math 0235 from professor lennard - honorscollege.pitt.edu · our fall 2020 text is: “calculus,...

MATH 0235 From Professor Lennard I will give out a more detailed Course Description on the first day of class. Also, any student interested in joining 0235 is very welcome to e-mail me directly at [email protected]. Due to the coronavirus, this course will almost surely have on-line aspects in Fall 2020. I will announce them later. _____________________________________ To students interested in Calculus at Pitt: There are many Fall Calculus options at Pitt: e.g. [1] regular Calc 1 (Math 0220); [2] going straight into regular Calc 2 (Math 0230), if you have high enough AP high school calculus scores; or [3] Honors Math 0235 I believe this is the most challenging option…; and hopefully it is a very interesting option… ! My philosophy and belief is that math is fun, creative and beautiful - both a science and an art. _____________________________ Math 0235. This Honors course covers all of 1-variable Calculus, and leads to (multi-variable) Calculus 3, and the beginning theoretical math analysis courses: Math 0413/0420 or Honors Math 0450. M0235 is designed for highly motivated students, and replaces M0220 and M0230 (regular Calc 1 and 2). Moreover we cover the same material - and more (and in the same depth - and more). There is too much material to cover it all, in detail, in lectures... A lot of the course homework will consist of assigned reading, noting and learning from our textbook - and/or elsewhere - of material not covered in class. Lecture notes: I recommend that you take clear and careful lecture notes; they will be the core of the class; and your main reference when studying for the in-term and final exams. I expect a lot from you... You are not competing against each other - since I do not "curve.'' You will be simply striving to learn the material to the best of your ability, knowing the absolute grade levels in advance. A previous serious encounter with Calculus - where you learned a lot – will be a great help in the class... A strong knowledge of pre-calculus algebra, trigonometry and plane geometry plays an important role in doing well in many math courses, and this course is no exception. I will assume that you have this knowledge. If this is not the case, then the first week of term is the best time to begin your revision. For example, I expect you to know all your trig formulas...

mailto:[email protected]

mailto:[email protected]

You will be required also to use and apply logic and algebra to prove certain mathematical statements and formulas in this course. Be aware of this from the very beginning. I have attached a sample of my 0235 Fall 2008 Lecture notes (lectures 12 through 21). My lecture notes may vary from year to year... Nevertheless, this will give you an idea of the type of material you will be expected to learn... Our Fall 2020 text is: “Calculus, 6th edition, Single and Multivariable", by D. Hughes-Hallett, A.M. Gleason, W.G. McCallum, et al; John Wiley & Sons, Inc.. Or, buy the 5th Edition, which is essentially the same… Both will work. I will cover much of the material in Chapters 1 through 10 of Hughes-Hallet et al – and possibly parts of some other sections; but not necessarily in the same manner or order. Also, the course will include: any extra topics or modifications to the material or approach in the text that I make in class. Note that regular calculus at Pitt uses a different textbook, written by Stewart...] PROBLEMS AND HOMEWORK: Homework Assignments are not for credit. They are given to help you prepare yourself for the exams. I will regularly assign problems. You should regularly solve these assigned problems... Please consult me or your Teaching Assistant regularly - in office hours or recitation - about problems you are "stuck on''; or when you are uncertain of your solutions. Also, you can go to the Engineering/Science/Math Library in Benedum Hall and consult the many books there about Calculus... Or, try "Google" to find mathematics resources: books, research papers, on-line encyclopedias, etc... Do not be surprised if you are spending many more hours per week on Math 0235 homework than you did on any previous math homework... I encourage students to form study groups, to solve problems together and learn from each other. ____________________________ Best wishes, Dr. Lennard.

FALL ’08 : 0235 HONORS CALCULUS : LECTURE NOTES :PART 2 : 14/OCT

CHRIS LENNARD

12. Math 0235 : Lecture 12 : Mon 22/September/08

“Power series” and “The Super-Fact”...

Henceforth, we will often abbreviate the word “respectively”to “resp.”.

Fix an arbitrary c ∈ R (respectively, c ∈ C) and also fix anarbitrary R with R ∈ (0,∞) or R = ∞. The open disc in R, withcenter c and radius R is simply the open interval

D(c, R) := (c−R, c+R) := {x ∈ R : c−R < x < c+R} , if R ∈ (0,∞) ;

and

(c−R, c + R) := (c−∞, c +∞) = (−∞,∞) = R , if R =∞ .

1

The open disc in C = R2, with center c = α + i β = (α, β) (whereα, β ∈ R) and radius R is the “solid interior of a circle (excluding thecircle itself)” given by

D(c, R) := {z ∈ C : |z − c| < R} , if R ∈ (0,∞) ;

andD(c, R) := C , if R =∞ .

Definition 12.1. Let (an)n∈N0 be a sequence in R (respectively, C), and sup-pose that c ∈ R (resp. C). Let x be a real (resp. complex) variable. We callthe sequence (

N∑n=0

an (x− c)n)N≥0

a power series, with center c . If the infinite sum

S(x) =

∞∑n=0

an (x− c)n := limN−→∞

N∑n=0

an (x− c)n

exists in R (resp. C), for all x in the open disc D(c, R); for some radiusR ∈ (0,∞], then we call S a power series function from D(c, R) into R (resp.C).

2

Example 12.2. In an earlier lecture, we learned that

S(x) = exp(x) := ex :=

∞∑n=0

xn

n!

=

[1 + x +

x2

2!+x3

3!+x4

4!+ · · · + xn

n!+ . . .

]exists in R (resp. C), for all x ∈ R (resp. x ∈ C). Let c = 0 and R =∞.This function S = exp is a power series function from D(0,∞) = R(resp. C) into R (resp. C). In this case,

a0 = 1 , a1 = 1 , a2 =1

2!, a3 =

1

3!, a4 =

1

4!;

and generally,

an =1

n!, for all n ≥ 0 .

3

Example 12.3. In an earlier lecture, we learned that

S(x) = sin(x) :=

∞∑n=0

(−1)n x2n+1

(2n + 1)!

=

[x− x3

3!+x5

5!− x7

7!+ · · · + (−1)n x2n+1

(2n + 1)!+ . . .

]exists in R (resp. C), for all x ∈ R (resp. x ∈ C). Let c = 0 and R =∞.This function S = sin is a power series function from D(0,∞) = R(resp. C) into R (resp. C). In this case,

a0 = 0 , a1 = 1 , a2 = 0 , a3 = − 1

3!, a4 = 0 , a5 =

1

5!, a6 = 0 , a7 = − 1

7!;

and generally,

a2 k = 0 and a2 k+1 =(−1)k

(2 k + 1)!, for all k ≥ 0 .

4

Example 12.4. Also, in an earlier lecture, we learned that

S(x) = cos(x) :=

∞∑n=0

(−1)n x2n

(2n)!

=

[1− x2

2!+x4

4!− x6

6!+ · · · + (−1)n x2n

(2n)!+ . . .

]exists in R (resp. C), for all x ∈ R (resp. x ∈ C). Let c = 0 and R =∞.This function S = cos is a power series function from D(0,∞) = R(resp. C) into R (resp. C). In this case,

a0 = 1 , a1 = 0 , a2 = − 1

2!, a3 = 0 , a4 =

1

4!, a5 = 0 , a6 = − 1

6!;

and generally,

a2 k =(−1)k

(2 k)!and a2 k+1 = 0 , for all k ≥ 0 .

5

Example 12.5. Consider the Power Series(N∑n=0

xn

)N≥0

,

where x varies over all of R (resp. C). For each integer N ≥ 0 and everyx ∈ R (resp. C), we define

SN(x) :=

N∑n=0

xn = 1 + x + x2 + x3 + · · · + xN .

In this case, we clearly have that the center c = 0;

a0 = 1 , a1 = 1 , a2 = 1 , a3 = 1 ;

and generally,

an = 1 , for all n ≥ 0 .

Moreover, in earlier work, we calculated that for all x 6= 1,

SN(x) :=

N∑n=0

xn =1− xN+1

1− x.

6

Further, when x ∈ R and |x| < 1 (i.e., −1 < x < 1), we know that

limN−→∞

xN+1 = 0 .

The argument we used to show this readily extends to show that for allcomplex numbers x = α + i β (where α, β ∈ R), with

|x| :=(α2 + β2

)1/2< 1, we also have that

limN−→∞

xN+1 = 0 .

Therefore, by the Algebra of Limits (A.o.L), for all x ∈ R (resp. C)with |x| < 1, we have that

S(x) =

∞∑n=0

xn := limN−→∞

N∑n=0

xn

= limN−→∞

SN(x) = limN−→∞

1− xN+1

1− x=

1− 0

1− x=

1

1− x.

7

In summary, for all real or complex numbers x with |x| < 1,1

1− x=[1 + x + x2 + x3 + x4 + · · · + xn + . . .

].

Note that for this power series function, the center c = 0 and the radiusR = 1. It can be shown that 1 is the largest possible radius R such thatthis power series (often called the “geometric (power) series”)(

N∑n=0

xn

)N≥0

converges in R (resp. C) at every point of D(0, R). [Aside: Note also thatthis Power Series does not converge at any point x on the circle of radius1; i.e., where |x| = 1 ].

Consider now the following (useful) addition to the Algebra ofLimits (A.o.L.).

8

[The Super-Fact.] Let (xn)n∈N and (an)n∈N0 be sequences in R (resp. C).Let c ∈ R (resp. C), and R ∈ (0,∞) or R =∞.

Suppose that the infinite sum

S(x) =

∞∑n=0


N∑n=0

an (x− c)n

exists in R (resp. C), for all x in the interval (c − R, c + R) (resp. the opendisc D(c, R)).

Further suppose thatlim

n−→∞xn = L

exists in (c−R, c + R) (resp. D(c, R)). Then,

limn−→∞

S(xn) = S(L) .

Let’s now use this Super-Fact to calculate the following limit.

Suppose that (xn)n∈N is a sequence in R (resp. C) with

limn−→∞

xn = 0 .

9

Also, assume xn 6= 0, for all n ∈ N.Note that the expression xkn, used below, is defined by xkn := (xn)k,

for all n ∈ N and for all integers k ≥ 0. Also, concerning thenotation used in the Super-Fact: note that c = 0 and L = 0.

Problem. Using the Super-Fact, algebra, and the power series for-mula for exp, calculate

limn−→∞

exp(xn)− 1

xn.

Solution. Fix an arbitrary n ∈ N. Note that xn 6= 0.

exp(xn)− 1

xn=

1

xn(exp(xn)− 1)

=1

xn

( ∞∑k=0

xknk!− 1

)

=1

xn

([1 + xn +

x2n

2!+x3n

3!+x4n

4!+ · · · + xkn

k!+ . . .

]− 1

).

10

Thus, by the distributive law,

exp(xn)− 1

xn=

1

xn

(xn +

x2n

2!+x3n

3!+x4n

4!+ · · · + xkn

k!+ . . .

)=

1

xnxn

(1 +

xn2!

+x2n

3!+x3n

4!+ · · · + xk−1

n

k!+ . . .

)=

[1 +

xn2!

+x2n

3!+x3n

4!+ · · · + xk−1

n

k!+ . . .

]= S(xn) ,

where S is the power series function defined by

S(x) :=

[1 +

x

2!+x2

3!+x3

4!+ · · · + xk−1

k!+ . . .

].

Note that the infinite sum S(x) exists in R (resp. C), because exp(x)does, for all x ∈ R (resp. C).

11

By the Super-Fact,

limn−→∞

exp(xn)− 1

xn= lim

n−→∞S(xn) = S(0)

=

[1 +

0

2!+

02

3!+

03

4!+ · · · + 0k−1

k!. . .

]:= lim

N−→∞

[1 +

0

2!+

02

3!+

03

4!+ · · · + 0N−1

N !

]= lim

N−→∞[1] = 1 .

13. Math 0235 : Lecture 13 : Wed 24/September/08

Homework Assignment 13 [HA.L13]

We will abbreviate the word “respectively” to “resp.”.

[1] Recall from Lecture 12 (Monday, 22/Sep/08) the followingaddition to the Algebra of Limits (A.o.L.):

12

[The Super-Fact.] Let (xn)n∈N and (an)n∈N0 be sequences in R (resp. C).Let c ∈ R (resp. C), and R ∈ (0,∞) or R =∞.

Suppose that the infinite sum

S(x) =

∞∑n=0


N∑n=0

an (x− c)n

exists in R (resp. C), for all x in the interval (c − R, c + R) (resp. the opendisc D(c, R)).

Further suppose thatlim

n−→∞xn = L

exists in (c−R, c + R) (resp. D(c, R)). Then,

limn−→∞

S(xn) = S(L) .

Remark: You may have encountered this idea before... It is one way ofsaying that the function S is continuous (on its given interval (resp. disc)domain).

13

Here, recall that (c − R, c + R) := {x ∈ R : c − R < x < c + R},if R ∈ (0,∞); and (c − R, c + R) = R, if R = ∞. Also recall thatD(c, R) := {z ∈ C : |z − c| < R}, if R ∈ (0,∞); and D(c, R) = C, ifR = ∞. Note that for the power series function S defined above,the value of S at the center c is S(c) = a0 .

Also, note that the expression xkn, used in the problems below,is defined by xkn := (xn)k, for all n ∈ N and for all integers k ≥ 0.

In the problems below, c = 0 and L = 0. In particular, we areassuming that

limn−→∞

xn = 0 .

Also, assume xn 6= 0, for all n ∈ N.

Using the A.o.L. Super-Fact, algebra, and known ways of express-ing some of the functions below as sums of certain Power Series,solve the following 3 problems.

14

Problem(1). Calculate

limn−→∞

sin(xn)− xn +x3n

6x5n

.


limn−→∞

exp(xn)− 1− xn −x2n

2x3n

.


limn−→∞

exp(xn)− 1

1− xnx2n

.

Next we will solve some example problems...

Example Problem(1). Calculate

limn−→∞

sin(xn)

xn.

15


sin(xn)

xn=

1

xnsin(xn)

=1

xn

∞∑k=0

(−1)k x2 k+1n

(2 k + 1)!

=1

xn

[xn −

x3n

3!+x5n

5!− x7

n

7!+ · · · + (−1)k x2 k+1

n

(2 k + 1)!+ . . .

].


sin(xn)

xn=

1

xnxn

[1− x2

n

3!+x4n

5!− x6

n

7!+ · · · + (−1)k x2 k

n

(2 k + 1)!+ . . .

]=

[1− x2

n

3!+x4n

5!− x6

n

7!+ · · · + (−1)k x2 k

n

(2 k + 1)!+ . . .

]= S(xn) ,


S(x) :=

[1− x2

3!+x4

5!− x6

7!+ · · · + (−1)k x2 k

(2 k + 1)!+ . . .

].

16

Note that the infinite sum S(x) exists in R (resp. C), because sin(x)does, for all x ∈ R (resp. C).By the Super-Fact, since limn−→∞ xn = 0,

limn−→∞

sin(xn)

xn= lim

n−→∞S(xn) = S(0)

=

[1− 02

3!+

04

5!− 06

7!+ · · · + (−1)k 02 k

(2 k + 1)!+ . . .

]:= lim

N−→∞

[1− 02

3!+

04

5!− 06

7!+ · · · + (−1)N 02N

(2N + 1)!

]= lim

N−→∞[1] = 1 .


limn−→∞

cos(xn)− 1

x2n

.

17


cos(xn)− 1

x2n

=1

x2n

(cos(xn)− 1)

=1

x2n

( ∞∑k=0

(−1)k x2 kn

(2 k)!− 1

)

=1

x2n

([1− x2

n

2!+x4n

4!− x6

n

6!+ · · · + (−1)k x2 k

n

(2 k)!+ . . .

]− 1

).


cos(xn)− 1

x2n

=1

x2n

(−x

2n

2!+x4n

4!− x6

n

6!+ · · · + (−1)k x2 k

n

(2 k)!+ . . .

)=

1

x2n

x2n

(− 1

2!+x2n

4!− x4

n

6!+ · · · + (−1)k x2 k−2

n

(2 k)!+ . . .

)=

[− 1

2!+x2n

4!− x4

n

6!+ · · · + (−1)k x2 k−2

n

(2 k)!+ . . .

]= S(xn) ,

18


S(x) :=

[− 1

2!+x2

4!− x4

6!+ · · · + (−1)k x2 k−2

(2 k)!+ . . .

].

Note that the infinite sum S(x) exists in R (resp. C), becausecos(x) does, for all x ∈ R (resp. C).By the Super-Fact, since limn−→∞ xn = 0,

limn−→∞

cos(xn)− 1

x2n

= limn−→∞

S(xn) = S(0)

=

[− 1

2!+

02

4!− 04

6!+ · · · + (−1)k 02 k−2

(2 k)!+ . . .

]:= lim

N−→∞

[− 1

2!+

02

4!− 04

6!+ · · · + (−1)N 02N−2

(2N)!

]= lim

N−→∞

[− 1

2!

]= −1

2.

19


limn−→∞

cos(xn)− 1

xn.

Solution. We can solve this problem similarly to Example Prob-lem (2)... Or we can use the following shortcut. By ExampleProblem(2),

limn−→∞

cos(xn)− 1

x2n

= −1

2.

For each n ∈ N,

cos(xn)− 1

xn= xn

(cos(xn)− 1

x2n

).

Recall thatlim

n−→∞xn = 0 .

Thus, from the Algebra of Limits,

limn−→∞

cos(xn)− 1

xn= (0)

(−1

2

)= 0 .

20

14. Math 0235 : Lecture 14 : Fri 26/September/08

Homework Assignment 14 [HA.L14][1] Carefully read and note Section 9.2 of our text [Stewart] on“Vectors”, pages 642-649. Then do problems 4, 5, 6, 7, 10, 11,14, 15, 18, 19, 20, 24, 26, 34, 37, 38 on pages 649-651.

[2] Fix an arbitrary sequence (xn)n∈N in R (resp. C). Let L ∈ R(resp. C). Suppose that

limn−→∞

xn = L ; and xn 6= L , for all n ∈ N .

Using the A.o.L. Super-Fact, algebra, and known ways of expressingsome of the functions below as infinite sums, solve the following 3problems.

Problem(1). Suppose that L = 0. Calculate

limn−→∞

exp(xn)−

(1 +

xn2

)(

1 −xn2

)

x3n

.

21

Problem(2). Suppose that L = π/2. Calculate

limn−→∞

(sin(xn)− 1 + 1

2

(xn − π

2

)2)(xn − π

2

)4 .

[Hint for solving Problem (2): Also use trigonometry...]

Problem(3). Suppose that L = 0. Calculate

limn−→∞

tan(xn)− xn −x3n

3x5n

.

Hint for Problem(3):

tan(x) =sin(x)

cos(x), for all x ∈ R with cos(x) 6= 0 .

End of [HA.L14].

We introduce here an extension of the Algebra of Limits.

22

Fact 14.1 (Part of the Algebra of Limits (A.o.L.)). Let (an)n∈N be a sequencein R, L ∈ R and w ∈ R. Suppose that

limn−→∞

an = L .

(1) If there exists k ∈ N such that

an ≥ w , for all integers n ≥ k ,

thenL ≥ w .

(2) Also, if there exists k ∈ N such that

an ≤ w , for all integers n ≥ k ,

thenL ≤ w .

We see that Fact 14.1 is related to the Squeeze Theorem. Note thatthere is no A.o.L. analogue of this fact for general sequences ofcomplex numbers (an)n∈N, because this fact deals with the naturalordering ≤ on R. Indeed, note that all of the parts of the A.o.L. stated forreal number sequences, still work for complex number sequences: exceptfor those parts that concern the ordering ≤ on R.

23

Recall the following useful fact:(#) [For all u, v > 0, the product u v > 0.]

Theorem 14.2. (1) For all x ∈ R with x > 0,

ex > 1 .

(2) For all x ∈ R,ex > 0 .

(3) For all x, z ∈ R with x < z,

ex < ez ;

i.e., the exponential function exp is strictly increasing on R.

Proof. (1) Fix an arbitrary x ∈ R with x > 0.

ex = exp(x) :=

∞∑n=0

xn

n!= lim

N−→∞

N∑n=0

xn

n!

Fix an arbitrary N ∈ N with N ≥ 2, and define

SN :=

N∑n=0

xn

n!= 1 + x +

x2

2!+x3

3!+x4

4!+x5

5!+ · · · + xN

N !.

24

By Fact(#) above, since x > 0, we see that x2 := x x > 0. Further, x3 :=x x2 > 0; and inductively we see that xk := x xk−1 > 0, for all integers k ≥ 2.

[Aside: Of course, for any real number x 6= 0, we have that x2 > 0, x4 >0, x6 > 0, etc... However, the odd powers of x are only positive when x itselfis positive...]

Since sums of positive numbers are also positive, we have that

x2

2!+x3

3!+x4

4!+x5

5!+ · · · + xN

N !> 0 ,

and consequently,

SN = 1 + x +x2

2!+x3

3!+x4

4!+x5

5!+ · · · + xN

N !> 1 + x + 0 = 1 + x .

In summary, for all integers N ≥ 2,

SN ≥ 1 + x .

By Fact 14.1 part (1),

ex := limN−→∞

SN ≥ 1 + x > 1 + 0 = 1 .

Hence, for all real x > 0, ex > 1.25

(2) Fix an arbitrary x ∈ R.[Case 1. x > 0.] By Part (1), ex > 1 > 0; and so ex > 0.

[Case 2. x = 0.] Clearly, e0 = exp(0) = 1 > 0. So, e0 > 0.

[Case 3. x < 0.] By the addition formula for exp,

ex e−x = ex+(−x) = e0 = 1 .

Recall that for every real number u > 0, we have that its reciprocal 1/u satisfies

1

u> 0 .

But −x > 0, and so from Case 1 above, e−x > 0. Hence,

ex =1

e−x> 0 .

In summary, we have shown that ex > 0.

(3) Fix arbitrary x, z ∈ R with x < z. Thus, z − x > 0.Further,

ez − ex = ex+(z−x) − ex = ex ez−x − ex

= ex (ez−x − 1) .

26

Now, from Part (2), u := ex > 0. Also, z − x > 0; and therefore Part (1)gives us that ez−x > 1. Hence, v := ez−x − 1 > 0. Applying Fact (#) above,we conclude that

ez − ex = ex (ez−x − 1) = u v > 0 .

In summary, ez − ex > 0, and consequently ez > ex; i.e., ex < ez. �

Theorem 14.2 enables us to roughly sketch the graph of expnear 0 ... The first graph below clearly illustrates all three partsof Theorem 14.2. As we will see later (and as you may havelearned before), there are other aspects of the graph of exp thatTheorem 14.2 does not tell us... In particular, exp is a functionthat is concave up; i.e., every straight line segment joining two pointson the graph (that is, every “chord”) lies above or on the graph. [Seethe second graph below...] It turns out that the second derivativeof exp is always positive, which implies that exp is concave up...Also, what does the graph of exp “do” when x is extremely large andpositive or x is very large in magnitude and negative? The following

27

limits, that we will discuss in more detail later, give us an answer:

limx−→∞

ex =∞ and limx−→−∞

ex = 0 .

graph H exp L

-2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5

1

2

3

4

28

3 chords

-2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5

1

2

3

4

29

“More thoughts related to our earlier discussion of true-or-falselogical sentences...”

Recall from an earlier lecture the following “nonsense questionand answer”:

Why is a duck ?Answer: Because one of its legs are both the same.During Lecture 14, I suggested that students Google for “an

answer” to the following question... :Why is a mouse when it spins?A common answer on the web seems to be:Answer: The higher, the fewer.

15. Math 0235 : Lecture 15 : Mon 29/September/08

Fix arbitrary real numbers a and b with a < b. Recall that theclosed and bounded interval [a, b] is given by

[a, b] := {x ∈ R : a ≤ x ≤ b} .Consider an arbitrary function f : [a, b] −→ R. We wish to define

the real number “I := the integral of f over [a, b]”, if it exists...30

Consider the following picture: Graph 1

R 1

R 2

graphH f L

1.5 2.0 2.5 3.0 3.5 4.0 4.5

-0.4

-0.2

0.2

0.4

0.6

The idea is that I should be given by:

I = “Area(Region R1)−Area(Region R2)” .

31

Put differently, I should equal “the signed (or net) area betweengraph(f ) and the horizontal or x-axis (i.e., the interval [a, b]).”In an attempt to carefully define “I := the integral of f over [a, b]”

in such a way that “a large and useful collection of functions have anintegral I”, we next make some helpful definitions...

Definition 15.1. (1) We define the set P [a, b] of all partitions−→P = (x0, x1, . . . , xn) of [a, b] by

P [a, b] :={

finite sequences−→P = (x0, x1, . . . , xn) :

n ∈ N , each xj ∈ R , and a = x0 < x1 < · · · < xn = b}.

(2) Fix an arbitrary partition−→P = (x0, x1, . . . , xn) ∈ P [a, b]. A finite se-

quence−→t = (t1, . . . , tn) is called a tag sequence for

−→P if

tk ∈ [xk−1, xk] , for all k ∈ {1, . . . , n} .We denote the set of all tag sequences

−→t for

−→P by

tag(−→P ) .

32

Two interesting examples of tag sequences are−→t left and

−→t right, defined by:

−→t left := (x0, . . . , xn−1) and

−→t right := (x1, . . . , xn) .

(3) Fix an arbitrary−→P ∈ P [a, b]. Next fix an arbitrary

−→t ∈ tag(

−→P ). We

define the (−→P ,−→t )-Riemann sum, S(f,

−→P ,−→t ), by

S(f,−→P ,−→t ) :=

n∑k=1

f (tk) (xk − xk−1) .

*** Note concerning Definition 15.1 (3): Fix an arbitrary k ∈{1, . . . , n}.If f (tk) > 0, then:f (tk) (xk − xk−1) = [the area of the shaded rectangle above the interval[xk−1, xk] in the picture “Graph 2” below].

If f (tk) < 0, then:f (tk) (xk − xk−1) = - [the area of the shaded rectangle below the interval[xk−1, xk] in the picture “Graph 2” below].

33

Thus, the (−→P ,−→t )-Riemann sum

S(f,−→P ,−→t ) :=

n∑k=1

f (tk) (xk − xk−1)

is [the signed area between the shaded bar graph in “Graph 2”, drawnbelow, and the horizontal axis].

Graph 2:34

t1 t2 t3 t4 tk tn-1 tn... ...

f H t2 L f H t4 L

f H tn-1 L

x0=a x1 x2 x3 x4 xk-1 xk xn-2 xn-1 xn=b

For any non-empty, finite set of real numbers S := {s1, . . . , sm},we define maxS to be the maximum value among all of the numberss1, . . . , sm. E.g., max{4,−3, 2, 1} = 4.

35

For all partitions−→P ∈ P [a, b], we define the mesh-size of

−→P ,

mesh(−→P ) by

mesh(−→P ) := max

{xk − xk−1 : k ∈ {1, . . . , n}

}.

The key idea underlying (Riemann) integration is that as we vary−→P over P [a, b] and also vary

−→t over tag(

−→P ) in such a way that

mesh(−→P ) −→ 0 , and simultaneously n −→∞ ,

the Riemann sums

S(f,−→P ,−→t ) :=

n∑k=1

f (tk) (xk − xk−1)

should converge to

I = [the signed area between graph(f ) and the horizontal axis] ,

if this limit exists in R...

36

16. Math 0235 : Lecture 16 : Wed 1/October/08

Definition 16.1. Let a, b ∈ R with a < b. Let f : [a, b] −→ R be a function.Let I ∈ R. We say that “f is Riemann integrable over [a, b]” with integralI if:for all ε > 0, there exists δ > 0 so that for all partitions

−→P ∈ P [a, b] with

mesh(−→P ) < δ, for all tag sequences

−→t ∈ tag(

−→P ),∣∣∣I − S(f,

−→P ,−→t )∣∣∣ < ε .

In this situation, we write

I =

∫ b

a

f (x) dx := limmesh(

−→P )−→0

S(f,−→P ,−→t ) .

We also denote the set of all Riemann-integrable functions f : [a, b] −→ Rby

R[a, b] .

37

Notes about Definition 16.1: [1] Recall that

S(f,−→P ,−→t ) :=

n∑k=1

f (tk) (xk − xk−1) .

[2] We also denote I =∫ ba f (x) dx by

I =

∫ x=b

x=a

f (x) dx =

∫ u=b

u=a

f (u) du =

∫ b

a

f (u) du .

[3] In the above integrals, x and u are (interchangeable) dummyvariables.

Definition 16.2. Let a, b ∈ R with a < b. Let f : [a, b] −→ R be a function.(1) We say that “f is increasing on [a, b]” if[

for all x < z in [a, b] , f (x) ≤ f (z)].

(2) We say that “f is decreasing on [a, b]” if[for all x < z in [a, b] , f (x) ≥ f (z)

].

38

(3) We say that “f is monotone on [a, b]” if[either f is increasing on [a, b] or f is decreasing on [a, b] ].

(4) We say that “f is piecewise monotone on [a, b]” if there exists a partition−→Q = (u0, . . . , um) of [a, b] such that[

for all k ∈ {1, . . . ,m} , f is monotone on [uk−1, uk]].

Notes about Definition 16.2: [1] If the function f is monotoneon [a, b], then f is piecewise monotone on [a, b]. (Let u0 = a andu1 = b...)[2] The functions exp, and [f (x) := xn, where n is an odd positiveinteger constant] are increasing functions on every interval [a, b].[3] The functions [f (x) := xn, where n is a negative integer con-stant] are decreasing functions on every interval [a, b] with a > 0.[4] The functions sin, cos and [f (x) := xn, where n is an even pos-itive integer constant] are piecewise monotone functions on everyinterval [a, b].[5] E.g., for the function [sin : [0, 2 π] −→ R], let u0 := 0, u1 := π/2,

u2 := 3 π/2 and u3 := 2 π. Then−→Q = (u0, u1, u2, u3) is a partition of

[0, 2 π] such that sin is monotone on each interval [uk−1, uk].39

1 2 3 4 5 6

-1.0

-0.5

0.5

1.0

[“Double-lightning-bolt”]Theorem Let a, b ∈ R with a < b. Letf : [a, b] −→ R be a piecewise monotone function. Then f is Riemannintegrable on [a, b].

40

For a proof: See future courses: e.g., Math 0420, 0450 and/or1530; or certain books in the Math Library...

So, all the piecewise monotone functions noted above: exp, sin, cos,etc, are Riemann integrable...

Note that when f ∈ R[a, b], we can calculate I =∫ ba f (x) dx in the

following way.

Consider any sequence of partitions (−→P (n))n∈N with

limn−→∞

mesh(−→P (n)) = 0 ,

and any sequence of tag sequences (−→t (n))n∈N with each

−→t (n) ∈ tag(

−→P (n)) .

Then we have ∫ b

a

f (x) dx = limn−→∞

S(f,−→P (n),

−→t (n)) .

For example, consider the sequence of equal width partitions (−→P (n))n∈N

and the corresponding sequence of right tag sequences (−→t

(n)right)n∈N de-

fined in this way.41

Fix an arbitrary integer n ∈ N. Define

δn :=b− an

.

Next, define

P (n) :=(x

(n)0 , . . . , x

(n)k , . . . , x(n)

n

),

where for all k ∈ {0, 1, . . . , n},

x(n)k := a + k δn .

In particular, x(n)0 = a, x

(n)1 = a + δn and

x(n)n = a + n δn = a + n

(b− an

)= a + (b− a) = b .

Hence, a = x(n)0 < x

(n)1 < · · · < x

(n)n = b. Also, for all k ∈ {1, . . . , n},

x(n)k − x

(n)k−1 = (a + k δn)− (a + (k − 1) δn) = δn .

It follows that

mesh(−→P (n)) = max

{x

(n)k − x

(n)k−1 : k ∈ {1, . . . , n}

}= δn :=

b− an−→ 0 ,

as n −→∞.42

Further, for all k ∈ {1, . . . , n}, define−→t

(n)right,k := x

(n)k := a + k δn ∈ [x

(n)k−1, x

(n)k ] .

Then, from above, we see that

I =

∫ b

a


S(f,−→P (n),

−→t

(n)right)

= limn−→∞

n∑k=1

f (−→t

(n)right,k)

(x

(n)k − x

(n)k−1

)= lim

n−→∞

n∑k=1

f (a + k δn) δn .

17. Math 0235 : Lecture 17 : Fri 3/October/08

“Calculating integrals...”Example[1] Fix arbitrary a, b ∈ R with a < b. Consider the piece-wise monotone function f : [a, b] −→ R given by

f (x) := x2 , for all x ∈ [a, b] .43

Recall these facts from class and a recent Homework Assignment:

n∑k=1

k =1

2n (n + 1) and

n∑k=1

k2 =1

6n (n + 1) (2n + 1) ,

for all n ∈ N. Also, recall an algebraic fact (that is easy to check):

(b− a)(b2 + a b + a2

)= b3 − a3 .

From above,∫ b

a

x2 dx =

∫ b

a


n∑k=1

f (a + k δn) δn = limn−→∞

n∑k=1

(a + k δn)2 δn ,

where each

δn :=b− an

.

Fix an arbitrary n ∈ N. Consider

Sn :=

n∑k=1

(a + k δn)2 δn = δn

n∑k=1

(a + k δn)2 = δn

n∑k=1

(a2 + 2 a k δn + (k δn)2

).

44

Thus,

Sn = δn

n∑k=1

(a2 + 2 a k δn + (k δn)2

)= δn

(n∑k=1

a2 +

n∑k=1

2 a k δn +

n∑k=1

k2 δ2n

)

= δn

(n a2 + 2 a δn

n∑k=1

k + δ2n

n∑k=1

k2

)= δn

(n a2 + 2 a δn

1

2n (n + 1) + δ2

n

1

6n (n + 1) (2n + 1)

)= δn n

(a2 + 2 a δn

1

2(n + 1) + δ2

n

1

6(n + 1) (2n + 1)

)= δn n

(a2 + 2 a

1

2(δn n + δn) +

1

6(δn n + δn) (2 δn n + δn)

).

45

Now,

δn n =

(b− an

)n = b− a .

Hence,

Sn = (b− a)

(a2 + a ((b− a) + δn) +

1

6((b− a) + δn) (2 (b− a) + δn)

).

But

limn−→∞

δn = limn−→∞

b− an

= 0 .

From the Algebra of Limits, we see that∫ b

a

x2 dx = limn−→∞

Sn

= (b− a)

(a2 + a ((b− a) + 0) +

1

6((b− a) + 0) (2 (b− a) + 0)

)= (b− a)

(a2 + a (b− a) +

1

6(b− a) 2 (b− a)

).

46

Therefore,∫ b

a


Sn

= (b− a)

(a2 + a b− a2 +

1

3(b− a)2

)=

1

3(b− a)

(3 a b + (b− a)2

)=

1

3(b− a)

(3 a b + b2 − 2 a b + a2

)=

1

3(b− a)

(b2 + a b + a2

)=

1

3(b3 − a3) .

The above example is the last lecture material that is directly eligiblefor Exam 1... Of course, there may be material in the rest of this lecture(Lecture 17) and in Lecture 18 that will help illuminate the Exam 1 material...

47

“Calculating integrals, continued...”Example[2] Fix arbitrary b ∈ R with 0 < b. Consider the piecewisemonotone function f : [0, b] −→ R given by

f (x) := cos(x) , for all x ∈ [0, b] .

In this example, the constant a := 0.

Let us calculate I :=∫ b

0 cos(x) dx ∈ R. Firstly, we recall Euler’sformulas for cos and sin.

((1)) cos(u) =ei u + e−i u

2and sin(u) =

ei u − e−i u

2 i,

for all u ∈ R (or C). Note that these two equations imply that

((2)) ei u = cos(u) + i sin(u) and e−i u = cos(u)− i sin(u) ,

for all u ∈ R (or C).Also, by the Addition Formula for exp: for all u ∈ C, for everyk ∈ N, we have that

((3)) ek u = e[ (1)1 u + ... + (1)k u ]

= e(1)1 u · . . . · e(1)k u

= (eu)k .48

Further, recall that for all w ∈ C with w 6= 1, for all n ∈ N,

((4))

n∑k=1

wk = w

n∑k=1

wk−1 = w

n−1∑j=0

wj

[ using : the change of variables j = k − 1 ⇐⇒ k = j + 1 ]

= w

(1− wn−1+1

1− w

)= w

(1− wn

1− w

).

From Lecture 16,

I :=

∫ b

0

cos(x) dx = limn−→∞

n∑k=1

cos(0 + k δn) δn = limn−→∞

Sn ,

where, for all n ∈ N,

δn :=b− 0

n=b

nand Sn :=

n∑k=1

cos(0 + k δn) δn .

49

Fix an arbitrary n ∈ N. By ((1)) above,

Sn :=

n∑k=1

cos(0 + k δn) δn = δn

n∑k=1

cos(k δn)

= δn

n∑k=1

ei k δn + e−i k δn

2= δn

n∑k=1

1

2

(ei k δn + e−i k δn

)= δn

1

2

n∑k=1

(ei k δn + e−i k δn

)=δn2

(n∑k=1

ei k δn +

n∑k=1

e−i k δn

).

From ((3)) above, we see that

Sn =δn2

(n∑k=1

(ei δn)k

+

n∑k=1

(e−i δn

)k).

18. Math 0235 : Lecture 18 : Mon 6/October/08

“Calculating integrals...”Example[2], continued...

50

Recall that 0 < δn = b/n −→ 0 as n −→∞. So, there exists m ∈ Nso that for all integer n ≥ m,

0 < δn <π

2.

From our knowledge of trigonometry it follows that for all n ≥ m,

0 < cos(δn) < 1 and 0 < sin(δn) < 1 .

Therefore, by ((2)) above, for all n ≥ m,

ei δn = cos(δn) + i sin δn 6= 1 + i 0 = 1 ;

i.e., ei δn 6= 1. Similarly, for all n ≥ m,

e−i δn = cos(δn)− i sin δn 6= 1 .

Note that

((5)) ei δn e−i δn = ei δn+(−i δn) = e0 = 1 .51

From ((4)), with w = ei δn and then w = e−i δn, we have that

Sn =δn2

(n∑k=1

(ei δn)k

+

n∑k=1

(e−i δn

)k)

=δn2

(ei δn

(1−

(ei δn)n

1− ei δn

)+ e−i δn

(1−

(e−i δn

)n1− e−i δn

))

=δn2

(1

e−i δn

(1− ei δn n

1− ei δn

)+

1

ei δn

(1− e−i δn n

1− e−i δn

)).

52

Further note that δn n = (b/n)n = b. Thus, by ((5)),

Sn =δn2

((1− ei b

e−i δn − e−i δn ei δn

)+

(1− e−i b

ei δn − ei δn e−i δn

))=δn2

(1− ei b

e−i δn − 1+

1− e−i b

ei δn − 1

)

=δn2

(1δn

)(

1δn

) ( 1− ei b

e−i δn − 1+

1− e−i b

ei δn − 1

)

=1

2

1− ei b(e−i δn −1

δn

) +1− e−i b(ei δn −1δn

)

=1

2

1− ei b

−i(e−i δn −1−i δn

) +1− e−i b

i(ei δn −1i δn

) .

53

Now, we know that 0 < δn := b/n −→ 0 as n −→ ∞. Thus, by theAlgebra of Limits,

limn−→∞

−i δn = 0 and limn−→∞

i δn = 0 .

Also, each −i δn 6= 0 and each i δn 6= 0. So, by earlier work, weknow that

limn−→∞

e−i δn − 1

−i δn= 1 and lim

n−→∞

ei δn − 1

i δn= 1 .

Hence, by the Algebra of Limits and ((1)) above,∫ b

0

cos(x) dx = limn−→∞

Sn

=1

2

(1− ei b

−i (1)+

1− e−i b

i (1)

)=

1

2 i

(1− ei b

−1+

1− e−i b

1

)=

1

2 i

(−(1− ei b) + 1− e−i b

)=

1

2 i

(−1 + ei b + 1− e−i b

)=

1

2 i

(ei b − e−i b

)= sin(b) .

54


“Calculating integrals...”Example[3].Fix arbitrary b ∈ R with 0 < b. Consider the piecewise monotone

function f : [0, b] −→ R given by

f (x) := x4 , for all x ∈ [0, b] .

In this example, the constant a := 0.Firstly, we recall the following “funny powers” formula.

((1))

n∑k=1

k〈4〉 =n〈5〉

5, for all n ∈ N .

From Lecture 16,

I :=

∫ b

0


n∑k=1

(0 + k δn)4 δn = limn−→∞

Sn ,

where, for all n ∈ N,

δn :=b− 0

n=b

nand Sn :=

n∑k=1

(0 + k δn)4 δn .

55

Fix an arbitrary n ∈ N. By ((1)) above,

Sn :=

n∑k=1

(0 + k δn)4 δn = δn

n∑k=1

(k δn)4

= δn

n∑k=1

k4 δ4n = δn δ

4n

n∑k=1

k4

= δ5n

n∑k=1

k4 .

Note that, for each k ∈ {1, . . . , n},

k4 ≤ k (k + 1) (k + 2) (k + 3) = k〈4〉 .

Let

Un := δ5n

n∑k=1

k〈4〉 .

56

Since δn = b/n > 0, it follows that

Sn = δ5n

n∑k=1

k4 ≤ δ5n

n∑k=1

k〈4〉 = Un .

Further, from ((1)) above,

Un = δ5n

n∑k=1

k〈4〉 = δ5n

n〈5〉

5

=

(b

n

)5n (n + 1) (n + 2) (n + 3) (n + 4)

5

=b5

5

(nn

) (n + 1

n

) (n + 2

n

) (n + 3

n

) (n + 4

n

)=b5

5(1)

(1 +

1

n

) (1 +

2

n

) (1 +

3

n

) (1 +

4

n

).

Since limn−→∞ 1/n = 0, we see by the Algebra of Limits that

limn−→∞

Un =b5

5(1) (1 + 0) (1 + 0) (1 + 0) (1 + 0) =

b5

5.

57

Next, fix n ∈ N with n ≥ 4. Note that, for each k ∈ {4, . . . , n},

k4 ≥ (k − 3) (k − 2) (k − 1) k = (k − 3)〈4〉 .

Let

Qn := δ5n

n∑k=4

(k − 3)〈4〉 .

Since δn = b/n > 0 and each k > 0, it follows that

Sn = δ5n

n∑k=1

k4 ≥ δ5n

n∑k=4

k4 ≥ δ5n

n∑k=4

(k − 3)〈4〉 = Qn .

58

Further, from ((1)) above, using the change of variables[j = k − 3 ⇐⇒ k = j + 3], we see that

Qn = δ5n

n∑k=4

(k − 3)〈4〉 = δ5n

n−3∑j=1

j〈4〉 = δ5n

(n− 3)〈5〉

5

=

(b

n

)5(n− 3) (n− 2) (n− 1)n (n + 1)

5

=b5

5

(n− 3

n

) (n− 2

n

) (n− 1

n

) (nn

) (n + 1

n

)=b5

5

(1− 3

n

) (1− 2

n

) (1− 1

n

)(1)

(1 +

1

n

).

Since limn−→∞ 1/n = 0, we see by the Algebra of Limits that

limn−→∞

Qn =b5

5(1− 0) (1− 0) (1− 0) (1) (1 + 0) =

b5

5.

Consequently, by the Squeeze Theorem, since [Qn ≤ Sn ≤ Un,∀n ≥ 4],∫ b

0


Sn =b5

5.

59

HA.L19 [1].

In a manner similar to the calculation of∫ b

0 cos(x) dx and∫ b

0 x4 dx

(where b > 0) in Lectures 17, 18 and 19, calculate

(1)∫ b

0 sin(x) dx, for all b > 0;

(2)∫ b

0 x5 dx, for all b > 0; and

(3)∫ b

0 xν dx, for all b > 0 and for all ν ∈ N.

20. Math 0235 : Lecture 20 : Fri 10/October/08

HA.L20.

[1] Using ideas from Lecture 20 (and earlier), calculate∫ ba sin(x) dx,

for all a, b ∈ R.

[2] Read and carefully note: Stewart, Section 9.3 “The Dot Prod-uct”, pages 651-655.

[3] Do questions 4, 5, 7, 11, 16, [17 (b) and (d)], 20, 27, 31, 35and 39: on pages 656-657 of Stewart.

60

[4] Do Question 43, p. 657 of Stewart: using only the algebraic“Properties of the Dot Product”, page 654 of Stewart”.Hint (1): Note that −→v • −→v ≥ 0, for all vectors −→v .

Hint (2): Expand (−→a −λ−→b )•(−→a −λ

−→b ) and choose λ appropriately...

[5] Do questions 44 and 45: on page 657 of Stewart.

Definition 20.1. Fix arbitrary a, b ∈ R with b < a. Suppose that thefunction f : [b, a] −→ R is Riemann-integrable on [b, a]. We define “the

integral of f from a to b”,∫ ba f (x) dx, by∫ b

a

f (x) dx := −∫ a

b

f (x) dx .

We also define ∫ a

a

f (x) dx := 0 .

Note: it follows that for all real numbers a, b ∈ R,

(‡)∫ b

a

f (x) dx = −∫ a

b

f (x) dx .

61

[A Useful Fact]. The formula

(#)

∫ b

a


n∑k=1

f (a + k δn) δn ,

where

δn :=b− an

,

which is true whenever a < b, remains true when b < a and whenb = a.Another useful fact is:

[¡WOW!] Proposition. Fix arbitrary u, v ∈ R with u < v. Let f :[u, v] −→ R be a Riemann-integrable function. Let a, b, c ∈ [u, v]. Then eachof the integrals below exists in R and∫ b

a

f (x) dx =

∫ c

a

f (x) dx +

∫ b

c

f (x) dx .

“Applications of Proposition [¡WOW!] and formula (#)”.62

Similarly to HA.L19 [1] (3) (where b > 0), we can show fromformula (#) that for all b ≤ 0,

(†)∫ b

0

xν dx =bν+1

ν + 1, for all ν ∈ N .

In summary, formula (†) is true for all real numbers b.Consequently, for all a, b ∈ R, for all ν ∈ N, using (‡), (†) and

[¡WOW!] with c := 0, we see that

∫ b

a

xν dx =

∫ 0

a

xν dx +

∫ b

0

xν dx

= −∫ a

0

xν dx +

∫ b

0

xν dx

= − aν+1

ν + 1+bν+1

ν + 1

=bν+1 − aν+1

ν + 1.

63

Next, recall that we calculated that∫ b

0

cos(x) dx = sin(b) ,

for all b > 0. Using formula (#) when b ≤ 0, a similar calculationreveals that ∫ b

0

cos(x) dx = sin(b) ,

for all b ≤ 0. In summary, this formula is true for all b ∈ R.Consequently, using (‡) and [¡WOW!], we see that for all a, b ∈ R,∫ b

a

cos(x) dx =

∫ 0

a

cos(x) dx +

∫ b

0

cos(x) dx

= −∫ a

0

cos(x) dx +

∫ b

0

cos(x) dx

= − sin(a) + sin(b)

= sin(b)− sin(a) .

The above results are special cases of “The Fundamental The-orem of Calculus”... Let’s now introduce the concept of the “thederivative...

64

Definition 20.2 (The Sequential Definition of Derivative). Let J be an inter-val and c ∈ J . Let f : J −→ R be a function. We say that f is differentiableat c, if there exists L ∈ R such that for every sequence (hn)n∈N with

limn−→∞

hn = 0 , [hn 6= 0 , for all n ∈ N] ,

and each c + hn ∈ J , it follows that

limn−→∞

f (c + hn)− f (c)

hn= L .

We denote this limit L by f ′(c), and call it the derivative of f at c.

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21. Math 0235 : Lecture 21 : Tues 14/October/08

“Calculating derivatives...”

Definition 21.1 (The Sequential Definition of Derivative). [Note: This defi-nition is equivalent to the usual one...] Let J be an interval and c ∈ J . Letf : J −→ R be a function. We say that f is differentiable or locally linear atc, if there exists L ∈ R such that for every sequence (hn)n∈N in R with

limn−→∞

hn = 0 , [hn 6= 0 , for all n ∈ N] ,

and each c + hn ∈ J , it follows that

limn−→∞

f (c + hn)− f (c)

hn= L .

We denote this common limit L by f ′(c), and call it the derivative of f at c.

Note that

Qn :=f (c + hn)− f (c)

hnis the slope of the chord C on graph(f ) joining the points (c, f (c))and (c+ hn, f (c+ hn). Extend this chord to a line ` with slope Qn.As n −→ ∞, this line approaches the line T with slope f ′(c) through

66

the point (c, f (c). We define this line to be the tangent to graph(f ) at(c, f (c)). Note that

T := {(x, y) ∈ R2 : y − f (c) = f ′(c) (x− c)} .

[In class, we drew a “typical graph of f”, to illustrate the aboveideas...]

Also,

Qn :=f (c + hn)− f (c)

hn

is the average rate of change of f over the interval [c, c+hn], if hn > 0;or [c + hn, c], if hn < 0.

In the above definition, we require that for all null sequences(hn)n∈N, as described above, Qn converges to the same limit L. Oth-erwise it is easy to draw the graph of functions f for which (forexample), for all such null sequences with each hn > 0, Qn tendsto a certain number L1; while for all such null sequences witheach hn < 0, Qn tends to a another number L2. Such functions fare not differentiable at c.

67

[In class, we drew a “typical graph of f”, to illustrate the aboveideas...]

Example [1]. Let J := R and the function f : J −→ R be definedby

f (x) := x3 , for all x ∈ J = R .

Fix an arbitrary point c ∈ J = R. Next, fix an arbitrary sequence(hn)n∈N in R with

limn−→∞

hn = 0 and [hn 6= 0 , for all n ∈ N] .

(Note that each c + hn ∈ J = R.) Further, fix an arbitrary n ∈ Nand consider

Qn :=f (c + hn)− f (c)

hn.

Using Pascal’s triangle to help us expand (c + hn)3, we see that

Qn =f (c + hn)− f (c)

hn=

1

hn

((c + hn)3 − c3

)=

1

hn

(c3 + 3 c2 hn + 3 c h2

n + h3n − c3

).

68

Thus, taking out a common factor of hn, we have

Qn =1

hn

(3 c2 hn + 3 c h2

n + h3n

)=

1

hnhn(3 c2 + 3 c hn + h2

n

)= 3 c2 + 3 c hn + h2

n .

Now,

limn−→∞

hn = 0 .

Consequently, by the Algebra of Limits (A.o.L.),

limn−→∞

Qn = limn−→∞

(3 c2 + 3 c hn + h2

n

)= 3 c2 + 3 c (0) + (0)2 = 3 c2 .

In summary, we have shown that for all c ∈ R,

f ′(c) = 3 c2 .

Example [2]. Let J := R, ν ∈ N and the function f : J −→ R bedefined by

f (x) := xν , for all x ∈ J = R .69

By similar arguments to above, using the Binomial Theorem, wecan show from the definition of the derivative that for all c ∈ R,

f ′(c) = ν cν−1 .


f (x) := ex = exp(x) , for all x ∈ J = R .


limn−→∞



Qn :=f (c + hn)− f (c)

hn.

70

Using the addition formula for exp to help us expand exp(c + hn), wesee that

Qn =f (c + hn)− f (c)

hn=

1

hn

(ec+hn − ec

)=

1

hn

(ec ehn − ec

).

Thus, taking out a common factor of ec, we have

Qn =1

hnec(ehn − 1

)= ec

(ehn − 1

hn

).

Now,lim

n−→∞hn = 0 and [hn 6= 0 , for all n ∈ N] .

By earlier work, using the A.o.L. Super Fact and the power seriesdefinition of exp, we know it follows that

limn−→∞

(ehn − 1

hn

)= 1 .

71


limn−→∞

Qn = limn−→∞

ec(ehn − 1

hn

)= ec (1) = ec .


f ′(c) = ec .


f (x) := sin(x) , for all x ∈ J = R .


limn−→∞



Qn :=f (c + hn)− f (c)

hn.

72

Using the addition formula for sin to help us expand sin(c+hn), we seethat

Qn =f (c + hn)− f (c)

hn=

1

hn(sin(c + hn)− sin(c))

=1

hn(sin(c) cos(hn) + cos(c) sin(hn)− sin(c))

=1

hn

(sin(c) (cos(hn)− 1) + cos(c) sin(hn)

)= sin(c)

(cos(hn)− 1

hn

)+ cos(c)

(sin(hn)

hn

).

Now,

limn−→∞


By earlier work, using the A.o.L. Super Fact and the power seriesdefinitions of sin and cos, we know it follows that

limn−→∞

(cos(hn)− 1

hn

)= 0 and lim

n−→∞

(sinhnhn

)= 1 .

73


limn−→∞

Qn = limn−→∞

[sin(c)

(cos(hn)− 1

hn

)+ cos(c)

(sin(hn)

hn

)]= sin(c) (0) + cos(c) (1) = cos(c) .


f ′(c) = cos(c) .

HA.L21 [1] Using the definition of the derivative, and in a similarmanner to Example [4] above: calculate f ′(c), for all c ∈ R; where

(1) f (x) := cos(x) , for all x ∈ J := R ;

and

(2) f (x) :=1

x, for all x ∈ J := (0,∞) .

We next began to discuss the natural logarithm function ln : (0,∞) −→R. We will re-cap and continue this discussion in Lecture 22...

74


“Calculating derivatives from the definition, continued...”

HA.L22 [1] Using the definition of the derivative, and in a similarmanner to the examples in Lecture 21: calculate f ′(c), for all c ∈ R;where

(1) f (x) := x5 , for all x ∈ J := R ;

(2) f (x) := x3/2 , for all x ∈ J := (0,∞) ;

and(3) f (x) := x1/5 , for all x ∈ J := (0,∞) .

Department of Mathematics, University of Pittsburgh, Pittsburgh, PA 15260, U.S.A.E-mail address: [email protected]

75

math 0235 from professor lennard - honorscollege.pitt.edu · our fall 2020 text is: “calculus,...

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