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Linear Functions

A function $f$ is linear if $f(a{\bf x} + b{\bf y}) = af({\bf x}) + 
bf({\bf y})$

Or equivalently $f$ is linear if\hfil \break
 1.)  $f(a{\bf x}) = af({\bf x})$ and 2.) $f({\bf x} + {\bf y}) = f({\bf 
x}) + f({\bf y})$


Theorem:  If $f$ is linear, then $f({\bf 0}) = {\bf 0}$

Proof:  $f({\bf 0}) = f(0 \cdot {\bf 0}) = 0 \cdot f({\bf 0}) = {\bf 0}$


Example 1.)  $f:R \rightarrow R$, $f(x) = 2x$

Proof:\hfil \break
  $f(ax + by) = 2(ax + by) = 2ax + 2by = af(x) + bf(y)$

Example 2.) $f:R^2 \rightarrow R^2$, 

\centerline{$f((x_1,x_2)) = (2x_1, x_1 + x_2)$}

Proof:  Let ${\bf x} = (x_1, x_2)$, ${\bf y} = (y_1, y_2)

$a{\bf x} + b{\bf y} = a(x_1, x_2) + b(y_1, y_2) = (ax_1, ax_2) + (by_1, by_2) 
= (ax_1 + by_1, ax_2 + by_2)$

\eject
$f(ax_1 + by_1, ax_2 + by_2) $

\hskip 0.4in $= (2(ax_1 + by_1), ax_1 + by_1 + ax_2 + by_2) $

\hskip 0.4in $= (2ax_1 + 2by_1, ax_1 + ax_2 + by_1 + by_2)   $

\hskip 0.4in $=(2ax_1, ax_1 + ax_2) + (2by_1, by_1 + by_2) $

\hskip 0.4in $= a(2x_1, x_1 + x_2) + b(2y_1, y_1 + y_2)$

\hskip 0.4in $= af((x_1, x_2)) + bf((y_1, y_2))$


Example 3.) $D:$ {set of all differential functions} $\rightarrow$ {set of all functions}, 
$D(f) = f'$

Proof: \hfil \break
 $D(af + bg) = (af + bg)' = af' + bg' = aD(f) + bD(g)$

Example 4.) Given $a, b$ real numbers,\hfil \break
 $I:$ {set of all integrable functions on [a, b]} $\rightarrow R$ , 
$I(f) = \int_a^b f$

Proof:  $I(sf + tg) = \int_a^b sf + tg = s \int_a^b f + t \int_a^b g = sI(f) + tI(g)$ 
\eject

Example 5.)  The inverse of a linear function is linear (when the inverse exists).

Suppose $f^{-1}(x) = c$, $f^{-1}(y) = d$. 

Then $f(c) = x$ and $f(d) = y$ and \hfil \break
$f(ac + bd) = af(c) + bf(d) = ax + by$.

 Hence $f^{-1}(ax + by) = ac + bd = a f^{-1}(x) + bf^{-1}(y)$.

Example 6.)  $D:$ {set of all twice differential functions} $\rightarrow$ {set of all functions}, 
$L(f) = af'' + bf' + cf$

Proof: \hfil \break
$L(sf + tg) = a(sf + tg)'' + b(sf + tg)' + c(sf + tg) $

\hskip 0.67in $= saf'' + tag'' + sbf' + tbg' + scf + tcg $

\hskip 0.67in $= s(af'' + bf' + cf) +t(ag'' + bg' + cg) $

\hskip 0.67in $= sL(f) + tL(g)$

\eject

Consequence 1:  If $\psi_1, \psi_2$ are  solutions to $af'' + bf' + cf = 0$, 
then $3 \psi_1 + 5 \psi_2$ is also a solution to \break
$af'' + bf' + cf = 0$,

Proof: Since $\psi_1, \psi_2$ are  solutions to $af'' + bf' + cf = 0$, 
$L(\psi_1) = 0$ and $L(\psi_2) = 0$.

Hence $L(3 \psi_1 + 5 \psi_2) = 3 L(\psi_1) + 5 L(\psi_2)$
\u
\hskip 1.24in$ = 3(0) + 5(0) 
= 
0$.  
\u
Thus $3 \psi_1 + 5 \psi_2$ is also a solution to $af'' + bf' + cf = 
0$

Consequence 2: \hfil \break
 If $\psi_1$ is a solution to $af'' + bf' + cf = h$ \hfil \break
and $\psi_2$ is a solution to $af'' + bf' + cf = k$, \hfil \break
then $3 \psi_1 + 5 \psi_2$ is  a solution to $af'' + bf' + cf = 3h + 5k$,

Since $\psi_1$ is a solution to $af'' + bf' + cf = h$, $L(\psi_1) = h$.
 
Since $\psi_2$ is a solution to $af'' + bf' + cf = k$, $L(\psi_2) = k$.

Hence $L(3 \psi_1 + 5 \psi_2) = 3 L(\psi_1) + 5 L(\psi_2)$
\u
\hskip 1.27in$ = 3h + 5k$.  
\u
Thus $3 \psi_1 + 5 \psi_2$ is also a solution to 

\centerline{$af'' + bf' + cf = 
3h + 5k$}

\end 

\v
\hrule

Calculus pre-requisites you must know.

Derivative = slope of tangent line = rate.

Integral = area between curve and x-axis (where area can be negative).

\eject
Integration by parts:

Derivative of a product:  $(uv)' = uv' + vu'$

\centerline{$uv' = (uv)' - vu'$}

\centerline{$\int uv' = \int (uv)' - \int vu'$}

\centerline{$\int uv' = (uv) - \int vu'$}


Example:
$\int e^{2x} sin(3x)$

Let $u = sin(3x)$, $dv = e^{2x}$

then $du = 3cos(3x)$, $v = {1 \over 2}e^{2x}$

then $d^2u = -9sin(3x)$, $\int v = {1 \over 4}e^{2x}$

$\int e^{2x} sin(3x) = {1 \over 2}sin(3x)e^{2x} - \int {3 \over 2}e^{2x}cos(3x)$

\hskip 0.4in $= {1 \over 2}sin(3x)e^{2x} - [{3 \over 4}cos(3x)e^{2x} - \int {-9 \over 4}sin(3x)e^{2x}$


$\int e^{2x} sin(3x) = {1 \over 2}sin(3x)e^{2x} - {3 \over 4}cos(3x)e^{2x} - {9 \over 4}\int sin(3x)e^{2x}$


${13 \over 4} \int e^{2x} sin(3x) = {1 \over 2}sin(3x)e^{2x} - {3 \over 4}cos(3x)e^{2x}$


$\int e^{2x} sin(3x) = {4 \over 13}[{1 \over 2}sin(3x)e^{2x} - {3 \over 4}cos(3x)e^{2x}]$
\vfill

Exercise:  Calculate $\int e^{x} cos(2x)$