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\Centerline{\bf MA106 Calculus I}
\medskip
\noindent{\bf Syllabus}
1. Real numbers; complex numbers:
\qquad representation, Argand diagram,
\qquad de Moivre's theorem, complex nth roots.
2. Real functions of one variable:
\qquad formulae, graphs, domain and range, inversion;
\qquad differentiation (formal rules, geometric interpretation,
tangents).
\qquad Local and global maxima and minima;
\qquad limits (formal rules and pictorial interpretations), l'H\^opital's
rule.
\qquad (Examples of functions to include $x^{\alpha}$, exp, sin, cos, tan, sinh,
cosh, tanh, polynomials, and their inverses.)
3. Sequences and series:
\qquad convergence of sequences, monotonic sequences,
\qquad summability of series, absolutely summable series, comparison and
ratio tests.
\qquad (Examples to include geometric and harmonic series, power series
for elementary functions.)
\qquad Power series. Radius of convergence
4. Intermediate and mean value theorems,
\qquad Taylor's theorem. Taylor's series;
\qquad Maclaurin's series.
\bigskip
\noindent{\bf Textbooks} There are no textbooks perfectly suited for this course. There is a shelf of approximately relevant books in section QA303 in the library. You should prefer those with lots of pictures and avoid any which use the Greek letters $\epsilon$ and $\delta$ in incomprehensible ways.
A book which you may find useful not only for this course but also for others is
\inset{E.Kreyszig, {\it Advanced Engineering Mathematics}, Wiley (QA37.K7).}
\noindent Two others which may be helpful are
\inset{E.W.Swokowski, M.Olinick, D.Pence \& J.A.Cole, {\it Calculus}, PWS Publishing (QA303.S7)}
\inset{J.Gilbert, {\it Guide to Mathematical Methods}, McMillan (QA37.G5).}
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D. As an example of something a little more complicated, consider the
semicircle function, given by saying
\qquad$\dom f=\{x:-1\le x\le 1\}=\{x:x\in\Bbb R,\,|x|\le 1\}$,
\qquad$f(x)=\sqrt{1-x^2}$,
\noindent the graph being
\def\Caption{$\sqrt{1-x^2}$}
\picture{ma1063a}{60pt}
\noindent The general rule is this: if you have some notion of what
rule-of-calculation you are going to use, pick for your domain the
largest set on which this rule works properly. Thus, if the
rule-of-calculation is $x\mapsto\sqrt{1-x^2}$, then the set of $x$ for
which this works is
\Centerline{$\{x:1-x^2\ge 0\}=\{x:x^2\le 1\}=\{x:|x|\le 1\}$,}
\noindent and this is likely to be the most sensible domain to choose.
But there might be circumstances in which you wanted (for instance) the
domain $\{x:-1f(x_0)$.
In general, we say that a function $f$ has a {\bf local maximum} at a
point $x_0\in\dom f$ if there is some interval $I$ around $x_0$ such
that $f(x)\le f(x_0)$ whenever $x\in I\cap\dom f$.
{\bf Local minima} are the same thing, upside down.
\medskip
{\bf True} or {\bf global} or {\bf overall} maxima: a real function $f$
has a {\bf global maximum} at $x_0\in\dom f$ if $f(x)\le f(x_0)$ for
every $x\in\dom f$. Thus $x\mapsto x^3-x$ has no global maximum
(supposing, that is, that we are taking its domain to be $\Bbb R$), but
$\cos$ has global maxima at $0$, $\pm 2\pi$, $\pm 4\pi,\ldots$ (with
value $1$) and global minima at $\pm\pi$, $\pm 3\pi,\ldots$ (with value
$-1$).
\inset{Remember: every global maximum is a local maximum; not every
local maximum is a global maximum.}
Another special kind of point is a {\bf stationary} point; this is a
point at which the graph of the function is running horizontally, that
is, the derivative exists and is zero.
In fact, when hunting for local and global maxima and minima of a
function $f$ the usual method is to start by finding its derivative $f'$
and solving the equation $f'(x)=0$. This is effective because
\inset{if $x$ is a local maximum or minimum, and $f'(x)$ is defined,
then $f'(x)=0$.}
\noindent So you can hope to pick up most of your maxima and minima this
way, at least when $f$ is differentiable. But
\inset{{\bf Warning!} Not every stationary point is a maximum or
minimum.}
\noindent For instance, consider the function $x\mapsto x^3$, with
domain $\Bbb R$:
\def\Caption{$x^3$}
\picture{ma1063c}{150pt}
%skets.for/ma1063c.for domain [-1.5,1.5] scale 2/2 cutoffs 1.5/1.5
\noindent This has a stationary point at $0$ but no local maximum or
minimum. What gives most people even more trouble is
\inset{{\bf Warning!} not every maximum or minimum is a stationary
point.}
\noindent For instance, try $f(x)=|x|$ (with domain $\Bbb R$):
\def\Caption{$|x|$}
\picture{ma1063d}{100pt}
\noindent This has a global minumum at $0$ which is not a stationary
point because there's no tangent there. (The function turns a sharp
corner.) Finally, you have to check sometimes at the endpoints of
domains, when functions stop suddenly:
\def\Caption{$\sqrt{1-x^2}$}
\picture{ma1063a}{60pt}
\noindent This has global minima at $\pm 1$, but the only stationary
point is at $0$, which is a global maximum.
So the best general rule I can give has to be
\inset{Find a formula for $f'$.
Look at all solutions of $f'(x)=0$; see if any of these are maxima or
minima.
Look at any odd points where $f'$ is undefined.
Look at the ends of the domain of $f$, if any.
{\it Draw the picture!} if you possibly can.}
\medskip
A word on what this course is about. I expect you find the rule above
a bit indefinite. In fact it actually asks you to use your brains. I
have to say that that is what I think mathematics is about. We do have
lots of formulae; places where all you have to do is type the numbers
in, press the right buttons and you'll get the right answer.
Understanding strict rules which, if exactly followed, will take you
safely to the end of a problem, is a very important part of mathematics.
But these techniques are important not because they enable us to escape
from thinking but because they liberate us to think about something more
interesting. I think the biggest difference between `school' and
`university' mathematics is that from now on we are going to be
exploring the edges where things don't quite work as we should like, and
finding out what might go wrong and what still goes right. This means
that quite a lot of the time -- in this course especially -- I am going
to have to ask you to reconsider something you've been confidently
doing, and warn you that in the wrong place it's chancy. But never
throw ideas away altogether. The things you have learnt are almost
always useful for something, even if they don't reach quite as far as
you imagined they did.
\bigskip
\noindent{\bf Differentiation} I've been talking about this for a couple
of lectures now. I am sure that you have all done quite a bit of
differentiating at one time or another, and I should like to believe
that you are all perfectly efficient. But I rather suspect that a bit
of revision will be useful, so I will run through the basic rules
carefully.
The {\bf derivative} of a real function $f$ at a point $x$ is the slope
of the tangent to the graph of $f$ at the point $(x,f(x))$, if there is
one ({\bf remember} $|x|$). Call it $f'(x)$.
\medskip
{\bf Basic formulae} $\sin'=\cos$, $\cos'=-\sin$, $\exp'=\exp$,
$\ln'(x)=\Bover1x$ (only for $x>0$!); if $f(x)=x^{\alpha}$ then
$f'(x)=\alpha x^{\alpha-1}$.
\medskip
{\bf Oops.} Are we quite sure we know what we mean by $x^{\alpha}$?
This is of course a matter of convention (just as it's a matter of
convention what we mean by the word `chair'). As with the argument of
a complex number, there are some real problems in the definition, which
mean that not all authors follow exactly the same rules. However, for
$x>0$ we have a general formula which works very reliably:
\Centerline{$x^{\alpha}=\exp(\alpha\ln x)$ if $x>0$, $\alpha\in\Bbb R$.}
\noindent (This is how most calculators do it. Can you think of a way
to find out whether yours does?) It works fine unless we want to look
at negative $x$. For integral $n$ we have a different way to do it:
\Centerline{$x^0=1$,
\quad$x^{n+1}=x^n\times x$ for integers $n\ge 0$,}
\Centerline{$x^{-n}=\Bover1{x^n}$ for integers $n\ge 1$, $x\ne 0$.}
\noindent (Can you find out whether your calculator uses this method?)
Of course this means that for things like $2^3$ we have two different
definitions:
\Centerline{$2^3=\exp(3\ln 2)$,\quad $2^3=2^2\times 2=2^1\times 2\times
2=2^0\times 2\times 2\times 2=1\times 2\times 2\times 2$}
\noindent and if I were giving proofs in this course, I should have to
take time out to prove that these two rules give the same answer.
Note in particular that on {\it my} rules, $0^0$ comes out as $1$. For
the things we shall want to do in this course, this rule is what makes
the formulae simplest, and I will stick to it.
\medskip
\noindent{\bf Rules for differentiation}
\medskip
A. Suppose $f$ and $g$ are two real functions. Then
\Centerline{$(f+g)'=f'+g'$;}
\noindent the derivative of the sum is the sum of the derivatives. So,
for instance,
\Centerline{$\bover{d}{dx}(x^2+x^3)
=\bover{d}{dx}x^2+\bover{d}{dx}x^3=2x+3x^2$.}
\noindent But what does this mean if, for instance, we are looking at
$\bover{d}{dx}(\ln x+\sqrt{1-x^2})$? In the next few pages I will give
conventions for dealing with combinations of partially-defined functions
which, while not universally used, seem to me to be closest to what
ordinary mathematicians and scientists actually do when they're thinking
about it at all.
Given two real functions $f$ and $g$ with domains $\dom f$, $\dom g$
respectively, define the function $f+g$ by saying
$$\eqalign{\dom(f+g)&=\{x:f(x)+g(x)\text{ is properly defined}\}\cr
&=\{x:x\in\dom f\text{ and }x\in\dom g\}
=\dom f\cap\dom g,\cr}$$
\noindent and write
\Centerline{$(f+g)(x)=f(x)+g(x)$ for $x\in\dom(f+g)=\dom f\cap\dom g$.}
\noindent Now if $x$ is a real number such that $f'(x)$ and $g'(x)$ are
both defined, $(f+g)'(x)$ is also defined and is equal to $f'(x)+g'(x)$.
\vfill\eject
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\noindent{\bf 5. Two famous theorems}
I want to spend a couple of lectures on two of the fundamental theorems of real analysis.
The first is a very straightforward fact:
\medskip
\inset{\noindent{\bf Intermediate Value Theorem: first form} Suppose that $f$ is a continuous real function defined (at least) on the whole of a closed interval $[a,b]$, where $a**0$. If we set $f(x)=\cos x$, $g(x)=x^2$, $a=0$ in Cauchy's Mean Value Theorem, we see that there is some $x\in\ooint{0,b}$ such that
\Centerline{$\Bover{\cos b-\cos 0}{b^2-0^2}=\Bover{\cos'x}{2x}
=-\Bover{\sin x}{2x}$,}
\noindent that is,
\Centerline{$\Bover{1-\cos b}{b^2}=\Bover{\sin x}{2x}$.}
\noindent Now we know from the result above that
\Centerline{$|\sin x-\sin 0|\le|x-0|$, that is, $|\sin x|\le|x|$,}
\noindent so that
\Centerline{$|\Bover{1-\cos b}{b^2}|=\Bover12|\Bover{\sin x}x|\le\Bover12$,}
\noindent and
\Centerline{$1-\cos b\le\Bover12b^2$, that is,
$\cos b\ge 1-\Bover12b^2$,}
\noindent as required.
If $b<0$, then $-b\ge 0$, so
\Centerline{$\cos b=\cos(-b)\le 1-\Bover12(-b)^2=1-\Bover12b^2$,}
\noindent while if $b=0$ then $\cos b=1=1-\Bover12b^2$. So the result is true for all $b\in\Bbb R$.
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\noindent{\bf Taylor's Theorem}
So far I've been talking of power series as if you just waited for an angel to reveal an interesting sequence $\sequence{r}{c_r}$ in a dream, and then worked out which function it represented. But of course there's a reverse approach which one uses quite as often. Recall the rule
\inset{if $f(x)=\sum_{r=0}^{\infty}c_rx^r$, then $c_r=\Bover1{r!}f^{(r)}(0)$}
\noindent (provided, at least, that the series $\sum_{r=0}^{\infty}c_rx^r$ has radius of convergence greater than $0$.) If we have a function $f$ which we understand well enough to get a formula for its $r$th derivative, we can turn this backwards:
\inset{$f(x)=\sum_{r=0}^{\infty}\Bover1{r!}f^{(r)}(0)x^r$}
\noindent for $x$ near $0$ (with a little bit of luck). This is {\bf Maclaurin's series}. For instance, starting with $f(x)=\sin x$, we have
\Centerline{$f'(x)=\cos x$,
\quad$f''(x)=-\sin x$,
\quad$f^{(3)}(x)=-\cos x$,
\quad$f^{(4)}(x)=\sin x=f(x)$}
\noindent and so on. This means that we can calculate $f^{(r)}(0)$ easily:
\Centerline{$f^{(0)}(0)=f(0)=\sin 0=0$,
\quad$f^{(1)}(0)=\cos 0=1$,}
\Centerline{$f^{(2)}(0)=-\sin 0=0$,
\quad$f^{(3)}(0)=-\cos 0=-1$,}
\Centerline{$f^{(4)}(0)=f(0)=0$,
\quad$f^{(5)}(0)=f'(0)=1$,
\quad$f^{(6)}(0)=f''(0)=0$}
\noindent and so on. Generally
$$\eqalign{f^{(r)}(0)&=0\text{ if }r\text{ is even},\cr
&=1\text{ if }r=4k+1,\cr
&=-1\text{ if }r=4k+3;\cr}$$
\noindent or, a little more manageably,
$$\eqalign{f^{(r)}(0)&=0\text{ if }r\text{ is even},\cr
&=(-1)^k\text{ if }r=2k+1.\cr}$$
\noindent So the corresponding Maclaurin series is
$$\eqalign{\sin x
&=\sum_{r=0}^{\infty}\Bover1{r!}f^{(r)}(0)x^r\cr
&=x-\Bover{x^3}{3!}+\Bover{x^5}{5!}-\Bover{x^7}{7!}+\ldots\cr
&=\sum_{k=0}^{\infty}\Bover{(-1)^k}{(2k+1)!}x^r.\cr
}$$
I gave this approach second place because there are theoretical problems with it. Recall once again the function $f(x)=e^{-1/x}$ for $x>0$, $f(x)=0$ for $x\le 0$; here all the derivatives $f^{(r)}(0)=0$ are perfectly well defined but tell us nothing about the function $f$. In practice, however, it's always worth forming the Maclaurin series first and thinking about what it means afterwards.
\medskip
\noindent{\bf Taylor series} All the power series I've discussed so far have been centered on $0$. This is much the easiest place to start and gives the prettiest formulae. But of course you can start anywhere else you like. If you want to understand the behaviour of a function $f$ near a point $x_0$, the natural approach, in the present context, is to look at the {\bf Taylor series}
\Centerline{$f(x_0)+(x-x_0)f'(x_0)+\Bover{(x-x_0)^2}{2!}f''(x_0)
+\Bover{(x-x_0)^3}{3!}f^{(3)}(x_0)+\ldots
+\Bover{(x-x_0)^r}{r!}f^{(r)}(x_0)+\ldots$,}
\noindent hoping that this will be summable and equal to $f(x)$ at least for $x$ close to $x_0$. (The Maclaurin series is just the special case $x_0=0$.) For instance, instead of writing
\Centerline{$\ln(1+x)=x-\Bover{x^2}2+\Bover{x^3}3-\Bover{x^4}4+\ldots$
for $-1R$; we still call $R$ the radius of convergence; we still have
\Centerline{$f'(x)=\sum_{r=1}^{\infty}rc_r(x-x_0)^{r-1}
=\sum_{r=0}^{\infty}(r+1)c_{r+1}(x-x_0)^r$}
\noindent when $|x-x_0|**