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MAE305-MAT301
Problem Set # 0, Self-test: Review of basic ideas
Assigned: 14 September 2016
Due: Never
This set of exercises is for review of elementary operations and ideas from calculus. It is
not to be turned in. The exercises also provide a few opportunities to gain familiarity
with elementary mathematical operations (including plotting) with Mathematica.
There are several topics from calculus that you should feel comfortable with before beginning this course. We will now provide a brief review of some of the more important topics
that will be of use either shortly or later in the course. These topics include changes
of variables and the chain-rule, standard integrals, integration by parts, partial di↵erentiation, Taylor series, and elementary operations involving complex numbers. It may
sound obvious, but is nevertheless worth stating, that it is wise to feel comfortable and
understand each of the operations, techniques and ideas discussed below. You should not
feel like you are being asked to memorize a menagerie of unrelated topics. All of these
themes are part of the toolbox of calculus that will appear in this course as well as other
quantitative courses you will take.
In addition, I suggest that you take this opportunity to learn a few basic commands using
Mathematica. On the course website we have provided a separate document with more
details including getting your own copy of the student version of Mathematica for your
computer. Here we only give brief instructions. To get started, it should be as easy as
(i) Open Mathematica
(ii) Elementary mathematical operations are +, , ⇤, / and powers are indicated with ˆ,
e.g. 2 ⇤ 3, x/6 and xˆ2. Also, Mathematica commands have a capitalized first letter
and arguments are contained within brackets, e.g. Cos[x], Sin[x], Exp[x2 ].
(iii) Important: To run a Mathematica command requires you to type “shift-return” at
the same time.
(iv) You integrate theR function f (x) = x2 with the command Integrate[x2 , x] and the
definite integral 01 x2 dx is obtained by typing Integrate[x2 , {x, 0, 1}].
(v) Similarly, you plot the function f (x) = x2 e x , from x = 0 to 2 with the command Plot[x2 e x ,{x,0,2}]. If you want to plot more than one function, then
you simply place the list of functions within curly brackets, as in Plot[{x2 e x ,
Sqrt[x]},{x,0,2}].
(vi) Try each of the above and then explore some of the Mathematica help pages.
0. Basic mathematical facts:
(a) ex+y = ex ey 6= ex + ey ; and for any constant a, we have ax+y = ax ay 6= ax + ay .
(b) ln e = 1; ln (eg ) = g; also, eln g = g; moreover, e
1
ln g
=g
1
=
1
g
6=
g.
(c) Show that lim t ln t ! 0. Hint: Use L’Hopital’s rule.
t!0
Explain why we then say that “t goes to zero faster than ln t goes to infinity”.
Solution: We can write
lim t ln t =
t!0
ln t
t 1
=
=
t 1
t 2
t
(1)
t ! 0.
after using L’Hopital’s rule. Thus, lim t ln t =
t!0
(d) Show that lim te t ! 0. Hint: Use the change of variables u = e
t!1
the given statement to item (c).
t
Explain why we then say that “e
t
to reduce
goes to zero faster than t goes to infinity”.
(e) An integral from calculus (that comes up later in the course): Use the identity
sin2 x = 1 cos(2x)
(familiar from calculus or high school) to show
2
Z ⇡/2
0
sin2 x dx =
⇡
.
4
(2)
Solution: Given the hint we write
Z ⇡/2
0
sin2 x dx =
1 Z ⇡/2
(1
2 0
cos(2x)) dx =
⇡
,
4
(3)
since the second term in parentheses integrates to sin(2x), which vanishes at
x = 0, ⇡/2.
If you want to test your skill further, then show
Z ⇡/2
0
sin4 x dx =
3⇡
.
16
(4)
Solution: We write
Z ⇡/2
0
sin4 x dx =
1 Z ⇡/2
(1
4 0
cos(2x))2 dx =
1 Z ⇡/2 ⇣
1
4 0
⌘
2 cos(2x) + cos2 (2x) dx,
(5)
which leads to (only giving the non-zero terms)
Z ⇡/2
0
sin4 x dx =
✓
1 ⇡ ⇡
+
4 2
4
◆
=
3⇡
.
16
(6)
If you feel energetic, show
Z ⇡/2
0
sin6 x dx =
5⇡
.
32
(7)
For perspective, these integrals were known already in 1792 by David Rittenhouse (1732-1796), a very early American intellectual, who apparently had no
2
formal education and never earned a degree (his method for evaluating the integrals was also di↵erent).1 In case you think that it is easy to spot a pattern,
R ⇡/2
note that 0 sin8 x dx = 35⇡
.
256
1. Changes of variables and the chain-rule: Recall the chain-rule
df (x(t))
dx(t) df (x)
=
.
dt
dt dx
(8)
df
as a function of t.
dt
with x = t2/3 , then
(a) Given f (x) = x2 e x . Let x = t2/3 . Evaluate
Solution: Given f (x) = x2 e
x
df (x)
= 2xe x x2 e x ,
dx
dx
2 1/3
=
t
.
dt
3
(9a)
(9b)
Therefore,
df (x(t))
dx df (x)
=
= (2x
dt
dt dx
x2 )e
x2
3
t
1/3
2
= (2t1/3
3
t)e
t2/3
.
(10)
(b) Where does the function f (x) = x2 e x have a maximum? Use your knowledge
of calculus, then plot the function and verify your answer.
(x)
= e x (2x x2 ) = 0 where x = 0 or x = 2. You
Solution: We calculate dfdx
can plot the function to verify the location of the maximum.
2. How ideas with simple changes of variables often arise in di↵erential equations:
Consider the function y(t), which satisfies the equation
dy
= ay 2
dt
with y(0) = y0 .
(11)
Here a, y0 and t0 are considered known numbers (constants). We will likely discuss
this equation the first day of class. Rescale the equation with the change of variables:
T =
t
,
tc
Y =
y
,
y0
(12)
where the function Y (T ) = y(t) and tc is a constant you are to identify. The key is to
choose tc to simplify the di↵erential equation to have all of the remaining constants
equal to unity.
1
See D.E. Zitarelli, “David Rittenhouse: Modern Mathematician”, Notices of the AMS, p. 11-14,
January 2015.
3
In the di↵erential equation, make the change of variables indicated. This question
tests your understanding of making a change of variables.
Show that Y (T ) satisfies the di↵erential equation
dY
=Y2 ,
dT
with Y (0) = 1.
(13)
What did you select for tc ?
Remark: Notice how equation (13) is “cleaner” (or we would say “nicer”) than the
original equation (11). The two equations are really structurally the same, but it
is easier to think about the equation that does not have extra algebraic quantities
that are cumbersome and do not influence how we think about the solution to the
equation. Moreover, in an actual physical problem, the quantity tc represents the
time scale over which significant changes occur, so with this identification you have
learned something before solving the problem.
Solution: We begin with the indicated change of variables. We write
dy
d
dT dY
y0 dY
=
(y0 Y (T )) =
=
,
dt
dt
dt dT
tc dT
(14)
where we have used the fact that y0 is a constant and dT
= t1c . The ODE can then
dt
be written
y0 dY
= ay02 Y 2 , with Y (0) = 1.
(15)
tc dT
Choose the constant tc to eliminate the other constants in the equation. Thus, we
choose tc = ay10 . The ODE then reduces to equation (13).
3. Integrals that yield logarithms:2
d ln x
d ln( x)
Reminder: Since
= 1/x for x > 0 and
= 1/x for x < 0, we use
dx
dx
the absolute value symbol to write
Z x
Z 2
(a) Evaluate
0
Solution:
Z 2
0
ds
= ln |x| + constant.
s
(16)
dx
. Are there any restrictions on the values of a?
x+a
dx
= [ln |x + a|]20 = ln |2 + a|
x+a
2
ln |a| = ln |1 + 2a 1 |.
(17)
In calculus you should beware that log x and ln x are often used interchangeably, as we do in this
document.
4
Notice that a 6= 0 and a 6= 2 for the interpretation based on the arguments of
the logarithm, which is discussed in advanced books as the meaning in terms
of the principal value. Otherwise, the integrand has a zero in the denominator
and is ill-defined for 2  a  0.
Z ⇡/2
cos x
(b) Evaluate
dx.
⇡/6 sin x
Solution: Recognize that the integrand is the derivative of ln | sin x| so that
Z ⇡/2
⇡/6
Z ⇡/2
cos x
d ln | sin x|
⇡/2
dx =
dx = [ln | sin x|]⇡/6 = ln(1)
sin x
dx
⇡/6
Z 2
ln(1/2) = ln 2.
(18)
dx
+ x2 )
1
Solution: Expand to obtain two straightforward integrals.
(c) Evaluate
Z 2
1
x(a2
✓
dx
1 Z2 1
=
x(a2 + x2 )
a2 1 x
=
"
◆

x
1
dx = 2 ln x
2
2
a +x
a
2
1
x
ln 2
2
2a
a + x2
#2
1
1
ln(a2 + x2 )
2
2
1
2
=
1
4(1 + a )
ln
.
2
2a
4 + a2
(19)
(d) Use a partial fraction expansion to evaluate
Z 1
0
dx
.
(x + a) (x + b)
(20)
Solution: Expanding the integrand we have
Z 1
0
✓
◆
dx
1 Z1
1
1
=
dx
(x + a)(x + b)
b a 0 x+a x+b
1
1
(1 + a)b
=
[ln |x + a| ln |x + b|]10 =
ln
(21)
.
b a
b a (1 + b)a
Note: For this problem, other problems here, and indeed standard problems in
calculus, this course, and other mathematics courses involving questions whose
answers require the use of an algorithm, you can check your answers by using
Maple or Mathematica, which are available to students at Princeton, or via the
Wolfram Alpha website, www.wolframalpha.com. We encourage you to use
these tools to aid in your learning.
4. Integration by parts:
5
Review the method of integration by parts:3
Z b
a
Z b
dv(x)
u(x)
dx = u(x)v(x)|ba
dx
a
v(x)
du(x)
dx.
dx
(23)
Z 1
(a) Evaluate
x log x dx. Can you a priori determine whether the integral is
0
positive or negative?
Solution: One application of integration by parts is necessary.
Z 1
0
"
x ln xdx =
x2
ln x
2
0
#
2 1
"
1 x
2 2
=
#1
1 Z 1 21
x dx
2 0
x
=
0
1
.
4
(24)
(Reminder: Use l’Hopital’s rule to show that limx!0 x2 ln x = 0.) Note that
the integral is clearly negative since on the interval 0 < x  1, ln x < 0.
(b) Evaluate
Z 1
0
x2 cos x dx.
Solution: Use integration by parts several times.
Z 1
2
x cos x dx =
0
=
=

sin x
sin
sin
x
2
2
+
1
"0 ✓
2
Z 1
x sin x dx
0
cos x
x
◆1
0
2 cos
2
2
3
+
1
Z 1
[sin x]10 =
cos x dx
0
sin
+
#
2 cos
2 sin
2
3
(. 25)
5. Partial di↵erentiation:
@f
@f
versus
.
@x
@y
Solution: In evaluating the partial derivative (@f /@x), the variable y is fixed,
i.e. it is regarded as a “constant” (independent of x). Similarly, the variable x is
regarded as a “constant” (independent of y) in evaluating the partial derivative
(@f /@y).
(a) Given a function f (x, y), state the meaning of
3
This equation is remembered easily by recalling the product rule for di↵erentiation
d
du(x)
dv(x)
(u(x)v(x)) =
v(x) + u(x)
,
dx
dx
dx
(22)
and then integrating both sides.The result was obtained first by the German philosopher and mathematician Gottfried Wilhelm Leibniz in 1675.
6
@f
@f
and
.
@x
@y
(b) Given f (x, y) = x2 exy log y, evaluate
Solution:
@f
= (2xexy + yx2 exy ) log y,
@x
!
@f
exy
2
xy
= x xe log y +
.
@y
y
(26a)
(26b)
(c) Suppose that u(x, y) is a specified function with x(t) and y(t). Write down an
equation to determine the derivative of u with respect to t.
Solution: Using the chain rule,
du(x(t), y(t))
dx @u dy @u
=
+
.
dt
dt @x dt @y
(27)
6. Changes of variables for special functions: Later in the course we will learn a variety
of special functions, which are defined in terms of integrals.
(a) Given the identify
evaluate
Z 1
0
Z 1
0
t2 e
2t
t2 e
t
dt = 2,
(28)
dt.
Solution: Start with equation (28) and make the change of variables u = 2t.
Then, we have
Z 1
0
e
2t 2
t dt =
Z 1
0
e
1
1Z 1 2
du =
ue
4 2
8 |0
{z
uu
2
u
=2
(b) Given the identify
Z 1
0
Z 1
sin x
⇡
dx = ,
x
2
1
du = .
4
}
(29)
(30)
sin(4x)
dx.
x
0
Solution: Start with equation (30) and make the change of variables u = 4x.
Then, as du = 4dx, we have
evaluate
Z 1
0
Z 1
sin(4x)
sin u
⇡
dx =
du = .
x
u
2
0
(31)
7. Some remarks about Taylor series:4 This exercise is more of a discussion then a set
of problems, but you may wish to write out some of the details to gain experience
with the ideas of a Taylor series.
4
Introduced by the English mathematician Brooke Taylor (1685-1731), though the idea was apparently
known earlier. Taylor series are one of the most useful ideas in calculus!
7
Why do we introduce the Taylor series? What does it tell you about functions? e.g.
the first term in the Taylor series? the second term? the third term? Draw sketch
of a typical smooth function and indicate the first few terms of a Taylor series of the
function near some point. If you have trouble thinking about these problems than
you did not pick one of the key intellectual features of calculus, which asks you to
think qualitatively about functions, as well as qualitatively about functions, such as
how they change, the rate of change, etc.
The Taylor (or MacLaurin) series of a function f (x) about the point x = 0 is defined
as
1
X
f (n) (0) n
dn f
f (x) ⌘
x
where f (n) ⌘ n .
(32)
n!
dx
n=0
Typically there is a range of x values, called the interval of convergence, for which
the series on the right-hand side converges to the value of f (x).
The generalization of the Taylor series for an expansion about the point x = a is
written
1
X
f (n) (a)
f (x) ⌘
(x a)n .
(33)
n!
n=0
Verify each of the Taylor series shown below (expansions about x = 0):
cos x = 1
sin x = x
1
X
x2 x4
x2n
+
+ ··· =
( 1)n
2!
4!
(2n)!
n=0
(34a)
1
X
x3 x5
x2n+1
+
+ ··· =
( 1)n
3!
5!
(2n + 1)!
n=0
(34b)
1
X
x2 x3
xn
+
+ ··· =
(34c)
2!
3!
n=0 n!
1
n(n + 1) 2 n(n + 1)(n + 2) 3
= 1 ⌥ nx +
x ⌥
x + · · · for |x| < 1(34d)
n
(1 ± x)
2
3!
1
X
1 2 1 3 1 4
( 1)n+1 n
ln(1 + x) = x
x + x
x + ··· =
x for 1 < x  1(34e)
2
3
4
n
n=1
ex = 1 + x +
Solution: From the basic di↵erentiation formula, we find
dn cos x
= ( 1)(n+1)/2 sin x (n odd)
dxn
dn cos x
= ( 1)n/2 cos x
(n even).
dxn
(35a)
(35b)
The Taylor series of cos x about x = 0 is therefore
1
X
xn
cos x =
n=0 n!
dn cos x
dxn
8
!
=
x=0
X
n=even
( 1)
n/2 x
n
n!
(36a)
1
X
=
( 1)m
m=0
x2m
,
(2m)!
(n = 2m).
(36b)
To see how well this idea works, plot the first few terms of the series and compare
with cos x.
We may follow a similar procedure as above to obtain the Taylor series for sin x
about x = 0. Or, by the identity
d cos x
=
dx
sin x,
we obtain
sin x =
=
d cos x
=
dx
1
X
( 1)m
m=1
Since
"
1
d X
x2m
( 1)m
dx m=0
(2m)!
(37a)
1
X
x2m 1
x2n+1
=
( 1)n
, (m = n + 1)
(2m 1)! n=0
(2n + 1)!
dn ex
dxn
#
(37b)
= [ex ]x=0 = 1,
x=0
then using the Taylor series formula we find
1
X
"
xn d n e x
e =
dxn
n=0 n!
x
#
=
x=0
1
X
xn
n=0 n!
(38)
Similarly,
d
1
n
=⌥
,
n
dx (1 ± x)
(1 ± x)n+1
(39)
then using the Taylor series formula we find
"
#
"
#
"
x d
1
x2 d 2
1
+
+
1! dx (1 ± x)n x=0 2! dx2 (1 ± x)n
x=0
x2
x3
= 1 ⌥ nx + n(n + 1) ⌥ n(n + 1)(n + 2) + · · ·
2
3!
1
=
(1 ± x)n
1
(1 ± x)n
#
x=0
+(40a)
···
(40b)
Since
d
1
ln(1 + x) =
,
dx
(1 + x)
(41)
then by the expansion in equation (40) (n = 1), we obtain
ln(1 + x) =
Z x
x2 x3
+
2
3
dt
=x
(1 + t)
9
1
X
x4
xn
+ ··· =
.
4
n=1 n
(42a)
8. Comparing functions when variables are small: In many practical problems we have
to compare the typical magnitude of two di↵erent terms. In a typical calculus course,
you learn the idea of a limit and so, for example, can evaluate (you might wish to
consult the Taylor series above for sin x)
sin2 (2x)
(2x)2
=
= 4.
(43)
x!0
x!0
x2
x2
It is helpful to think about these limits as being taken for x ⌧ 1. Hence, it is generally most useful to write lim sin x ⇡ x and to think about this as the approximation
x⌧1
of sin x when the argument is small compared to 1. You do not simply substitute
x = 0 in the formula since we seek an approximation to the original function.
lim sin x ⇡ x ! 0
or
lim
Using the idea of a Taylor series, approximate the following ratios when an argument
is small. Report your answer as a function of x.
sin(4x)
(a) Approximate
for x ⌧ 1. Is this ratio much smaller or much larger
x3/2
than unity?
Solution: We have
sin(4x)
4x
4
lim
= 3/2 = 1/2
1.
(44)
3/2
x⌧1 x
x
x
For x ⌧ 1 the ratio is clearly a large number.
2x2
(b) Approximate 1 ex2 for x ⌧ 1.
Solution: For small arguments, e
1
u
e
x2
⇡1
2x2
u, then we have
2x2
= 2 = 2.
x
(45)
9. Elementary properties of complex numbers: (there are no exercises here)
We only require a few elementary operations in this course but the subject is studied
in more detail in MAE306.
p
(a) We begin with the definition i2 = 1 or i =
1. A complex number is
written c = a + ib where the pair (a, b) are real numbers. We refer to a as
the real part of the complex number, a = Re c, and b as the imaginary part of
the complex number, b = Im c.
(b) The usual rules of algebra apply, with the caveat that i2 =
ib)(c + id) = (ac bd) + i(ad + bc).
1. Thus, (a +
(c) Euler’s formula:5 (This is one of many formulas associated with Euler’s name.)
eix = cos x + i sin x
5
(49)
We note that if z is a complex number, then we define ez according to equation (3) so that
ez = 1 + z +
z2
z3
z4
+
+
+ ···.
2!
3!
4!
10
(46)
(d) imaginary powers: Recall that y = x↵ may be written as y = e↵ ln x where we
have used the identity c = eln c . Thus, we see that if the power is a complex
number, we may write
xa+ib = xa xib = xa eib ln x = xa [ cos (b ln x) + i sin (b ln x) ] .
(50)
Note: This idea will reappear early in the course when we study the Euler equidimensional di↵erential equation (remember we said Euler’s name appears a lot).
Thus, letting z = ix we have
x2
x3
x4
i +
+ ··· ,
2!
3!
4!
which may be written by grouping real and imaginary parts as
✓
◆
x2
x4
x3
x5
ix
e =1
+
+ ··· + i x
+
+ · · · = cos x + i sin x.
2!
4!
3!
5!
eix = 1 + ix
The final identity follows by recognizing the Taylor series for cos x and sin x.
11
(47)
(48)
Figure 1: Reference: D. J. Struik, A Concise History of Mathematics, Dover (1948).
Historical note: Leonard Euler was one of the great mathematicians of all time. Born and
raised in Basle Switzerland, Euler studied under Johann Bernoulli, one of the Bernoulli
clan of mathematicians that spanned more than four generations (Euler’s father, a clergyman, in fact studied under Jakob Bernoulli who was Johann’s older brother). Euler moved
to the St. Petersburg Academy in Russia in 1727, the academy having been founded some
years earlier by Peter the Great. In 1741 Euler moved to the Berlin Academy only to
return to St. Petersburg in 1766. Euler lost one eye in 1735 and the other eye in 1766 but
somehow this did not deter the enormous breadth and productivity of his mathematics.
Euler’s notation is often the notation we use today. For example, notation in
trigonometry, the exponential e, and the definitive use of ⇡ for the ratio of the circumference
to the diameter of a circle are largely due to Euler, as are the use of i
p
for
1 and ⌃ to designate summation. Euler introduced the Gamma and Beta functions, which we will see later in this course, as well as the Euler constant defined as
1 1
1
lim ( + + . . . +
ln n) = 0.577216 . . . ⌘ . In one of Euler’s many textbooks we
n!1 1
2
n
find a section on di↵erential equations that distinguishes linear, exact and homogeneous
equations and serves as a model for textbooks today (hence the first week of this class).
References: D. J. Struik, A Concise History of Mathematics, Dover (1948); C. B. Boyer, A
History of Mathematics, as revised by U. C. Merzbach, John Wiley & Sons (1991).
12