Übersetzung für "summaus" auf englisch

Summaus

• summing
• summation

Übersetzungsbeispiele

summation

Näiden vektorien summaus on suoritettava komponenteittain.
Summation of these vectors is being done componentwise.

Borelin mukaan on nimetty seuraavat matemaattiset käsitteet: Borelin algebra, Borelin lemma, Borelin suurten lukujen laki Borel-joukko Borel-mitta, Borelin–Kolmogorovin paradoksi, Borelin–Cantellin lemma, Borelin–Carathéodoryn lause Heinen–Borelin lause, Borelin summaus, Borelin jakauma.
Besides the Centre Émile Borel at the Institut Henri Poincaré in Paris and a crater on the Moon, the following mathematical notions are named after him: Borel algebra, Borel's lemma, Borel's law of large numbers, Borel measure, Borel–Kolmogorov paradox, Borel–Cantelli lemma, Borel–Carathéodory theorem, Heine–Borel theorem, Borel summation, Borel distribution.

Tässä tapauksessa summaus johtaa tulokseen: A ⋅ B = − A 0 B 0 + A 1 B 1 + A 2 B 2 + A 3 B 3 {\displaystyle \mathbf {A\cdot B} =-A^{0}B^{0}+A^{1}B^{1}+A^{2}B^{2}+A^{3}B^{3}} matriisimuodossa: A ⋅ B = ( A 0 A 1 A 2 A 3 ) ( − 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ) ( B 0 B 1 B 2 B 3 ) {\displaystyle \mathbf {A\cdot B} =\left({\begin{matrix}A^{0}&A^{1}&A^{2}&A^{3}\end{matrix}}\right)\left({\begin{matrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{matrix}}\right)\left({\begin{matrix}B^{0}\\B^{1}\\B^{2}\\B^{3}\end{matrix}}\right)} On huomattava, että tässä tapauksessa yhdessä vertailujärjestelmässä pätee: A ⋅ B = − A 0 B 0 + A 1 B 1 + A 2 B 2 + A 3 B 3 = − C {\displaystyle \mathbf {A} \cdot \mathbf {B} =-A^{0}B^{0}+A^{1}B^{1}+A^{2}B^{2}+A^{3}B^{3}=-C} toisessa sen sijaan: A ′ ⋅ B ′ = − A ′ 0 B ′ 0 + A ′ 1 B ′ 1 + A ′ 2 B ′ 2 + A ′ 3 B ′ 3 = − C ′ {\displaystyle \mathbf {A} '\cdot \mathbf {B} '=-{A'}^{0}{B'}^{0}+{A'}^{1}{B'}^{1}+{A'}^{2}{B'}^{2}+{A'}^{3}{B'}^{3}=-C'} joten: − C = − A 0 B 0 + A 1 B 1 + A 2 B 2 + A 3 B 3 = − A ′ 0 B ′ 0 + A ′ 1 B ′ 1 + A ′ 2 B ′ 2 + A ′ 3 B ′ 3 {\displaystyle -C=-A^{0}B^{0}+A^{1}B^{1}+A^{2}B^{2}+A^{3}B^{3}=-{A'}^{0}{B'}^{0}+{A'}^{1}{B'}^{1}+{A'}^{2}{B'}^{2}+{A'}^{3}{B'}^{3}} ja näin ollen C saadaan edellisen kanssa yhtä­pitävästi A.n ja B:n avulla.
Evaluating the summation with this signature: A ⋅ B = − A 0 B 0 + A 1 B 1 + A 2 B 2 + A 3 B 3 {\displaystyle \mathbf {A\cdot B} =-A^{0}B^{0}+A^{1}B^{1}+A^{2}B^{2}+A^{3}B^{3}} while the matrix form is: A ⋅ B = ( A 0 A 1 A 2 A 3 ) ( − 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ) ( B 0 B 1 B 2 B 3 ) {\displaystyle \mathbf {A\cdot B} =\left({\begin{matrix}A^{0}&A^{1}&A^{2}&A^{3}\end{matrix}}\right)\left({\begin{matrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{matrix}}\right)\left({\begin{matrix}B^{0}\\B^{1}\\B^{2}\\B^{3}\end{matrix}}\right)} Note that in this case, in one frame: A ⋅ B = − A 0 B 0 + A 1 B 1 + A 2 B 2 + A 3 B 3 = − C {\displaystyle \mathbf {A} \cdot \mathbf {B} =-A^{0}B^{0}+A^{1}B^{1}+A^{2}B^{2}+A^{3}B^{3}=-C} while in another: A ′ ⋅ B ′ = − A ′ 0 B ′ 0 + A ′ 1 B ′ 1 + A ′ 2 B ′ 2 + A ′ 3 B ′ 3 = − C ′ {\displaystyle \mathbf {A} '\cdot \mathbf {B} '=-{A'}^{0}{B'}^{0}+{A'}^{1}{B'}^{1}+{A'}^{2}{B'}^{2}+{A'}^{3}{B'}^{3}=-C'} so that: − C = − A 0 B 0 + A 1 B 1 + A 2 B 2 + A 3 B 3 = − A ′ 0 B ′ 0 + A ′ 1 B ′ 1 + A ′ 2 B ′ 2 + A ′ 3 B ′ 3 {\displaystyle -C=-A^{0}B^{0}+A^{1}B^{1}+A^{2}B^{2}+A^{3}B^{3}=-{A'}^{0}{B'}^{0}+{A'}^{1}{B'}^{1}+{A'}^{2}{B'}^{2}+{A'}^{3}{B'}^{3}} which is equivalent to the above expression for C in terms of A and B. Either convention will work.