course notes math 146

7/29/2019 course notes math 146

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1 Overview of M147

1. Completeness property ofR.

Definition 1. R is an ordered field containing the N and for every non-empty subset that is boundedabove contains a least upper bound.

S is bounded above means there exists a c R such that x C for all x S. Any such C is anupper bound. C is a least upper bound for S if C is an upper bound and if D is any upper boundfor S than C D.People usually called the LUB is also called supremum or sup. The greatest lower bound is calledthe infimum or inf.All properties ofR follows from defn.

Decimal representation Rationals are Dense Archimedean Property: Given any r Rn N such that n > r. Continuum Property

2. Sequences

Definition 2. Say (xn) converges to a real number L if N N such that |xn L| < for alln N.Theorem 1.1. (Monotone Convergence) Every bounded monotone sequence converges.

Theorem 1.2. (Bolzano-Weierstrass) Every bounded sequence has a convergent subsequence.

Definition 3. (Cauchy) > 0N N such that |xn xm| < n, m N.

Theorem 1.3. Every Cauchy sequence converges.

3. Limits + Continuity.At a A Domain(f) if for all > 0 such that if x A and |x a| < , then |f(x) f(a)| < .If A = R or A = (c, d) then this definition is equivalent to limxa f(x) = f(a).

Sequential characterization of continuity. Intermediate Value Theorem Extreme Value Theorem

4. Differentiability.

MVT: Iff : [a, b] R continuous and is differentiable on (a, b) then there exists c (a, b) suchthat f(c) =

f(b)f(a)ba .

Cauchy MVT.

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2 Integration Theory

2.1 Definite Integral

Suppose f : [a, b] R is bounded. Partition the interval [a, b] into n subintervals [ti1, ti] for i [1, n].For f : [ti1, ti] R is a bounded function. Let

Mi = sup{

f(t) : t

[ti1, ti]}

(1)

mi = inf{f(t) : t [ti1, ti]} (2)(3)

In particular if f is continuous then Mi and mi is attained.

Definition 4 (Riemann Sums). Let ti = ti ti1.

U(f, P) =ni=1

Miti (4)

L(f, P) =n

i=1

miti (5)

We say U(f, P) is the upper riemann sum and L(f, P) is lower Riemann sum.

Theorem 2.1. L(f, P) Area U(f, P).Definition 5. Let Q be a refinement if Q is a partition which includes all the points in the partition P.

Theorem 2.2. If Q is a refinment of P then U(f, P) U(f, P) and L(f, P) L(f, Q). Furthermore,for any two partitions P1, P2, take Q = P1 P2 then L(f, P1) L(f, Q) U(f, Q) U(f, P2).

Look at {L(f, P) : P partition}. Every U(f, P) is an upper bound for this set. This set is boundedabove and therefore has a lowest upper bound. Let S = sup {L(f, P)}. Thus, S U(f, P). Similarly,

{U(f, P)

}is a set bounded below let I = inf

{U(f, P)

}. Thus, S

I.

Now,L(f, P) S I U(f, Q)

Definition 6. We say f is integrable on [a, b] if sup {L(f, P)} = inf{U(f, P)} and we write ba f =common value. If f 0 then we define area under f between a and b as ba f.Example 1. f(x) = c in [a, b]. Then U(f, P) = c(b a) = L(f, P) = ba f.Example 2. f(x) =

1, x Q0, x Q . sup = 1 and inf = 0. Thus not integrable.

Theorem 2.3. Suppose f : [a, b] R is bounded. Then f is integrable over [a, b] if and only if for every > 0 there is a partition P such that U(f, P) L(f, P) < .Proof. Assume f is integrable. That means I = S. Fix > 0. Consider S 2 < S. So there must exista partition P1 such that L(f, P1) > S 2 . Similarly there is a partition P2 such that U(f, P2) < I + 2 .Let P = P1 P2. Then,

U(f, P) L(f, P) U(f, P2) L(f, P1) I + 2

(S 2

) = I S+

Conversely, assume we have such a partition P. Suppose I = S, let = IS > 0. Then I U(f, P), S L(f, P). Then = U(f, P)L(f, P) < but this is a contradiction. Therefore I = S and f is integrable.za

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Example 3. f(x) = x2 on [0, 1].Choose a uniform partition Pn : 0t0 < t1 =

1n < t2 =

2n .

ti1 =i 1

n, ti =

i

n(6)

Mi = t2i (7)

mi = t2i1 (8)

U(f, P) =

(i

n)2(

i

n i 1

n) =

1

n3

ni2 =

1

n3n(n + 1)(2n + 1)

6(9)

L(f, P) =

(i 1

n)2(

i

n i 1

n) =

1

n3

n1i2 =

1

n3(n 1)n(2(n 1) + 1)

6(10)

U L = 1n

(11)

Let > 0. Pick N > 1 . For all n N we have U L = 1n < . Thus, f is integrable.L(f, P), U(f, P), 23 . Thus

10 x

2 = 23 .

3 Uniform Continuity

Recall:

Definition 7. f is continuous at x0 D(f) if > 0 > 0 such that if x D(f) aand |x x0| < ,then |f(x) f(x0)| < .

depends on both x0 and !

Definition 8. f is uniformly continuous if > 0 > 0 such that if x, y D(f) and |x y| < then|f(x) f(y)| < .

Uniform continuity implies continuity, but continuity uniform cont.Example 4. f(x) = x2 is contiuous but not uniformly continuous.

Proof. Let = 1. Say > 0 Take x = a + 2 , y = a.

|x y| = 2

<

|f(x) f(y)| = |(a + 2

)2 a2| = |a2 + a+ 2

4 a2| a

But if a > 1 . This is greater than 1.

Example 5. f(x) = 1x on (0, 1). Let > 0. Say works. Take x = 1N, y = 2N.

Example 6. f : [a, b] R or (a, b), with a bounded derivative, then f is uniformly continuous.Proof. Let > 0 and = C. If x, y D(f) by MVT

|f(x) f(y)| = |f(t)| |x y| C|x y| < C = C C

= .

for some t (x, y).Big theorem:

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Theorem 3.1. If f : [a, b] R is continuous then f is uniformly continuous.Proof. (By Contradiction) Suppose this is false. There is a choice of such that no works. In particular, = 1n fails for each n. Hence for each n N there are points xn, yn [a, b] such that |xn yn| < but|f(x) f(y)| .Consider (x, n), (yn) in [a, b].(xn) is bounded sequence. Bolzano-W thm, it has a convergent subsequence (xnk)

x0

[a, b].

Look at (ynk) with the same indices. This has convergent subsequence that converges to y0 [a, b]. Callseq (ynkj). (xnkj) x0.f is continuous at x0, y0. So f(xnkj) f(x0 and f(ynkj) f(y0).|f(xnkj f(ynkj)| |f(x0) f(y0)||xnkj ynkj | < 1nkj 0 as j . Hence 0 = |y0 x0| x0 = y0. f(x0) = f(y0). Contradiction.Proof. Suppose not. There is some such that no works.In particular, = 1n fails for each n. We willbe able to pick xn, yn D(f) such that |xn yn| < = 1n but |f(xn) f(yn)| . The sequence (xn)from [a, b] has a convergent subsequence (xnk), say limit x0. Consider (ynk). We have |xnk ynk | < 1nk .By squeeze theorem, (ynk)

x0.

Now the closure of the interval becomes important when we have f(xnk) f(x0) since x0 [a, b] and fis continuous. Similarly for f(ynk) f(x0). But |f(xnk) f(ynk)| |f(x0) f(y0)| = 0.

4 Integration Theory Contd

Theorem 4.1. If f : [a, b] R is continuous, then f is integrable.Proof. Note that f is bounded by the EVT and f is uniformly continuous. We will use the criteria thatf is integrable if for every > 0 there exists a parition P such that U(f, P) L(f, P) < .Let > 0. Pick from the defn of uniform continuity such that |x y| < so |f(x) f(y)| < ba . TakeP such that ti

ti1 < .

Assume Mi = sup f|[ti1,ti] = f(di) for di in[ti1, ti]. Similarly for mi = f(ci). These points are attainedbecause of the EVT.

U(f, P) L(f, P) =

(f(di) f(ci))(ti ti1)Since ci, di [ti1, ti] thus, |ci di| < . By uniform continuity, |f(ci) f(di)| < ba . So,

U(f, P) L(f, P) <

b a (ti ti1) =

Theorem 4.2. Every monotonic function f : [a, b] R.Proof. Assume f is increasing. f is bounded above by f(b) and below by f(a). Take a uniform partitionPn such that a = t0 < < tn = b where ti ti1 = ban .

U(f, P) =

Mi(ti ti1) =

f(ti)b a

n

L(f, P) =

mi(ti ti1) =

f(ti1)b a

n

U(f, P) L(f, P) =

(f(ti) f(ti1))( b an

) =b a

n(f(b) f(a))

Let > 0. Choose n = (ba)(f(b)f(a)) . Therefore U(f, P) L(f, P) < .4


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Exercise 1. Prove that a monotionic function on [a.b] can have countable many discontinuities.

Theorem 4.3. f : [a, b] R is bounded and continuous except at finitely many points c1, . . . , cs.Proof. We begin by surrounding the discontinuities with very small sub intervals. r0 = a , . . . , r2s+1 = bsuch that cj (r2j1, rj) and

sj=1 (r2j r2j1) < 2sup |f| .

Look at [r2j , r2j+1]. Since no ci [r2j , r2j+1] and thus f|[r2j ,r2j+1] is uniformly continuous. Pick j > 0from uniform continuouty such that if x, y [r2j , r2j+1] and |x y| < j then |f(x) f(y)| 0 there are countable many openintervals Ij , j = 1, 2, . . . such that E j=1Ij and

j=1 lengthIj < .

Example 7. Finite set Q Any countable set has measure zero. E = [0, 1] is NOT measure zero

E is irrationals in [0, 1] are not of measure zero.

Theorem 4.4. (Lebesgue) Bounded function f : [a, b] R is integrable if and only if the set of disconti-nuities of f is of measure zero.

5 Arc Length Derivation

Definition 10. Arc Length of a continuously differentiable f : [a, b] R. We mean sup {||f(tj) f(tj1||}where ||f(tj)f(tj1)|| is the length of the line segment joining (tj1, f(tj1)), (tj , f(tj)) =

(tj tj1)2 + (f(

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f(tj) f(tj1) = f(cj)(tj tj1)for some cj (tj1, tj) by MVT. Thus,

tj =

(tj tj1)2 + f(cj)2(tj tj1)2 = (tj tj1)

1 + (f(cj))2

ni=1

Lj =n

j=1

1 + (f(cj))2(tj tj1 (22)

Take F(x) =

1 + (f(x))2 then,Have observed

f(tj) f(tj1) = R(F, P):

L(F, P) R(F, P) U(F, P)

If Q is a refinement of P then:

R(F, P) = j L(P)

j i L(Q)

i R(F, Q)

Let P, P be any parition and Q is a common refinement,

R(F, P) R(F, Q) U(f, Q) U(F, P)

supP

L(F, P) supP

R(F, P) U(F, P) P

Therefore,supP

R(F, P) infP

U(F, P)

Thus, supP R(F, P) = ba F which means that the arclength R(F, P) is ba 1 + (f(x))2.6 Improper Integrals

Definition 11. a

f(x)dx = lims

sa

f(x)dx

f =

a

f +

a

f

Example 8.

1

dx

x3/2 = limss

1 x

3/2

dx = lims(2s1/2

+ 2) = 2

Exercise 2. a

xpdx

converges if p > 1 and diverges if p 1, provided a > 0.Theorem 6.1. (Comparison Test) Suppose 0 f(x) g(x)x a. If a g converges, then a fconverges.

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Proof. For n N, let n > a. Let

In =

na

f(x)dx

In+1 In =n+1a

f na

f =

n+1n

f 0

(In)n=n is an increasing sequence.

In na

g a

g R

sincea g converges.

So (In) is bounded above, hence it converges by MCT. Still must check limssa f converges. Given s,

pick n such that n < s n + 1.

In =

na

f sa

f n+1a

f = In+1

By squeeze theorem, since In, In+1 L, then sa f L.Example 9.

0ex

2

dx =

2

Proof it converges: Note: ex2 ex for x 1.

limss0 e

x2dx.

s0

=

10

ex2

C+

s1

ex2

dx

Using the comparison test since ex2 ex. Check if1 ex converges.

lims

s1

exdx = lims

exs1

= limx

ex + e1 = e1

Example 10. Find the volume of the infinitely llong horn generated by rotating y = 1x , to the firght ofx = 1 around the x-axis.

V = lims

s

1

1

x2dx = lim

s

x1

s

1

= lims

(

1

s+ 1) =

What is its surface area?

SA = lims

s0

2f(x)

1 + f(x)2

Definition 12. Say f is unbounded at a. Defineba

f = lim0+

ba+

f

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Example 11. 11

dx1 x2

7 Series

Definition 13.n=1

an = limN

Nn=1

an.

Theorem 7.1. (Geometric Series) Sn =N

1 rn =

r(1rn)1r . If |r| < 1 then rn 0 as N hence

Sn . Therefore:1

rn =r

1 r

If |r| 1,N1 an diverges.Example 12. (Telesoping Sum)

1

1

n 1

n(n + 1)= 1

n(n + 1)

Sn =N1

1

n 1

n + 1

= 1 1

2+

1

2 1

3+

1

3 1

4+ + 1

N 1

N + 1= 1 1

N + 1

Therefore Sn 1 as n .Facts:

1. If1 an and

1 bn both converges then so does

1 an

bn.

Proof. Write S(a)n =

an, S

(b)n =

bn. Say Sn(a) S(a) and sim for S(b).

N1

an + bn = S(a)n + S

(b)n S(a) + S(b)

2. If

an converges then so does

can = c

an.

3. Ifan 0 then

an converges if and only if (Sn)n=1 is bounded.

Proof. If an converges then by defn (Sn) converges (Sn) bounded.Suppose (Sn) bounded. We have Sn+1 Sn = an+1 0 Sn+1 Sn. Therefore (Sn) is increasing. Hence by MCT. (Sn) converges. So be defn.

an converges.

4. If

an converges then an 0.

Proof. Since (Sn) is convergent, its Cauchy. So given > 0. We can find N such that |an+1| =|Sn Sn+1| < if n N an 0.

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Example 13. (Harmonic Series)n=1

1

n

Let us compare with1

dxx . (Pictorially, the rectangles all sit above the curve in the discrete sum).

Formally, the discrete sum is the upper riemann sum when taking the partition of evenly spaced outalong the integers. 1

n dx

x= ln(N + 1)

So hence Sn is unbounded and hence 1

n diverges.

Example 14.1

1

n1+

for > 0, this sum converges. Compare with1

dxx1+

= x

Sn

= Area of Rect under curve = 1 +1

21++

+

1

N1+

N1

1

x1++ 1 =

x

N1

+ 1 =!

1

N+ 1 1

+ 1

So (Sn) is bounded and thus 1

n1+ converges.

Theorem 7.2. (Integral Test) If f is positive and decreasing on [1, ) and f(n) = an, then

anconverges if and only if

1 f converges.

Proof. Proof by contrapositive, assume f diverges then an diverges.Example 15. 1

np

converges if and only if p > 1.

Theorem 7.3. (Comparison Test) If 0 an bnn n0. If

bn converges then so does

an.

Proof. Write S(a)n and for b as well. For large N, S

(a)n =

n01 an +

Nn0

an A + S(b)n C for all N sincebn converges. So (S

(a)n ) is bounded and thus

an converges.

Example 16.

1

2n + 1

12n+1 13n and 13n diverges.Example 17. 2n 1

n3 + 1 3

n2

Theorem 7.4. (Ratio Test) If 0 an, bn and lim anbn = c = 0. Then

an converges if and only if

bnconverges.

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8 Area

Definition 14. If f 0 continuously differentiable then SA < V ol < .Proof. (Sketch)

SA = 2

a

f(x)1 + f(x)2dx 2

a

f(x)

Claim 1. For all nxn n such that f(xn) 1. If not, then n0 st f(x) 1x n0.

2

a

f 2n0

2(N n0)

Proof. (Real) Sn = solid by rotating f on [a, xn] around xaxis.

V(Sn) Vol of sphere that has the same SA (By Isoperimetric property)

SA(Sn) = f(a)2 + f(xn)

2 + f(xn)2 + 2

a

f(x)

1 + f(x)2dx

f(a)2 + 1 + 2 a

f(x)

1 + f(x)2dx =

f(a)2 + 1

+ SA = C <

V olSn V ol sphere which has surface area = C.V ol Solid of revn = lim V(Sn) Vol sphere which has surface area = C < .

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