Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 4 pptx

0 <  < 2 there is no value of δ > 0 such that |sin(1/x
1
) − sin(1/x
2
)| <  for all x
1
, x
2
∈ (0, 1) and |x
1
− x
2
| < δ.
Thus sin(1/x) is not uniformly continuous in the open interval (0, 1).
Solution 3.6
First consider the function
√
x. Note that the function
√
x + δ −
√
x is a decreasing function of x and an increasing
function of δ for positive x and δ. Thus for any fixed δ, the maximum value of
√
x + δ −
√
x is bounded by
√
δ.
Therefore on the interval (0, 1), a sufficient condition for |

√
x −
√
ξ| <  is |x −ξ| < 
2
. The function
√
x is uniformly
continuous on the interval (0, 1).
Consider any positive δ and . Note that
1
x
−
1
x + δ
> 
for
x <
1
2


δ
2
+
4δ

− δ

.

Thus there is no value of δ such that




1
x
−
1
ξ




< 
for all |x − ξ| < δ. The function
1
x
is not uniformly continuous on the interval (0, 1).
Solution 3.7
Let the function f(x) be continuous on a closed interval. Consider the function
e(x, δ) = sup
|ξ−x|<δ
|f(ξ) − f(x)|.
Since f(x) is continuous, e(x, δ) is a continuous function of x on the same closed interval. Since continuous functions
on closed intervals are bounded, there is a continuous, increasing fun ction (δ) satisfying,
e(x, δ) ≤ (δ),
for all x in the closed interval. Since (δ) is continuous and increasing, it has an inverse δ(). Now note that
|f(x) − f(ξ)| <  for all x and ξ in the closed interval satisfying |x − ξ| < δ(). Thus the function is uniformly
continuous in the closed interval.

94
Solution 3.8
1. The statement
lim
n→∞
a
n
= L
is equivalent to
∀  > 0, ∃ N s.t. n > N ⇒ |a
n
− L| < .
We want to show that
∀ δ > 0, ∃ M s.t. m > M ⇒ |a
2
n
− L
2
| < δ.
Suppose that |a
n
− L| < . We obtain an upper bound on |a
2
n
− L
2
|.
|a
2
n

− L
2
| = |a
n
− L||a
n
+ L| < (|2L|+ )
Now we choose a value of  such that |a
2
n
− L
2
| < δ
(|2L| + ) = δ
 =
√
L
2
+ δ − |L|
Consider any fixed δ > 0. We see that since
for =
√
L
2
+ δ − |L|, ∃ N s.t. n > N ⇒ |a
n
− L| < 
implies that
n > N ⇒ |a
2

n
− L
2
| < δ.
Therefore
∀ δ > 0, ∃ M s.t. m > M ⇒ |a
2
n
− L
2
| < δ.
We conclude that lim
n→∞
a
2
n
= L
2
.
2. lim
n→∞
a
2
n
= L
2
does not imply that lim
n→∞
a
n

= L. Consider a
n
= −1. In this case lim
n→∞
a
2
n
= 1 and
lim
n→∞
a
n
= −1.
95
3. If a
n
> 0 for all n > 200, and lim
n→∞
a
n
= L, then L is not necessarily positive. Consider a
n
= 1/n, which
satisfies the two constraints.
lim
n→∞
1
n
= 0
4. The statement lim

x→∞
f(x) = L is equivalent to
∀  > 0, ∃ X s.t. x > X ⇒ |f(x) − L| < .
This implies that for n > X, |f(n) −L| < .
∀  > 0, ∃ N s.t. n > N ⇒ |f(n) − L| < 
lim
n→∞
f(n) = L
5. If f : R → R is continuous and lim
n→∞
f(n) = L, then for x ∈ R, it is not necessarily true that lim
x→∞
f(x) = L.
Consider f(x) = sin(πx).
lim
n→∞
sin(πn) = lim
n→∞
0 = 0
lim
x→∞
sin(πx) does not exist.
Solution 3.9
a.
d
dx
(x
n
) = lim
∆x→0


(x + ∆x)
n
− x
n
∆x

= lim
∆x→0



x
n
+ nx
n−1
∆x +
n(n−1)
2
x
n−2
∆x
2
+ ···+ ∆x
n

− x
n
∆x



= lim
∆x→0

nx
n−1
+
n(n −1)
2
x
n−2
∆x + ··· + ∆x
n−1

= nx
n−1
96
d
dx
(x
n
) = nx
n−1
b.
d
dx
(f(x)g(x)) = lim
∆x→0

f(x + ∆x)g(x + ∆x) − f(x)g(x)

∆x

= lim
∆x→0

[f(x + ∆x)g(x + ∆x) − f(x)g(x + ∆x)] + [f(x)g(x + ∆x) −f(x)g(x)]
∆x

= lim
∆x→0
[g(x + ∆x)] lim
∆x→0

f(x + ∆x) − f(x)
∆x

+ f(x) lim
∆x→0

g(x + ∆x) −g(x)
∆x

= g(x)f

(x) + f(x)g

(x)
d
dx
(f(x)g(x)) = f(x)g


(x) + f

(x)g(x)
c. Consider a right triangle with hypotenuse of length 1 in the first quadrant of the plane. Label the vertices A, B,
C, in clo ckwis e order, starting with the vertex at the origin. The angle of A is θ. The length of a circular arc of
radius cos θ that connects C to the hypotenuse is θ cos θ. The length of the side BC is sin θ. The length of a
circular arc of radius 1 that connects B to the x axis is θ. (See Figure 3.20.)
Considering the length of these three curves gives us the inequality:
θ cos θ ≤ sin θ ≤ θ.
Dividing by θ,
cos θ ≤
sin θ
θ
≤ 1.
Taking the limit as θ → 0 gives us
lim
θ→0
sin θ
θ
= 1.
97
B
θ
sin
A
C
θ
θ
θ

cosθ
Figure 3.20:
One more little tidbit we’ll need to know is
lim
θ→0

cos θ − 1
θ

= lim
θ→0

cos θ − 1
θ
cos θ + 1
cos θ + 1

= lim
θ→0

cos
2
θ − 1
θ(cos θ + 1)

= lim
θ→0

−sin
2

θ
θ(cos θ + 1)

= lim
θ→0

−sin θ
θ

lim
θ→0

sin θ
(cos θ + 1)

= (−1)

0
2

= 0.
98
Now we’re ready to find the derivative of sin x.
d
dx
(sin x) = lim
∆x→0

sin(x + ∆x) − sin x
∆x


= lim
∆x→0

cos x sin ∆x + sin x cos ∆x −sin x
∆x

= cos x lim
∆x→0

sin ∆x
∆x

+ sin x lim
∆x→0

cos ∆x −1
∆x

= cos x
d
dx
(sin x) = cos x
d. Let u = g(x). Consider a nonzero increment ∆x, which induces the increments ∆u and ∆f. By definition,
∆f = f(u + ∆u) − f(u), ∆u = g(x + ∆x) − g(x),
and ∆f, ∆u → 0 as ∆x → 0. If ∆u = 0 then we have
 =
∆f
∆u
−

df
du
→ 0 as ∆u → 0.
If ∆u = 0 for some values of ∆x then ∆f also vanishes and we define  = 0 for theses values. In either case,
∆y =
df
du
∆u + ∆u.
99
We divide this equation by ∆x and take the limit as ∆x → 0.
df
dx
= lim
∆x→0
∆f
∆x
= lim
∆x→0

df
du
∆u
∆x
+ 
∆u
∆x

=

df

du

lim
∆x→0
∆f
∆x

+

lim
∆x→0



lim
∆x→0
∆u
∆x

=
df
du
du
dx
+ (0)

du
dx

=

df
du
du
dx
Thus we see that
d
dx
(f(g(x))) = f

(g(x))g

(x).
Solution 3.10
1.
f

(0) = lim  → 0
|| −0

= lim  → 0||
= 0
The function is differentiable at x = 0.
100
2.
f

(0) = lim  → 0

1 + || − 1


= lim  → 0
1
2
(1 + ||)
−1/2
sign()
1
= lim  → 0
1
2
sign()
Since the limit does not exist, the function is not differentiable at x = 0.
Solution 3.11
a.
d
dx
[x sin(cos x)] =
d
dx
[x] sin(cos x) + x
d
dx
[sin(cos x)]
= sin(cos x) + x cos(cos x)
d
dx
[cos x]
= sin(cos x) −x cos(cos x) sin x
d
dx

[x sin(cos x)] = sin(cos x) −x cos(cos x) sin x
b.
d
dx
[f(cos(g(x)))] = f

(cos(g(x)))
d
dx
[cos(g(x))]
= −f

(cos(g(x))) sin(g(x))
d
dx
[g(x)]
= −f

(cos(g(x))) sin(g(x))g

(x)
d
dx
[f(cos(g(x)))] = −f

(cos(g(x))) sin(g(x))g

(x)
101
c.

d
dx

1
f(ln x)

= −
d
dx
[f(ln x)]
[f(ln x)]
2
= −
f

(ln x)
d
dx
[ln x]
[f(ln x)]
2
= −
f

(ln x)
x[f(ln x)]
2
d
dx


1
f(ln x)

= −
f

(ln x)
x[f(ln x)]
2
d. First we write the expression in terms exponentials and logarithms,
x
x
x
= x
exp(x ln x)
= exp(exp(x ln x) ln x).
Then we differentiate using the chain rule and the produ ct rule.
d
dx
exp(exp(x ln x) ln x) = exp(exp(x ln x) ln x)
d
dx
(exp(x ln x) ln x)
= x
x
x

exp(x ln x)
d
dx

(x ln x) ln x + exp(x ln x)
1
x

= x
x
x

x
x
(ln x + x
1
x
) ln x + x
−1
exp(x ln x)

= x
x
x

x
x
(ln x + 1) ln x + x
−1
x
x

= x
x

x
+x

x
−1
+ ln x + ln
2
x

d
dx
x
x
x
= x
x
x
+x

x
−1
+ ln x + ln
2
x

102
e. For x > 0, the expression is x sin x; for x < 0, the expression is (−x) sin(−x) = x sin x. Thus we see that
|x|sin |x| = x sin x.
The first derivative of this is
sin x + x cos x.

d
dx
(|x|sin |x|) = sin x + x cos x
Solution 3.12
Let y(x) = sin x. Then y

(x) = cos x.
d
dy
arcsin y =
1
y

(x)
=
1
cos x
=
1
(1 −sin
2
x)
1/2
=
1
(1 −y
2
)
1/2
d

dx
arcsin x =
1
(1 −x
2
)
1/2
103
Let y(x) = tan x. Then y

(x) = 1/ cos
2
x.
d
dy
arctan y =
1
y

(x)
= cos
2
x
= cos
2
(arctan y)
=

1
(1 + y

2
)
1/2

=
1
1 + y
2
d
dx
arctan x =
1
1 + x
2
Solution 3.13
Differentiating the equation
x
2
+ [y(x)]
2
= 1
yields
2x + 2y(x)y

(x) = 0.
We can solve this equation for y

(x).
y


(x) = −
x
y(x)
To find y

(1/2) we need to find y(x) in terms of x.
y(x) = ±
√
1 −x
2
Thus y

(x) is
y

(x) = ±
x
√
1 −x
2
.
104
y

(1/2) can have the two values:
y


1
2


= ±
1
√
3
.
Solution 3.14
Differentiating the equation
x
2
− xy(x) + [y(x)]
2
= 3
yields
2x −y(x) − xy

(x) + 2y(x)y

(x) = 0.
Solving this equation for y

(x)
y

(x) =
y(x) −2x
2y(x) −x
.
Now we differentiate y


(x) to get y

(x).
y

(x) =
(y

(x) −2)(2y(x) −x) −(y(x) − 2x)(2y

(x) −1)
(2y(x) −x)
2
,
y

(x) = 3
xy

(x) −y(x)
(2y(x) −x)
2
,
y

(x) = 3
x
y(x)−2x
2y(x)−x
− y(x)

(2y(x) −x)
2
,
y

(x) = 3
x(y(x) −2x) − y(x)(2y(x) − x)
(2y(x) −x)
3
,
y

(x) = −6
x
2
− xy(x) + [y(x)]
2
(2y(x) −x)
3
,
y

(x) =
−18
(2y(x) −x)
3
,
105
Solution 3.15
a.

f

(x) = (12 −2x)
2
+ 2x(12 − 2x)(−2)
= 4(x −6)
2
+ 8x(x − 6)
= 12(x −2)(x − 6)
There are critical points at x = 2 and x = 6.
f

(x) = 12(x −2) + 12(x − 6) = 24(x −4)
Since f

(2) = −48 < 0, x = 2 is a local maximum. Since f

(6) = 48 > 0, x = 6 is a local minimum.
b.
f

(x) =
2
3
(x −2)
−1/3
The first derivative exists and is nonzero for x = 2. At x = 2, the derivative does not exist and thus x = 2 is a
critical point. For x < 2, f

(x) < 0 and for x > 2, f


(x) > 0. x = 2 is a local minimum.
Solution 3.16
Let r be the radius and h the height of the cylinder. The volume of the cup is πr
2
h = 64. The radius and height are
related by h =
64
πr
2
. The surface area of the cup is f(r) = πr
2
+ 2πrh = πr
2
+
128
r
. The first derivative of the surface
area is f

(r) = 2πr −
128
r
2
. Finding the zeros of f

(r),
2πr −
128
r

2
= 0,
2πr
3
− 128 = 0,
r =
4
3
√
π
.
The second derivative of the surface area is f

(r) = 2π +
256
r
3
. Since f

(
4
3
√
π
) = 6π, r =
4
3
√
π
is a local minimum of

f(r). Since this is the only critical point for r > 0, it must be a global minimum.
The cup has a radius of
4
3
√
π
cm and a height of
4
3
√
π
.
106
Solution 3.17
We define the function
h(x) = f(x) −f(a) −
f(b) −f(a)
g(b) −g(a)
(g(x) −g(a)).
Note that h(x) is differentiable and that h(a) = h(b) = 0. Thus h(x) satisfie s the conditions of Rolle’s theorem and
there exists a point ξ ∈ (a, b) such that
h

(ξ) = f

(ξ) −
f(b) −f(a)
g(b) −g(a)
g


(ξ) = 0,
f

(ξ)
g

(ξ)
=
f(b) −f(a)
g(b) −g(a)
.
Solution 3.18
The first few terms in the Taylor series of sin(x) ab out x = 0 are
sin(x) = x −
x
3
6
+
x
5
120
−
x
7
5040
+
x
9
362880
+ ··· .

The seventh derivative of sin x is −cos x. Thus we have that
sin(x) = x −
x
3
6
+
x
5
120
−
cos x
0
5040
x
7
,
where 0 ≤ x
0
≤ x. Since we are considering x ∈ [−1, 1] and −1 ≤ cos(x
0
) ≤ 1, the approximation
sin x ≈ x −
x
3
6
+
x
5
120
has a maximum error of

1
5040
≈ 0.000198. Using this polynomial to approximate sin(1),
1 −
1
3
6
+
1
5
120
≈ 0.841667.
107
To see that this has the required accuracy,
sin(1) ≈ 0.841471.
Solution 3.19
Expanding the terms in the approximation in Taylor series,
f(x + ∆x) = f(x) + ∆xf

(x) +
∆x
2
2
f

(x) +
∆x
3
6
f


(x) +
∆x
4
24
f

(x
1
),
f(x −∆x) = f(x) − ∆xf

(x) +
∆x
2
2
f

(x) −
∆x
3
6
f

(x) +
∆x
4
24
f


(x
2
),
where x ≤ x
1
≤ x + ∆x and x −∆x ≤ x
2
≤ x. Substituting the expansions i nto the formula,
f(x + ∆x) − 2f(x) + f(x −∆x)
∆x
2
= f

(x) +
∆x
2
24
[f

(x
1
) + f

(x
2
)].
Thus the error in the approximation is
∆x
2
24

[f

(x
1
) + f

(x
2
)].
Solution 3.20
Solution 3.21
a.
lim
x→0

x −sin x
x
3

= lim
x→0

1 −cos x
3x
2

= lim
x→0

sin x

6x

= lim
x→0

cos x
6

=
1
6
108
lim
x→0

x −sin x
x
3

=
1
6
b.
lim
x→0

csc x −
1
x


= lim
x→0

1
sin x
−
1
x

= lim
x→0

x −sin x
x sin x

= lim
x→0

1 −cos x
x cos x + sin x

= lim
x→0

sin x
−x sin x + cos x + cos x

=
0
2

= 0
lim
x→0

csc x −
1
x

= 0
109
c.
ln

lim
x→+∞

1 +
1
x

x

= lim
x→+∞

ln

1 +
1
x


x

= lim
x→+∞

x ln

1 +
1
x

= lim
x→+∞

ln

1 +
1
x

1/x

= lim
x→+∞


1 +
1
x


−1

−
1
x
2

−1/x
2

= lim
x→+∞


1 +
1
x

−1

= 1
Thus we have
lim
x→+∞

1 +
1
x


x

= e.
110
d. It takes four successive applications of L’Hospital’s rule to evaluate the limit.
lim
x→0

csc
2
x −
1
x
2

= lim
x→0
x
2
− sin
2
x
x
2
sin
2
x
= lim
x→0
2x −2 cos x sin x

2x
2
cos x sin x + 2x sin
2
x
= lim
x→0
2 −2 cos
2
x + 2 sin
2
x
2x
2
cos
2
x + 8x cos x sin x + 2 sin
2
x −2x
2
sin
2
x
= lim
x→0
8 cos x sin x
12x cos
2
x + 12 cos x sin x −8x
2

cos x sin x −12x sin
2
x
= lim
x→0
8 cos
2
x −8 sin
2
x
24 cos
2
x −8x
2
cos
2
x −64x cos x sin x −24 sin
2
x + 8x
2
sin
2
x
=
1
3
It is easier to use a Taylor series expansion.
lim
x→0


csc
2
x −
1
x
2

= lim
x→0
x
2
− sin
2
x
x
2
sin
2
x
= lim
x→0
x
2
− (x − x
3
/6 + O(x
5
))
2
x

2
(x + O(x
3
))
2
= lim
x→0
x
2
− (x
2
− x
4
/3 + O(x
6
))
x
4
+ O(x
6
)
= lim
x→0

1
3
+ O(x
2
)


=
1
3
111
Solution 3.22
To evaluate the first limit, we use the identity a
b
=
e
b ln a
and then apply L’Hospital’s rule.
lim
x→∞
x
a/x
= lim
x→∞
e
a ln x
x
= exp

lim
x→∞
a ln x
x

= exp

lim

x→∞
a/x
1

=
e
0
lim
x→∞
x
a/x
= 1
We use the same method to evaluate the second limit.
lim
x→∞

1 +
a
x

bx
= lim
x→∞
exp

bx ln

1 +
a
x


= exp

lim
x→∞
bx ln

1 +
a
x

= exp

lim
x→∞
b
ln(1 + a/x)
1/x

= exp


lim
x→∞
b
−a/x
2
1+a/x
−1/x
2



= exp

lim
x→∞
b
a
1 + a/x

lim
x→∞

1 +
a
x

bx
=
e
ab
112
3.11 Quiz
Problem 3.1
Define continuity.
Solution
Problem 3.2
Fill in the blank with necessary, sufficient or necessary and sufficient.
Continuity is a condition for differentiability.
Differentiability is a condition for continuity.

Existence of lim
∆x→0
f(x+∆x)−f(x)
∆x
is a condition for differentiability.
Solution
Problem 3.3
Evaluate
d
dx
f(g(x)h(x)).
Solution
Problem 3.4
Evaluate
d
dx
f(x)
g(x)
.
Solution
Problem 3.5
State the Theorem of the Mean. Interpret the theorem physically.
Solution
Problem 3.6
State Taylor’s Theorem of the Mean.
Solution
Problem 3.7
Evaluate lim
x→0
(sin x)

sin x
.
Solution
113
3.12 Quiz Solutions
Solution 3.1
A function y(x) is said to be continuous at x = ξ if lim
x→ξ
y(x) = y(ξ).
Solution 3.2
Continuity is a necessary condition for differentiability.
Differentiability is a sufficient condition for continuity.
Existence of lim
∆x→0
f(x+∆x)−f(x)
∆x
is a necessary and sufficient condition for differentiability.
Solution 3.3
d
dx
f(g(x)h(x)) = f

(g(x)h(x))
d
dx
(g(x)h(x)) = f

(g(x)h(x))(g

(x)h(x) + g(x)h


(x))
Solution 3.4
d
dx
f(x)
g(x)
=
d
dx
e
g(x) ln f (x)
=
e
g(x) ln f (x)
d
dx
(g(x) ln f(x))
= f(x)
g(x)

g

(x) ln f(x) + g(x)
f

(x)
f(x)

Solution 3.5

If f(x) is continuous in [a b] and differentiable in (a b) then there exists a point x = ξ such that
f

(ξ) =
f(b) −f(a)
b −a
.
That is, there is a point where the instantaneous velocity is equal to the average velocity on the interval.
114
Solution 3.6
If f(x) is n + 1 times continuously differentiable in (a b) then there exists a point x = ξ ∈ (a b) such that
f(b) = f(a) + (b − a)f

(a) +
(b −a)
2
2!
f

(a) + ··· +
(b −a)
n
n!
f
(n)
(a) +
(b −a)
n+1
(n + 1)!
f

(n+1)
(ξ).
Solution 3.7
Consider lim
x→0
(sin x)
sin x
. This is an indeterminate of the form 0
0
. The limit of the logarithm of the expression
is lim
x→0
sin x ln(sin x). This is an indeterminate of the form 0 · ∞. We can rearrange the expression to obtain an
indeterminate of the form
∞
∞
and then apply L’Hospital’s rule.
lim
x→0
ln(sin x)
1/ sin x
= lim
x→0
cos x/ sin x
−cos x/ sin
2
x
= lim
x→0
(−sin x) = 0

The original limit is
lim
x→0
(sin x)
sin x
=
e
0
= 1.
115
Chapter 4
Integral Calculus
4.1 The Indefinite Integral
The opp osite of a derivative is the anti-derivative or the indefinite integral. The indefinite integral of a function f(x)
is denoted,

f(x) dx.
It is defined by the property that
d
dx

f(x) dx = f(x).
While a function f(x) has a unique derivative if it is differentiable, it has an infinite number of indefinite integrals, each
of which differ by an additive constant.
Zero Slope Implies a Constant Function. If the value of a function’s derivative is identically zero,
df
dx
= 0,
then the function is a constant, f(x) = c. To prove this, we assum e that there exists a non-constant differentiable
function whose derivative is zero and obtain a contradiction. Let f(x) be such a function. Since f(x) is non-constant,

there exist points a and b such that f(a) = f(b). By the Mean Value Theorem of differential calculus, there exists a
116
point ξ ∈ (a, b) such that
f

(ξ) =
f(b) −f(a)
b −a
= 0,
which contradicts that the derivative is everywhere zero.
Indefinite Integrals Differ by an Additive Constant. Suppose that F(x) and G(x) are indefinite integrals
of f(x). Then we have
d
dx
(F (x) − G(x)) = F

(x) −G

(x) = f(x) −f(x) = 0.
Thus we see that F (x) − G(x) = c and the two indefinite integrals must differ by a constant. For example, we have

sin x dx = −cos x + c. While every function that can b e expressed in terms of elementary functions, (the exponent,
logarithm, trigonometric functions, etc.), has a derivative that can be written explicitly in terms of elementary functions,
the same is not true of integrals. For example,

sin(sin x) dx cannot be written explicitly in terms of elementary
functions.
Properties. Since the derivative is linear, so is the indefinite integral. That is,

(af(x) + bg(x)) dx = a


f(x) dx + b

g(x) dx.
For each derivative identity there is a corresponding integral identity. Consider the power law identity,
d
dx
(f(x))
a
=
a(f(x))
a−1
f

(x). The corresponding integral identity is

(f(x))
a
f

(x) dx =
(f(x))
a+1
a + 1
+ c, a = −1,
where we require that a = −1 to avoid division by zero. From the derivative of a logarithm,
d
dx
ln(f(x)) =
f


(x)
f(x)
, we
obtain,

f

(x)
f(x)
dx = ln |f(x)| + c.
117