Translation examples
noun
Probleemat 21–23 ovat täydennysprobleemoja, jota nykyisin ratkaistaisiin yksinkertaisesti murtolukujen vähennyslaskulla.
Problems 21–23 are problems in completion, which in modern notation are simply subtraction problems.
Lukusuoraa käytetään apuvälineenä yhteen- ja vähennyslaskun opetuksessa, etenkin negatiivisten lukujen osalta.
It is often used as an aid in teaching simple addition and subtraction, especially involving negative numbers.
Helpoimmillaan esimerkiksi 2×3-tapauksessa (vaikeusaste 0) ne sopivat esimerkiksi koululaisille yhteen- ja vähennyslaskun harjoitustehtäviksi.
In their simplest form, as in the following 2 × 3 case (degree of difficulty 0) Survo puzzles are suitable exercises in addition and subtraction.
Niinpä riittää todeta, että jos e i G 1 ⋅ R = 1 {\displaystyle e^{i\mathbf {G} _{1}\cdot \mathbf {R} }=1} ja e i G 2 ⋅ R = 1 {\displaystyle e^{i\mathbf {G} _{2}\cdot \mathbf {R} }=1} summa ja erotus G 1 ± G 2 {\displaystyle \mathbf {G} _{1}\pm \mathbf {G} _{2}} toteuttavat saman yhtälön. e i ( G 1 + G 2 ) R = ( e i G 1 ⋅ R ) ( e i G 2 ⋅ R ) = ( 1 ) ( 1 ) = 1 e i ( G 1 − G 2 ) ⋅ R = e i G 1 ⋅ R e i G 2 ⋅ R = 1 {\displaystyle {\begin{aligned}e^{i\left(\mathbf {G} _{1}+\mathbf {G} _{2}\right)\mathbf {R} }&=\left(e^{i\mathbf {G} _{1}\cdot \mathbf {R} }\right)\left(e^{i\mathbf {G} _{2}\cdot \mathbf {R} }\right)=\left(1\right)\left(1\right)=1\\e^{i\left(\mathbf {G} _{1}-\mathbf {G} _{2}\right)\cdot \mathbf {R} }&={\frac {e^{i\mathbf {G} _{1}\cdot \mathbf {R} }}{e^{i\mathbf {G} _{2}\cdot \mathbf {R} }}}=1\end{aligned}}} Tämä osoittaa, että käänteishilakin on sulkettu vektorien yhteen- ja vähennyslaskun suhteen, toisin sanoen kahden käänteishilaan kuuluvan vektorin summa ja erotus kuuluvat siihen myös.
Thus it is sufficient to say that if we have e i G 1 ⋅ R = 1 {\displaystyle e^{i\mathbf {G} _{1}\cdot \mathbf {R} }=1} and e i G 2 ⋅ R = 1 {\displaystyle e^{i\mathbf {G} _{2}\cdot \mathbf {R} }=1} then the sum and difference G 1 ± G 2 {\displaystyle \mathbf {G} _{1}\pm \mathbf {G} _{2}} satisfy the same. e i ( G 1 + G 2 ) R = ( e i G 1 ⋅ R ) ( e i G 2 ⋅ R ) = ( 1 ) ( 1 ) = 1 e i ( G 1 − G 2 ) ⋅ R = e i G 1 ⋅ R e i G 2 ⋅ R = 1 {\displaystyle {\begin{aligned}e^{i\left(\mathbf {G} _{1}+\mathbf {G} _{2}\right)\mathbf {R} }&=\left(e^{i\mathbf {G} _{1}\cdot \mathbf {R} }\right)\left(e^{i\mathbf {G} _{2}\cdot \mathbf {R} }\right)=\left(1\right)\left(1\right)=1\\e^{i\left(\mathbf {G} _{1}-\mathbf {G} _{2}\right)\cdot \mathbf {R} }&={\frac {e^{i\mathbf {G} _{1}\cdot \mathbf {R} }}{e^{i\mathbf {G} _{2}\cdot \mathbf {R} }}}=1\end{aligned}}} Thus we have shown the reciprocal lattice is closed under vector addition and subtraction.
Niitä voidaan laskea yhteen tavalliseen tapaan: A + B = ( A 0 , A 1 , A 2 , A 3 ) + ( B 0 , B 1 , B 2 , B 3 ) = ( A 0 + B 0 , A 1 + B 1 , A 2 + B 2 , A 3 + B 3 ) {\displaystyle \mathbf {A} +\mathbf {B} =(A^{0},A^{1},A^{2},A^{3})+(B^{0},B^{1},B^{2},B^{3})=(A^{0}+B^{0},A^{1}+B^{1},A^{2}+B^{2},A^{3}+B^{3})} ja ne voidaan kertoa skalaarilla λ komponenteittain: λ A = λ ( A 0 , A 1 , A 2 , A 3 ) = ( λ A 0 , λ A 1 , λ A 2 , λ A 3 ) {\displaystyle \lambda \mathbf {A} =\lambda (A^{0},A^{1},A^{2},A^{3})=(\lambda A^{0},\lambda A^{1},\lambda A^{2},\lambda A^{3})} Samoin vähennyslasku on nelivektoreillakin yhteenlaskun käänteistoimitus, joka määritellään komponenteittain: A + ( − 1 ) B = ( A 0 , A 1 , A 2 , A 3 ) + ( − 1 ) ( B 0 , B 1 , B 2 , B 3 ) = ( A 0 − B 0 , A 1 − B 1 , A 2 − B 2 , A 3 − B 3 ) {\displaystyle \mathbf {A} +(-1)\mathbf {B} =(A^{0},A^{1},A^{2},A^{3})+(-1)(B^{0},B^{1},B^{2},B^{3})=(A^{0}-B^{0},A^{1}-B^{1},A^{2}-B^{2},A^{3}-B^{3})} Katso myös: Intervalli (fysiikka) Kahden nelivektorin A ja B sisätulo eli skalaaritulo määritellään Einsteinin notaatiota käyttäen seuraavasti: A ⋅ B = A μ η μ ν B ν {\displaystyle \mathbf {A} \cdot \mathbf {B} =A^{\mu }\eta _{\mu \nu }B^{\nu }} missä η {\displaystyle \eta } on Minkowskin metriikka.
They can be added in the usual entrywise way: A + B = ( A 0 , A 1 , A 2 , A 3 ) + ( B 0 , B 1 , B 2 , B 3 ) = ( A 0 + B 0 , A 1 + B 1 , A 2 + B 2 , A 3 + B 3 ) {\displaystyle \mathbf {A} +\mathbf {B} =(A^{0},A^{1},A^{2},A^{3})+(B^{0},B^{1},B^{2},B^{3})=(A^{0}+B^{0},A^{1}+B^{1},A^{2}+B^{2},A^{3}+B^{3})} and similarly scalar multiplication by a scalar λ is defined entrywise by: λ A = λ ( A 0 , A 1 , A 2 , A 3 ) = ( λ A 0 , λ A 1 , λ A 2 , λ A 3 ) {\displaystyle \lambda \mathbf {A} =\lambda (A^{0},A^{1},A^{2},A^{3})=(\lambda A^{0},\lambda A^{1},\lambda A^{2},\lambda A^{3})} Then subtraction is the inverse operation of addition, defined entrywise by: A + ( − 1 ) B = ( A 0 , A 1 , A 2 , A 3 ) + ( − 1 ) ( B 0 , B 1 , B 2 , B 3 ) = ( A 0 − B 0 , A 1 − B 1 , A 2 − B 2 , A 3 − B 3 ) {\displaystyle \mathbf {A} +(-1)\mathbf {B} =(A^{0},A^{1},A^{2},A^{3})+(-1)(B^{0},B^{1},B^{2},B^{3})=(A^{0}-B^{0},A^{1}-B^{1},A^{2}-B^{2},A^{3}-B^{3})} Applying the Minkowski tensor ημν to two four-vectors A and B, writing the result in dot product notation, we have, using Einstein notation: A ⋅ B = A μ η μ ν B ν {\displaystyle \mathbf {A} \cdot \mathbf {B} =A^{\mu }\eta _{\mu \nu }B^{\nu }} It is convenient to rewrite the definition in matrix form: A ⋅ B = ( A 0 A 1 A 2 A 3 ) ( η 00 η 01 η 02 η 03 η 10 η 11 η 12 η 13 η 20 η 21 η 22 η 23 η 30 η 31 η 32 η 33 ) ( B 0 B 1 B 2 B 3 ) {\displaystyle \mathbf {A\cdot B} ={\begin{pmatrix}A^{0}&A^{1}&A^{2}&A^{3}\end{pmatrix}}{\begin{pmatrix}\eta _{00}&\eta _{01}&\eta _{02}&\eta _{03}\\\eta _{10}&\eta _{11}&\eta _{12}&\eta _{13}\\\eta _{20}&\eta _{21}&\eta _{22}&\eta _{23}\\\eta _{30}&\eta _{31}&\eta _{32}&\eta _{33}\end{pmatrix}}{\begin{pmatrix}B^{0}\\B^{1}\\B^{2}\\B^{3}\end{pmatrix}}} in which case ημν above is the entry in row μ and column ν of the Minkowski metric as a square matrix.
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