\magnification 1800
\parskip 7pt
\parindent 0pt
\hoffset -0.3truein
\hsize 7truein

\def\N{{\bf N}}
\def\S{\Sigma_{i=1}^n}
\def\Sm{\Sigma_{i=1}^m}

\def\R{{\bf R}}
\def\C{{\bf C}}
\def\x{{\bf x}}
\def\a{{\bf a}}
\def\y{{\bf y}}
\def\e{{\bf e}}
\def\i{{\bf i}}
\def\j{{\bf j}}
\def\k{{\bf k}}
\def\l{{\it l}}
\def\ep{\epsilon}
\def\v{\vskip 50pt}

\def\f{\vskip 7pt}
\def\u{\vskip -5pt}
\def\h{\hskip 10pt}

\def\hb{\hfil \break}
\def\hr{\vskip 5pt \hrule \vskip -5pt}

A {\it scalar field} on $\R^n$ is a function $f: \R^n \rightarrow \R$.
\u
\h\h Visualized by drawing graph in $\R^n \times \R$ or 
\u\u
\h\h  by drawing level sets in domain $\R^n$.

A {\it vector field} on $\R^n$ is a function $F: \R^n \rightarrow \R^n$.
\u
\h\h Visualized by drawing vectors in domain $\R^n$.

\hr
Vector Fields can represent many things

{\bf Example 1:  multiple paths.}

Vectors represent velocity (tangent) vectors of paths.

Defn:  A {\it flow line} of a vector field $F: \R^n \rightarrow \R^n$ is a differentiable path  $\x: I \subset \R 
\rightarrow \R^n$ such that $$\x'(t) = F(\x(t))$$
\u
That is that tangent vector to $\x$ at time $t$ is $\x'(t) = F(\x(t))$.
\u
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
\u

Ex 8 (p. 212):  $F: \R^2 \rightarrow \R^2$, $F(x, y) = (-y, x)$


To find flow line(s) need   $\x: \R \rightarrow \R^n$ such that  $\x'(t) = F(\x(t))$
\f
\centerline{I.e, need $(x'(t), y'(t)) = F( x(t), y(t))$}
\f
\centerline{I.e, need $(x'(t), y'(t)) = (-y(t), x(t))$}


That is need to solve $x'(t) = -y(t)$ and $y'(t) = x(t)$.

Solution:  $\x(t) = (acos t - b sint, a sint + b cos t)$

Check:   $\x'(t) = (-asin t - b cost, a cost - b sin t)$

$F(\x(t)) =$ \hb
$ F(acos t - b sint, a sint + b cos t) = (-asin t - b cost, a cost - b sin t)$

\eject

{\bf Example 2:  A vector field can represent a gradient field.}

Vectors = $\nabla f$, for some scalar field $f$.

Defn:  A {\it gradient field} on $\R^n$ is a  vector field $F: \R^n \rightarrow \R^n$  such that $F(x, y) = \nabla f$, for 
some scalar field $f:  \R^n \rightarrow \R$   

$f$ is called the {\it potential function} for $F$.  

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Ex:  $f:  \R^2 \rightarrow \R$, ~~~~$f(x, y) = x^2 + 3y^2$, ~~~~$\nabla f = (2x, 6y)$.

$F:  \R^2 \rightarrow \R^2$, $F(x, y) =  (2x, 6y)$ is a gradient field with potential function $f$.  

{\it Equipotential set} of $F$ = level set of $f$.  

Thus the vector $F(x, y)$ is perpendicular to an equipotential set of $F$ = level set of $f$  since $F(x, y) = \nabla f$ 
points in the direction of steepest ascent for the terrain described by the graph of $f$ in $\R^2 \times \R$. 

Magnitude of the vector $F(x, y)$ indicate the steepness of the slope 



\eject

\def\R{{\bf R}}
\def\x{{\bf x}}
\def\a{{\bf a}}
\def\S{\Sigma_{i=1}^n}
\def\Sk{\Sigma_{i, j=1}^n}
\def\Sjk{\Sigma_{i_1,..., i_k=1}^n}

Calc 1 Review:

Taylor's Thm for $f: \R \rightarrow \R$.  Suppose   $f \in C^k$,

Let $p_k(x) = f(a) + f'(a) (x - a) + ... + {f^{(k)}(a) \over k!}(x - a)^k$

Then $f(x) = p_k(x) + R_k(x, a)$ where $lim_{x \rightarrow a} { R_k(x, a) 
\over (x- a)^k} = 0$.

\vskip 5pt
\hrule
\vskip -5pt

Prop 1.2:  If $f^{(k+1)}$ exists, then there exists $c$ between $a$ and 
$x$ such that  $R_k(x, a) = {f^{(k+1)}(c) \over (k+1)!}(x - a)^{k+1}$

\vskip 5pt
\hrule
\vskip -5pt

Estimate $ln(2)$ using degree 3 Taylor polynomial for $f(x) = ln(x)$ about 
$a = 1$

 $f(x) = ln(x)$ 
\hfill
 $f'(x) = x^{-1}$ 
\hfill
$f''(x) = -x^{-2}$ 
\hfill
$f'''(x) = 2x^{-3}$ 

\vskip 5pt
$p_3(x) = $
\vskip 25pt


Thus $ln(2) \sim p_3(2) = $
\vskip 25pt

$ln(x) = p_3(x) + R_3(x-1)$

$f^{(4)}(x) = -6x^{-4} = -{6 \over x^4}$

 
$R_3(x-1) = {f^{(4)}(c) \over (4)!}(x - 1)^{4} = $ 

\vskip 10pt
\rightline{\hfill where $c$ is btwn 1 and 2.}

%%$R_3(x, 1) = {f^{(4)}(c) \over (4)(3)(2)}(x - 1)^{4}$

\vfill

$ln(2) = p_3(2) + R_3(2, 1) = $ \hfill where $c \in (1, 2)$


\eject

Multivariable version:

Taylor's Thm for $f: \R \rightarrow \R$.  Suppose   $f \in C^k$,

Let $p_k(\x) = f(\a) + \S f_{x_i}(\a) (x_i - a_i) + ... }$

\vskip 10pt

\rightline{$+  {1 \over k!} \Sjk f_{i_1 ...i_k}(\a) (x_{i_1} - a_{i_1}) 
... (x_{i_k} - a_{i_k})  $}

Then $f(\x) = p_k(\x) + R_k(\x, \a)$ where $lim_{x \rightarrow a} { R_k(x, 
a) \over ||x- a||^k} = 0$.

\vskip 5pt
\hrule
\vskip -5pt


If $f:\R^n \rightarrow \R \in C^2$, then

$R_1({\bf x}, {\bf a}) = {1 \over 2} \Sigma_{i, j = 1}^n f_{x_i x_j} ({\bf 
c}) (x_i - a_i)(x_j - a_j)$  

for some {\bf c}  on the line segment joining {\bf a} and {\bf x}.




If $f:\R^n \rightarrow \R \in C^{k+1}$, then

%%$R_k({\bf x}, {\bf a}) = $


%%${1 \over (k+1)!} \Sigma_{i_1,...,  i_{k+1} = 1}^n 
%%f_{x_{i_1}... x_{i_{k+1}} }({\bf c}) (x_{i_1} - a_{i_1}) ... 
(x_{i_{k+1}} - a_{i_{k+1}})$  
\vskip 10pt

\line{$R_k({\bf x}, {\bf a}) = $
${1 \over (k+1)!} \Sigma_{i_1,...,  i_{k+1} = 1}^n f_{x_{i_1}... 
x_{i_{k+1}} }({\bf c}) (x_{i_1} - a_{i_1}) ... (x_{i_{k+1}} - 
a_{i_{k+1}})$  }

for some {\bf c}  on the line segment joining {\bf a} and {\bf x}.











  
\end
