# properties of vector-valued functions

If  $F=(f_{1},\,\ldots,\,f_{n})$  and  $G=(g_{1},\,\ldots,\,g_{n})$  are vector-valued and $u$ a real-valued function of the real variable $t$, one defines the vector-valued functions $F\!+\!G$ and $uF$ componentwise as

 $F\!+\!G\;:=\;(f_{1}\!+\!g_{1},\,\ldots,\,f_{n}\!+\!g_{n}),\quad uF\;:=\;(uf_{1% },\,\ldots,\,uf_{n})$

and the real valued dot product as

 $F\cdot G\;:=\;f_{1}g_{1}\!+\ldots+\!f_{n}g_{n}.$

If  $n=3$,  one my define also the vector-valued cross product function as

 $F\!\times\!G\;:=\;\left(\left|\begin{matrix}f_{2}&f_{3}\\ g_{2}&g_{3}\end{matrix}\right|\!,\,\left|\begin{matrix}f_{3}&f_{1}\\ g_{3}&g_{1}\end{matrix}\right|\!,\,\left|\begin{matrix}f_{1}&f_{2}\\ g_{1}&g_{2}\end{matrix}\right|\right)\!.$

It’s not hard to verify, that if $F$, $G$ and $u$ are differentiable on an interval, so are also $F\!+\!G$, $uF$ and $F\cdot G$, and the formulae

 $(F\!+\!G)^{\prime}\;=\;F^{\prime}\!+\!G^{\prime},\quad(uF)^{\prime}\;=\;u^{% \prime}F\!+\!uF^{\prime},\quad(F\cdot G)^{\prime}\;=\;F^{\prime}\cdot G+F\cdot G% ^{\prime}$

are valid, in $\mathbb{R}^{3}$ additionally

 $(F\!\times\!G)^{\prime}\;=\;F^{\prime}\!\times\!G+F\!\times\!G^{\prime}.$

Likewise one can verify the following theorems.

Theorem 1.  If $u$ is continuous in the point $t$ and $F$ in the point $u(t)$, then

 $H\;=\;F\!\circ\!u\;:=\;(f_{1}\!\circ\!u,\,\ldots,\,f_{n}\!\circ\!u)$

is continuous in the point $t$.  If $u$ is differentiable in the point $t$ and $F$ in the point $u(t)$, then the composite function $H$ is differentiable in $t$ and the chain rule

 $H^{\prime}(t)\;=\;F^{\prime}(u(t))\,u^{\prime}(t)$

is in .

Theorem 2.  If $F$ and $G$ are integrable on  $[a,\,b]$,  so is also $c_{1}F\!+\!c_{2}G$, where $c_{1},\,c_{2}$ are real constants, and

 $\int_{a}^{b}\!(c_{1}F\!+\!c_{2}G)\,dt\;=\;c_{1}\int_{a}^{b}\!F\,dt+c_{2}\int_{% a}^{b}\!G\,dt.$

Theorem 3.  Suppose that $F$ is continuous on the interval $I$ and  $c\in I$.  Then the vector-valued function

 $t\;\mapsto\;\int_{c}^{t}\!F(\tau)\,d\tau\;:=\;G(t)\quad\forall\,t\in I$

is differentiable on $I$ and satisfies  $G^{\prime}=F$.

Theorem 4.  Suppose that $F$ is continuous on the interval  $[a,\,b]$  and $G$ is an arbitrary function such that  $G^{\prime}=F$  on this interval.  Then

 $\int_{a}^{b}\!F(t)\,dt\;=\;G(b)\!-\!G(a).$

Theorem 2 may be generalised to

Theorem 5.  If $F$ is integrable on  $[a,\,b]$  and  $C=(c_{1},\,\ldots,\,c_{n})$  is an arbitrary vector of $\mathbb{R}^{n}$, then dot product $C\cdot F$ is integrable on this interval and

 $\int_{a}^{b}\!C\cdot F(t)\,dt\;=\;C\cdot\!\int_{a}^{b}\!F(t)\,dt.$
Title properties of vector-valued functions PropertiesOfVectorvaluedFunctions 2013-03-22 19:02:42 2013-03-22 19:02:42 pahio (2872) pahio (2872) 9 pahio (2872) Topic msc 26A42 msc 26A36 msc 26A24 ProductAndQuotientOfFunctionsSum