Methods Of Real Analysis, R. Goldberg Solutions-1 PDF

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Solutions Part-1 to the text. Covering Real Numbers, Sequences and little part of Limits....

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Real Analysis Solution Set

Methods of Real Analysis Richard R. Goldberg

Anish Sachdeva DTU / 2K16 / MC / 13

Contents 1 Sets and Functions 1.1 Exercise 1.1 . . . 1.2 Exercise 1.2 . . . 1.3 Exercise 1.3 . . . 1.4 Exercise 1.4 . . . 1.5 Exercise 1.5 . . . 1.6 Exercise 1.6 . . . 1.7 Exercise 1.7 . . . 2 Sequences Of Real 2.1 Exercise 2.1 . . 2.2 Exercise 2.2 . . 2.3 Exercise 2.3 . . 2.4 Exercise 2.4 . . 2.5 Exercise 2.5 . . 2.6 Exercise 2.6 . . 2.7 Exercise 2.7 . . 2.8 Exercise 2.8 . . 2.9 Exercise 2.9 . . 2.10 Exercise 2.10 . 2.11 Exercise 2.11 . 2.12 Exercise 2.12 .

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Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1 2 3 4 5 6 7 8 8 10 18 20 24 26 33 44 46 48 49 50

3 Limits and Metric Spaces 51 3.1 Exercise 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

at10.95ptSETS AND FUNCTIONS

1

Exercise 1.1

Sets and Functions at12.0pt at12.0pt1.1

Exercise 1.1

1. Describe the following sets of real numbers geometrically : (a) A = {x| x < 7} (b) B = {x| |x| ≥ 2} (c) C = {x| |x| = 1} 2. Describe the following sets of points in the plane geometrically:   (a) A = hx, yi | x2 + y2 = 1 (b) B = {hx, yi | x ≤ y}

(c) C = {hx, yi | x + y = 2} 3. Let P be the set of prime Integers, which of the following are true ? (a) 7 ∈ P (b) 9 ∈ P (c) 11 6∈ P

1

SETS AND FUNCTIONS

1.2

Exercise 1.2

Exercise 1.2

2

SETS AND FUNCTIONS

1.3

Exercise 1.3

Exercise 1.3

3

SETS AND FUNCTIONS

1.4

Exercise 1.4

Exercise 1.4

4

SETS AND FUNCTIONS

1.5

Exercise 1.5

Exercise 1.5

5

SETS AND FUNCTIONS

1.6

Exercise 1.6

Exercise 1.6

6

SETS AND FUNCTIONS

1.7

Exercise 1.7

Exercise 1.7

7

SEQUENCE OF REAL NUMBERS

2

Exercise 2.1

Sequences Of Real Numbers

2.1

Exercise 2.1

1. Let {sn }∞ n=1 be the sequence defined by

s1 = 1 s2 = 1

sn+1 = sn + sn−1 (n = 3, 4, 5, · · · ) Find s8

s3 = s2 + s1 s3 = 1 + 1 s3 = 2 s4 = s3 + s2 s4 = 2 + 1 s4 = 3 s5 = s4 + s3 s5 = 3 + 2 s5 = 5 s6 = s5 + s4 s6 = 5 + 3 s6 = 8 and so on we get · · ·

s8 = 21

2. Write a formula or formulae for sn for each of the following sequences. (a) 1,0,1,0... sn = 1 ∀ N ∈ I where n = 2N − 1 sn = 0 ∀ N ∈ I where n = 2N (b) 1,3,6,10,15... sn = sn−1 + n ∀n ∈ N (c) 1,-4,9,-16,25,-36... sn = (−1)n+1 n2 ∀n ∈ N (d) 1,1,1,2,1,3,14,1,5,1,6... sn = 1 ∀ N ∈ I where n = 2N − 1 sn = n2 ∀ N ∈ I where n = 2N 8

SEQUENCE OF REAL NUMBERS

Exercise 2.1

3. Which of the following sequences (a), (b), (c) and (d) in the previous exercise are subsequences of {n}∞n=1 ? The sequences (a), (b) and (d) are subsequences of {n}∞ n=1   ∞ ∞ 2 ∞ 4. If S = {sn } n=1 = {2n − 1}∞ n=1 and N = {ni } n=1 = i i=1 . Find s5 , s9 , n2 , sn3 . Is N a subsequence of ∞ ? {k}k=1

s5 = 2 · 5 − 1

s5 = 9

s9 = 2 · 9 − 1

s9 = 17

n 2 = 22 = 4 sn3 = 2 · n3 − 1

We know n3 = 32 = 9 sn3 = s9 = 17

 ∞ ∞ = {1, 2, 3, 4 · · · } and the sequence N = i2 i=1 = {1, 4, 9 · · · } is clearly a subsequence. Now {k}k=1

9

SEQUENCE OF REAL NUMBERS

2.2

Exercise 2.2

Exercise 2.2

∞ is a sequence of real numbers, if sn 6 M (n ∈ I) and if lim = L . Prove L 6 M 1. If {sn }n=1 n→∞

We Know sn 6 M ∀ n ∈ I

and lim sn = L n→∞

Now, L 6 max (sn ) and max (sn ) 6 M So, L 6 M

2. If L ∈ R, M ∈ R and L ≤ M +ǫ for every ǫ ≥ 0, prove that L ≤ M We Know, |sn | ≤ M ∀n ∈ N L ≤ |sn | ∀n ∈ N

Now we know that, L ≯ |sn | Hence, L ≤ M

3. If {sn }∞ n=1 is a sequence of real numbers and if, for every ǫ > 0, |sn − L| < ǫ (n ≥ N) where N does not depend on ǫ, prove that all but a finite number of terms of {sn }∞ n=1 are equal to L. 4. (a) find N ∈ I such that    1  2n  < (n ≥ N )  − 3  5 n + 3     2n    n + 3 − 3 <    2n − 3(n + 3)   <   n+3    −n − 9     n+3 <   n + 9   n + 3 <   −1  n + 9  <

0. 32 1 n+1 > 0.09 n > 11.11 − 1 n > 10.11

11

(1)

(2)

SEQUENCE OF REAL NUMBERS

Exercise 2.2

So, for any N ∈ I where N is positive and N > 10.11 and n > N, the equation (1) is satisfied 1 =0 (b) Prove that lim √n+1 n→∞ To prove: limit L = 0 Let the limit of the function as n → ∞ be L = 0 Now,

   1  √   n + 1 − L  < ǫ (ǫ > 0)      √ 1  n + 1 − 0 < ǫ    1  √  2 ǫ 1 n > 2 −1 ǫ Now let us take a N such that N > Hence the limit of { √ 1 }∞ n=1 is 0.

1 ǫ2

(3)

− 1. So for all n ≥ N the equation (1) is satisfied.

n+1

6. If θ is a rational number prove that the sequence {sin nθπ }∞ n=1 has a limit. 7. For each of the following sequences, prove either that the sequence has a limit or that the sequence does not have a limit. 2

n (a) { n+5 }∞ n=1

2

n }=L Let the sequence have a limit L such that lim { n+5 n→∞ Now if the function has a limit, it would also satisfy the equation :

  2   n  < ǫ (∀n ≥ N ) and (ǫ > 0)  − L  n + 5     n  1 1 0     3  7n 1 1 < 7n 1 + 7n 3 3 < 7n 1 + 7n

(3)

Substituting equation (3) into (2)

  3     7n − L < ǫ

We can see that this resembles the series

1 n

and hence can assume L=0 in equation (4)

   3    7ǫ

3 for any ǫ > 0. Now let us take N ∈ I where N > 7ǫ So for n ≥ N the equation (6) is satisfied and in conclusion equation (1) is also satisfied. So the 3n exists. limit for the function n+7 n2

8. (a) Prove that the sequence {107 /n}∞ n=1 has a limit 0. To prove that the function has a limit L =0 7 lim 10n = 0 n→∞

Let the limit of the function exist and let L = 0 14

(4)

(5)

(6)

SEQUENCE OF REAL NUMBERS

Exercise 2.2

So the function would satisfy the equation

  7   10  < ǫ (∀n ≥ N)  − L   n

(1)

where ǫ > 0 and N ∈ I and N > 0  7  10     n  ǫ

(2)

(3)

7

If we take N > 10ǫ , we will get ∀ n ≥ N and the equation (1) will be satisfied Hence the limit for the function exists 7 lim 10n = 0 n→∞

(b) Prove that {n/107 }∞ n=1 does not have a limit. ∞ doesn’t exist. To prove: We have to prove that limit for { 10n7 }n=1 Let us assume that the limit of the function exists and it is L. As teh Limit L exists the function will satisfy the equation :

  n    7 − L < ǫ ∀n ≥ N 10

(1)

Here L is the limit and N>0. N ∈ I and ǫ is an arbitrary positive rational number. Now as the limit exists we would be able to find some N for any arbitrary ǫ where equation (1) is satisfied. Let ǫ = 1  n     7 − L < 1 10 n ∈ (L − 1, L + 1) 107 n ∈ (107 (L − 1), 107 (L + 1)) For any Limit value L that we take, we can find a value for n where equation (2) is not satisfied. Hence our assumption that this function converges was wrong. n ∞ The series { 10 7 }n=1 clearly diverges

(2)

(c) Note that the first 107 terms of the sequence in (a) are greater than the corresponding terms in sequence (b. This emphasizes that the existence of a limit for a sequence does not depend on the first few (’few’ = any finite number of terms) terms. 15

SEQUENCE OF REAL NUMBERS

Exercise 2.2

9. Prove that {n − 1/n}∞ n=1 does not have a limit. ∞ is divergent To prove : that series {n − 1/n}n=1 We know that:

n − 1/n >

n (∀n ∈ N) and n > 1 2

(1)

∞ is divergent then the given sequence {n − 1/n}∞ Now from (1) we can infer that if the sequence { n2 }n=1 n=1 will also diverge. We have previously proved that the sequence {n}∞ n=1 is divergent. n ∞ diverges using the property that Hence we can also state that { 2 }n=1

lim {c · sn } = lim c {sn }

n→∞

(2)

n→∞

Hence we can state from (2) that { n2 }∞ n=1 is divergent. ∞ We can now state that sequence {n − 1/n}n=1 is also divergent Hence proved 10. If sn = 5n /n! show that lim sn = 0. n→∞ e can write sn as: 5 · 5 · 5··· sn = 1 · 2 · 3 · 4···             5 5 5 5 5 5 sn = · · · · · ··· 1 2 3 4 5 6 We can write this as: sn =

55 5 Π 5! n

So,

sn <

55 5 · 5! n

(1) 5

5 ∞ 5 From equation (1) we can infer that {sn }∞ n=1 will be convergent if the series { 5! · n }n=1 is convergent. ∞ is a convergent series. (Proved above) Now we know that {1/n}n=1 we also know that

lim c · sn = c · lim sn

n→∞

n→∞

5

using (2) we can state that the series { 55! · n5 }∞ n=1 is convergent We can then state that the series {sn }∞ n=1 Hence proved 16

(2)

SEQUENCE OF REAL NUMBERS

Exercise 2.2

11. If P is a polynomial function of the third degree P (x) = ax3 + bx2 + cx + d (a, b, c, d, x ∈ R) Prove that lim

n→∞

P (n + 1) =1 P (n)

To prove : That limit exists and Limit L = 1. We know that if the limit for any given function exists then it will satisfy the equation:

    P (n + 1)  < ǫ (∀n ≥ N )  − L   P (n)

(1)

Here ǫ is an arbitrary constant and L is the value of the limit. Here Limit, L =1. N ∈ I and N > 0 Now,

    P (n + 1)  0. Now we know that ||a| − |b|| ≤ |a − b| It is also given that :

||sn | − 0| < ǫ (given)

(2)

From equation (2) and the above mentioned identity we can infer that:

||sn | − |0|| ≤ |sn − 0|

(3)

Plugging equation (3) in (1) we get that :

|sn − 0| < ǫ (Standard Form of Limit Equation) Hence we can say that the series {sn }∞ n=1 is convergent and converges to 0.

18

(4)

SEQUENCE OF REAL NUMBERS

Exercise 2.3

∞ 4. Can you find a sequence of real numbers {sn }∞ n=1 which has no convergent subsequence and yet {|sn |} n=1 converges? ∞ If the sequence {sn }∞ n=1 has no convergent subsequence, then that implies that {sn } n=1 is divergent to either positive or negative Infinity. ∞ If the sequence {sn }∞ n=1 is divergent to positive or negative infinity then {|sn |} n=1 will also diverge to positive Infinity. ∞ Hence such a case is not possible where {sn }∞ n=1 has no convergent subsequence but {|sn |} n=1 converges.

5. If {sn }∞ n=1 is a sequence of real numbers and if lim s2m = L

m→∞

lim s2m−1 = L

m→∞

prove that sn → L as n → ∞. It is given that terms with even subscripts converge to L as n → ∞ and terms with odd terms also converge to L as n → ∞. We can say that lim sn as lim s2m where n = 2m for all even numbers. n→∞

n→∞

Similarly we can say that lim sn as lim s2m−1 where n = 2m − 1 for all odd numbers. n→∞

n→∞

We know that:

lim s2m = L

m→∞

lim s2m−1 = L

m→∞

So, we can conclusively say that lim sn = L n→∞ Hence proved.

19

SEQUENCE OF REAL NUMBERS

2.4

Exercise 2.4

Exercise 2.4

1. Label each of the following sequences either (A) convergent. (B) divergent to infinity, (C) divergent to -Infinity, or (D) Oscillating (a) {sin nπ/2}∞ n=1 (D) Oscillating ∞ (b) {sin nπ}n=1 (A) Convergent

(c) {en }∞ n=1 (B) Divergent to +∞ ∞ (d) {e1/n }n=1 (A) Convergent

(e) {n sin(π/n)}∞ n=1 (A) Convergent (f) {(−1)n tan(π/2 − 1/n)}∞ n=1 (D) Oscillating ∞  (g) 1 + 12 + 31 + 14 + · · · + 1n n=1 (B) Divergent to +∞ (h) {−n2 }∞ n=1 (C) Divergent to −∞

√ ∞ diverges to Infinity. 2. Prove that { n}n=1 √ To prove that the sequence { n}∞ n=1 diverges to +∞ we must prove that for any given M > 0 : sn > M (n ≥ N )

(1)

√ n>M

(2)

Here N ∈ I and N > 0 Now,

n>M

2

(3)

So for any N > M 2 will satisfy the equation (3) and hence satisfy (1). So we can conclusively say that √ the sequence { n}∞ n=1 diverges to +∞. √ √ ∞ 3. Prove that { n + 1 − n}n=1 is convergent. First let us simplify the expression √ √ n+1− n (n + 1) − n √ √ n+1+ n 20

(1)

SEQUENCE OF REAL NUMBERS

Exercise 2.4

Rationalizing numerator and denominator Further simplifying

1 √ √ n+1+ n (2) We know that : √ √ √ √ n+1+ n > n+ n √ √ √ n+1+ n > 2 n 1 1 √ √ < √ 2 n n+1+ n √ Now if the equation in (3) is convergent we can say that { n}∞ n=1 will also be convergent.

(3)

√1 { √1n }∞ > √1n ∀n ∈ N. n=1 is a bounded above series by 0. It is also monotonically decreasing as n−1 As this series is both monotonically non-increasing as well as bounded above it is also convergent. We also know that:

lim c · sn = c · lim sn

n→∞ 1 √ n→∞ 2 n

From (4) we can infer that lim

=

n→∞

1 lim √1 2 n→∞ n

(4)

=0

√ √ ∞ Now as { 2√1 n }∞ n=1 is convergent we also conclude that { n + 1 − n}n=1 is convergent. ∞ 4. Prove that if the sequence of real numbers {sn }∞ n=1 diverges to infinity, then {−sn } n=1 diverges to minus infinity. It is given that the sequence {sn }∞ n=1 diverges to +∞. So we can write it mathematically as:

sn > M (n ≥ N )

(1)

Here M > 0 is an arbitrary constant and N ∈ I where N > 0 is a term subscript where sn surpasses the value of M. Solving further :-

−sn < −M (n ≥ N )

(2)

Now let the negative sequence be denoted by {−sn }∞ n=1 . Let the sequence be represented by S . ∞ S = {−sn }n=1

S < −M (n ≥ N )

(3)

Equation (3) is the standard form of a divergent series that diverges to −∞. Hence {−sn }∞ n=1 diverges to ∞ −∞ when {sn }n=1 diverges tp +∞. 21

SEQUENCE OF REAL NUMBERS

Exercise 2.4

n ∞ 5. Suppose {sn }∞ n=1 converges to 0. Prove that {(−1) sn }n=1 converges to 0. ∞ It is given that {sn }∞ n=1 converges to 0. We can also then state that {−sn }n=1 converges to 0. As

lim c · sn = c · lim sn

(1)

sn = {s1 , s2 , s3 , s4 · · · }

(2)

−sn = {−s1 , −s2 , −s3 · · · }

(3)

n→∞

n→∞

Using (1) lim −sn = − lim sn = 0 n→∞ n→∞ Now, we can represent {sn }∞ n=1 as

And we can represent{−sn }∞ n=1 as

Taking sub-sequences from (2) and (3) we get

{s2 , s4 , s6 · · · } and {−s1 , −s3 , −s5 · · · }

(4)

respectively We know that sub-sequences of a convergent sequence are also convergent and they converge to the same value as their parent sequence. So limits for the sub-sequences (4) are 0. On combining the sub-sequences that we have created we create the sequence:

sn = {−s1 , s2 , −s3 · · · }

(5)

Hence {(−1)n sn }∞ n=1 = {−s1 , s2 , −s3 · · · }. This sequence has been formed by the combination of 2 convergent sub-sequences that are convergent to the same value 0, and hence {(−1)n sn }∞ n=1 is also convergent. n ∞ 6. Suppose {sn }∞ n=1 converges to L 6= 0. Prove that {(−1) sn } n=1 oscillates. Similar to the above example we can state that the series {−sn }∞ n=1 will also converge. We know that:

lim c · sn = c · lim sn

n→∞

n→∞

From (1) we can deduce that lim)n → ∞ − sn = − lim sn = −L n→∞

Both these series can now be represented as :22

(1)

SEQUENCE OF REAL NUMBERS

Exercise 2.4

{sn }∞ n=1 = {s1 , s2 , s3 · · · }

{−sn }∞ n=1

(2)

= {−s1 , −s2 , −s3 · · · }

(3)

We know that sub-sequences of convergent sequences are convergent hence both the sub-sequences (2) and (3) will be convergent. Sequence (2) will converge to L and sequence (3) will converge to -L. Now let us create a sequence by the combination of these sub-sequences:

{(−1)n sn }∞ n=1 = {−s1 , s2 , −s3 · · · }

(4)

Clearly (4) will oscillate as it’s sub-sequences are converging to different points L 6= -L. n ∞ 7. Suppose {sn }∞ n=1 diverges to infinity. Prove that {(−1) sn } n=1 oscillates. lim = L where From example 6 above it can be proved that any sequence {(−1)n sn }∞ n=1 oscillates if n→∞ L6...