Traduction de "lineaarisuus" à anglaise
Lineaarisuus
Exemples de traduction
Tämä z-muunnoksen ominaisuus on lineaarisuus.
It can be shown that the dependence of z on x is linear.
Sana lineaarisuus tulee latinankielisestä sanasta linearis, joka tarkoittaa viivoista tehtyä.
The word linear comes from the Latin word linearis, which means "created by lines".
Esimerkkinä geometrisen sarjan arviointi (aina, kun r ≠ 1,), G ( r , c ) = ∑ k = 0 ∞ c r k = c + ∑ k = 0 ∞ c r k + 1  (stabilisuus)  = c + r ∑ k = 0 ∞ c r k  (lineaarisuus)  = c + r G ( r , c ) ,  jossa  G ( r , c ) = c 1 − r , {\displaystyle {\begin{aligned}G(r,c)&=\sum _{k=0}^{\infty }cr^{k}&&\\&=c+\sum _{k=0}^{\infty }cr^{k+1}&&{\mbox{ (stabilisuus) }}\\&=c+r\sum _{k=0}^{\infty }cr^{k}&&{\mbox{ (lineaarisuus) }}\\&=c+r\,G(r,c),&&{\mbox{ jossa }}\\G(r,c)&={\frac {c}{1-r}},&&\\\end{aligned}}} kun r on yhtä suurempi reaaliluku osasummat kasvavat rajatta ja keskiarvomenetelmät lähenevät ääretöntä.
For instance, whenever r ≠ 1, the geometric series G ( r , c ) = ∑ k = 0 ∞ c r k = c + ∑ k = 0 ∞ c r k + 1  (stability)  = c + r ∑ k = 0 ∞ c r k  (linearity)  = c + r G ( r , c ) ,  hence  G ( r , c ) = c 1 − r ,  unless it is infinite {\displaystyle {\begin{aligned}G(r,c)&=\sum _{k=0}^{\infty }cr^{k}&&\\&=c+\sum _{k=0}^{\infty }cr^{k+1}&&{\text{ (stability) }}\\&=c+r\sum _{k=0}^{\infty }cr^{k}&&{\text{ (linearity) }}\\&=c+r\,G(r,c),&&{\text{ hence }}\\G(r,c)&={\frac {c}{1-r}},{\text{ unless it is infinite}}&&\\\end{aligned}}} can be evaluated regardless of convergence.
Ristitulo on anti­kommuta­tiivinen: a × b = − ( b × a ) . {\displaystyle \mathbf {a} \times \mathbf {b} =-(\mathbf {b} \times \mathbf {a} ).} Ristitulo noudattaa osittelulakia yhteen­laskun suhteen: a × ( b + c ) = ( a × b ) + ( a × c ) , {\displaystyle \mathbf {a} \times (\mathbf {b} +\mathbf {c} )=(\mathbf {a} \times \mathbf {b} )+(\mathbf {a} \times \mathbf {c} ),} Jos ristitulon jompikumpi tekijä kerrotaan skalaarilla, myös ristitulon arvo tulee kerrotuksi samalla skalaarilla: ( r a ) × b = a × ( r b ) = r ( a × b ) . {\displaystyle (r\mathbf {a} )\times \mathbf {b} =\mathbf {a} \times (r\mathbf {b} )=r(\mathbf {a} \times \mathbf {b} ).} Ristitulo ei ole liitännäinen, mutta sille pätee Jacobin identiteetti: a × ( b × c ) + b × ( c × a ) + c × ( a × b ) = 0 . {\displaystyle \mathbf {a} \times (\mathbf {b} \times \mathbf {c} )+\mathbf {b} \times (\mathbf {c} \times \mathbf {a} )+\mathbf {c} \times (\mathbf {a} \times \mathbf {b} )=\mathbf {0} .} Osoittelulaki, lineaarisuus ja Jacobin identiteetti osoittavat, että vektoriavaruus R 3 {\displaystyle \mathbb {R} ^{3}} varustettuna vektorien yhteen­laskulla ja risti­tulolla on Lien algebra, tarkemmin sanottuna kolmi­ulotteisen reaalisen orto­gonaalisen ryhmän Lien algebra SO(3).
The self cross product of a vector is the zero vector: a × a = 0 {\displaystyle \mathbf {a} \times \mathbf {a} =\mathbf {0} } The cross product is anticommutative, a × b = − ( b × a ) , {\displaystyle \mathbf {a} \times \mathbf {b} =-(\mathbf {b} \times \mathbf {a} ),} distributive over addition, a × ( b + c ) = ( a × b ) + ( a × c ) , {\displaystyle \mathbf {a} \times (\mathbf {b} +\mathbf {c} )=(\mathbf {a} \times \mathbf {b} )+(\mathbf {a} \times \mathbf {c} ),} and compatible with scalar multiplication so that ( r a ) × b = a × ( r b ) = r ( a × b ) . {\displaystyle (r\mathbf {a} )\times \mathbf {b} =\mathbf {a} \times (r\mathbf {b} )=r(\mathbf {a} \times \mathbf {b} ).} It is not associative, but satisfies the Jacobi identity: a × ( b × c ) + b × ( c × a ) + c × ( a × b ) = 0 . {\displaystyle \mathbf {a} \times (\mathbf {b} \times \mathbf {c} )+\mathbf {b} \times (\mathbf {c} \times \mathbf {a} )+\mathbf {c} \times (\mathbf {a} \times \mathbf {b} )=\mathbf {0} .} Distributivity, linearity and Jacobi identity show that the R3 vector space together with vector addition and the cross product forms a Lie algebra, the Lie algebra of the real orthogonal group in 3 dimensions, SO(3).
How many English words do you know?
Test your English vocabulary size, and measure how many words you know.
Online Test