Zero Derivative Theorem: Suppose f is continuous on [a, b] and differentiable on a, b.. Label as x1and x2any two distinct numbers in the interval a, b where x1< x2... Equal Derivatives T
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B f A
f
x
B A
f
x
Figure C.4
The equation of the secant line through A and B is
y− f (a) = m(x − a), where m =f (b)b− f (a)
− a or
y=f (b)b− f (a)
− a (x− a) + f (a).
Therefore the vertical displacement v is given by
v(x)= f (x) − f (b) − f (a)
b− a (x− a) + f (a)
for x∈ [a, b]
v(a)= f (a) − f (b) − f (a)
b− a (a− a) + f (a)
= f (a) − f (a) = 0
v(b)= f (b) − f (b) − f (a)
b− a (b− a) + f (a)
= f (b) − f (b) + f (a) − f (a) = 0 Because f is continuous on [a, b] and differentiable on (a, b), we know that v
is continuous on [a, b] and differentiable on (a, b) Therefore, we can apply Rolle’s Theorem to v; there is a number c∈ (a, b) such that
v(c)= 0
We know that the derivative of a linear function is just the slope of the line, so
v(x)= f(x)−f (b)− f (a)
b− a . Therefore,
v(c)= f(c)−f (b)b− f (a)
The statement v(c)= 0 is equivalent to
f(c)=f (b)− f (a)
b− a .
5 Zero Derivative Theorem: Suppose f is continuous on [a, b] and differentiable on
(a, b) If f(x)= 0 for all x ∈ (a, b), then f is constant on (a, b)
Proof Label as x1and x2any two distinct numbers in the interval (a, b) where x1< x2
f is differentiable on (a, b) so it must be differentiable on (x , x )and continuous on
Trang 21092 APPENDIX C The Theoretical Basis of Applications of the Derivative
[x1, x2] Therefore, we can apply the Mean Value Theorem to f on [x1, x2] to say that there exists c∈ (x1, x2)such that
f(c)=f (xx2)− f (x1)
2− x1
But f(c)= 0 by hypothesis (f(x)= 0 for all x ∈ (a, b)), so
f (x2)− f (x1)= 0
f (x1)= f (x2)
Since x1and x2were chosen arbitrarily in (a, b), we conclude that f (x) is constant throughout the interval (a, b)
6 Equal Derivatives Theorem: Suppose f is continuous on [a, b] and differentiable on
(a, b) If f(x)= g(x)for all x∈ (a, b), then f (x) and g(x) differ by a constant on (a, b) That is, there exists a constant C such that
f (x)= g(x) + C for all x∈ (a, b)
Proof Let j (x)= f (x) − g(x)
j(x)= f(x)− g(x)= 0 for all x∈ (a, b)
Therefore, by the Zero Derivative Theorem, j (x)= C for some constant C for all
x∈ (a, b)
f (x)− g(x) = C or
f (x)= g(x) + C for all x ∈ (a, b)
7 Increasing/Decreasing Function Theorem: Suppose f is continuous on [a, b] and
differentiable on (a, b)
If f(x) >0 for all x∈ (a, b), then f is increasing on (a, b);
if f(x) <0 for all x∈ (a, b), then f is decreasing on (a, b)
Proof We will show that if f(x) >0 for all x∈ (a, b), then f is increasing on (a, b) Label as x1and x2any two distinct numbers in the interval (a, b), where x1< x2 Our goal is to show that
f (x1) < f (x2),
or equivalently,
f (x2)− f (x1) >0
To do this, we apply the Mean Value Theorem to f on the interval [x1, x2] There exists
a number c∈ (x1, x2)such that
f(c)=f (xx2)− f (x1)
2− x1
or f(c)(x2− x1)= f (x2)− f (x1)
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By hypothesis f(x) >0 on (a, b); therefore, f(c) >0 We know that x2− x1>0 because x1< x2 From sign analysis it follows that
f (x2)− f (x1) >0
f is increasing
We leave it as an exercise to show that if f(x) <0 for all x∈ (a, b), then f is decreasing on (a, b)
Trang 5A P P E N D I X
Proof by Induction
Suppose we arrange a collection of dominos in a line so that if any one domino falls, we are sure that the domino next to it will also fall Then, if we knock over the first domino
in the line, all the rest of the dominos will also be knocked down Proof by mathematical induction works under very much the same principle, except that it works with an infinite number of dominos
Suppose we do some calculations and notice a pattern We may wonder whether this pattern will hold indefinitely For example, consider the sum of the first n odd numbers
1= 1
1+ 3 = 4
1+ 3 + 5 = 9
1+ 3 + 5 + 7 = 16
1+ 3 + 5 + 7 + 9 = 25
1+ 3 + 5 + 7 + 9 + 11 = 36
Do you notice a pattern? The numbers on the right are 12, 22, 32, 42, 52, and 62 We might conjecture that the sum of the first n odd integers is n2 But how can we prove this? Let’s call our conjecture C This conjecture encompasses infinitely many assertions We’ll name them C1, C2, C3, et cetera.
C1: 1= 12
C2: 1+ 3 = 22
C3: 1+ 3 + 5 = 32
et cetera
We will show that the conjecture is true using mathematical induction, working on the domino principle
First we show that the statement holds for n= 1
1095
Trang 61096 APPENDIX D Proof by Induction
That is, we show C1is true
We show that the first domino will fall.
Then we show that if Ckis true, where k is a positive integer, then Ck +1must be true
as well
We show that if any one domino falls, it will knock down the domino after it.
Then we have completed our proof Ckholds true for all positive integers
Let’s apply it to our conjecture that the sum of the first n odd integers is n2
Proof First, we show that C1holds 1= 12 (Not hard to show!)
Now we show that if Ckis true, then Ck +1must be true
Suppose that Ckis true Then
1+ 3 + 5 + 7 + 9 + · · · + (2k − 1) = k2
We must show that
1+ 3 + 5 + 7 + 9 + · · · + (2k − 1) + (2(k + 1) − 1) = (k + 1)2
1+ 3 + 5 + 7 + 9 + · · · + (2k − 1) + (2(k + 1) − 1)
= [1 + 3 + 5 + 7 + 9 + · · · + (2k − 1)] + (2k + 2 − 1)
= [k2]+ (2k + 1) (by the induction hypothesis)
= k2+ 2k + 1
= (k + 1)2 Therefore, if Ck is true, then Ck +1 must be true This completes the induction proof The sum of the first n odd integers is n2
Below we give a second example
EXAMPLE D.1 Prove that
12+ 22+ 32+ 42+ · · · + n2=n(n+ 1)(2n + 1)
Proof We will show that this statement is true using mathematical induction.
First we’ll show that the statement holds for n= 1
1=? 1(1+ 1)(2 · 1 + 1)
6
1=? 6 6
1= 1 yes, indeed!
Now we want to show that if the statement holds for n= k, then it must hold for
n= k + 1 If we show this, then we know that because the statement holds for n = 1, it must hold for n= 2, and because it holds for n = 2, it must hold for n = 3, and so on, ad infinitum
Therefore, we need to show that if 12+ 22+ 32+ 42+ · · · + k2=k(k+1)(2k+1)6 , then
it follows that
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12+ 22+ 32+ 42+ · · · + k2+ (k + 1)2=(k+ 1)[(k + 1) + 1][2(k + 1) + 1]
The right-hand side of this equation can be written as
(k+ 1)(k + 2)(2k + 3)
By assumption,
12+ 22+ 32+ 42+ · · · + k2=k(k+ 1)(2k + 1)
From this it follows that
12+ 22+ 32+ 42+ · · · + k2+ (k + 1)2=k(k+ 1)(2k + 1)
=k(k+ 1)(2k + 1)
2
6
=k(k+ 1)(2k + 1) + 6(k + 1)
2
6
=(k+ 1)[k(2k + 1) + 6(k + 1)]
6
=(k+ 1)[2k
2+ k + 6k + 6]
6
=(k+ 1)[2k
2+ 7k + 6]
6
=(k+ 1)(k + 2)(2k + 3)
We have proven what is known as the induction step The statement holds for n= 1, so it holds for n= 2, and because it holds for n = 2, it holds for n = 3, and so on, n = 1, 2, 3, 4
We have shown that
12+ 22+ 32+ 42+ · · · + n2=n(n+ 1)(2n + 1)
6 holds for n any positive integer
Later in our study of calculus we will find ourselves trying to calculate the area under the graph of y= x2on the interval [0, 1] by evaluating the limit of a Riemann sum without using the Fundamental Theorem of Calculus Just when we need it, the identity
12+ 22+ 32+ 42+ · · · + n2=n(n+ 1)(2n + 1)
6 will drop like manna from heaven Below are some other bits of manna
1+ 2 + 3 + 4 + · · · + n =n(n+ 1)
2
13+ 23+ 33+ 43+ · · · + n3=n
2(n+ 1)2
4 They too can be proven by mathematical induction
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P R O B L E M S F O R A P P E N D I X D
For the problems below, use mathematical induction to prove that each statement is true for all positive integers.
1 Prove
1+ 2 + 3 + 4 + · · · + n =n(n+ 1)
2 for n any positive integer
2 Prove
13+ 23+ 33+ 43+ · · · + n3=n
2(n+ 1)2
4 for n any positive integer
3 Prove that the sum of the first n nonzero even integers is n(n− 1)
4 Prove
1+ 2 + 22+ 23+ · · · + 2n= 2n+1− 1 for n any positive integer
5 Prove
2(1+ 3 + 32+ 33+ · · · + 3n)= 3n+1− 1 for n any positive integer
6 Prove
1+1 1
·
1+1 2
·
1+1 3
· · ·
1+1 n
= n + 1 for n any positive integer
Trang 9A P P E N D I X
Conic Sections
Conic sections get their name from the fact that they can be thought of as the intersection
of a plane with a pair of circular cones with vertices joined in an hourglass configuration
Depending upon the orientation of the plane relative to the cones we obtain a circle, an ellipse, a parabola, or a hyperbola.
Parabola Ellipse
Figure E.1 Conic sections
If the plane passes through the vertex of the cones, then the resulting figure, a point, a line,
or a set of two intersecting lines, is a “degenerate conic.”
Figure E.2 Degenerate conics
1099
Trang 101100 APPENDIX E Conic Sections
We can describe conics in several ways: as “slices” of the cones, as a set of points with a given geometric property, or as the graph of the second degree equation in x and y of the form
Ax2+ Bxy + Cy2+ Dx + Ey + F = 0, where A, B, C, D, E, and F are constants
Below we present the characteristics of the conics from a geometric point of view
E.1 CHARACTERIZING CONICS FROM A
GEOMETRIC VIEWPOINT
A circle
A circle is the set of points (x, y) equidistant from a fixed point C (the center)
C d
d
d
(x, y)
d is constant
To sketch a circle, tack one end
of a string at C, and attach
a pen point to the other Holding the string taut, you can trace out
a circle with the pen.
Figure E.3 Circle
A parabola
A parabola is the set of points (x, y) equidistant from a fixed line L and a fixed point
F, where F does not lie on L
Lis called the directrix, and F is called the focus.
The vertex, V, of the parabola lies midway between F and L.
The axis of symmetry runs
through F and perpendicular
to L.
F
V
L
d1
d1
d2
d2
d3
d3
Axis of symmetry
Figure E.4
An Ellipse
An ellipse is the set of points (x, y), the sum of whose distances from two fixed points
F and F is constant