*386*
*120*
*330KB*

*English*
*Pages 83*
*Year 2008*

- Author / Uploaded
- Marian Muresan

Marian Mure¸san

A Concrete Approach to Classical Analysis Solutions to Exercises February 29, 2012

Springer

Contents

1

Sets and Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

2

Vector Spaces and Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3

Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4

Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

5

Diﬀerential Calculus on R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

6

Integral Calculus on R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

7

Diﬀerential Calculus on Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

8

Double Integrals, Triple Integrals, and Line Integrals . . . . . . 75

1 Sets and Numbers

1.1. (i) Suppose that the set in the left-side contains an element, say x. Then x belongs to A, not to B, and to B. We get a contradiction, hence the set in the left-side contains no element. (ii) Obviously, A ∩ B ⊂ A \ (A \ B). If there exists an x ∈ A \ (A \ B), then x ∈ A and x ∈ / A \ B. It follows that x ∈ B, so that x ∈ A ∩ B. (iii) Both sides of this identity contain elements belonging either to A or to B. (iv) Both sides of the identity contain elements in A and neither in B nor in C. (v) The three members of this double identity contain elements belonging to A and C, but not to B. 1.2. Suppose that A \ B ⊂ C and pick up an x ∈ A. If x ∈ / B, then x ∈ A \ B and thus x ∈ C. If x ∈ B, then x ∈ B ∪ C. Conversely, suppose that A ⊂ B ∪ C and pick up an x ∈ A \ B. Then x ∈ A and x ∈ / B. Thus x ∈ (B ∪ C) \ B ⊂ C. 1.3. (i) We have x ∈ (A ∪ B) \ C ⇐⇒ (x ∈ A or x ∈ B) and x ∈ /C ⇐⇒ (x ∈ A and x ∈ / C) or (x ∈ B and x ∈ / C) ⇐⇒ x ∈ (A \ C) ∪ (B \ C). (ii) We have x ∈ (A ∩ B) \ C ⇐⇒ (x ∈ A and x ∈ B) and x ∈ /C ⇐⇒ (x ∈ A and x ∈ / C) and (x ∈ B and x ∈ / C) ⇐⇒ x ∈ (A \ C) ∩ (B \ C).

6

1 Sets and Numbers

(iii) By (i) we have (A ∪ C) \ B = (A \ B) ∪ (C \ B) ⊂ (A \ B) ∪ C. 1.4. (i) One can easily show that Y = (B \ C) \ A and Y ⊂ X. If C = ∅ and A = B ̸= ∅, then ∅ = Y ̸= A = X. We conclude that Y ⊂ X. 1.5. (i) x ∈ ∪∞ n=0 Bn ⇐⇒ ∃n ∈ N, x ∈ Bn ⇐⇒ ∃n ∈ N, i ∈ {0, . . . , n}, x ∈ Ai ⇐⇒ x ∈ ∪∞ n=0 An . (ii) x ∈ ∩∞ n=0 Bn ⇐⇒ ∀n ∈ N, x ∈ Bn ⇐⇒ ∀n ∈ N, i ∈ {0, . . . , n}, x ∈ Ai ⇐⇒ x ∈ ∩∞ n=0 An . 1.6. (i) ∞ ∞ x ∈ ∪∞ m=0 (∩n=0 Am,n ) =⇒ ∃m0 ∈ N, x ∈ ∩n=0 Am0 ,n

=⇒ ∃m0 ∈ N, ∀n ∈ N, x ∈ Am0 ,n ∞ ∞ ∞ =⇒ x ∈ ∩∞ n=0 (∪m=m0 Am,n ) ⊂ ∩n=0 (∪m=0 Am,n ). (ii) Not true. 1.7. (i) {X ∅ = X \ ∅ = X and {X X = X \ X = ∅. (ii) {(A \ B) = {(A ∩ {B)) = ({A) ∩ ({({B)) = ({A) ∪ B. (iii) (A ∪ B) \ (A ∩ B) = (A ∪ B) ∩ ({(A ∩ B)) = (A ∪ B) ∩ ({A ∪ {B) = (A ∩ ({A ∪ {B)) ∪ (B ∩ ({A ∪ {B)) = ((A ∩ {A) ∪ (A ∩ {B) ∪ ((B ∩ {A) ∪ (B ∩ {B) = (A ∩ {B) ∪ ((B ∩ {A). (iv) x ∈ {B =⇒ x ∈ / B =⇒ x ∈ / A =⇒ x ∈ {A. 1.8. Successively we have {(X ∪ A) ∪ (X ∪ {A) = B ⇐⇒ (({X ∩ {A) ∪ {A) ∪ X = B ⇐⇒ (({X ∪ {A) ∩ ({A ∪ {A)) ∪ X = B ⇐⇒ (({X ∪ {A) ∩ {A) ∪ X = B ⇐⇒ {A ∪ X = B ⇐⇒ A ∩ ({A ∪ X) = A ∩ B ⇐⇒ A ∩ X = A ∩ B. Thus X is any subset of U satisfying A ∩ X = A ∩ B.

1.1 Solutions

7

1.9. (i) X ∈ P(A) ∪ P(B) =⇒ X ∈ P(A) or X ∈ P(B) =⇒ X ⊂ A or X ⊂ B =⇒ X ⊂ A ∪ B =⇒ X ∈ P(A ∪ B). (ii) Pick a nonempty set X ∈ P(A ∪ B). Then X ⊂ A ∩ B. If X ⊂ A or X ⊂ B, the equality is true. Suppose that X = X1 ∪ X2 , with X1 ̸= ∅, X1 ⊂ A, X2 ̸= ∅, X2 ⊂ B, and X1 ̸= X2 . Then X ∈ / P(A) ∪ P(B). (iii) and (iv) are straightforward. 1.10. (i) One-to-one. (ii) Onto. (iii) One-to-one and onto. 1.11. (a) (i) =⇒ (ii) Consider f∗ (M1 ) = f∗ (M2 ). This is equivalent to f (M1 ) = f (M2 ). We show that M1 = M2 . Pick an arbitrary x ∈ M1 . Then f (x) ∈ f (M1 ) = f (M2 ). It means that there exists y ∈ M2 with f (x) = f (y). Since the function f is one-to-one, it follows that x = y and x ∈ M2 . Thus M1 ⊂ M2 . In a similar way we show that M2 ⊂ M1 , and therefore M1 = M2 . Thus the function f∗ is one-to-one. (ii) =⇒ (iii) Let M ∈ P(A) be arbitrary. We show that there exists N ∈ P(B) with f ∗ (N ) = M, that is, for M ⊂ A there exists N ⊂ B with f −1 (N ) = M. Deﬁne N = f (M ). From the injectivity of f we indeed have that M = f −1 (N ) = f ∗ (N ). (iii) =⇒ (iv) The inclusion f (M ∩ N ) ⊂ f (M ) ∩ f (N ) is obvious. We now show its reverse. Let N1 , N2 ∈ P(B) be with f −1 (N1 ) = M1 and f −1 (N2 ) = M2 be arbitrary. We have f −1 (N1 ∩ N2 ) = f −1 (N1 ) ∩ f −1 (N2 ). For y ∈ f (M1 ) ∩ f (M2 ) there exists x1 ∈ M1 , x2 ∈ M2 with f (x1 ) = f (x2 ) = y. Since f (x1 ) ∈ N1 , f (x2 ) ∈ N2 , it follows that y ∈ N1 ∩ N2 . Thus f −1 (y) ∈ f −1 (N1 ∩N2 ) = f −1 (N1 )∩f −1 (N2 ) = M1 ∩M2 , and there exists x ∈ M1 ∩M2 with f (x) = y. So f (M ) ∩ f (N ) ⊂ f (M ∩ N ). (iv) =⇒ (v) Taking M1 = M and M2 = {M, we have that f (M ) ∩ f ({M ) = f (M ∩ {M ) = f (∅) = ∅ and thus f ({M ) ⊂ {f (M ). (v) =⇒ (i) Consider x, y ∈ A with f (x) = f (y) and deﬁne M = A \ {x}. Then {M = {x} and f (x) = f (y) ∈ f ({M ) ⊂ {f (M ). Thus f (y) ∈ / f (M ) and y ∈ / M. Therefore y = x and the function f is one-to-one. (b) The proof is similar to (a). (c) Follows from (a) and (b). 1.12. (a) (i) =⇒ (ii) For every ξ ∈ C we have that (f ◦ g)(ξ) = (f ◦ h)(ξ), i.e., f (g(ξ)) = f (h(ξ)). Because of f is one-to-one, it follows that g(ξ) = h(ξ) and since ξ was taken arbitrarily in C, we have that g = h. (ii) =⇒ (i) Suppose that f is not one-to-one. Then there exist x, y ∈ A, with x ̸= y and f (x) = f (y). Consider C = {u, v} and the functions g, h : C → A

8

1 Sets and Numbers

satisfying g(u) = x, g(v) = y, and h(u) = h(v) = y. Then we have f ◦g = f ◦g and g ̸= h. (b) (i) =⇒ (ii) Let y ∈ B be arbitrary. Since f is onto, there exists x ∈ A so that f (x) = y. We have the sequence of implications g(f (x)) = h(f (x)) =⇒ g(y) = h(y) =⇒ g = h. (ii) =⇒ (i) Suppose f is not onto. Then there exists y ∈ B so that for every x ∈ A, f (x) ̸= y. Consider C = {0, 1} and g(z) = 0, for all z ∈ B, h(z) = 0, for all z ∈ B \ {y} and h(y) = 1. Then g ◦ f = h ◦ f and g ̸= h. 1.13. (i) There exists an injection N∗ ∋ n → 2n ∈ N∗ . (ii) The set is at least countable since the coordinates a given vertice runs on the countable set Q × Q. The same set is a subset of the countable set (Q × Q) × (Q × Q) × (Q × Q), Theorem 2.17. (iii) The set coincides with Q × Q which is countable. (iv) It follows from Theorem 2.17. 1.14. (i) Denote N = {a1 , a2 , . . . , an } and M = {b1 , b2 , . . . , bm } Then to a1 one can assign an arbitrary element from M, i.e., there are m possibilities, to a2 one can assign an arbitrary element from M, i.e., there are m possibilities, and so on. (ii) Follows from (iii). (iii) To a1 one can assign an arbitrary element from M, let it be {bi }, i.e., there are m possibilities, to a2 one can assign an arbitrary element from M \ {bi }, let it be {bj }, i.e., there are m − 1 possibilities, to a3 one can assign an arbitrary element from M \ {bi , bj }, let it be {bk }, i.e., there are m − 2 possibilities, and so on till to each element from M is assigned one from N. n , m ≥ n. For (iv) Denote the set of onto functions from M into N by Sm each i = 1, . . . , n, let Ai be the set of functions from M into N such that n bi does not belongs to the image of any such function. Then the set Sm of onto functions from M into N coincides with the set of all functions from M into N except the functions belonging to some of the sets Ai . Then n |Sm | = nm − |A1 ∪ A2 ∪ · · · ∪ An |. By the ﬁrst equality in Exercise 1.18, we expand |A1 ∪ A2 ∪ · · · ∪ An |, and remark that |Ai | = (n − 1)m ,

|Ai ∩ Aj | = (n − 2)m , . . . , ∩ni=1 Ai = ∅. ( ) ( ) We also note that there are n1 sets Ai , n2 sets Ai ∩ Aj , . . . . Now the result follows. 1.15. First approach. One proves them by induction. Second approach. One proves them by combinatorial argues. For the ﬁrst identity a combinatorial argue is the next one. Choose a set A of n ∈ N elements and a set B with two elements 0 and 1. We identify a subset M ⊂ A with a mapping f : M → B deﬁned as f (x) = 1 if x ∈ M and f (x) = 0

1.1 Solutions

9

otherwise. Thus we obtain a bijection between the family of subsets of A and the set of functions from M into B. The cardinal of the later set is 2n , according to (i) in Exercise 1.14. For the next two identities consider C and D a partition of family of subsets of A so that C contains subsets of odd cardinal whereas D contains subsets of even cardinals. We already saw that |C ∪ D| = 2n . Pick up an element in A, denote it x. We deﬁne a bijection f from C onto D such that f (X) = X ∪ {x} if x ∈ / X and f (X) = X \ {x} otherwise. It follows that the sets C and D are of the same cardinal and thus the last two identities are proved. One can try using the next Mathematica r commands. TraditionalForm[Sum[Binomial[n, k], {k, 0, n},Assumptions \[RightArrow] n\[Element]Integers &&n\[GreaterSlantEqual] 0]] TraditionalForm[Sum[Binomial[n,2*k],{k,0,n/2},Assumptions \[RightArrow] n\[Element]Integers &&n\[GreaterSlantEqual] 0]] TraditionalForm[Sum[Binomial[n,2*k+1],{k,0,n/2},Assumptions \[RightArrow] n\[Element]Integers &&n\[GreaterSlantEqual] 0]] 1.16. Let A be the family of partitions of S. Then S = {1} =⇒ A = {S}, S = {1, 2} =⇒ A = {S, {{1}, {2}}}, S = {1, 2, 3} =⇒ A = {S, {{1}, {2}, {3}}, {{1, 2}, {3}}, {{1, 3}, {2}}, {{2, 3}, {1}} }, S = {1, 2, 3, 4} =⇒ A = {S, {{1}, {2}, {3}, {4}}, {{1, 2, 3}, {4}}, {{1, 2, 4}, {3}}, {{1, 3, 4}, {2}}, {{2, 3, 4}, {1}}, {{1, 2}, {3, 4}}, {{1, 2}, {3}, {4}}, {{1, 3}, {2, 4}}, {{1, 3}, {2}, {4}}, {{1, 4}, {2, 3}}, {{1, 4}, {2}, {3}}, {{2, 3}, {1}, {4}}, {{2, 4}, {1, 3}}, {{3, 4}, {1}, {2}}}. Thus the equalities in (1.11) of the statement hold. Some results can be obtained by Mathematica r , Figure 1.1. One can also try using the next Mathematica r command. Table[BellB[n], {n, 0, 4}] Now suppose S = {a1 , a2 , . . . , an , an+1 }. Pick up( an ) arbitrary element in S, say an+1 . Select k elements in S \ {an+1 } (in nk ways) and denote by A the set of remaining n − k elements. By A one has Bn−k partitions. Thus we get (1.12). 1.17. Every x ∈ A belongs to at least a set Ai . The set of indices of the sets Ai containing x is a nonempty subset to {1, 2, . . . , n}. So, there are 2k − 1 possibilities for x belonging to at least a set Ai . Since each x ∈ A belongs to at least a set Ai , the conclusion follows.

10

1 Sets and Numbers

H* Set partitions*L SetPartitions@4D 8881, 2, 3, 4 1. Hence M = {1}. 3.18. Since the set {a1 , a2 , . . . , an } is ﬁnite, max{a1 , a2 , . . . , an } is attained and is a member of the set. Denote ak = max{a1 , . . . , an }. If a1 = · · · = an , then max{a1 , a2 , . . . , an } = a1 .

3.1 Solutions

33

Suppose not all the members are equal. Suppose k ∈ {2, . . . , n − 1} and a1 < ak , ak > an . Then a2k ≤ ak−1 ak+1 ≤

max am · max am = a2k

m=1,...,n

m=1,...,n

and thus a2k = ak−1 ak+1 . One has ak ≥ ak−1 and ak ≥ ak+1 . Thus we get ak−1 = ak = ak+1 . Using the same argumentation, we ﬁnd that all the elements of the set are equal, which is a contradiction. Thus the maximum is equal to a1 or an . Both results are possible. Suppose a1 = 1, a2 = 2, a3 = 3. Then max{a1 , a2 , a3 } = a3 . Suppose now a1 = 3, a2 = 2, a3 = 1. Then max{a1 , a2 , a3 } = a1 . 3.19. From xn+1 = xn (xn −1)+1 we have that xn > 1, for all n ≥ 1 and from xn+1 −xn = (xn −1)2 > 0 we conclude that the sequence is increasing. Suppose it is bounded. Then it converges and let limn→∞ xn = α > 1. From the recurrence we ﬁnd out that α = 1, a contradiction. Thus (xn ) monotonically diverges to ∞. Since xn+1 − 1 = xn (xn − 1), then 1 xn+1 − 1

=

1 xn (xn − 1)

=

1 1 1 1 1 − =⇒ = − . xn − 1 xn xn xn − 1 xn+1 − 1

Thus the general term of the sequence of partial sums can be written as 1 1 1 + + ··· + x1 x2 xn 1 1 1 1 1 1 = − + − + ··· + − x1 − 1 x2 − 1 x2 − 1 x3 − 1 xn − 1 xn+1 − 1 1 1 = − . x1 − 1 xn+1 − 1

sn =

n→∞

Since (xn ) diverges to ∞, we conclude sn −−−−→ 1/(a − 1) and therefore ∞ ∑

1/xn = 1/a − 1.

n=1

3.20. The sequence (xn ) is convergent. Let x = lim xn . The general term of the series is nonnegative. Thus the series is convergent if and only if the sequence of partial sums is bounded. The general terms of the sequence of partial sums can be written as sn = (1 − x1 /x2 ) + (1 − x2 /x3 ) + · · · + (1 − xn /xn+1 )

34

3 Sequences and Series

=n−

n ∑

xk /xk+1

v u n ( ) √ u∏ n n ≤n− t (xk /xk+1 ) = n 1 − x1 /xn+1

k=1

=−

k=1

e(1/n) · ln(x1 /xn+1 ) − 1 (1/n) · ln(x1 /xn+1 )

· ln

n→∞ x1 x −−−−→ ln . xn+1 x1

3.21. (i) It diverges because is the generalized harmonic series with p = 1/2. (ii) It converges. One can check it by the root of ratio tests. (iii) It diverges. (iv) It diverges because is the generalized harmonic series with p = 1/3. (v) It converges based on the comparison, ratio, or root test. Also try the Mathematica r command Sum[(n + 1)/( n^2

3^n), {n, 1, Infinity}]

(vi) It converges to 5 sin(x)/(26−10 cos x). Try the Mathematica r command Sum[Sin[n * x]/5^n, {n, 1, Infinity}]

(vii) It converges to 1. Try the Mathematica r command Sum[2/3^n, {n, 1, Infinity}]

(viii) The series converges. (ix) It converges to 1 because the partial sum is 1 − 1/(n + 1). Also try the Mathematica r command Sum[1/(n (n + 1)), {n, 1, Infinity}]

(x) It converges to 1. Also try the Mathematica r command Sum[((-1)^(n - 1)/(n + 1)) (2 + 1/n), {n, 1, Infinity}]

(xi) It converges. Try the Mathematica r command Sum[ArcTan[1/(2 n^2)], {n, 1, Infinity}]

√ 3 (xii) Write the partial sum of the series to conclude that it converges to 1− 2. r (xiii) It converges to ln 2. Try the Mathematica command Sum[Log[(n^2 + 5 n + 6)/(n^2 + 5 n + 4)], {n, 1, Infinity}]

∑∞ −√n (xiv) The series n=1 e√n has positive terms and since e−x < x−3/2 for all ∑ −2 positive x, is majorized by the generalized harmonic series n . (xv) The series diverges for p ≤ 0, converges for p ∈ ]0, 1], and converges absolutely for p > 1. (xvi) The series converges by the root test. (xvii) The series diverges. 3.22. (a) Its domain of convergence is the set ]1, ∞[ , according to Theorem 3.10. On this set the series converges even absolutely. Try the Mathematica r commands

3.1 Solutions

35

SumConvergence[1/n^x, n] $Assumptions = {x \[Element] Reals}; SumConvergence[1/n^x, n] $Assumptions = {x > 1}; SumConvergence[1/n^x, n]

(b) For x > 1, the series converges absolutely by (a). For 0 < x ≤ 1, it converges by Theorem 3.22 of Leibniz. For x ≤ 0 the necessary condition (Theorem 3.2) is not satisﬁed, hence the series diverges. We conclude that the domain of convergence is ]0, ∞[ . Try the Mathematica r commands $Assumptions = {x > 0}; SumConvergence[(-1)^n/n^x, n]

(c) Since it is periodic, we restrict our discussion to ]−π/2, π/2[ . By Theorem 3.8, the series converges absolutely on ] − π/4, π/4[ . Otherwise it diverges. Try also the Mathematica r command SumConvergence[(Tan[x])^n, n]

(d) For x = 0, it converges obviously. By the root test we have / |xn+1 | n lim sup = |x|. n| n + 1 |x n→∞ So, for |x| < 1, the series converges absolutely. For x = −1, the series converges by Leibniz Theorem. For x = 1, the series is the harmonic series, thus it diverges. For |x| > 1, the necessary condition is not satisﬁed, thus it diverges. (e) We have lim sup n→∞

√ |x − 3| n |(x − 3)n /nn | = lim sup = 0. n n→∞

Thus the given series converges for all x ∈ R. Try also the Mathematica r command SumConvergence[(x - 3)^n/n^n, n]

(f) By the ratio test we have |x|, n+1 2n 2n x (1 + x ) 1 + x 1, = |x| · lim lim sup = n→∞ 1 + x2n+2 (1 + x2n+2 )xn 1 n→∞ , |x|

|x| < 1, x ̸= 0 |x| = 1 |x| > 1.

For x = 0, the series converges. It also converges absolutely for all x ̸= ±1. ∑ For x = ±1, the series diverges since it looks like (−1)n /2. Try also the Mathematica r command SumConvergence[x^n/(1 + x^(2 n)), n]

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3 Sequences and Series

(g) For x = 0, the series diverges. The ration test gives no information regarding the convergence. By the Raabe–Duhamel test, i.e., Theorem 3.14, we have ( ) n! (x2 + 1)(x2 + 2) · · · (x2 + n + 1) lim inf n · − 1 n→∞ (x2 + 1)(x2 + 2) · · · (x2 + n) (n + 1)! = x2 . Thus the series converges for x2 > 1, that is x < ∑ −1 or x > 1. It diverges for −1 < x < 1. If x2 = 1, the series looks like 1/(n + 1), so diverges. Hence the series converges if and only if x2 > 1. Try also the Mathematica r commands $Assumptions = {x \[Element] Reals}; SumConvergence[n!/Product[x^2 + i, {i, 1, n}], n]

3.23. (i) x ∈ [−1, 1[ . (ii) x ∈ ] − 1, 1]. (iii) x ∈ ] − 5, 5[ . (iv) x = {0}. (v) x ∈ ] − 3, √ 3[ . √ (vi) x ∈ [−5/ 3, 5/ 3]. (vii) x ∈ [−1, 3[ . √ √ (viii) x ∈ ] − 2 − 5, −2 + 5 [ . (ix) x ∈ ] − 1, 5[ . (x) x ∈ ] − 1/ max{a, b}, 1/ max{a, b} [ . √ 3.24. For the ﬁrst series we apply the root test. Then limn→∞ n | sinn x|/nα = | sin x|. Thus if | sin x| < 1, i.e., for all x ∈ R \ {2kπ ± π/2 | k ∈ Z}, the series converges ∑ absolutely. Suppose x is of the form 2kπ + π/2. Then the series becomes 1/nα , which converges if and only if α > 1. For of the form ∑ 2kπ − π/2, the series is (−1)n /nα , which converges for α > 0. Hence if α > 1, the domain of convergence is R; if 0 < α ≤ 1, the domain of convergence is R \ {2kπ + π/2 | k ∈ Z}; if α ≤ 0, the domain of convergence is R \ {2kπ ± π/2 | k ∈ Z}. For the second series we apply the ratio test. Then we get as limit of the ratio 2 in Theorem 3.18 the fraction |1 − x2 |/(1 + x For x ̸= 0, the series converges ∑). ∞ n absolutely. For x = 0, we get the series n=2 (−1) / ln n, which converges by Theorem 3.22 of Leibniz. Hence the domain of convergence is R. Try as well the Mathematica r command SumConvergence[(-1)^n/Log[n] ((1 - x^2)/(1 + x^2))^n, n]

For the third series we apply Theorem 3.22 of Leibniz for an arbitrary real x. Because √ √ x2 + 1/n2 ≤ |x| + 1/n and 0 ≤ x2 + 1/n2 − |x| ≤ 1/n,

3.1 Solutions

37

we conclude that the series converges on R. 3.25. The sequence (an (x)), where an (x) = 1/(x2 + n), has positive terms and converges monotonically to 0. Since |an (x) − 0| ≤ 1/n → 0 as n → ∞, we conclude that the sequence (an (x)) converges uniformly to 0 on R. By Theorem 4.3 we have that the series converges uniformly on R. The series ∑∞ 2 1/(x + n) diverges for each real x, since comparing with the harmonic n=1 series, the comparison test supplies the answer.

4 Limits and Continuity

4.1. Consider p > 0 since otherwise the necessary condition for convergence is not satisﬁed. Denote f (x) = x−p , x > 0. We have that ex · f (ex )/f (x) = xp /e(p−1)x . Thus for p > 1, limx→∞ ex · f (ex )/f (x) = 0 < 1, and we conclude the series converges. For 0 < p < 1, ex · f (ex )/f (x) ≥ 1, and we conclude the series diverges. The case p = 1 has been already discussed. 4.2. The function is continuous at each point in R2 \{(0, 0)} as composition of continuous functions. Consider an arbitrary small neighborhood of the origin. Then for x = ρ cos θ and y = ρ sin θ we have { ρ(cos3 θ + sin3 θ), ρ ̸= 0 f (x, y) = f (ρ cos θ, ρ sin θ) = 0, ρ = 0. Further |ρ(cos3 θ + sin3 θ)| ≤ 2ρ. Hence the function is continuous on the real plane. 4.3. Note the null function and the identity function are solutions to (4.54). Thus the set of solutions is nonempty. Note also that a solution necessarily is an odd function. We are looking for the whole set of solutions. Taking x = y = 0, we ﬁnd φ(0) = 2φ(0), i.e., φ(0) = 0. By induction one can show that φ(n) = nφ(1), n ∈ N∗ . Then from φ(1) = φ (m/m) = mφ (1/m) it follows φ (1/m) = 1/mφ(1) for m ∈ N∗ . By induction one can further show that φ(p/q) = (p/q)φ(1), p ∈ Z, q ∈ Z \ {0}. Now we know the representation of the solution if we restrict the exercise to the system of rational numbers. Consider an arbitrary but ﬁxed x ∈ R \ Q. Then there exists a sequence (qn ) of rational numbers such that qn → x. By the continuity hypothesis one can conclude that

40

4 Limits and Continuity

φ(x) = lim φ(qn ) = lim qn φ(1) = (lim qn )φ(1) = xφ(1). Denote a = φ(1). Then a solution of the Cauchy functional equation (4.54) is any function φ(x) = ax, for every x ∈ R. 4.4. The constant function φ(x) = −a, x ∈ R, is a solution. Set f (t) = φ(t) + a. Then we get the Cauchy functional equation for function f. Thus we conclude that φ(x) = cx − a, x ∈ R. 4.5. Remark that φ(x) = ln x, x > 0, is a solution to (4.56). Set u = ex , v = ey , f (t) = φ(et ), and substitute them in (4.56). Then we have that f is continuous and satisﬁes f (u + v) = f (x) + f (v), for u, v ∈ R. From Exercise (4.3) we get that f (u) = au, u ∈ R. Hence φ(x) = loga x, x > 0. 4.6. Note that function φ(x) = exp(x), x ∈ R, is a solution to (4.57). Take y = 0 in (4.57). It results that φ(x) = φ(x)φ(0) for x ∈ R. If φ(0) = 0, φ is the zero function, contrary to our assumption. So φ(0) ̸= 0, more precisely, φ(0) = 1. Take y = x in (4.57). Then φ(2x) = φ2 (x) ≥ 0, for all x ∈ R. So, φ is a non-negative function. Suppose that there is an x0 such that φ(x0 ) = 0. Then for any real x we have φ(x) = φ(x − x0 + x0 ) = φ(x − x0 )φ(x0 ) = 0, contradicting that φ is a non-zero function. Thus we get that φ is a positive function. We have the following equalities φ(n) = φ(1 + 1 + · · · + 1) = φn (1),

n ∈ N∗ ;

1 = φ(0) = φ(1)φ(−1); φ(−n) = φn (−1), n ∈ N∗ ; φ(1) = φ (1/n + · · · + 1/n) =⇒ φ (1/n) = (φ(1)) φ (±m/n) = (φ(1))

±m/n r

φ(r) = (φ(1)) ,

,

1/n

,

n ∈ N∗ ;

m, n ∈ N∗ ;

r ∈ Q.

From the last equality taking into account the continuity assumption we infer x that φ(x) = (φ(1)) , for any x ∈ R. If we denote φ(1) = a(> 0), we get that a solution to our exercise is a function of the form φ(x) = ax , x ∈ R. 4.7. Remark φ(x) = x, x ∈ R, is a solution to (4.58). Set x = eu , y = ev , and f (t) = φ(et ), t ∈ R. Then we get f (u + v) = f (u)f (v), u, v ∈ R. From Exercise (4.6) we obtain that f (t) = at , for any t ∈ R. Thus φ(x) = f (ln x) = aln x . If a = eα , φ(x) = xα , for all x > 0.

4.1 Solutions

41

4.8. Note the null function is a solution to (4.59). Consider the auxiliary function et − e−t f : R → R \ {−1, 1}, f (t) = t . e + e−t It easy to check that function f is one-to-one and onto. Then for any x, y ∈ R \ {−1, 1} there exist u, v ∈ R such that x = f (u) and y = f (v). Moreover, 1 + xy = 1 + f (u)f (v), with 1 + xy ̸= 0. Consider the function g : R → R deﬁned by g = φ ◦ f. Then (φ ◦ f )(u) + (φ ◦ f )(v) = (φ ◦ f )(u + v). Since this is precisely the functional equation, we write (φ ◦ f )(t) = ct, for a constant c. Now we conclude that φ(x) = cf −1 (x) =

c 1+x ln . 2 1−x

4.9. Suppose a function f : R → R satisﬁes the ﬁrst relation. Then f (xy + x + y) = f (xy) + f (x + y) = f (xy) + f (x) + f (y),

x, y ∈ R,

that is, f satisﬁes the second relation. Now suppose that a function f : R → R satisﬁes the second relation. Set y = u + v + uv. Then f (x + u + v + xu + xv + uv + xuv) = f (x) + f (u + v + uv) + f (xu + xv + xuv) or f (x + u + v + xu + xv + uv + xuv) = f (x) + f (u) + f (v) + f (uv) + f (xu + xv + xuv).

(4.1)

In (4.1) we interchange x by u and get f (x + u + v + xu + xv + uv + xuv) = f (x) + f (u) + f (v) + f (xv) + f (xu + uv + xuv).

(4.2)

From (4.1) and (4.2) we can write f (uv) + f (xu + xv + xuv) = f (xv) + f (xu + uv + xuv). In (4.3) we take x = 1 and it follows f (uv) + f (u + v + uv) = f (v) + f (u + 2uv) or f (uv) + f (u) + f (v) + f (uv) = f (v) + f (u + 2uv).

(4.3)

42

4 Limits and Continuity

From here f (u) + 2f (uv) = f (u + 2uv).

(4.4)

Setting u = 0 in (4.4), we get f (0) = 3f (0), so f (0) = 0.

(4.5)

Setting v = −1 in (4.4), we get f (−u) = f (u) + 2f (−u), and f (−u) = −f (u).

(4.6)

Setting v = −1/2 in (4.4), we get f (0) = f (u) + 2f (−u/2) . From (4.5) and (4.6), we have f (2u) = 2f (u).

(4.7)

From (4.7) and (4.4), we have f (u + 2uv) = f (u) + f (2uv), and by the substitution 2v = t, it follows f (u + ut) = f (u) + f (ut).

(4.8)

By (4.5), equality f (x + y) = f (x) + f (y) holds for x = 0. Suppose x ̸= 0. Then by (4.8), we have ( ( y) y) f (x + y) = f x + x · = f (x) + f x · = f (x) + f (y). x x 4.10. The null function is a solution to (4.60). Setting x = y = 0, yields f (0) = 0. For x = y, we obtain f 2 (2x) = 4f 2 (x) and then, since f (x) ≥ 0, f (2x) = 2f (x). One idea is to show that f (nx) = nf (x), for all n ∈ N∗ , and n ≥ 1. Assume that f (kx) = kf (x), for all k = 1, 2, . . . , n. Take y = nx in (4.62) to get f 2 ((n + 1)x) − f 2 ((n − 1)x) = 4f (nx)f (x), and f 2 ((n + 1)x) = ((n − 1)2 + 4n)f 2 (x). Hence f ((n + 1)x) = (n + 1)f (x) and the proof by induction is completed. We follow the way in the Cauchy functional equation, Exercise 4.3, to show that p, q are positive integers, then q · f (p/q) = f (p) = pf (1), so f (p/q) = (p/q)f (1) and thus f (r) = rf (1), for all r ∈ Q, r ≥ 0. Take x = 0 in (4.62) to get f 2 (y) − f 2 (−y) = 0, for all y ∈ R, that is f (y)−f (−y) = 0, for y ∈ R. Thus f is an even function. Then f (r) = |r|f (1), for r ∈ Q. Now we show that f (x) = |x| · f (1), for all x ∈ R. Set an arbitrary real x and set (rn ) a sequence of rational numbers such that limn→∞ rn = x. Since rn ∈ Q, for all n, we have f (rn ) = |rn |f (1), for all n ∈ N∗ . Function f is continuous, so we pass to the limit in the last equality to get f (x) = |x|f (1).

4.1 Solutions

43

Note that a = f (1) ≥ 0, therefore the set of solutions contains exactly the functions f (x) = a|x|, for some nonnegative a. 4.11. Deﬁne g(x) = f (x) − f (tx), for all x ∈ R. Remark that g is continuous at 0 and g(0) = 0. We have x2 = f (x)−f (tx)−(f (tx)−f (t2 x)) = g(x)−g(tx). From g(x) − g(tx) = x2 ,

g(tx) − g(t2 x) = t2 x2 ,

...,

g(tn−1 x) − g(tn x) = t2(n−1) x2 , we have g(x) − g(tn x) = x2 (1 + t2 + · · · + t2(n−1) ) = x2 (1 − t2n )(1 − t2 ). We pass to the limit as n → ∞ and since t ∈ ]0, 1[ , it follows g(x) − g(0) = x2 /(1 − t2 ) ⇐⇒ f (x) − f (tx) = x2 /(1 − t2 ). From x2 t2 x2 2 , f (tx) − f (t x) = , 1 − t2 1 − t2 t2(n−1) x2 f (tn−1 x) − f (tn x) = , 1 − t2 f (x) − f (tx) =

...,

we have f (x) − f (tn x) =

x2 x2 2 2(n−1) (1 + t + · · · + t ) = (1 − t2n ). 1 − t2 (1 − t2 )2

We pass to the limit as n → ∞ and since t ∈ ]0, 1[ , it follows f (x) − f (0) = x2 /(1 − t2 )2 , for every x ∈ R. So f (x) = k + x2 /(1 − t2 )2 ,

x ∈ R.

Conversely, each function of the form f (x) = k + x2 /(1 − t2 )2 , x ∈ R is a solution of the Exercise. 4.12. Exists m > 0 such that |f (x)| ≤ m, for any x ∈ R. Consider the auxiliary function g(x) = f (x) − x, for x ∈ R. Then g is continuous. From g(m) = f (m) − m ≤ 0, g(−m) = f (−m) + m ≥ 0, and the continuity of g, it follows that there exists a ξ ∈ R such that 0 = g(ξ) = f (ξ) − ξ. 4.13. The ”only if” part is clear. Now suppose that (4.61) is true. Then we have a sequence in a compact interval. It has at least a limit point. Suppose that the sequence (xn ) does not converge. Then it has at least two limit points. Choose p, q two limit points with p < q. There must be points from the interval ]p, q[ in the sequence. There is an x ∈ ]p, q[ such that f (x) ̸= x. Set ε = |f (x) − x|/2. Then from the continuity of f we get that for some

44

4 Limits and Continuity

δ > 0 for all y ∈ ]x − δ, x + δ [ , one has |f (y) − y| > ε. On the other hand for n large enough |xn+1 − xn | < 2δ and |f (xn ) − xn | = |xn+1 − xn | < ε. So the sequence cannot come into the interval ]x − δ, x + δ[, but also cannot jump over this interval. Then all limit points have to be at most x−δ (contradicting that q is a limit point), or at least x+δ (contradicting that p is a limit point). 4.14. Consider g(x) = x3 +x, for all x ∈ R. Function g is a third degree polynomial, so it is continuous and onto. Moreover, it is strictly increasing. Then the its inverse g −1 is strictly increasing and continuous on R, by Theorem 2.20. Suppose that f is a solution to the exercise. Then f (g(x)) ≤ x ≤ g(f (x)), for all x ∈ R. Applying g −1 to the right hand side inequality, we have g −1 (x) ≤ f (x), for all x ∈ R.

(4.9)

Take y = g(x). Then from the left hand side inequality, we have f (y) ≤ g −1 (y), for all y ∈ R.

(4.10)

From (4.9) and (4.10) we conclude that f (x) = g −1 (x), x ∈ R. Easy to see that f satisﬁes the assumption. For, take x ∈ R, y ∈ f (x), f −1 (y) = x, and g −1 (x) = y. Then g(y) = x and f (g(x)) ≤ x ≤ g(f (x)) ⇐⇒ x ≤ x ≤ x, for all x ∈ R. 4.15. Since f is continuous, f ([a, b]) is a compact set. Thus the sequence (xn ) is included in a compact set and thus it is bounded. Function f being increasing, the sequence (xn ) is monotone. So we have established that the sequence (xn ) converges, let x∗ ∈ [a, b] be its limit. From the estimates 0 ≤ |f (xn ) − x∗ | = |xn+1 − x∗ | → 0 as n → ∞, we conclude that f (x∗ ) = x∗ . 4.16. (a) Yes. Indeed, let A = {x ∈ [0, 1] | f (x) > x}. If f (0) = 0, then x = 0 supplies the answer. If not, A in nonempty ( 0 ∈ A ) and bounded, so it has a supremum, say a. Then b = f (a). There are two cases. I case: a < b. Then, since f is monotonic and a is a supremum, we get b = f (a) ≤ f ((a + b)/2) ≤ (a + b)/2, which contradicts a < b. II case: b < a. Then we get b = f (a) ≥ f ((a + b)2) > (a + b)2, a contradiction. Therefore we must have a = b. (b) No. Here is an example

4.1 Solutions

45

{ 1 − x/2, x ≤ 1/2; f (x) = 1/2 − x/2, x > 1/2. 4.17. Let ]x − α, x + α ] ⊂ [0, 1] be an arbitrary nonempty open interval. The function is not monotone in the intervals [x − α, x] and [x, x + α], thus there exist some real numbers x − α ≤ p < q ≤ x and x ≤ r < s ≤ x + α so that f (p) > f (q) and f (r) < f (s). By Theorem 2.10, f has a global minimum in the interval [p, s]. The values f (p) and f (q) are not the minimum, because they are greater than f (q) and f (s), respectively. Thus the minimum is in the interior of the interval, i.e., it is a local minimum. So each nonempty open interval ]x − α, x + α ] ⊂ [0, 1] contains at least one local minimum. 4.18. Suppose that at least one such function exists. Then it is one-to-one since f (x) = f (y) =⇒ f (f (x)) = f (f (y)) =⇒ −x = −y =⇒ x = y. Since f is one-to-one and continuous, it is strictly monotonic. Then f ◦ f is strictly increasing. But this is impossible since (f ◦ f )(x) = −x, x ∈ R. Thus, there is no continuous function satisfying (4.62). 4.19. Note that we have a solution, namely the function f (x) = x, x ∈ R. We show that there is no other solution. Suppose there is a solution g : R → R and a number a ∈ R such that g(a) ̸= a. If g(a) < a, then since g is strictly increasing, g(g(a)) < g(a). By assumption g(g(a)) = a, so we get a < g(a), contradiction. If g(a) > a, then since g is strictly increasing, g(g(a)) > g(a). By assumption g(g(a)) = a, so we get a > g(a), contradiction. Now we conclude the only one solution is f (x) = x, x ∈ R. 4.20. Suppose function f is a solution. Since f is strictly increasing, 0 ≤ f (0) < f (1) < f (2) = 2, from where f (0) = 0 and f (1) = 1. Similarly we have f (4) = f (2 · 2) = f (2) · f (2) = 4, 2 < 3 < 4 =⇒ 2 = f (2) < f (3) < f (4) = 4 =⇒ f (3) = 3, f (6) = f (2 · 3) = f (2) · f (3) = 6, 4 < 5 < 6 =⇒ 4 = f (4) < f (5) < f (6) = 6 =⇒ f (5) = 5. We prove by induction that f (n) = n, for all n ∈ N∗ .

46

4 Limits and Continuity

Let p1 < p2 < . . . be the set of all primes. It is obvious that f (2k ) = 2k , for all k ∈ N∗ , f (3k ) = 3k , for all k ∈ N∗ , and f (2k 3m ) = 2k 3m , k, mN∗ . mn−1 mn−1 m1 m2 1 m2 Suppose that f (pm 1 p2 . . . pn−1 ) = p1 p2 . . . pn−1 , for all m1 , m2 , ∗ . . . , mn−1 ∈ N , n ≥ 4. Then pn − 1 are pn + 1 are composed. Their prime decomposition contain primes from {p1 , p2 , . . . , pn−1 } at nonnegative powers. Then pn − 1 < pn < pn + 1 implies pn − 1 = f (pn − 1) < f (pn ) < f (pn + 1) = pn +1. From here one has f (pn ) = pn . Immediately one can see that f (n) = n for all natural n such that their prime decomposition contain primes from {p1 , p2 , . . . , pn−1 , pn } at nonnegative powers. Thus f (n) = n, for all natural n. 4.21. Take an arbitrary but ﬁxed x ∈ R \ Q and two rational sequences (un ) and (vn ) such that un → x, vn → x, and un < x < vn , for all n ∈ N∗ . We show that f (x) = g(x). Suppose that function g is increasing. Then g(un ) ≤ g(x) ≤ g(vn ), f (un ) = g(un ) and f (vn ) = g(vn ). Thus f (un ) ≤ g(x) ≤ f (vn ).

(4.11)

Passing to the limit in (4.11) we get that f (x) = g(x). 4.22. From Exercise 4.3 we have that f (p) = p · f (1), for all p ∈ Q. Consider x an arbitrary irrational number, two rational sequences (an ) and (bn ) converging to x and satisfying an < x and x < bn , for all n. Then f (1)an = f (an ) ≤ f (x) ≤ f (bn ) = f (1)bn . Passing to the limit, we get f (1)x = f (x) ≤ f (x) ≤ f (x) = f (1)x. Thus f (x) = f (1)x, x ∈ R. So, f (x) = ax, for all x ∈ R. 4.23. From |f (x) − f (y)| < |x − y| for all x, y ∈ [a, b], we have that f is continuous on [a, b]. Inspired by recurrence we deﬁne g(x) = (x + f (x))/2, x ∈ [a, b]. It is easy to see that g([a, b]) ⊂ [a, b]. Moreover, g is increasing. So the sequence (an ) is bounded and monotonic. Therefore it is convergent. Let a0 be its limit. Since f is continuous we pass to the limit in an+1 = (an + f (an ))/2 and ﬁnd that f (a0 ) = a0 . 4.24. We may assume that a ≥ b ≥ c. Then a − c ≥ b − c ≥ 0, a − b ≥ 0 and f (a) ≥ f (b) =⇒ f (a)(a − b)(a − c) ≥ f (b)(b − c)(a − b) =⇒ f (a)(a − b)(a − c) + f (b)(b − a)(b − c) ≥ 0. Also, f (c)(c−a)(c−b) ≥ 0. Summing up these two inequalities, we get (4.65). 4.25. First approach. We introduce the function

4.1 Solutions

47

f (y) = ⌊y⌋ + ⌊y + 1/n⌋ + · · · + ⌊y + n − 1/n⌋ − ⌊ny⌋ . Function f is deﬁned on R. Moreover, it is periodic, since f (y) = f (y + 1/n) for all y ∈ R. Then it is enough to consider the case x = k + α, where k is the integer part of x, while 0 ≤ α < 1/n. Thus f (x) = 0 and equality (4.66) is established. Second approach. Take y = n in Exercise 1.19. 4.26. There is a positive p such that f (x + p) = f (x), for all x ∈ R. Then on [0, p] function f is bounded by a constant, let it be, m ≥ 0. It follows that function f is bounded on R by the same constant m ≥ 0. Now we recall Exercise 4.12 and thus this exercise is solved. 4.27. By induction we get that f (x) = f (x + m/n), for all x ∈ Q and m, n ∈ N∗ . Taking successively x = 0 and x = −m/n, it follows that f (±m/n) = f (0), for all m, n ∈ N∗ . It means that f (x) = f (0), for all x ∈ Q. It is enough to consider an arbitrary real x0 and a sequence of rationals converging to it. 4.28. (a) First examine the case a = 0. In this case (4.67) √ is a second degree equation in f (x), having two real roots, namely (2 ± 2)/4. Only one root √ is acceptable, that is (2 + 2)/4. For the other we have a negative √ number under square root. So, we get the constant function f (x) = (2 + 2)/4, for all x ∈ R. This function is obviously periodic (of any period) as a constant function. Now assume that a ̸= 0. Suppose there exists f a solution. Note that 1/2 ≤ f (x) ≤ 1, for all x ∈ R. Deﬁne g(x) = f (x) − 1/2, for all x ∈ R. Then √ √ g(x + a) = 1/4 − g 2 (x) and g(x + 2a) = 1/4 − g 2 (x + a) = g(x). So, f (x + 2a) = f (x), for all x ∈ R. (b) We supply three examples f (x) = (1/2) (|sin πx/2| + 1) , x ∈ R; { 1/2, 2n ≤ x < 2n + 1, f (x) = n ∈ Z, x ∈ R; 1, 2n + 1 ≤ x < 2n + 2

x 1 2n ≤ x ≤ 2n + 1, + − n, 2 f (x) = 12 √ n ∈ Z, ( ) 2 + x − n − x − n , 2n + 1 ≤ x ≤ 2n + 2 2 2 2

x ∈ R.

4.29. First approach. The set J = f (I) is an interval and f is strictly monotonic, by Corollary 4.9. Suppose f is strictly increasing. Take u, v ∈ J arbitrary, but ﬁxed, such that u < v; Then let x = f −1 (u) and y = f −1 (v) be

48

4 Limits and Continuity

their counter images and let λ be a real between them. Since f −1 is strictly monotonic, we have x < λ < y. Then z = f (λ) ∈ ] u, v [ and thus f −1 (z) = λ. Second approach. By Corollary 4.9, f is strictly monotonic, while by Theorem 2.20, the conclusion follows. 4.30. From (4.30) we have (xn − x1 )f (xk ) ≤ (xk − x1 )f (xn ) + (xn − xk )f (x1 ),

k = 1, 2, . . . , n.

From here for k = 1, . . . , n − 1, we have (xk+1 − xk )(f (xk ) − f (xk+1 )) ≤

(x2k+1

−

x2k )(f (xn )

− f (x1 )) + 2(xn f (x1 ) − x1 f (xn ))(xn − x1 ) . xn − x1

Let A be the right-hand side of the inequality in the statement of Exercise 4.30. Then adding side by side the previous inequality we get A ≤ (xn − x1 )(f (x1 ) + f (xn )). 4.31. For k = 0 function f has a discontinuity of the second kind at 0. By Corollary 7.22, f is not of ﬁnite variation. For k = 1, similar to Example 7.1, consider the partition P = ∑ {0, 1/(nπ+π/2), 1/(nπ), . . . , 1/(π+π/2), 1/π, 2/π} n and note that V (f, P ) = 2 m=1 1/(mπ +π/2)+2/π. This sum tend to ∞ as n → ∞. So the function is not of ﬁnite variation. For k ≥ 2, f has a bounded derivative on the whole interval, thus is Lipschitz. By (b) in Remarks 7.1, f is of bounded variation. 4.32. The partial sum is sn = nx/(1 + n2 x2 ). Since for each x ∈ [0, 1] limn→∞ sn (x) = 0, the series converges pointwise to the null function (which is continuous). Since for xn = 1/n we have |sn (xn ) − 0| = 1/2, the series converges non uniformly. 4.33. It is clear that each term of the series is a discontinuous function. The partial sum is sn = f /(n + 1). Since |sn (x) − 0| ≤ |f (x)|/(n + 1), for each x ∈ R, we conclude that the series converges uniformly to the null function.

5 Diﬀerential Calculus on R

5.1. We write dn as a convex combination of the form f (βn ) − f (αn ) f (βn ) − f (0) βn f (0) − f (αn ) −αn = · + · , βn − αn βn βn − αn −αn βn − αn βn −αn λ1 = , λ2 = , λ1 , λ2 > 0, λ1 + λ2 = 1. βn − αn βn − αn It follows

{

} f (βn ) − f (0) f (0) − f (αn ) f (βn ) − f (αn ) min , ≤ βn −αn βn − αn { } f (βn ) − f (0) f (0) − f (αn ) ≤ max , . βn −αn

The assumptions guarantee that lim (f (βn ) − f (0))/βn = lim (f (0) − f (αn ))/(−αn ) = f ′ (0).

n→∞

n→∞

Now invoking Theorem 1.18 from page 81, we get the conclusion. (b) Consider the identity ( ) f (βn ) − f (αn ) f (0) − f (αn ) βn f (βn ) − f (0) f (0) − f (αn ) − = − . βn − αn −αn βn − αn βn −αn The sequence (βn /(βn − αn ))n is bounded and since the diﬀerence in the parenthesis tends to 0, the conclusion follows. (c) Invoking the mean value Theorem 2.3, we write dn = (f (βn ) − f (αn ))/(βn − αn ) = f ′ (γn ),

50

5 Diﬀerential Calculus on R

for a sequence (γn ) satisfying αn < γn < βn . Then γn → 0 and the conclusion follows since f ′ is continuous. (d) Consider the function given in (b) of Examples 1.1 and the sequences (αn ) and (βn ) deﬁned as βn = 2/((4n + 1)π), respectively αn = 1/(2nπ). 5.2. Suppose there exists a point u ∈ ] a, b[ so that g(u) = 0. Since g is nonconstant, there exists v ∈ ]a, b[ so that g(v) ̸= 0. We may suppose that u < v. Deﬁne w = sup{x ∈ [u, v] | g(x) = 0}. Since g is continuous on ]a, b[ , g(w) = 0 and g(x) ̸= 0 for every x ∈ ] w, v [ . On the interval ]w, v] deﬁne h = f /g. Then h′ (x) = 0 for all x ∈ ]w, v [ . By (b) of Theorem 2.4 and continuity of g we have that h(x) = f (v)/g(v) and f (x) = g(x)f (v)/g(v) for all x ∈ ] w, v] and f (w) = lim f (x) = lim x↓w

x↓w

f (v) g(x) = 0. g(v)

Thus we get that f (w) + g(w) = 0, a contradiction. Therefore g has no zero on ]a, b[ and h is deﬁned and constant on ]a, b[ . 5.3. The inequality is equivalent with ln a/a ≥ ln x/x. So, we are led to consider the function f (x) = ln x/x, for x > 0. This function has a unique maximum at x = e. Thus the answer follows for a = e. 5.4. Deﬁne g(x) = f (x) · e−x on [0, 1]. Since g(0) = g(1), we ﬁnd a point ξ ∈ ]0, 1[ so that g ′ (ξ) = 0. Then 0 = g ′ (ξ) = (f ′ (ξ) − f (ξ))e−ξ , that is f (ξ) = f ′ (ξ). 5.5. It is clear that all the terms are positive if x1 is so. It is also clear that if x1 = 0, all the terms are null. Deﬁne the sequence of polynomials P0 (x) = x,

Pn (x) = Pn−1 (x) (Pn−1 + 1/n) ,

n = 1, 2, . . . .

We have xn = Pn−1 (x1 ). Also Pn (0) = 0, Pn (1) > 1 for n = 1, 2, 3, . . . . Since Pn (x) has nonnegative coeﬃcients, it is strictly increasing in the interval I = [0, 1]. Hence one ﬁnds (unique) solutions an and bn to Pn (an ) = 1 − 1/n,

Pn (bn ) = 1,

n = 1, 2, 3, . . . .

Since Pn+1 (an ) = Pn (an ) (Pn (an ) + 1/n) = 1 − 1/n < 1 − 1/(n + 1), we have an < an+1 . Similarly Pn+1 (bn ) = Pn (bn ) (Pn (bn ) + 1/n) = 1 + 1/n > 1 + 1/(n + 1),

5.1 Solutions

51

so bn > bn+1 . Thus (an ) is increasing, (bn ) is decreasing, and an < bn , for all n > 1. Then there exists at least one point x1 satisfying an < x1 < bn , for all n > 1. Hence 1 − 1/n < Pn (x1 ) < 1, for all n > 1. But Pn (x1 ) = xn+1 . So xn+1 < 1, for all n > 1. Also xn > 1 − 1/n implies xn+1 = xn (xn + 1/n) > xn . So, sequence (xn ) satisﬁes all the required conditions. It remained to consider uniqueness. Pick an x1 satisfying the given conditions. Then 1 − 1/n < Pn (x1 ) < 1, for all n > 1. So we have for all n > 1 that an < x1 < bn . The uniqueness of x1 will follow by showing that bn − an → 0, as n → ∞. Since all the coeﬃcients of Pn (x) are nonnegative, it has increasing derivative. But Pn (0) = 0 and Pn (bn ) = 1, so for any x ∈ [0, bn ] we have Pn (x) ≤ x/bn . Then we have 1 − 1/n < an /bn and n→∞

0 < bn − an ≤ bn − bn (1 − 1/n) = bn /n < 1/n −−−−→ 0.

5.6. The inequality has been already proved geometrically by Exercise 1.51 at page 39. Here we prove it using Theorem 2.4. We only show that sin x < x, for every x ∈ ]0, π/2[ . Consider the function f deﬁned by f (x) = x−sin x on x ∈ [0, π/2[ . Then f (0) = 0 and f ′ (x) = 1−cos x > 0, for every x ∈ ]0, π/2[ . Hence sin x < x on 0 < x < π/2. 5.7. (a) One might apply directly Theorem 4.1. Instead of it we choose writing the limit as ( ( 2 )) xln x ln x lim = exp lim x − ln ln x . x→∞ (ln x)x x→∞ x By Theorem 4.1 we have lim ln2 x/x = lim 2 ln x/x = lim 2/x = 0.

x→∞

(

x→∞

x→∞

)

Therefore limx→∞ ln2 x/x − ln ln x = −∞ and so the limit is equal to 0. We can try ﬁnding the answer by the Mathematica r commands. f[x_] := x^(Log[x])/(Log[x])^x Limit[f[x], x -> Infinity]

(b) We write ( ) ln x lim xx = exp(lim x ln x) = exp lim . x↓0 x↓0 x↓0 1/x By Theorem 4.1 we further have that the previous limit is equal to 0. We conclude that the limit is equal to 1. We can try ﬁnding the answer by the Mathematica r command. Limit[x^x, x -> 0]

52

5 Diﬀerential Calculus on R

(c) Since ( ) lim ln(ex + x)/x = lim 1 + ln(1 + xe−x )/x = 1

x→∞

x→∞

we successively have (√ ) √ ln(ex + x) 3 lim x3 + x2 + x + 1 − x2 + x + 1 · x→∞ x (√ ) √ 3 = lim x3 + x2 + x + 1 − x2 + x + 1 x→∞ (√ ) √ 1 1 1 1 1 3 = lim x 1+ + 2 + 3 − 1+ + 2 x→∞ x x x x x (1 + t + t2 + t3 )1/3 − (1 + t + t2 )1/2 t↓0 t ( ) 2 1 + 2t + 3t 1 + 2t 1 √ = lim − √ =− . t↓0 6 2 1 + t + t2 3 3 (1 + t + t2 + t3 )2 = lim

5.8. A more general result will be proved, namely f (t) ≥ 0, for all t ∈ ]0, π]. Suppose f (x) ≤ 0 for all x ∈ [0, t]. We show that in this case f (x) = 0 for all x ∈ [0, t]. Indeed, if f (u) < 0 for some u ∈ ]0, t [ , by the mean value Theorem of Lagrange there exists v ∈ ]0, u[ so that f ′ (v) = (f (u) − f (0))/u < 0. We apply the mean value Theorem of Lagrange to the function af ′ to get (a · f ′ )′ (w) = (a(v)f ′ (v) − a(0)f ′ (0))/v < 0 for some w ∈ ]0, v [ . In this cases (a(w)f ′ (w))′ + f (w) < 0, a contradiction. Suppose f (p) > 0 for some p ∈ ]0, t [ . Deﬁne ξ1 = sup{x | x ∈ [0, p], f (x) ≤ 0},

ξ2 = inf{x | x ∈ [p, t], f (x) ≤ 0}.

We have 0 ≤ ξ1 < ξ2 < t, f (ξ1 ) = f (ξ2 ) = 0, and f (x) > 0, for all x ∈ ]ξ1 , ξ2 [ . On the interval ] ξ1 , ξ2 [ deﬁne g(x) = a(x)f ′ (x)/f (x). Then g ′ (x) =

(a(x)f ′ (x))′ g 2 (x) − ≥ 1 − g 2 (x). f (x) a(x)

(5.1)

We note that function g has arbitrary large values in a neighborhood of ξ1 and arbitrary small values in a neighborhood of ξ2 . Indeed, let ε > 0. Let η be the point of maximum to f on [ξ1 , ξ1 + ε]. By the mean value Theorem of Lagrange we ﬁnd a point θ ∈ ]ξ1 , η [ fulﬁlling

5.1 Solutions

f ′ (θ) =

f (η) − f (ξ1 ) f (η) f (η) = ≥ . η − ξ1 η − ξ1 ε

g(θ) =

a(θ)f ′ (θ) f ′ (θ) f ′ (θ) 1 ≥ ≥ ≥ . f (θ) f (θ) f (η) ε

Then

53

Similarly, one can show that on [ξ2 − ε, ξ2 ], function g is less than −1/ε. From (5.1) we have that g ′ (x) ≥ −1 − g 2 (x) =⇒ (arctan g(x))′ =

g ′ (x) ≥ −1 1 + g 2 (x)

(5.2)

for x ∈ ]ξ1 , ξ2 [ . One the other side, for arbitrary positive δ there are θ1 , θ2 ∈ ] ξ1 , ξ2 [ , with ξ1 < ξ2 , so that arctan g(ξ1 ) >

π −δ 2

and

arctan g(ξ2 ) > −

π + δ. 2

By the mean value Theorem of Lagrange we ﬁnd a point ξ ∈ ]θ1 , θ2 [ fulﬁlling (arctan g)′ (ξ) =

(arctan g)(θ2 ) − (arctan g)(θ1 ) −π + 2δ < . θ2 − θ1 ξ2 − ξ1

For suﬃciently small δ the last fraction is less than −1, contradicting (5.2). This contradiction shows that the assumption that f (p) > 0 for some p ∈ ]0, t [ is false. Now the conclusion follows. 5.9. From the given equality we may write that for y ̸= 0 we have f ′ (x) =

f (x + y) − f (x − y) , 2y

x ∈ R.

The right hand side is diﬀerentiable at x, so ( ) f ′ (x + y) − f ′ (x − y) 1 f (x + 2y) − f (x) f (x − 2y) − f (x) ′′ f (x) = = − 2y 2y 2y −2y f (x + 2y) + f (x − 2y) − 2f (x) = . 4y 2 Diﬀerentiating again we ﬁnd f ′ (x + 2y) + f ′ (x − 2y) − 2f ′ (x) 1 = 2· 2 4y 4y ( ) f (x + 4y) − f (x) f (x − 4y) − f (x) f (x + 4y) − f (x − 4y) + − = 0. 4y (−4y) 4y f ′′′ (x) =

Thus f ′′′ (x) = 0, for all x ∈ R. Then

54

5 Diﬀerential Calculus on R

f ′′ (x) = f ′′ (0),

f ′ (x) = f ′′ (0)x + f ′ (0),

f (x) = f ′′ (0)

x2 + f ′ (0)x + f (0). 2

We note that each second degree polynomial f (x) = ax2 + bx + c, x ∈ R, satisﬁes all the requirements of the exercise. ⊓ ⊔ 5.10. First approach. Setting x = y = 0, we ﬁnd that f (0)(f (0) − 1) = 0. So, either f (0) = 0 or f (0) = 1. Assume f (0) = 0 and take into account the identity f (0)f (x) = f (x), for all x ∈ R. Then f (x) = 0, for all real x, that is, f is the null function, contrary to our assumption. We suppose that f (0) = 1. We show that f is diﬀerentiable at every point on the real axis. Let x be an arbitrary point of the real axis. From f (x + y) = f (x)f (y) we have lim (f (x + y) − f (x))/y = f (x) lim (f (y) − f (0))/y.

y→0

y→0

The limit in the left hand side there exists and we have f ′ (x) = f (x)f ′ (0), for all x ∈ R. Set f ′ (0) = a. Then f ′ (x) = a · f (x), for all x ∈ R. Thus f is indeﬁnite diﬀerentiable at each x ∈ R and f (n) (x) = an f (x), for all x ∈ R and n ∈ N∗ . Second approach with an extra assumption of continuity. By Exercise 4.6, under the assumption of continuity we have that f (x) = ax , a > 0. This function is diﬀerentiable on R. In this case we need not the assumption that f is diﬀerentiable at x = 0. 5.11. Consider g(x) = f (x) − x2 /2. Obviously, g is diﬀerentiable and g ′ (0) = 1, g ′ (1) = −1. Then g attains its maximum on a point in int [0, 1]. Let ξ be this point of maximum, that is, g(ξ) = maxx∈[0,1] g(x), ξ ∈ ]0, 1[ . Then g ′ (ξ) = 0, that is f ′ (ξ) = ξ. 5.12. Set

g(x) = [f (x) + f ′ (x) + · · · + f (n) (x)]e−x .

From the assumption one gets g(a) = g(b). Then there exists c ∈ ]a, b [ such that g ′ (c) = 0. Substituting in the last equality g ′ (x) = (f (n+1) (x)−f (x))e−x , we ﬁnish the proof. 5.13. Consider the convex function [0, ∞ [ ∋ x 7→ f (x) = xm . 5.14. We prove the ﬁrst inequality. We have that A, B, C ∈ ] 0, π [ and A + B + C = π. Because the sine function is concave on ] 0, π [ , we may write

5.1 Solutions

55

√ 3 sin A + sin B + sin C A+B+C ≤ sin = . 3 3 2 5.15. Since |e−n sin(nx)| ≤ e−n for every n ∈ N and x ∈ R, the series converges uniformly and absolutely on R. Its sum, denoted by f, is deﬁned ∑∞ on the whole real axis. The series n=1 ne−n cos(nx) of derivatives converges uniformly and absolutely on R. Pick up an arbitrary x0 ∈ R and an interval [a, b] so that x0 ∈ [a, b]. Then by Theorem 7.3, f is diﬀerentiable and its derivative is continuous as a uniform convergent series of continuous functions. Since x0 is arbitrary on R, the conclusion follows. 5.16. First approach. Function f may be written as { { sin x/x, x ̸= 0, 0, x ̸= 0, f (x) = f1 (x) = + f2 (x) = 1 x=0 −1 x = 0. Function f1 is continuous on R, so it has a primitive. Function f2 is not a Darboux function, so it has no primitive. Hence f has no primitive. Second approach. We note that function f is discontinuous at x = 0, since limx→0 f (x) = 1 ̸= 0 = f (0). More precisely, x = 0 is a point of discontinuity of the ﬁrst kind. Based on Corollary 10.11, function f has no primitive. Third approach. We show that f is not a Darboux function. Because limx→0 f (x) = 1, for all V ∈ V(1) exists ] − δ, δ [ ∈ V(0) such that for all x ∈ ]0, δ [ , f (x) ∈ V. Choose V = ]1/2, 3/2 [ . So f (x) > 1/2 for every x ∈ ]0, δ [ . Then we conclude that f ( ] − δ, δ [ ) = {1/2 < f (x) < 3/2 | x ∈ ]0, δ [ } ∪ {0}, which is not an interval. Hence the function f has no primitive on R.

6 Integral Calculus on R

6.1. We can write n ∑ k=1

∑ ∑ n 1 1 1 = = 2 2 2 n +k 1 + (k/n) n 1 + (k/n)2 n

n

k=1

k=1

(

k k−1 − n n

) .

Consider the function f (x) = 1/(1 + x2 ) on x ∈ [0, 1] and the partition 0 = x0 < x1 < · · · < xn = 1, where xk = k/n, k = 0, 1, 2, . . . , n. Then based on Theorem 1.4, we have lim

n→∞

n ∑ k=1

n = n2 + k 2

∫ 0

1

dx π = arctan 1 − arctan 0 = . 1 + x2 4

(6.1)

We can approach this exercise by the Mathematica r commands Limit[Sum[n/(n^2 + i^2), {i, 1, n}], n -> Infinity] TraditionalForm[%] (*getting*) 1/2 (\[Pi]+I (log(1+I)-log(1-I))). With some elementary background on complex functions we write (1/2)(π + ı ln((1 + ı)/(1 − ı))) = (1/2)(π − π/2) = π/4. This result agrees with (6.1). 6.2. First we note that arctan(b/a)+arctan(a/b) = sign (ab)π/2. Successively we have ∫ π/4 ∫ π/2 dx dx I(a, b) = + 2 2 2 2 a2 cos2 x + b2 sin2 x 0 π/4 a cos x + b sin x

6 Integral Calculus on R

58

∫ = 0

1

b/a a/b ∫ 1 du du 1 1 + = arctan t arctan t + 2 + a2 u2 a2 + b2 u2 b ab ab 0 0 0 ( ) 1 b a π 1 arctan + arctan = . = ab a b 2 |ab|

We can approach this exercise by the Mathematica r commands Integrate[1/(a^2 (Cos[x])^2 + b^2 (Sin[x])^2), {x, 0, Pi/2}] TraditionalForm[%] (*getting*) (\[Pi] Sqrt[b^2/a^2])/(2 b^2) 6.3. One has ∫ 1√ ∫ f (x) dx = 0

1 0

f (x) √ dx ≥ f (x)

∫

1

0

f (x) √ dx = a2/3 . a2/3

6.4. Since all ak > 0, we denote gk (x) = fk (x)/ak . Then for all k, ∫1 g (x) dx = 1. By the arithmetic-geometric inequality (Exercise 1.28 at 0 k page 35 of the book), we have ∫

1

∫ √ n g1 (x)g2 (x) · · · gn (x) dx ≤

1 0

0

g1 (x) + g2 (x) + · · · + gn (x) dx = 1. n

√ Then there exists ξ ∈ [0, 1] so that g1 (ξ)g2 (ξ) · · · gn (ξ) ≤ 1, that is g1 (ξ)g2 (ξ) · · · gn (ξ) ≤ 1. The last inequality is equivalent to f1 (ξ)f2 (ξ) · · · fn (ξ) ≤ a1 a2 · · · an . 6.5. For 0 ≤ x < 1, the sequence (xn )n tends monotonically to zero as n → ∞. Therefore ∫ 1 lim cos(xn ) dx = 1. n→∞

0

∫

We evaluate

π

cos(xn ) dx.

lim

n→∞

1

Substitute xn = y. Then ∫

π 1

∫ n 1 π d sin y cos y 1−1/n = n 1 y 1−1/n y 1 πn ∫ πn sin y n−1 sin y dy = + . 2 1−1/n n ny y 2−1/n 1 1

1 cos(x ) dx = n n

∫

πn

dy

6.1 Solutions

Obviously,

59

πn n→∞ sin y −−−−→ 0. 1−1/n ny 1

For n → ∞, we have n − 1 ∫ πn sin y dy n − 1 ∫ πn dy n→∞ 2(n − 1) = (1 − π −n/2 ) −−−−→ 0. 2 ≤ 2 2 2−1/n 3/2 n n n y y 1 1 ∫

Thus

π

cos(xn ) dx = 1.

lim

n→∞

0

6.6. Consider an integer n large enough, at least n > T. Function g(nx) is of period Tn = T /n. Consider the partition of [0, 1] by {x0 = 0, x1 = Tn , x2 = 2Tn , . . . , xm = mTn }, where m is the largest integer satisfying mTn ≤ 1. We write ∫

1

f (x)g(nx) dx = 0

m−1 ∑ ∫ xk+1

∫

xk

k=0

1

f (x)g(nx) dx +

f (x)g(nx) dx. xm

First suppose that g takes only positive values. By (h) in Theorem 1.7, we write for some ξk ∈ [xk , xk+1 ], ∫ xk+1 ∫ xk+1 f (x)g(nx) dx = f (ξk ) g(nx) dx. xk

Further ∫ xk+1 xk

1 g(nx) dx = n

xk

∫

∫ ∫ 1 (k+1)T 1 T g(t) dt = g(t) dt = g(t) dt, n kT n 0 n→∞ f (x)g(nx) dx −−−−→ 0.

nxk+1 nxk 1

∫

xm

Thus ∫

1

f (x)g(nx) dx = 0

m−1 ∑

f (ξk )

k=0

1 = T

∫

1 n

∫

g(t) dt + o(1) 0

T

g(t) dt · 0

n→∞

−−−−→

1 T

T

m−1 ∑

f (ξk )(xk+1 − xk ) + o(1)

k=0

∫

T

∫

g(t) dt · 0

1

f (x) dx. 0

6 Integral Calculus on R

60

Now consider g is not positive. Then there exists a positive constant c such that g + c is positive. By the earlier arguments one can write ∫

1

1 T

n→∞

f (x)(g(nx) + c) dx −−−−→

0

We subtract from the both sides c ·

∫1 0

∫

∫

T

1

(g(t) + c) dt · 0

f (x) dx. 0

f (x) dx and get the result.

6.7. Since |f ′ (x)| ≤ 1 on [a, b], we have |f (x) − f (ξk )| ≤ |x − ξk |. Therefore ∫ n ∫ xk n ∑ b ∑ f (x) dx − f (ξk )(xk − xk−1 ) ≤ |f (x) − f (ξk )| dx a k=1 xk−1 k=1 n ∫ xk ∑ ≤ |x − ξk | dx. k=1

xk−1

Since ∫

xk

|x − ξk | dx =

xk−1

(ξk − xk−1 )2 (xk − xk−1 )2 (xk − ξk )2 + ≤ , 2 2 2

we have ∫ n n b ∑ ∑ (xk − xk−1 )2 f (x) dx − f (ξk )(xk − xk−1 ) ≤ a 2 k=1

k=1

1 ∑ xk − xk−1 1 ≤ = . b−a 2 2 n

k=1

The estimate is sharp since for f (x) = x we get equality. 6.8. Consider n a positive integer large enough, a partition a = x0 < · · · < xn = b, satisfying xk − xk−1 = xk−1 − xk−2 , for k = 2, 3, . . . , n, and Mk =

sup

f (x),

mk =

x∈[xk−1 ,xk ]

inf

f (x), for all k = 1, 2, . . . , n.

x∈[xk−1 ,xk ]

Then by Theorem 1.2 one has that n ∑

lim

n→∞

(Mk − mk )(xk − xk−1 ) = 0.

k=1

Thus for n large enough we have n ∑

(Mk − mk )(xk − xk−1 ) < ε/2 .

k=1

(6.2)

6.1 Solutions

{ mk , h(x) = f (b),

Deﬁne

61

x ∈ [xk−1 , xk [ , k = 1, 2, . . . , n x = b.

Then 0 ≤ h(x) ≤ f (x) ≤ 1, for every x ∈ [a, b]. By (6.2), we still have ∫

∫

β

0≤

b

(f (x) − h(x))dx ≤ α

(f (x) − h(x))dx < a

ε . 2

Now, we look for a function g : [a, b] → {0, 1} such that on every [α, β] ⊂ [a, b] one has ∫ β ε (h(x) − g(x))dx < . (6.3) α 2 For it, on every interval [xk , xk+1 ] there exists ηk satisfying ∫ xk+1 ηk − x k = h(x) dx = mk (xk+1 − xk ). xk

Deﬁne

1, g(x) = 0, 0,

Then for each k,

∫

x ∈ [xk , ηk [ , x ∈ [ηk , xk+1 [ , x = b. ∫

xk+1

xk+1

g(x) dx = xk

h(x) dx.

(6.4)

xk

We show that this function g satisﬁes (6.3). If the boundary points of the interval [α, β] coincide with some xk ′ , then by (6.4) the left-hand side of (6.3) is zero. Suppose there exists an interval [xk , xk+1 ] so that it contains [α, β]. Then ∫ β b−a (h(x) − g(x))dx ≤ β − α ≤ α n and (6.3) holds whenever n is large enough. In the general case (not covered by the previous two cases) we argue as follows. We write the integral in (6.3) as ∫ ∫ β xk (h(x) − g(x))dx ≤ (h(x) − g(x))dx α α ∫ ∫ p β x ∑ k+m+1 + (h(x) − g(x))dx + (h(x) − g(x))dx . xk+m xk+p+1 m=0

6 Integral Calculus on R

62

The sum of integrals is zero since each term is so. The sum of the two extreme integrals is less then ε/2, whenever n is large enough. 6.9. For k = 0 or k = p, the inequality is trivial. We show the inequality by induction. If p = 1, the inequality is trivial. Suppose p = 2 and k = 1, that is √ |f ′ (x)| ≤ 2M0 M2 , for all x ∈ R. Suppose there exists x0 ∈ R so that |f ′ (x0 )| >

√

2M0 M2 .

We suppose that f (x0 ) ≥ 0 and f ′ (x0 ) ≥ 0, since if f (x0 ) ≥ 0 and f ′ (x0 ) < 0, we may consider instead of f, the function R ∋ x 7→ f (2x0 − x) and if f (x0 ) < 0, we consider the function R ∋ x 7→ −f (x). Thus √ f ′ (x0 ) > 2M0 M2 . For x ≥ x0 , we have ∫ x f ′′ (u) du ≥ f ′ (x0 ) + (−M2 ) du x0 x0 √ = f ′ (x0 ) − M2 (x − x0 ) > 2M0 M2 − M2 (x − x0 )

f ′ (x) ≥ f ′ (x0 ) +

∫

x

and ∫

x

f (x) = f (x0 ) +

∫

′

x

f (u)du > f (x0 ) + x0

(

√ 2M0 M2 − M2 (x − x0 ))du

x0

√ 1 = f (x0 ) + 2M0 M2 (x − x0 ) − M2 (x − x0 )2 2 √ 1 ≥ 2M0 M2 (x − x0 ) − M2 (x − x0 )2 . 2 The √ point of maximum to the right-hand side is attained at x = x0 + 2M0 M2 /M2 . The √ value of the right-hand side on this point is M0 . Then we get that f (x0 + 2M0 M2 /M2 ) > M0 , which is a contradiction. Thus the inequality is proved in this case. Suppose now that p > 2 and 0 < k < p are integers. Denote 1−k/p

ak = Mk /2k(p−k)/2 M0 Then ak ≤

√

ak−1 ak+1 ,

Mpk/p .

a0 = ap = 1.

By Exercise 3.18 at page 142, we have that max{a0 , a1 , . . . , ap } = max{a0 , ap } = 1. Then all ak ≤ 1 and thus the exercise is solved.

6.1 Solutions

63

6.10. Note that f (x) = x2 , x ∈ R, is a solution. Thus the set of solutions is nonempty. We are looking for other solutions. Setting x = y = 0, we ﬁnd that f (0) = 0. For each x ∈ R from f (x + y) − f (x) = f (y) + 2xy, it follows f (x + y) − f (x) f (y) + 2xy f (y) − f (0) = lim = 2x + lim y→0 y→0 y y y = 2x + f ′ (0).

f ′ (x) = lim

y→0

So we are looking for functions f satisfying ∫ x f (x) = f (x) − f (0) = f ′ (y) dy = x2 + f ′ (0)x. 0

We note for every real a the function f (x) = x2 + ax, x ∈ R satisﬁes all the requirements of the exercise. 6.11. The idea is to write the double sum as an integral and then to use the Cauchy-Buniakovski-Schwarz inequality for integrals, (i) in Theorem 1.7. Successively we have ∫ 1 n ∑ kl ak al = kak · lal tk+l−2 dt k+l−1 0 k,l=1 k,l=1 )2 ∫ 1 (∑ ∫ 1 ∑ n n k−1 k−1+l−1 dt = kak · t dt. kak · lal · t = n ∑

0

0

k,l=1

k=1

Applying the Cauchy-Buniakovski-Schwarz inequality for integrals, we get ∫ 0

1

(

n ∑ k=1

)2 kak · t

k−1

(∫

1

dt ≥ 0

(

n ∑ k=1

) kak · t

k−1

)2 dt

( =

n ∑

)2 ak

.

k=1

6.12. We use the rearrangement inequality (Exercise 1.38 at page 38 of the book). For each positive integer n we deﬁned the step function { fn (0) = f (0), fn (x) = inf x∈(k/n,(k+1)/n] f (x), whenever x ∈ ] k/n, (k + 1)/n]. Function fn has all the properties of f, but continuity. First suppose α is a rational, that is α = p/n, for an integer p. We show

6 Integral Calculus on R

64

∫

∫

2

2

fn (x)fn (x + α) dx ≥

fn (x)fn (x + 1) dx.

0

(6.5)

0

Denote ak = inf x∈(k/n,(k+1)/n] f (x). Then ∫

2

fn (x)fn (x + α) dx = 0

2n−1 1 ∑ ak ak+p n k=0

and (6.5) is equivalent to 2n−1 ∑ k=0

ak ak+p ≥

2n−1 ∑

ak ak+n .

(6.6)

k=0

System of numbers {ak+p } is obtained from the system {ak+n }. By hypothesis we have that if ai ≤ aj , then ai+n ≥ aj+n . Therefore (6.6) follows from the rearrangement inequality. Suppose now that α is arbitrary. We show that ∫ 2 ∫ 2 f (x)f (x + α) dx ≥ f (x)f (x + 1) dx. 0

0

For every positive ε we can ﬁnd a rational number p/n such that |p/n − α| < ε. Suppose n is large enough, at least n > 1/ε. Since f is uniformly continuous, for ε → 0, we have ∫ 2 ∫ 2 f (x)f (x + α) dx = f (x)f (x + p/n) dx + o(1) 0

∫

0 2

∫

fn (x)fn (x + p/n) dx + o(1) ≥

= 0

∫ =

2

fn (x)fn (x + 1) dx + o(1) 0

2

f (x)f (x + 1) dx + o(1), 0

and the inequality follows. 6.13. It is suﬃcient to show that ) ∫ ∞( 1 1 − dx < ∞, u u + u′ 1 since the sum of (6.65) and (6.7) supplies the answer. We have 1 u′ u′ 1 − = ≤ 2. ′ ′ u u+u u(u + u ) u Then

(6.7)

6.1 Solutions

∫

∞

1

(

1 1 − u u + u′

)

∫

∞

dx ≤ 1

65

u′ dx 1 1 = − lim . 2 u u(1) x→∞ u(x)

Since the last limit ﬁnite (u is increasing), we conclude that (6.7) holds and the exercise is proved. 6.14. Since the general term of the sequence of partial sums is sn = 2 2 2n2 xe−n x , limn→∞ sn = 0, for every x ∈ [0, 1]. So the series converges pointwise to the function f (x) = 0, x ∈ [0, 1]. The series does not converges n→∞ uniformly since for xn = 1/n, we have sn (xn ) = 2n/e −−−−→ ∞. On the ∫1 other side 0 f (x)dx = 0 whereas ∞ ∫ ∑ n=1

1

( ) 2 2 2x n2 e−n2 x2 − (n − 1)2 e−(n−1) x dx

0

=

∞ ∑

(1/e−(n−1) − 1/e−n ) = 1. 2

2

n=1

Hence the answer is no. 6.15. We have 3x/(x2 + 5x + 6) = 9/(x + 3) − 6/(x + 2). Also ∞ ∑ 3 9 xn = =3 (−1)n n , for |x| < 3, x+3 1 + x/3 3 n=0 ∞ ∑ −6 −3 xn = = −3 (−1)n n , for |x| < 2. x+2 1 + x/2 2 n=0

Since both series converge absolutely for |x| < 2, we write ) ( ∞ ∑ 1 3x 1 n+1 = 3 − xn , (−1) n n x2 + 5x + 6 2 3 n=0

|x| < 2.

6.16. Denote arctan x, x ∈ R. Then g ′ (x) = 1/(1 + x2 ) and therefore ∑∞ g(x) n= 2n ′ g (x) = n=0 (−1) x , |x| < 1. We integrate the last equality from 0 to x and it follows ∞ ∑ x2n+1 arctan x = (−1)n , |x| < 1. 2n + 1 n=0 Substituting x = −1 and x = 1 in the right-hand side of the last equality we get two convergent series. So the series of arctan x converges on |x| ≤ 1. Squaring the series of arctan x, we get

66

6 Integral Calculus on R

) 2n+2 ∞ ( ∑ 1 1 x (arctan x) = 1 + + ··· + , 3 2n + 1 n+1 n=0 2

|x| < 1.

Easily can be checked that the last identity holds also for x = ±1. 6.17. First we check that the integral converges. We xn e−x = xn e−x/2 · ∫ ∞write −x/2 n −x/2 −x/2 e and note that limx→∞ x e = 0 and 0 e dx is convergent. Then by Theorem 3.7 of Abel we have that our integral converges. By Theorem 3.9, we have ∫ ∞ ∞ In = xn e−x dx = −xn e−x|0 + n · In−1 = n · In−1 . 0

∫∞

Since 0 e−x dx = 1, we have In = n!. One can try using one of the Mathematica r command. Integrate[x^n*Exp[-x], {x, 0, Infinity}, Assumptions -> n \[Element] Integers && n > 0]

6.18. Since the integrand is not deﬁned at x = 1, we ∫have to take into ac1 count several subcases. We show that the integral I1 = 0 x ln x dx/(1 − x2 )2 ∫1 diverges. But instead of it we discuss the integral I = 1/e x ln x dx/(1 − x)2 , showing its divergence. Deﬁne f (x) = −2x ln x = x − 1, for x ∈ I = [1/e, √1]. We have f (1/e) = (3−e)/e > 0, f (1) = 0, f is nondecreasing on [1/e, 1/ e], √ and f is nonincreasing on [1/ e, 1]. Therefore f is nonnegative on I. Thus −2x ln x ≥ 1 − x, x ∈ I. Immediately we have ∫ 1 ∫ 1 −2x ln x 1 −2x ln x dx dx ≥ =⇒ ≥ = +∞. 2 2 (1 − x) 1−x 1/e (1 − x) 1/e 1 − x Thus I diverges and hence I1 diverges, too. We conclude that the initial integral diverges. 6.19. The ∫ a integrand is not deﬁned at x = 1. Consider 0 < a < 1 and I(a) = 0 (ln(1/(1 − x)))dx. We get I(a) = (1 − ln(1 − a))(1 − a) + 1. The integral is equal to lima↑1 I(a) = 1, so it converges. We can approach this exercise by the Mathematica r command. Integrate[Log[1/(1 - x)], {x, 0, 1}] (*getting*) 1

6.20. The integral is improper since at 0 and π/2 the integrand might be undeﬁned. Consider x = sin u. Then ∫ π/2 ∫ 1 sina−1 u cosb−1 u du = xa−1 (1 − x2 )(b−1)/2 (1 − x2 )−1/2 dx 0

0

6.1 Solutions

∫ =

1

67

t(a−1)/2 (1 − t)b/2−1 (1/2)t−1/2 dt = (1/2)B(a/2, b/2).

0

We can approach this exercise by the Mathematica r command. Integrate[{Sin[x]}^(a - 1) {Cos[x]}^(b - 1), {x, 0, Pi/2}, Assumptions -> a > 0 && b > 0] TraditionalForm[%] (*getting*) (\[CapitalGamma](a/2) \[CapitalGamma](b/2)) /(2\[CapitalGamma]((a+b)/2))

6.21. For every x ∈ [0, 1] we have limy→0 f (x, y) = 0. For x = y, we have y→0

f (y, y) −−−−→ ∞. So the convergence to the null function is not uniform in ∫1 ∫1 y. We have limy→0 0 f (x, y)dx = 1/2 whereas 0 limy→0 f (x, y)dx = 0.

7 Diﬀerential Calculus on Rn

7.1. If a1 = · · · = an = 0, f is the null functional and its norm is equal to 0. Suppose that m = a21 + · · · + a2n > 0. It is clear that f is a linear mapping. Let B be the closed unit ball in Rn . To ﬁnd the norm of f, we use formula (7.1). From one side by the Cauchy inequality, Exercise 1.35 in Chapter 1, we have v v u n u n u∑ u∑ √ a2k t x2k ≤ m. ∥f ∥ = max |f (x)| ≤ t x∈B

k=1

k=1

For the reverse inequality choose a particular point in B, namely x0 = √ 0 0 0 0 (x1 , x2 , . . . , xn ) by xk = ak / m, k = 1, . . . , n. Clearly, x0 ∈ B and √ √ ∥f ∥ ≥ |a1 x01 + · · · + an x0n | = m/ m = m. Thus ∥f ∥ = m. 7.2. Denote t = x2 + y 2 . Then we have ∂u ∂t ∂u ∂t = φ′t = φ′t · 2x and = φ′t = φ′t · 2y. ∂x ∂x ∂y ∂y Then y

∂u ∂u −x = φ′t · (y(2x) − x(2y)) = 0. ∂x ∂y

7.3. Pick an arbitrary but ﬁxed a ∈ A. Denote L = f ′ (a). We show that ∥L∥ ≤ M. There exists a function r(a, · −a) such that for every x ∈ A, f (x) = f (a) + L(x − a) + r(a, x − a). We may write r(a, x − a) = ∥x − a∥ω(a, x − a) such that limx→a ω(a, a − x) = 0 and consider that ω is continuous at zero.

70

7 Diﬀerential Calculus on Rn

For an x of norm one there exists η > 0 such that a + tx ∈ A for all t with |t| < η. Then f (a + tx) = f (a) + L(tx) + r(a, tx) = f (a) + tL(x) + ∥tx∥ω(a, tx) =⇒ ∥tL(x)∥ ≤ ∥f (a + tx) − f (a) − |t|ω(a, tx)∥ ≤ ∥f (a + tx) − f (a)∥ + |t| · ∥ω(a, tx)∥ ≤ M ∥tx∥ + |t| · ∥ω(a, tx)∥. From here we have ∥L(x)∥ ≤ M + ∥ω(a, tx)∥, for all 0 < t < η. Letting t ↓ 0, by the continuity of ω, we ﬁnd out ∥L(x)∥ ≤ M. By Theorem 1.3 we write ∥L∥ ≤ M. 7.4. Consider an arbitrary point (x, y) ∈ R2 and deﬁne the function g(t) = f (tx, ty) for t ∈ [0, 1]. Then g is of class C 1 , and g ′ (t) = x

∂f (tx, ty) ∂f (tx, ty) +y . ∂x ∂y

By Theorem 1.5 at page 318, we have that ∫ f (x, y) = f (x, y) − f (0, 0) = g(1) − g(0) =

1

g ′ (t)dt.

0

7.5. If α = 0, the conclusion is clear. Otherwise, pick up an arbitrary but ﬁxed a ∈ Rn . Since limx→a ∥f (x) − f (a)∥/∥x − a∥ = 0, we have that f is diﬀerentiable at a and df (a) = 0. The conclusion follows by Theorem 2.8. 7.6. We show that f ′ (a) = 0, for all a ∈ Rn . Suppose that this is not the case, i.e., there is an a ∈ Rn so that f ′ (a) ̸= 0. Then for some h ∈ Rn the scalar f ′ (a)h is nonzero. Admit f ′ (a)h > 0. Pick a real t and for a certain ξ on the segment deﬁned by a and a + th it holds f (a + th) = f (a) + f ′ (a)(th) + (1/2)f ′′ (ξ)(th, th), Immediately we have that f (a + th) ≤ f (a) + tf ′ (a)(h). Letting t → −∞, we get a contradiction. If f ′ (a)h < 0, we let t → ∞ and again we get a contradiction. Hence f ′ (a) = 0 for all a ∈ Rn . 7.7. From the system

∂z = 3x2 − 3y = 0, ∂x ∂z = 3y 2 − 3x = 0, ∂y

we ﬁnd two stationary points M1 = (0, 0) and M2 = (1, 1). Since

7.1 Solutions

f ′′ (x, y)(h, h) =

71

∂2z 2 ∂2z ∂2z h1 + 2 h1 h2 + 2 h22 = 6xh21 − 6h1 h2 + 6yh22 , 2 ∂x ∂x∂y ∂y

we have that f ′′ (0, 0)(h, h) = −6h1 h2 and f ′′ (1, 1)(h, h) = 6(h21 − h1 h2 + h22 ). Thus M1 is not an extremum point since f ′′ (0, 0) is neither positive nor negative deﬁned, while M2 is a minimum point since f ′′ (1, 1) is positively deﬁned. We have xmin = 1, ymin = 1, and zmin = −1. We may use equally well Theorem 1.9 of Sylvester to study if the stationary points are extreme points or not. For M1 we have [ ] 0 −3 A= =⇒ A1 = 0, A2 = −9. −3 0 Thus M1 is not an extreme point. For M2 we have [ ] 6 −3 A= =⇒ A1 = 6, −3 6

A2 = 27.

Thus M2 is a point of local minimum. 7.8. The Lagrangian function F is F (x, y, z, λ1 , λ2 ) = xyz + λ1 (x2 + y 2 + z 2 − 1) + λ2 (x + y + z). The set of stationary points is given by the system of nonlinear equations ∂F = yz + 2λ1 x + λ2 = 0, ∂x ∂F = xz + 2λ1 y + λ2 = 0, ∂y ∂F = xy + 2λ1 z + λ2 = 0, ∂z ∂F = x2 + y 2 + z 2 − 1 = 0, ∂λ1 ∂F = x + y + z = 0. ∂λ2

(7.1) (7.2) (7.3) (7.4) (7.5)

Add (7.1), (7.2), and (7.3) and take into account (7.5) to get xy + yz + zx + 3λ2 = 0.

(7.6)

From (7.4) and (7.5) we get xy + yz + zx = −1/2. Together with (7.6) we ﬁnd that

72

7 Diﬀerential Calculus on Rn

λ2 = 1/6.

(7.7)

Subtracting (7.2) from (7.1), we have (x − y)(z − 2λ1 ) = 0.

(7.8)

Therefore we have two cases. (i) Suppose z = 2λ1 . Substitute it in (7.2). We get the equation xz + yz + λ2 = 0.

(7.9)

By (7.7) we have that z = 0 is not a solution to (7.9). Therefore we multiply (7.5) by z to get xz + yz + z 2 = 0.

(7.10)

√ 2 From (7.9) and (7.10) we √ get that z − λ2 = 0. Thus z1,2 = ±1/ 6. (i1 ) Suppose z1 = 1/ 6. Then we have the stationary point √ √ √ √ M0 = (−2/ 6, 1/ 6, 1/ 6), λ1 = 1/2 6, λ2 = 1/6. √ (i2 ) Suppose z1 = −1/ 6. Then we have the stationary point √ √ √ √ M1 = (2/ 6, −1/ 6, −1/ 6), λ1 = −1/2 6, λ2 = 1/6. (ii) Suppose x = y. Substituting in (7.4) and (7.5), we get the system { 2x2 + z 2 = 1, 2x + z = 0. √ Then x = ±1/ 6. √ (ii1 ) Suppose x = 1/ 6. Then we have the stationary point √ √ √ √ M2 = (1/ 6, 1/ 6, −2/ 6), λ1 = 1/2 6, λ2 = 1/6. √ (ii2 ) Suppose x = −1/ 6. Then we have the stationary point √ √ √ √ M3 = (−1/ 6, −1/ 6, 2/ 6), λ1 = −1/2 6, λ2 = 1/6. We have ∂2F ∂2F 2 ∂2F 2 ∂2F ∂2F 2 2 dx + dy + dz + dλ + dλ22 1 ∂x2 ∂y 2 ∂z 2 ∂λ21 ∂λ22 ∂2F ∂2F ∂2F +2 dx dy + 2 dy dz + 2 dz dx ∂x∂y ∂y∂z ∂z∂x

d2 F =

7.1 Solutions

73

∂2F ∂2F ∂2F dx dλ1 + 2 dy dλ1 + 2 dz dλ1 ∂x∂λ1 ∂y∂λ1 ∂z∂λ1 ∂2F ∂2F ∂2F ∂2F +2 dx dλ2 + 2 dy dλ2 + 2 dz dλ2 + 2 dλ1 dλ2 ∂x∂λ2 ∂y∂λ2 ∂z∂λ2 ∂λ1 ∂λ2 (7.11) +2

It is clear that

∂2F ∂2F ∂2F = = = 0. ∂λ1 ∂λ2 ∂λ21 ∂λ22

From (7.65) by diﬀerentiation we get x dx + y dy + z dz = 0 and dx + dy + dz = 0.

(7.12)

We have ∂2F ∂2F = 2λ1 , = 2λ1 , 2 ∂x ∂y 2 ∂2F ∂2F = z, = x, ∂x∂y ∂y∂z ∂2F ∂2F = 2x, = 2y, ∂x∂λ1 ∂y∂λ1 ∂2F ∂2F = 1, = 1, ∂x∂λ2 ∂y∂λ2

∂2F = 2λ1 , ∂z 2 ∂2F = y, ∂z∂x ∂2F = 2z, ∂z∂λ1 ∂2F = 1. ∂z∂λ2

Substituting the last partial derivatives and (7.12) in (7.11) we write d2 F = 2λ1 (dx2 + dy 2 + dz 2 ) + 2zdx dy + 2xdy dz + 2ydz dx.

(7.13)

Now we have to determine if the quadratic mapping (7.13) is positive deﬁnite, negative deﬁnite, or none at each point Mk , k = 0 . . . , 3. If y = z ̸= x, then from (7.12) we have that dx = 0 and dz = −dy. Thus d2 F |M0 or M1 = 2λ1 (2dy 2 ) − 2x dy 2 = 2(2λ1 − x)dy 2 √ √ √ M0 = 2(1/ 6 + 2/ 6)dy 2 = 6 dy 2 > 0 =⇒ M0 is a point of minimum √ √ √ M1 = − 2(1/ 6 + 2/ 6dy 2 = − 6 dy 2 < 0 =⇒ M1 is a point of maximum. If x = y ̸= z, then from (7.12) we have that dz = 0 and dy = −dx. Thus d2 F |M2 or M3 = 2λ1 (2dx2 ) − 2z dx2 = 2(2λ1 − z)dx2 √ √ √ M2 = 2(1/ 6 + 2/ 6)dx2 = 6 dx2 > 0 =⇒ M2 is a point of minimum √ √ √ M3 = − 2(1/ 6 + 2/ 6dx2 = − 6 dx2 < 0 =⇒ M3 is a point of maximum. There are other two cases similar to (7.8), namely

74

7 Diﬀerential Calculus on Rn

(y − z)(x − 2λ1 ) = 0 and (z − x)(y − 2λ1 ) = 0.

8 Double Integrals, Triple Integrals, and Line Integrals

8.1. By Theorem 1.5, we have ∫ ∫ 1 (∫ x2 ydx dy = D

0

∫

2

) x2 ydy dx

0

(∫

1 2

=

x

2

)

∫

1

x2 dx = 2/3.

ydy dx = 2

0

0

0

One can try using the Mathematica r commands f[x_, y_] := x^2*y Integrate[Integrate[f[x, y], {y, 0, 2}], {x, 0, 1}]

8.2. By Theorem 1.5, we have ) ∫ ∫ 1 (∫ 2 ∫ (x2 + y)dx dy = (x2 + y)dy dx = D

0

∫ =

0 1

(

1 0

(

y=2 ) (x2 y + y 2 /2) y=0 dx

)

2x2 + 2 dx = 8/3.

0

One can try using the Mathematica r commands f[x_, y_] := x^2 + y Integrate[Integrate[f[x, y], {y, 0, 2}], {x, 0, 1}]

8.3. See Figure 8.1. The two curves intersect at (0, 0) and (1, 1). Therefore by Theorem 1.6, we have ) ∫ 1( ∫ ∫ 1 (∫ √x y=√x ) (x + y)dy dx = (xy + y 2 /2) y=x2 dx (x + y)dx dy = D

0

x2

0

76

8 Double Integrals, Triple Integrals, and Line Integrals (1,1)

(0,0)

Fig. 8.1. Figure for Exercise 8.3

∫

1

=

Fig. 8.2. Figure for Exercise 8.4

( √ ) x x + x/2 − (x3 + x4 /2) dx = 3/10.

0

One can try using the Mathematica r commands f[x_, y_] := x + y Integrate[Integrate[f[x, y], {y, x^2, Sqrt[x]}], {x, 0, 1}]

8.4. See Figure 8.2 and apply Theorem 1.9. We transform the unit closed ball in a rectangle by x = ρ cos θ and y = ρ sin θ and get T (D) = {(ρ, θ) | 0 ≤ ρ ≤ 1,

0 ≤ θ ≤ 2π}.

cos θ −ρ sin θ = ρ. det T ′ (x, y) = sin θ ρ cos θ

Then

By equality (8.18), we have ∫

∫

∫

D

T

∫

2π

ρ2 · ρ dρ dθ =

(x2 + y 2 )dx dy =

1

dθ ·

ρ3 dρ =

0

0

π . 2

One can try using the Mathematica r commands f[x_, y_] := x^2 + y^2 Integrate[Integrate[f[x, y], {y, -Sqrt[1 - x^2], Sqrt[1 - x^2]}], {x, -1, 1}]

8.5. Similarly to the previous Exercise we have x = ρ cos θ, y = ρ sin θ, for 0 ≤ ρ ≤ 1, π/6 ≤ θ ≤ π/3. Then ∫

∫ D

∫

T

∫

π/3

1

dθ ·

ρ2 · ρ dρ dθ =

(x2 + y 2 )dx dy =

π/6

One can try using the Mathematica r commands

ρ3 dρ = π/24. 0

8.1 Solutions

77

f[x_, y_] := x^2 + y^2 Integrate[Integrate[f[x, y], {y, Sqrt[3]*x/3, Sqrt[3]*x}], {x, 0, 0.5}] + Integrate[Integrate[f[x, y], {y, Sqrt[3]*x/3, Sqrt[1 - x^2]}], {x, 0.5, Sqrt[3]/2}]

8.6. The boundary curves intersect at (0, 0) and (2p, 2p). Consider x = √ u2 · v √ 2 3 3 and y = u · v . The new domain is {(u, v) | 0 ≤ u ≤ 2p, 0 ≤ v ≤ 2p} whereas the determinant of the transformation is 2uv u2 2 2 2 v 2uv = 3u v . Denote I the integral in the statement. Thus ∫ I=

e T

=

1 3

(

u3 +v 3

∫

√ 3

2p

∫

u3 2

√ 3 2p

e u du ·

2 2

3u v du dv = 3

3

ev v 2 du

0

0

) 3 √ 2p 2 )2 1 ( 2p u3 e = e −1 . 3 0

8.7. By Theorem 2.5, we have ∫

∫

1

(∫

)

2

2

x yzdx dy dz = D

x yzdy dz ∫

0

[0,2]×[0,1]

(∫

1

=

x ∫

(∫

2

2

y

0

0

∫

2

x2 dx · 0

One can try using the Mathematica

r

∫

1

ydy ·

0

)

zdz dy dx

0 1

=

)

1

dx

zdz = 0

1 . 3

commands

f[x_, y_, z_] := x^2*y*z Integrate[Integrate[Integrate[f[x, y, z], {z, 0, 1}], {y, 0, 2}], {x, 0, 1}]

8.8. Denote I the integral in (8.44) in the statement. By Theorem 2.5, we successively have ) ) ∫ 1 (∫ 2 (∫ 1 I= (x2 + y + z)dz dy dx 0

∫

1

(∫

0

0

2

) z=1 (x z + yz + z /2) z=0 dy dx 2

= 0

∫

1

(∫

0

2

)

2

∫

1

2

=

(x + y + 1/2)dy dx = 0

0

0

y=2 (x2 y + y 2 /2 + y/2) y=0 dx

78

8 Double Integrals, Triple Integrals, and Line Integrals

∫

1

(2x2 + 3)dx = 11/3.

= 0

One can try using the Mathematica r commands f[x_, y_, z_] := x^2 + y +z Integrate[Integrate[Integrate[f[x, y, z], {z, 0, 1}], {y, 0, 2}], {x, 0, 1}]

8.9. One approach based on Theorem 2.6 is the following. We have ) (∫ 1 ∫ ∫ x · y · zdz dx dy (x · y · z)dx dy dz = D x2 +y 2 ≤1 x2 +y 2 ∫ 1 = xy(1 − (x2 + y 2 )2 )dx dy 2 x2 +y2 ≤1 ∫ ∫ 2π 1 1 3 = ρ (1 − ρ4 )dρ sin θ cos θdθ = 0. 2 0 0 One can try using the Mathematica r commands f[x_, y_, z_] := x*y*z Integrate[Integrate[Integrate[f[x, y, z], {z, x^2 + y^2, 1}], {y, -Sqrt[1 - x^2], Sqrt[1 - x^2]}], {x, -1, 1}]

8.10. Apply Theorem 2.8. We transform the unit closed ball into a parallelepiped T by x = ρ cos α sin β,

y = ρ sin α sin β,

z = ρ cos β,

where ρ ∈ [0, 1], α ∈ [0, 2π], and β ∈ [0, π]. Then cos α sin β −ρ sin α sin β ρ cos α cos β det T ′ (x, y) = sin α sin β ρ cos α sin β ρ sin α cos β = −ρ2 sin β. cos β 0 −ρ sin β By equality (8.35), we have ∫ ∫ 2 2 2 (x + y + z )dx dy dz = ρ4 sin βdρ dα dβ D

T 2π

∫ =

∫

0

∫

π

dα 0

One can try using the Mathematica r commands f[x_, y_, z_] := x^2 + y^2 + z^2

1

ρ4 dρ = 4π/5.

sin βdβ 0

8.1 Solutions

79

Integrate[Integrate[Integrate[f[x, y, z], {z, -Sqrt[1 - x^2 - y^2], Sqrt[1 - x^2 - y^2]}], {y, -Sqrt[1 - x^2], Sqrt[1 - x^2]}], {x, -1, 1}]

8.11. We have to evaluate

∫ D

dx dy dz, where D is given as

{(x, y, z) | x2 + y 2 + z 2 ≤ 1}. Then

∫

∫ D

∫

2π

dα ·

dx dy dz = 0

∫

π

1

sin βdβ · 0

ρ2 dρ = 4π/3. 0

One can try using the Mathematica r commands f[x_, y_, z_] := 1 Integrate[Integrate[Integrate[f[x, y, z], {z, -Sqrt[1 - x^2 - y^2], Sqrt[1 - x^2 - y^2]}], {y, -Sqrt[1 - x^2], Sqrt[1 - x^2]}], {x, -1, 1}]

8.12. Use the Taylor formula and a given transformation of variables. ∫ ∫ π/2 √ 8.13. C f (x, y)ds = 2 0 1 − cos2 t dt = 2. One can try using the Mathematica r command Integrate[Sqrt[2 (1 + Cos[t])]*Sqrt[(1 - Cos[t])^2 + (Sin[t])^2], {t, 0, Pi/2}]

8.14. The curve is continuous diﬀerentiable on its domain and the function is continuous on the image∫ of the curve. Denote √ by I the √ integral. Then after 1 3√ 2 substitution we have I = 0 t 2 + t dt = 8 2/15 − 3/5. One can try using the Mathematica r command Integrate[t^3 *Sqrt[2 + t^2], {t, 0, 1}]

8.15. The curve γ is continuous diﬀerentiable. The function γ ∋ (x, y) 7→ 1/(x3 + y 3 ) is also continuous diﬀerentiable along γ. Then ∫

π/2

I=− 0

sin t dt −1 = 2 3 2 3 a a (sin t + cos t)

∫ 0

∞

√ u du −2π 3 = . 1 + u3 9a2

∫ 8.16. We write I = I1 + I2 + I3 , where Ik = γk y dx + z dy + x dz and γk is the segment Ak Ak+1 , with A4 = A1 . Then γ1 is parameterized as x = 1 − t,

80

8 Double Integrals, Triple Integrals, and Line Integrals

y = t, z = 0, with t ∈ [0, 1]; γ2 is parameterized as x = 0, y = 1 − t, z = t, with t ∈ [0, 1]; γ3 is parameterized as x = t, y = 0, z = 1 − t, with t ∈ [0, 1]. Then ∫ ∫ ∫ 1

I=

1

(−t) dt + 0

1

0

8.17. The area is given by A = 1/2 a cos t, y = b cos t, t ∈ [0, 2π]. Then 1 A= 2

∫

(−t) dt = −3/2.

(−t) dt + 0

∫ γ

x dy − y dx, where γ is given as x =

2π

ab(cos2 t + sin2 t) dt = πab. 0

One can try using the Mathematica r commands (* by the double integral *) f[x_, y_] := 1 Integrate[ Integrate[ f[x, y], {y, -b*Sqrt[1 - x^2/a^2], b*Sqrt[1 - x^2/a^2]}], {x, -a, a}, Assumptions -> b \[Element] Reals && a \[Element] Reals && a > 0 && b > 0] (* or by the line integral*) x[t_] := a*Cos[t] y[t_] := b*Sin[t] Integrate[(x[t]*D[y[t], t] - y[t]*D[x[t], t])*(1/2), {t, 0, 2*Pi}]

8.18. If |a| = |b|, then immediately follows that F (a, a) = π/(2a4 ). Suppose that |a| ̸= |b|. By Exercise 6.3 in Chapter 6 we showed that ∫

π/2

dx/(a2 cos2 x + b2 sin2 x) = π/(2|ab|).

I(a, b) = 0

There are satisﬁed the requirements of Theorem 5.3. Then taking the partial derivatives from I(a, b) in respect to a and b we get π = 4a3 b

∫

π/2 0

cos2 x dx , (a2 cos2 x + b2 sin2 x)2

π = 4ab3

∫ 0

π/2

sin2 x dx . (a2 cos2 x + b2 sin2 x)2

We add the previous equalities getting F (a, b) = π(a2 + b2 )/(4|ab|3 ). One can try using the Mathematica r commands f[x_] := 1/(a^2 {Cos[x]}^2 + a^2 {Sin[x]}^2)^2 Integrate[f[x], {x, 0, Pi/2}] g[x_] := 1/(a^2 {Cos[x]}^2 + b^2 {Sin[x]}^2)^2 Integrate[g[x], {x, 0, Pi/2}, Assumptions -> b \[Element] Reals && a \[Element] Reals && a != 0 && b != 0]

8.1 Solutions

81

8.19. We may suppose that a, b > 0. Note that F (1, 1) = 0. There are satisﬁed the requirements of Theorem 5.3. Then taking partial derivatives from F (a, b) in respect to a and b, we get ∂F (a, b)/∂a = π/(a + b) and ∂F (a, b)/∂b = π/(a + b). From dF =

∂F ∂F da + db, ∂a ∂b

we ﬁnd that dF = π d(a + b)/(a + b). Then F (a, b) = π ln(a + b) + C, where C is a constant. Substituting a = b = 1, we get C = −π ln 2. Thus F (a, b) = π ln((a + b)/2). If we remove the assumption that a, b > 0, then F (a, b) = π ln((|a| + |b|)/2). ∫b 8.20. Function f is continuous on [0, 1]. We have that f (x) = a xu du, for all x ∈ [0, 1]. The integrand is continuous on [0, 1] × [a, b], supposing that a < b. Then we apply Theorem 5.5. Hence ) ) ∫ 1 ∫ 1 (∫ b ∫ b (∫ 1 ∫ b du u f (x) dx = x du dx = xu dx du = 0 0 a a 0 a u+1 = ln

b+1 . a+1

One can try using the Mathematica r commands f[x_] := (x^b - x^a)/Log[x] f[0] := 0 f[1] := b - a Integrate[f[x], {x, 0, 1}, Assumptions -> a > 0 && b > 0] TraditionalForm[%]

8.21. The integrand is continuous, if we deﬁne it as b − c for x = 0. Then we have to estimate ∫ α −ax e (sin(bx) − sin(cx)) lim dx. α→∞ 0 x We have

∫

b

e−ax cos ux du =

c

e−ax (sin(bx) − sin(cx)) . x

The function [0, ∞ [ ×[c, b] ∋ (x, u) 7→ e−ax cos ux is continuous. Therefore ) ) ∫ b (∫ α ∫ b ∫ α (∫ b −ax −ax g(u) du, e cos ux dx du = e cos ux du dx = 0

where

c

c

0

c

82

8 Double Integrals, Triple Integrals, and Line Integrals

g(u) =

a + e−au (u sin(αu) − a cos(αu)) . a2 + u2

The function g tends uniformly to a/(a2 + u2 ) for α → ∞. Therefore ∫

∞ 0

e−ax (sin(bx) − sin(cx)) dx = x

∫ c

b

a du b−c = arctan 2 . 2 +u a + bc

a2

One can try using the Mathematica r commands f[x_] := Exp[-a*x] (Sin[b*x] - Sin[c*x])/x Integrate[f[x], {x, 0, Infinity}, Assumptions -> a > 0 && b \[Element] Reals && c \[Element] Reals]

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