right)^{2}}{2m}}+{\tilde {V}}\quad (2)} ( 1 ) {\displaystyle (1)} and ( 2 ) {\displaystyle (2)} can be combined into a single complex equation by introducing the complex function ψ = ρ e i S ℏ {\displaystyle \psi ={\sqrt {\rho }}e^{\frac {iS}{\hbar }}} , then the two equations are equivalent to i ℏ ∂ ψ ∂ t = ( − ℏ 2 2 m ∇ 2 + V ~ − Q ) ψ {\displaystyle i\hbar {\frac {\partial \psi }{\partial t}}=\left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+{\tilde {V}}-Q\right)\psi \quad } with Q = − ℏ 2 2 m ∇ 2 ρ ρ . {\displaystyle Q=-{\frac {\hbar ^{2}}{2m}}{\frac {\nabla ^{2}{\sqrt {\rho }}}{\sqrt {\rho }}}.} The time dependent Schrödinger equation is obtained if we start with V ~ = V + Q {\displaystyle {\tilde {V}}=V+Q} , the usual potential with an extra quantum potential Q {\displaystyle Q} .
Klassisen mekaniikan Hamiltonin–Jacobin muotoilussa nopeuden ilmaisee lauseke v ( x , t ) = ∇ S ( x , t ) m {\displaystyle v(x,t)={\frac {\nabla S(x,t)}{m}}} , missä S ( x , t ) {\displaystyle S(x,t)} on Hamiltonin–Jacobin yhtälön ratkaisu − ∂ S ∂ t = ( ∇ S ) 2 2 m + V ( 2 ) {\displaystyle -{\frac {\partial S}{\partial t}}={\frac {\left(\nabla S\right)^{2}}{2m}}+V\quad (2)} Yhtälöt ( 1 ) {\displaystyle (1)} ja ( 2 ) {\displaystyle (2)} voidaan yhdistää yhdeksi kompleksiseksi yhtälöksi ottamalla käyttöön kompleksinen funktio ψ = ρ e i S ℏ {\displaystyle \psi ={\sqrt {\rho }}e^{\frac {iS}{\hbar }}} .
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\,.} .
It must be combined with accepting YAHUSHUA as MESSIAH.
Sen täytyy olla yhdistettynä YAHU'SHUAn Messiaana hyväksymisen kanssa.
The correct diet regimen ought to be combined with good prepared routine exercise sessions
Oikea ruokavalio kuuri pitäisi olla yhdistettynä hyvään valmis rutiini harjoituskertoja
There can be combined graceful sha
Ei voi olla yhdistettynä siro muodot ja puhtaat linjat, tavallinen Kukkapuutarhan värikäs kukkapenkit, sanalla, luoda sellainen puutarha fantasian ei ole mitään rajaa.
Several product of physical fitness muscle building use the Anavar to be combination with a majority of features.
Suurin osa tuotteen fyysinen kunto lihaksen rakennuksen käyttää Anavar olla yhdistettynä useita toimintoja.
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