\magnification=1100 \overfullrule0pt \input amssym.def \input prepictex \input pictex \input postpictex % ********************* Definitions ************************************ %\def\widetilde{\mathaccent"0365 } \def\CC{{\Bbb C}} \def\FF{{\Bbb F}} \def\HH{{\Bbb H}} \def\NN{{\Bbb N}} \def\OO{{\Bbb O}} \def\QQ{{\Bbb Q}} \def\RR{{\Bbb R}} \def\ZZ{{\Bbb Z}} \def\cA{{\cal A}} \def\cB{{\cal B}} \def\cR{{\cal R}} \def\cZ{{\cal Z}} \def\cC{{\cal C}} \def\cR{{\cal R}} \def\fb{\frak{b}} \def\fg{\frak{g}} \def\fh{\frak{h}} \def\fn{\frak{n}} \def\Card{\hbox{Card}} \def\End{\hbox{End}} \def\Hom{\hbox{Hom}} \def\Ind{\hbox{Ind}} \def\Id{\hbox{Id}} \def\codim{\hbox{codim}} \def\diag{\hbox{diag}} \def\id{\hbox{id}} \def\im{\hbox{im}} \def\tr{\hbox{tr}} \def\Tr{\hbox{Tr}} \def\fbox{} % ********************* FONTS ************************************ \font\smallcaps=cmcsc10 \font\titlefont=cmr10 scaled \magstep1 \font\titlefontbold=cmbx10 scaled \magstep1 \font\titlesubfont=cmr10 scaled \magstep1 \font\sectionfont=cmbx10 \font\tinyrm=cmr10 at 8pt % ******************** SECTION HEADERS *************************** \newcount\sectno \newcount\subsectno \newcount\resultno \def\section #1. #2\par{ \sectno=#1 \resultno=0 \bigskip\noindent{\sectionfont #1. #2}~\medbreak} \def\subsection #1\par{\bigskip\noindent{\it #1} \medbreak} %******************* MATHEMATICAL LABELS ************************** \def\prop{ \global\advance\resultno by 1 \medskip\noindent{\bf Proposition \the\sectno.\the\resultno. }\sl} \def\lemma{ \global\advance\resultno by 1 \medskip\noindent{\bf Lemma \the\sectno.\the\resultno. } \sl} \def\fact{ \global\advance\resultno by 1 \medskip\noindent{\bf Fact \the\sectno.\the\resultno. } \sl} \def\remark{ \global\advance\resultno by 1 \medskip\noindent{\bf Remark \the\sectno.\the\resultno. }} \def\example{ \global\advance\resultno by 1 \medskip\noindent{\bf Example \the\sectno.\the\resultno. }\sl} \def\cor{ \global\advance\resultno by 1 \medskip\noindent{\bf Corollary \the\sectno.\the\resultno. }\sl} \def\thm{ \global\advance\resultno by 1 \medskip\noindent{\bf Theorem \the\sectno.\the\resultno. }\sl} \def\defn{ \global\advance\resultno by 1 \medskip\noindent{\it Definition \the\sectno.\the\resultno. }\slrm} \def\endthm{\rm\medskip} \def\thmend{\rm\medskip} \def\endlemma{\rm\medskip} \def\endfact{\rm\medskip} \def\endexample{\rm\medskip} \def\endprop{\rm\medskip} \def\endcor{\rm\medskip} \def\pf{\rm\smallskip\noindent{\it Proof. }} \def\endpf{\qed\hfil\medskip} \def\pfend{\qed\hfil\medskip} \def\note{\smallbreak\noindent{Note:}} \def\enddefn{\rm\medskip} %Homemade Struts: \newbox\strutAbox \setbox\strutAbox=\hbox{\vrule height 12pt depth6pt width0pt} \def\strutA{\relax\copy\strutAbox} \newbox\strutBbox \setbox\strutBbox=\hbox{\vrule height 10pt depth5pt width0pt} \def\strutB{\relax\copy\strutBbox} \newbox\strutDbox \setbox\strutDbox=\hbox{\vrule height 11pt depth5pt width0pt} \def\strutD{\relax\copy\strutDbox} %high strut: \newbox\strutHbox \setbox\strutHbox=\hbox{\vrule height 11pt depth1pt width0pt} \def\strutH{\relax\copy\strutHbox} %low strut: \newbox\strutLbox \setbox\strutLbox=\hbox{\vrule height 1pt depth5pt width0pt} \def\strutL{\relax\copy\strutLbox} % hack to ignore lots of typed stuff.... \def\ignore#1{\relax} % ****************** QED SIGNS ********************************* \def\qed{\hbox{\hskip 1pt\vrule width4pt height 6pt depth1.5pt \hskip 1pt}} \def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt \hbox{\vrule width.#2pt height#1pt \kern#1pt \vrule width.2pt} \hrule height.2pt}}}} \def\square{\mathchoice\sqr54\sqr54\sqr{3.5}3\sqr{2.5}3} \def\whiteslug{\bf $ \square $ \rm} % open square %*************** EQUATIONS WITH NUMBERS ************** \def\formula{\global\advance\resultno by 1 \eqno{(\the\sectno.\the\resultno)}} \def\formulano{\global\advance\resultno by 1 (\the\sectno.\the\resultno)} \def\tableno{\global\advance\resultno by 1 \the\sectno.\the\resultno. } \def\lformula{\global\advance\resultno by 1 \leqno(\the\sectno.\the\resultno)} %************Commutative diagrams********************** \def\mapright#1{\smash{\mathop {\longrightarrow}\limits^{#1}}} \def\mapleftright#1{\smash{\mathop {\longleftrightarrow}\limits^{#1}}} \def\mapsrightto#1{\smash{\mathop {\longmapsto}\limits^{#1}}} \def\mapleft#1{\smash{ \mathop{\longleftarrow}\limits^{#1}}} \def\mapdown#1{\Big\downarrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \def\lmapdown#1{{\hbox{$\scriptstyle#1$}} \llap {$\vcenter{\hbox{\Big\downarrow}}$} } \def\mapup#1{\Big\uparrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \def\mapne#1{\Big\nearrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \def\mapse#1{ %{$\vcenter{ \hbox{$\scriptstyle#1$} %$} \rlap{ $\vcenter{\hbox{$\searrow$}}$ } } \def\mapnw#1{\Big\nwarrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \def\mapsw#1{ %\Big \swarrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} %********** 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 \divide\hours by60 % NOTE: \divide only gives integer answers. \ifnum\hours>12 \advance\hours by-12 \number\hours:\ifnum\mins<10 0\fi\number\mins\ P.M.\else \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{} \def\eoheadline{\ifodd\pageno\oddheadline\else\evenheadline\fi} \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