Brief Calculus

2005·6.1 MB·English

by  Benjamin Crowell

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Fullerton, California
www.lightandmatter.com
copyright 2005 Benjamin Crowell
rev. November 10, 2015
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1 Rates of Change 7 3.6 Generalizations of
1.1 Changeindiscretesteps 7 l’Hoˆpital’srule . . . . . . 65
Two sides of the same coin, Multiple applications of the
7.—Some guesses, 9. rule, 66.—The indeterminate
1.2 Continuouschange . . 10 form ∞/∞, 66.—Limits at
infinity, 66.
A derivative, 13.—Properties
Problems. . . . . . . . 68
of the derivative, 14.—
Higher-order polynomials,
14.—The second derivative, 4 Integration
15.
4.1 Definite and indefinite
1.3 Applications . . . . . 17
integrals . . . . . . . . 71
Maxima and minima, 17.—
4.2 Thefundamentaltheorem
Propagation of errors, 19.
ofcalculus . . . . . . . 74
Problems. . . . . . . . 21
4.3 Propertiesoftheintegral 75
4.4 Applications . . . . . 76
2 To infinity — and
Averages, 76.—Work, 77.—
beyond! Probability, 77.
Problems. . . . . . . . 83
2.1 Infinitesimals. . . . . 25
2.2 Safeuseofinfinitesimals 30
5 Techniques
2.3 Theproductrule . . . 35
2.4 Thechainrule . . . . 37 5.1 Newton’smethod . . . 85
2.5 Exponentials and 5.2 Implicitdifferentiation . 86
logarithms . . . . . . . 39 5.3 Methodsofintegration . 87
The exponential, 39.—The Change of variable, 87.—
logarithm, 40. Integration by parts, 89.—
Partial fractions, 91.—
2.6 Quotients . . . . . . 42
Integrals that can’t be done,
2.7 Differentiation on a
95.
computer. . . . . . . . 43
Problems. . . . . . . . 98
Problems. . . . . . . . 47
6 Improper integrals
3 Limits and continuity
6.1 Integrating a function that
3.1 Continuity. . . . . . 53 blowsup . . . . . . . . 101
The intermediate value 6.2 Limits of integration at
theorem, 54.—The extreme infinity . . . . . . . . . 102
value theorem, 56.
Problems. . . . . . . . 104
3.2 Limits . . . . . . . 58
3.3 L’Hoˆpital’srule . . . . 61 7 Sequences and
3.4 Another perspective on
Series
indeterminateforms . . . 63
3.5 Limitsatinfinity. . . . 64 7.1 Infinitesequences. . . 105
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7.2 Infiniteseries . . . . 105 151.—Derivativeofex,151.—
7.3 Testsforconvergence . 106 Proofs of the generalizations
7.4 Taylorseries. . . . . 108 of l’Hoˆpital’s rule, 152.—
Problems. . . . . . . . 114 Proofofthefundamentalthe-
orem of calculus, 154.—The
intermediate value theorem,
8 Complex number
156.—Proof of the extreme
techniques value theorem, 159.—Proof
of the mean value theorem,
8.1 Review of complex
161.—Proofofthefundamen-
numbers . . . . . . . . 119
tal theorem of algebra, 162.
8.2 Euler’sformula . . . . 122
8.3 Partialfractionsrevisited 124
B Answers and solutions
Problems. . . . . . . . 126
165
9 Iterated integrals C Photo Credits 199
9.1 Integralsinsideintegrals 129
9.2 Applications . . . . . 131 D References and Fur-
9.3 Polarcoordinates . . . 133 ther Reading 201
9.4 Spherical and cylindrical
Further Reading, 201.—
coordinates. . . . . . . 135
References, 201.
Problems. . . . . . . . 137
E Reference 203
A Detours 139
E.1 Review. . . . . . . 203
Formal definition of the tan-
Algebra, 203.—Geometry,
gent line, 139.—Derivatives
area, and volume, 203.—
of polynomials, 140.—Details
Trigonometry with a right
of the proof of the deriva-
triangle, 203.—Trigonometry
tive of the sine function,
with any triangle, 203.
141.—Formal statement of
E.2 Hyperbolicfunctions. . 203
the transfer principle, 143.—
Isthetransferprincipletrue?, E.3 Calculus . . . . . . 204
144.—The transfer principle Rules for differentiation,
applied to functions, 149.— 204.—Integral calculus,
Proof of the chain rule, 204.—Table of integrals, 204.
1 Rates of Change
1.1 Change in
discrete steps
Toward the end of the eighteenth
century, a German elementary
school teacher decided to keep his
pupils busy by assigning them a
long, boring arithmetic problem:
to add up all the numbers from
one to a hundred.1 The chil- b/A trick for finding the
sum.
dren set to work on their slates,
and the teacher lit his pipe, con-
ing the area of the shaded region.
fident of a long break. But al-
Roughly half the square is shaded
most immediately, a boy named
in, so if we want only an approxi-
Carl Friedrich Gauss brought up
mate solution, we can simply cal-
his answer: 5,050. culate 72/2=24.5.
But, as suggested in figure b, it’s
not much more work to get an ex-
act result. There are seven saw-
teethstickingoutoutabovethedi-
agonal, with a total area of 7/2,
so the total shaded area is (72 +
7)/2 = 28. In general, the sum of
the first n numbers will be (n2 +
n)/2, which explains Gauss’s re-
a/Adding the numbers sult: (1002+100)/2=5,050.
from1to7.
Figure a suggests one way of solv- Twosidesofthesamecoin
ing this type of problem. The
filled-in columns of the graph rep- Problems like this come up fre-
resent the numbers from 1 to 7, quently. Imagine that each house-
and adding them up means find- hold in a certain small town sends
atotalofonetonofgarbagetothe
1I’m giving my own retelling of a dump every year. Over time, the
hoary legend. We don’t really know the
garbage accumulates in the dump,
exactproblem,justthatitwassupposed
tohavebeensomethingofthisflavor. taking up more and more space.
7
8 CHAPTER 1. RATES OF CHANGE
rate of change accumulated
result
13 13n
n (n2+n)/2
The rate of change of the function
x can be notated as x˙. Given the
function x˙, we can always deter-
mine the function x for any value
of n by doing a running sum.
Likewise, if we know x, we can de-
termine x˙ by subtraction. In the
c/Carl Friedrich Gauss example where x = 13n, we can
(1777-1855), alongtime
find x˙ = x(n)−x(n−1) = 13n−
aftergraduatingfromele-
13(n − 1) = 13. Or if we knew
mentaryschool.
that the accumulated amount of
garbage was given by (n2 +n)/2,
Let’s label the years as n = 1, 2,
3,..., and let the function2 x(n) we could calculate the town’s pop-
ulation like this:
represent the amount of garbage
that has accumulated by the end
of year n. If the population is
n2+n (n−1)2+(n−1)
constant, say 13 households, then −
garbage accumulates at a constant 2 2
rate, and we have x(n)=13n. n2+n−(cid:0)n2−2n+1+n−1(cid:1)
=
2
But maybe the town’s population
=n
isgrowing. Ifthepopulationstarts
out as 1 household in year 1, and
then grows to 2 in year 2, and so
on, then we have the same kind
of problem that the young Gauss
solved. After 100 years, the accu-
mulatedamountofgarbagewillbe
5,050tons. Thepileofrefusegrows
morequicklyeveryyear;therateof
change of x is not constant. Tabu-
lating the examples we’ve done so
far, we have this:
2Recallthatwhenxisafunction,the
notation x(n) means the output of the d/x˙ istheslopeofx.
function when the input is n. It doesn’t
representmultiplicationofanumberxby
anumbern. The graphical interpretation of
1.1. CHANGE IN DISCRETE STEPS 9
this is shown in figure d: on a of n.
graph of x=(n2+n)/2, the slope
of the line connecting two succes-
sivepointsisthevalueofthefunc- Someguesses
tion x˙.
Even though we lack Gauss’s ge-
Inotherwords,thefunctionsxand nius, we can recognize certain pat-
x˙ arelikedifferentsidesofthesame terns. One pattern is that if x˙ is a
coin. Ifyouknowone,youcanfind function that gets bigger and big-
the other — with two caveats. ger, it seems like x will be a func-
tion that grows even faster than
First, we’ve been assuming im-
x˙. In the example of x˙ = n and
plicitly that the function x starts x=(n2+n)/2,considerwhathap-
out at x(0) = 0. That might
pens for a large value of n, like
not be true in general. For in-
100. At this value of n, x˙ = 100,
stance, if we’re adding water to a
which is pretty big, but even with-
reservoir over a certain period of
outpawingaroundforacalculator,
time, the reservoir probably didn’t
weknowthatxisgoingtoturnout
start out completely empty. Thus,
really really big. Since n is large,
if we know x˙, we can’t find out n2 is quite a bit bigger than n, so
everything about x without some
roughly speaking, we can approxi-
further information: the starting mate x ≈ n2/2 = 5,000. 100 may
value of x. If someone tells you
be a big number, but 5,000 is a lot
x˙ = 13, you can’t conclude x =
bigger. Continuing in this way, for
13n, butonlyx=13n+c, wherec
n = 1000 we have x˙ = 1000, but
is some constant. There’s no such
x≈500,000 — now x has far out-
ambiguity if you’re going the op-
strippedx˙. Thiscanbeafungame
posite way, from x to x˙. Even
to play with a calculator: look at
if x(0) (cid:54)= 0, we still have x˙ =
which functions grow the fastest.
13n+c−[13(n−1)+c]=13.
Forinstance,yourcalculatormight
have an x2 button, an ex button,
Second,itmaybedifficult,oreven
and a button for x! (the factorial
impossible, to find a formula for
function,definedasx!=1·2·...·x,
the answer when we want to de-
e.g., 4! = 1·2·3·4 = 24). You’ll
termine the running sum x given
find that 502 is pretty big, but e50
a formula for the rate of change x˙.
is incomparably greater, and 50! is
Gauss had a flash of insight that
so big that it causes an error.
led him to the result (n2 + n)/2,
but in general we might only be All the x and x˙ functions we’ve
abletouseacomputerspreadsheet seen so far have been polynomials.
to calculate a number for the run- Ifxisapolynomial,thenofcourse
ning sum, rather than an equation we can find a polynomial for x˙ as
that would be valid for all values well, because if x is a polynomial,
10 CHAPTER 1. RATES OF CHANGE
thenx(n)−x(n−1)willbeonetoo.
It also looks like every polynomial
we could choose for x˙ might also
correspond to an x that’s a poly-
nomial. And not only that, but it
looks as though there’s a pattern
in the power of n. Suppose x is a
polynomial, and the highest power e/A pyramid with a vol-
of n it contains is a certain num- umeof12+22+32.
ber — the “order” of the polyno-
22+...+n2,andapplyingtheresultof
mial. Then x˙ is a polynomial of
theprecedingparagraph,wefindthat
that order minus one. Again, it’s
the volume of such a pyramid is ap-
fairly easy to prove this going one proximately (1/3)Ah, where A = n2 is
way,passingfromxtox˙,butmore
theareaofthebaseandh = n isthe
difficulttoprovetheoppositerela- height.
tionship: that if x˙ is a polynomial
When n is very large, we can get as
of a certain order, then x must be
good an approximation as we like to
a polynomial with an order that’s
a smooth-sided pyramid, and the er-
greater by one.
rorincurredinx(n)≈(1/3)n3+...by
We’d imagine, then, that the run- omittingthelower-orderterms...can
ning sum of x˙ = n2 would be a bemadeassmallasdesired.
polynomial of order 3. If we cal- We therefore conclude that the vol-
culate x(100) = 12 + 22 + ... + ume is exactly (1/3)Ah for a smooth-
1002 on a computer spreadsheet, sidedpyramidwiththeseproportions.
we get 338,350, which looks sus-
This is a special case of a theorem
piciously close to 1,000,000/3. It
lookslikex(n)=n3/3+..., where first proved by Euclid (propositions
XII-6 and XII-7) two thousand years
the dots represent terms involving
lower powers of n such as n2. The beforecalculuswasinvented.
fact that the coefficient of the n3
1.2 Continuous
term is 1/3 is proved in problem
21 on p. 23. change
Did you notice that I sneaked
Example1 somethingpastyouintheexample
Figure e shows a pyramid consisting ofwaterfillingupareservoir? The
of a single cubical block on top, sup- x and x˙ functions I’ve been using
ported by a 2×2 layer, supported in asexampleshaveallbeenfunctions
turnbya3×3layer. Thetotalvolume defined on the integers, so they
is12+22+32,inunitsofthevolumeof represent change that happens in
asingleblock.
discretesteps,buttheflowofwater
Generalizing to the sum x(n) = 12 + intoareservoirissmoothandcon-