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%********** DATING ******************************************
\def\monthname {\ifcase\month\or January\or February\or March\or April\or
May\or June\or
July\or August\or September\or October\or November\or December\fi}

\newcount\mins  \newcount\hours  \hours=\time \mins=\time
\def\now{\divide\hours by60 \multiply\hours by60 \advance\mins by-\hours
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       \number\hours:\ifnum\mins<10 0\fi\number\mins\ A.M.\fi}
\def\today {\monthname\ \number\day, \number\year}


%**************** PAGE HEADERS *************************

\nopagenumbers
\def\runningtitle{\smallcaps taylor's theorem and the limit formula}
\headline={\ifnum\pageno>1\eoheadline\else\firstheadline\fi}
\def\names{\smallcaps a.\ ram}
%\def\firstheadline{\noindent Preliminary Draft \hfill  \today}
\def\firstheadline{}
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\def\oddheadline{\tenrm\hfil\runningtitle\hfil\folio}
\def\evenheadline{\tenrm\folio\hfil{\names}\hfil}


%**************** TITLE *************************
\vphantom{$ $}  %My kludge to get the first page to move down a bit
\vskip.75truein
\centerline{\titlefont Taylor's theorem and the limit formula}
\bigskip
\centerline{\rm Arun Ram}
%${}^\ast$ 
\centerline{Department of Mathematics}
\centerline{University of Wisconsin, Madison}
\centerline{Madison, WI 53706 USA}
\centerline{{\tt ram@math.wisc.edu}}
\medskip
\centerline{Version: \today}

%\footnote{}{\tinyrm 
%${}^\ast$ 
%Research partially supported by the National Security Agency
%and by EPSRC Grant GR K99015 at the Newton Institute for
%Mathematical Sciences.}
%\footnote{}{\tinyrm
%\noindent {Keywords:} exponential functions, trigonometric functions}

\bigskip

%**************** ABSTRACT *************************
%\noindent{\bf Abstract.}



\bigskip\bigskip
The derivative of $f$ with respect to $x$ is 
$\displaystyle{ {df\over dx} }$.
It is common to write $f'(x)$ in place of 
$\displaystyle{ {df\over dx} }$.
$$f'(x) = {df\over dx}.$$
The {\it second derivative of $f$ with respect to $x$} is 
$$f''(x) = {d^2f\over dx^2}={d\over dx}\left({df\over dx}\right),$$
the derivative of the derivative of $f$.  Both
$\displaystyle{
{d^2f\over dx^2} }$ and $f''(x)$ are notations for the same thing,
the second derivative of $f$.
\smallskip\noindent
The {\it third derivative of $f$ with respect to $x$} is 
$$f'''(x) = {d^3f\over dx^3}={d\over dx}\left({d^2f\over dx^2}\right),$$
the derivative of the second derivative of $f$.  Use the notations
$\displaystyle{
{d^3f\over dx^3} }$ and $f'''(x)$ interchangably for the third
derivative of $f$.
\smallskip\noindent
The {\it fourth derivative of $f$ with respect to $x$} is 
$$f^{(4)}(x) = {d^4f\over dx^4}={d\over dx}\left({d^3f\over dx^3}\right),$$
the derivative of the third derivative of $f$.  
\smallskip\noindent
Let $a$ be a number.  Then {\it $f$ evaluated at $a$} is 
$$f(a) = f\big\vert_{x=a} = c_0+c_1a+c_2a^2+c_3x^3+\cdots,$$
if $f(x) = c_0+c_1x+c_2x^2+c_3x^3+\cdots$.
Use both notations, $f(a)$ and $f\big\vert_{x=a}$, interchangably, for
$f$ evaluated at $a$.

\bigskip\noindent
{\bf Example:} If $f(x) = 7x^3+3x^2+5x+12$ and $a=3$
then
$$\eqalign{
f(3)\ \  & = 7\cdot 3^3+3\cdot 3^2+5\cdot 3+12 = 8\cdot 3^3+27= 9\cdot 3^3
=3^5, \cr
f\big\vert_{x=3}
& = 7\cdot 3^3+3\cdot 3^2+5\cdot 3+12 = 8\cdot 3^3+27= 9\cdot 3^3
=3^5. \cr
}$$
\medskip
$$\matrix{
\eqalign{
\displaystyle{ {df\over dx} } &= 21x^2+6x+5, \cr
f'\ \  &= 21x^2+6x+5, \cr
\cr
\displaystyle{ {d^2f\over dx^2} } &= 42x+6, \cr
f''\ \  &= 42x+6, \cr
\cr
\displaystyle{ {d^3f\over dx^3} } &= 42, \cr
f'''\ \  &= 42, \cr
\cr
\displaystyle{ {d^4f\over dx^4} } &= 0, \cr
f^{(4)}\ \  &= 0, \cr
}
&\qquad 
&
\eqalign{
\displaystyle{ {df\over dx}\Big\vert_{x=3} } 
&= 21\cdot 3^2+6\cdot 3+5 = 189+23 = 202, \cr
f'(3)\ \ \  &= 21\cdot 3^2+6\cdot 3+5 = 189+23 = 202, \cr
\cr
\displaystyle{ {d^2f\over dx^2}\Big\vert_{x=3} } 
&= 42\cdot 3+6 = 132, \cr
f''(3)\ \ \  &= 42\cdot 3+6 = 132, \cr
\cr
\displaystyle{ {d^3f\over dx^3}\Big\vert_{x=3} } &= 42, \cr
f'''(3)\ \ \  &= 42, \cr
\cr
\displaystyle{ {d^4f\over dx^4}\Big\vert_{x=3} } &= 0, \cr
f^{(4)}(3)\ \ \  &= 0. \cr
} \cr
}$$



\bigskip\bigskip
\noindent
{\it Taylor's and Macluarin's theorems 
and the limit formula for the derivative}
\bigskip
\bigskip

If
\qquad 
$f(x) = c_0 + c_1(x-a)+c_2(x-a)^2+c_3(x-a)^3+c_4(x-a)^4+c_5(x-a)^5+\cdots$
\medskip\noindent
then
$$\eqalign{
f(a) &= c_0, \cr
\cr
{df\over dx}\Big\vert_{x=a} 
&= \big(c_1+2c_2(x-a)+3c_3(x-a)^2+4c_4(x-a)^3
+5c_5(x-a)^4+\cdots\big)\Big\vert_{x=a}
\ =\ c_1, \cr
\cr
{d^2f\over dx^2}\Big\vert_{x=a} 
&= \big(2c_2+3\cdot 2c_3(x-a)+4\cdot 3c_4(x-a)^2
+5\cdot 4c_5(x-a)^3+\cdots\big)\Big\vert_{x=a}
\ =\ 2c_2, \cr
\cr
{d^3f\over dx^3}\Big\vert_{x=a} 
&= \big(3\cdot 2c_3+4\cdot 3\cdot 2c_4(x-a)
+5\cdot 4\cdot 3c_5(x-a)^2+6\cdot 5\cdot 4c_6(x-a)^3
%+7\cdot 6\cdot 5c_7(x-a)^4
+\cdots\big)\Big\vert_{x=a}
\ =\ 3\cdot 2c_3, \cr
\cr
{d^4f\over dx^4}\Big\vert_{x=a} 
&= \big(4\cdot 3\cdot 2c_4+5\cdot 4\cdot 3\cdot 2c_5(x-a)
+6\cdot 5\cdot 4\cdot 3c_4(x-a)^2
%+7\cdot 6\cdot 5\cdot 4c_5(x-a)^3
%+8\cdot 7\cdot 6\cdot 5c_6(x-a)^4
+\cdots\big)\Big\vert_{x=a}
\ =\ 4\cdot 3\cdot 2c_4, \cr
}$$
\medskip\noindent
and we can continue this process to find
$${d^kf\over dx^k}\Big\vert_{x=a} 
= k! c_k,
\qquad\hbox{for $k=1,2,3,\ldots$}.$$
\medskip\noindent
Dividing both sides by $k!$ gives
$$c_k = 
{1\over k!}\left({d^kf\over dx^k}\Big\vert_{x=a} \right).$$
\medskip\noindent
So 
$$f(x) = f(a) 
+ \left({df\over dx}\Big\vert_{x=a}\right)(x-a)
+ {1\over 2!}\left({d^2f\over dx^2}\Big\vert_{x=a}\right)(x-a)^2
+ {1\over 3!}\left({d^3f\over dx^3}\Big\vert_{x=a}\right)(x-a)^3
%+ {1\over 4!}\left({d^4f\over dx^4}\Big\vert_{x=a}\right)(x-a)^4
+\cdots,$$
\smallskip\noindent
or, equivalently,
$$f(x) = f(a) 
+f'(a)(x-a)
+{1\over 2!}f''(a)(x-a)^2
+{1\over 3!}f'''(a)(x-a)^3
+{1\over 4!}f^{(4)}(a)(x-a)^4
%+{1\over 5!}f^{(5)}(a)(x-a)^5
+\cdots.$$
\medskip\noindent
Now subtract $f(a)$ from both sides:
$$f(x)- f(a) =
f'(a)(x-a)
+{1\over 2!}f''(a)(x-a)^2
+{1\over 3!}f'''(a)(x-a)^3
+{1\over 4!}f^{(4)}(a)(x-a)^4
%+{1\over 5!}f^{(5)}(a)(x-a)^5
+\cdots.$$
\medskip\noindent
Divide both sides by $x-a$.
$${f(x)- f(a)\over x-a} =
f'(a)
+{1\over 2!}f''(a)(x-a)
+{1\over 3!}f'''(a)(x-a)^2
+{1\over 4!}f^{(4)}(a)(x-a)^3
%+{1\over 5!}f^{(5)}(a)(x-a)^4
+\cdots.$$
\medskip\noindent
Evaluate both sides at $x=a$.
$${f(x)- f(a)\over x-a}\Big\vert_{x=a} =
f'(a)+0+0+0+0+\cdots. \qquad\qquad\qquad\qquad \qquad\qquad\qquad$$
\bigskip\noindent
So \qquad $\displaystyle{
f'(a) = {f(x)- f(a)\over x-a}\Big\vert_{x=a} }$.
\bigskip\noindent
Let $x = a+h$.  Then\qquad
$\displaystyle{
f'(a) = {f(a+h)- f(a)\over a+h-a}\Big\vert_{a+h=a} }$.
\bigskip\noindent
So \qquad $\displaystyle{
{df\over dx}\Big\vert_{x=a} 
= {f(a+h)- f(a)\over h}\Big\vert_{h=0} }$.
\bigskip\noindent
Another way to write this is
$${df\over dx}\Big\vert_{x=a} 
= \lim_{h\to 0} {f(a+h)- f(a)\over h}. $$

\bigskip\bigskip\noindent
{\bf Example:}  Suppose you want to know what $f$ I'm thinking of
and I refuse to tell you.
\bigskip\noindent
You ask me what $f(0)$ is and I say ``$6$''.
\smallskip\noindent
You ask me what $f'(0)$ is and I say ``$10$''.
\smallskip\noindent
You ask me what $f''(0)$ is and I say ``$31$''.
\smallskip\noindent
You ask me what $f'''(0)$ is and I say ``$5$''.
\smallskip\noindent
You ask me what $f^{(4)}(0)$ is and I say ``$7$''.
\smallskip\noindent
You ask me what $f^{(5)}(0)$ is and I say ``$0$''.
\smallskip\noindent
You ask me what $f^{(6)}(0)$ is and I say ``$0$''.
\smallskip\noindent
You ask me what $f^{(7)}(0)$ is and I say ``$0$, they are all
coming out to $0$ now.''.
\medskip\noindent
At this point you win because you know that 
$$\eqalign{
f(x) 
&= f(0)+f'(0)(x-0)+{1\over 2!}f''(0)(x-0)^2+
{1\over 3!}f'''(0)(x-0)^3+\cdots \cr
&=6+10(x-0)+{1\over 2!}31(x-0)^2+{1\over 3!}5(x-0)^3
+{1\over 4!}7(x-0)^4 \cr
&\qquad\qquad\qquad
+{1\over 5!}\cdot0(x-0)^5 
+{1\over 6!}\cdot 0(x-0)^6+{1\over 7!}\cdot0(x-0)^7+0+0+\cdots \cr
&= 6+10x+{31\over 2}x^2+{5\over 6}x^3+{7\over 24}x^4, \cr
}$$
and so you have found out what $f$ is.


\bigskip\bigskip\noindent
{\bf Example:}  Suppose you want to know what $f$ I'm thinking of
and I refuse to tell you.
\bigskip\noindent
You ask me what $f(0)$ is and I say 
``I won't tell you, but $f(3)=4$''.
\smallskip\noindent
You ask me what $f'(0)$ is and I say 
``I won't tell you, but 
$\displaystyle{ {df\over dx}\Big\vert_{x=3}=2 }$''.
\smallskip\noindent
You ask me what $f''(0)$ is and I say 
``I won't tell you, but 
$\displaystyle{ {d^2f\over dx^2}\Big\vert_{x=3}=5 }$''.
\smallskip\noindent
You ask me what $f'''(0)$ is and I say 
\smallskip
``I won't tell you, but 
$\displaystyle{ {d^3f\over dx^3}\Big\vert_{x=3}=0 }$ and
all the rest of the
$\displaystyle{ {d^kf\over dx^k}\Big\vert_{x=3} }$ are
coming out to $0$''.
\medskip\noindent
At this point you win because you know that 
$$\eqalign{
f(x) 
&= f\big\vert_{x=3}
+\left({df\over dx}\Big\vert_{x=3}\right)(x-3)
+{1\over 2!}\left({d^2f\over dx^2}\Big\vert_{x=3}\right)(x-3)^2
+{1\over 3!}\left({d^3f\over dx^3}\Big\vert_{x=3}\right)(x-3)^3
%+{1\over 4!}\left({d^4f\over dx^4}\Big\vert_{x=3}\right)(x-3)^4
+\cdots \cr
&= 2+5(x-3) + {1\over 2!}5(x-3)^2+{1\over 3!}\cdot 0(x-3)^3+0+0+\cdots
\cr
&= 2+5x-15+{5\over2}(x^2-6x+9)+0+0+\cdots \cr
&=-13+5x+{5\over2}x^2-15x+{45\over 2} \cr
&= {5\over 2}x^2-10x+{19\over 2}, \cr
}$$
and so you know what $f$ is.



\vfill\eject
\end

