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 $u F$ 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 $F\!+\!G$ , $u F$ 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
Canonical name	PropertiesOfVectorvaluedFunctions
Date of creation	2013-03-22 19:02:42
Last modified on	2013-03-22 19:02:42
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	9
Author	pahio (2872)
Entry type	Topic
Classification	msc 26A42
Classification	msc 26A36
Classification	msc 26A24
Related topic	ProductAndQuotientOfFunctionsSum