Translation for "osittaisderivaatat" to english

Osittaisderivaatat

• partial derivatives

Translation examples

partial derivatives

Reaalisen ja kompleksisen derivoituvuuden välinen suhde voidaan esittää seuraavasti: Jos kompleksifunktio ƒ(x + i y) = u(x, y) + i v(x, y) on holomorfinen, funktioilla u ja v on ensimmäisen kertaluvun osittaisderivaatat x:n ja y:n suhteen.
The relationship between real differentiability and complex differentiability is the following. If a complex function f(x + i y) = u(x, y) + i v(x, y) is holomorphic, then u and v have first partial derivatives with respect to x and y, and satisfy the Cauchy–Riemann equations:[6

Ratkaisu löytyy asettamalla osittaisderivaatat nolliksi.
We can now solve this by setting the partial derivatives to zero.

Yksinkertaisimmillaan voidaan osoittaa, että jos funktioilla u ja v on jatkuvat ensimmäisen kertaluvun osittaisderivaatat, jotka toteuttavat Cauchyn–Riemannin yhtälöt, ƒ on holomorfinen.
A simple converse is that if u and v have continuous first partial derivatives and satisfy the Cauchy–Riemann equations, then f is holomorphic.

Tällöin f {\displaystyle \scriptstyle f} :n Jacobin matriisin muodostavat u {\displaystyle \scriptstyle u} :n ja v {\displaystyle \scriptstyle v} :n neljä osittaisderivaattaa x {\displaystyle \scriptstyle x} :n ja y {\displaystyle \scriptstyle y} :n suhteen.
The Jacobian matrix of f {\displaystyle f} is formed by the four partial derivatives of u {\displaystyle u} and v {\displaystyle v} with respect to x {\displaystyle x} and y {\displaystyle y} .

Tyydyttävämpi lause, joka on vaikeampi todistaa, on Looman–Menchoffin lause: jos funktio ƒ on jatkuva ja funktioilla u ja v on ensimmäisen kertaluvun jatkuvat osittaisderivaatat, jotka toteuttavat Cauchyn–Riemannin yhtälöt, ƒ on holomorfinen.
A more satisfying converse, which is much harder to prove, is the Looman–Menchoff theorem: if f is continuous, u and v have first partial derivatives (but not necessarily continuous), and they satisfy the Cauchy–Riemann equations, then f is holomorphic.

Käytettäessä standardikantaa indeksi- ja lyhennysmerkintöineen neligradientin kontravariantit komponentit ovat: ∂ = ( ∂ ∂ x 0 , − ∂ ∂ x 1 , − ∂ ∂ x 2 , − ∂ ∂ x 3 ) = ( ∂ 0 , − ∂ 1 , − ∂ 2 , − ∂ 3 ) = e 0 ∂ 0 − e 1 ∂ 1 − e 2 ∂ 2 − e 3 ∂ 3 = e 0 ∂ 0 − e i ∂ i = e α ∂ α = ( 1 c ∂ ∂ t , − ∇ ) = e 0 1 c ∂ ∂ t − ∇ {\displaystyle {\begin{aligned}{\boldsymbol {\partial }}&=\left({\frac {\partial }{\partial x_{0}}},\,-{\frac {\partial }{\partial x_{1}}},\,-{\frac {\partial }{\partial x_{2}}},\,-{\frac {\partial }{\partial x_{3}}}\right)\\&=(\partial ^{0},\,-\partial ^{1},\,-\partial ^{2},\,-\partial ^{3})\\&=\mathbf {e} _{0}\partial ^{0}-\mathbf {e} _{1}\partial ^{1}-\mathbf {e} _{2}\partial ^{2}-\mathbf {e} _{3}\partial ^{3}\\&=\mathbf {e} _{0}\partial ^{0}-\mathbf {e} _{i}\partial ^{i}\\&=\mathbf {e} _{\alpha }\partial ^{\alpha }\\&=\left({\frac {1}{c}}{\frac {\partial }{\partial t}},\,-\nabla \right)\\&=\mathbf {e} _{0}{\frac {1}{c}}{\frac {\partial }{\partial t}}-\nabla \\\end{aligned}}} On otettava huomioon, että kantavektorit on asetettu komponenttien, jotta tämä ei sekaantuisi kantavektorien derivaattoihin tai yksinkertaisesti sen osoittamiseksi, että osittaisderivaatta on tämän nelivektorin komponentti.
Using the standard basis, in index and abbreviated notations, the contravariant components are: ∂ = ( ∂ ∂ x 0 , − ∂ ∂ x 1 , − ∂ ∂ x 2 , − ∂ ∂ x 3 ) = ( ∂ 0 , − ∂ 1 , − ∂ 2 , − ∂ 3 ) = E 0 ∂ 0 − E 1 ∂ 1 − E 2 ∂ 2 − E 3 ∂ 3 = E 0 ∂ 0 − E i ∂ i = E α ∂ α = ( 1 c ∂ ∂ t , − ∇ ) = ( ∂ t c , − ∇ ) = E 0 1 c ∂ ∂ t − ∇ {\displaystyle {\begin{aligned}{\boldsymbol {\partial }}&=\left({\frac {\partial }{\partial x_{0}}},\,-{\frac {\partial }{\partial x_{1}}},\,-{\frac {\partial }{\partial x_{2}}},\,-{\frac {\partial }{\partial x_{3}}}\right)\\&=(\partial ^{0},\,-\partial ^{1},\,-\partial ^{2},\,-\partial ^{3})\\&=\mathbf {E} _{0}\partial ^{0}-\mathbf {E} _{1}\partial ^{1}-\mathbf {E} _{2}\partial ^{2}-\mathbf {E} _{3}\partial ^{3}\\&=\mathbf {E} _{0}\partial ^{0}-\mathbf {E} _{i}\partial ^{i}\\&=\mathbf {E} _{\alpha }\partial ^{\alpha }\\&=\left({\frac {1}{c}}{\frac {\partial }{\partial t}},\,-\nabla \right)\\&=\left({\frac {\partial _{t}}{c}},-\nabla \right)\\&=\mathbf {E} _{0}{\frac {1}{c}}{\frac {\partial }{\partial t}}-\nabla \\\end{aligned}}} Note the basis vectors are placed in front of the components, to prevent confusion between taking the derivative of the basis vector, or simply indicating the partial derivative is a component of this four-vector.