The JoM Journal - Tolaso...(C) and suppose that Nis nilpotent. Show that if A,N commute then det(A+ N) = detA Solution. Since C is algebraically closed and A,Ncommute this means that

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3

Editor

Tolaso J. Kos , Farsala, Larisa, Greece , [email protected]

The JoM Journal is an electronic mathematical journal which aims at

giving the chance to the readers, and the editor himself, to work on interest-

ing mathematical problems or find information about various mathematical

topics. The problems presented here are basically a collection of the problems

posted on the JoM Blog ( hosted at math.tolaso.com.gr ) . The level of the

topics is undergraduate and beyond. However, there is a section dedicated

to inequalities and general mathematics sometimes including mathematical

competitions. The JoM journal is consisted of 6 parts:

� Algebra

� Calculus

� Real Analysis

� Inequalities

� JoM ... proposes

� JoM ... study

The JoM ... proposes column contains problems that extend the ideas

already seen in the previous 4 columns. The JoM study, on the other hand,

studies several mathematical concepts. Examples are included whenever nec-

essary. At the end of this part the reader will find problems to exercise

himself.

If you want to submit an article at the JoM ... study please contact the

author at [email protected].

This journal has been typeset using the known LTEX system. All drawings

are produced by the TIkZ package. All links appearing in this journal are

clickable.

Should you notice any typographical errors , please contact the author so

that can be fixed.

JoM Journal • Tolaso Network™Tolaso Network © 2019


math.tolaso.com.gr


4

Editor’s Note

The editor would like to invite the interested reader to visit the new and

improved mathematical forum

located at the address mathimatikoi.org. Undoubtedly , he will find in-

teresting material on university mathematics. The main branches the forum

focuses on are:

� Algebra

� Analysis

� Geometry

� Number Theory

Other fora include Topology, Category Theory, Differential Equations ,

Foundations as well as General Mathematics.

We suggest you to take a look. You might find something interesting

during your search.

https://www.mathimatikoi.org

5

Contents1 Algebra 7

2 Calculus 21

3 Analysis 33

4 General Mathematics - Inequalities 53

5 Jom ... Proposes 83

6 JoM ... Study 87

6

1

1PART

Algebra

7

9

1 Prove that

max1≤j≤p

√√q∑i=1

a2

ij ≤

∥∥∥∥∥∥∥∥∥a

11· · · a

1p...

. . ....

aq1 · · · aqp

∥∥∥∥∥∥∥∥∥2

≤

√√ q∑i=1

p∑j=1

a2

ij

Solution. Fix j. Apply the matrix A on ej thus:

‖Aej‖2 ≤ ‖A‖2‖ej‖2 = ‖A‖2

Since Aej is exactly the j - th column of A the previous equality can bererwritten as √√

q∑i=1

a2

ij ≤ ‖A‖2

Since this holds for all j = 1, 2, . . . , n we get max and the left inequalityfollows.

For a random unit vector x = (x1, x

2, . . . , xn) the i coordinate of the

vector Ax is∑jaijxj. It follows from Cauchy - Schwartz that

∣∣∣∣∣∣∣∑j aijxj∣∣∣∣∣∣∣2

≤∑j

|aij|2

∑j

|xj|2 =

∑j

|aij|2

Summing over all i ’s till we find ‖Ax‖2

we conclude that, for every unitvector x , it holds that ‖Ax‖

2is less than the right hand side. Taking

supremum with respect to all x the right hand side inequality follows.

�

Exercise lies in https://www.math.tolaso.com.gr/?p=2439.

2 Let A, B ∈ Mn(C) be two idempotent matrices such that A − B is in-

vertible and let α, � ∈ C. Let I ∈ Mn(C) be the identity matrix. Show

that:

(i) if α < {0,−1}Â then I + αAB is not necessarily invertible.

(ii) if α ∈ {0,−1} then I + αAB is invertible.

(iii) A + B − AB is invertible.

(iv) if α� , 0 then αA + �B is invertible.

https://www.math.tolaso.com.gr/?p=2439

10

Solution. (i) Pick

A =

(1 0

0 0

), B =

(− 1

α − 1

α1 + 1

α 1 + 1

α

)and note that A2 = A , B2 = B and A − B is invertible but I + αABis not.

(ii) Itâ™s clear for α = 0. For α = −1, suppose that ABv = v for somev ∈ Cn . We need to show that v = 0. We have ABv = A2Bv = Avthus Bv − v ∈ kerA. But since B2 = B we also have Bv − v ∈ kerBand hence

Bv − v ∈ kerA ∩ kerB ⊆ ker(A − B) = {0}

because A−B is invertible. So Bv = v and therefore Av = ABv = v.So v ∈ ker(A − B) = {0}.

(iii) Let A′ = I−A, B′ = I−B. Since A′, B′ are idempotents we concludeby (ii) that I−A′B′ is invertible since A′−B′ = −(A−B) is invertible.The result now follows because

I − A′B′ = I − (I − A) (I − B) = A + B − AB

(iv) Let γ = −�α , 0 and suppose that Av = γBv for some v ∈ Cn . We are

done if we show that v = 0. Well, we have BAv = γBv = Av = A2vand thus (A − B)Av = 0 implying that Av = 0 because A − B isinvertible. Hence γBv = Av = 0 which gives Bv = 0 because γ , 0.Thus (A − B)v = 0 and therefore v = 0.

�


3 Let n ≥ 1 and let

pn(x) = x2n

+ x2n−1

+ 1 ∈ Z[x]

Find all irreducible factors of pn(x).

Solution. Setting qn(x) = x2n− x2

n−1+ 1 we note that

pn(x) = pn−1(x)qn−1(x) , n ≥ 2

Hence


11

pn(x) = p1(x)q

1(x)q

2(x) · · · qn−1(x) (1)

It’s clear that p1(x) = x2 + x + 1 is irreducible over Z. Now, for n ≥ 1 let

Φn(x) be the n-th cyclotomic polynomial. Using well-known propertiesof Φn, we have

Φ3·2n (x) =

Φ2n (x3)

Φ2n (x)

=x3·2n−1 + 1

x2n−1

+ 1

= x2n− x2

n−1+ 1 = qn(x)

Thus qn is irreducible over Z because cyclotomic polynomials are irre-ducible over Z. Hence, by ( 1 ) pn has exactly n irreducible factors andthey are p

1, q

1, q

2, . . . , qn−1.

�


4 Let A, N ∈ Mn(C) and suppose that N is nilpotent. Show that if A, Ncommute then

det(A + N) = detA

Solution. Since C is algebraically closed and A, N commute this meansthat A, N are simultaneously triangularizable i.e there exists an invertibleelement P ∈ Mn(C) such that both PNP−1 and PAP−1 are triangular. SincePNP−1 is both nilpotent and triangular, all its diagonal entries are zeroand so the diagonal entries of P(A+N)P−1 = PAP−1 +PNP−1 are the sameas the diagonal entries of PAP−1. Thus,

det(P(A + N)P−1) = det(PAP−1)

because P(A + N)P−1 , PAP−1 are both triangular and the determinant ofa triangular matrix is the product of its diagonal entries. So,

det(A + N) = det(P(A + N)P−1) = det(PAP−1) = det(A)

�




12

5 Let G be a finite group and suppose that H , K are two subgroups of

G such that H , G and K , G. Show that

|H ∪ K| ≤3

4

|G|

Solution. Recall that |HK| =|H||K|

|H ∩ K|and thus

|H||K|

|H ∩ K|≤ |G|. Hence

|H ∩ K| ≥|H||K|

|G|and so

|H ∪ K| = |H| + |K| − |H ∩ K|

≤ |H| + |K| −|H||K|

|G|

= (a + b − ab)|G| (1)

where a =|H|

|G|and b =

|K|

|G|.

Now, since H , G and K , G we have [G : H] ≥ 2 and [G : K] ≥ 2

that is a ≤ 1

2and b ≤ 1

2. So if we let a′ = 1 − 2a and b′ = 1 − 2b then

a′, b′ ≥ 0 and thus

a + b − ab =3

4

−a′ + b′ + a′b′

4

≤3

4

due to (1).

�


6 Let P, Q be nilpotent matrices such that PQ + P +Q = 0 . Evaluate the

determinant

∆ = det (I + 2P + 3Q)

Solution. We state a lemma:

Lemma

If A and B are nilpotent matrices that commute and a, b arescalars, then aA + bB is nilpotent.


13

Proof. Since A and B commute, they are simultaneously triangulariz-

able. Let S be an invertible matrix such that A = STS−1 and B = SUS−1,where T and U are upper triangular. Note that since A and B are nilpo-

tent, T and U must have zeros down the main diagonal. Hence aT+bUis upper triangular with zeros along the main diagonal which means

that it’s nilpotent. Finally aA + bB = S(aT + bU )S−1 and so aA + bB is

nilpotent. �

We have (P + I) (Q + I) = I =⇒ (Q + I) (P + I) = I. Then equating thetwo left hand sides and simplifying gives us PQ = QP . Thus by thelemma we know that 2P + 3Q is nilpotent, i.e., it’s eigenvalues are allzero. It follows that the eigenvalues of I + 2P + 3Q are all one and so

∆ = det (I + 2P + 3Q) = 1

�


7 Let A ∈ Mn (C) with n ≥ 2. If

det (A + X ) = detA + detX

for every matrix X ∈ Mn (C) then prove that A = On .

Solution. Suppose that A , 0, say Aij , 0 for some i, j. Let P be anypermutation matrix with Pij = 1 and let Q be the matrix obtained from Pby changing its ij-entry to 0. Finally let X = xQ where x ∈ C.

We have that detX = 0 and that

detX = det(A + X ) − detA

is a polynomial in x . Furthermore, the coefficient of xn−1 of this polyno-mial is ±Aij depending on the sign of the corresponding permutation. Sothe polynomial is not identically zero, a contradiction.

�


8 Consider the matrices A ∈ Mm×n and B ∈ Mn×m . If AB+Im is invertible

prove that BA + In is also invertible.



14

Solution. So we have to answer the question if −1 is a zero of the es-sentially same characteristic polynomials. AB and BA have quite similarcharacteristic polynomials. In fact if p(x) denotes the polynomial of AB,then the polynomial of BA will be q(x) = xn−mp(x). It is easy to see that−1 cannot be an eigenvalue of the AB matrix, otherwise it wouldn’t beinvertible. Now, let us assume that BA is not invertible. Then it musthave an eigenvalue of −1 and let x be the corresponding eigenvector.Hence:

(BA) x = −x⇒ AB (Ax) = −Ax

meaning that AB has an eigenvalue of −1 which is a contradiction. Theresult follows.

�


9 Let a = 2πn . Prove that the matrix

An =

1 1 · · · 1

cosa cos 2a · · · cosnacos 2a cos 4a · · · cos 2na...

.... . .

...cos(n − 1)a cos 2(n − 1)a · · · cos(n − 1)na

is invertible.

Solution. For a = 2πn , n ∈ N and for every k = 0, 1, . . . , n − 1 the

numbers

em2kπn i , m = 1, 2, . . . n − 1

are different n-th roots of unity.

So

n−1∑k=0

em2kπn i = 0 ⇒

n−1∑k=0

<(em 2kπ

n i)

= 0

⇒

n−1∑k=0

cos(m 2kπ

n

)= 0


15

⇒

n−1∑k=0

cos(mka) = 0 (1) , m = 1, 2, . . . n − 1

We have that

|An | =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 1 · · · 1 1

cosa cos 2a · · · cos(n − 1)a cosnacos 2a cos 4a · · · cos 2(n − 1)a cos 2na...

.... . .

......

cos(n − 2)a cos 2(n − 2)a... cos(n − 2)(n − 1)a cos(n − 2)na

cos(n − 1)a cos 2(n − 1)a · · · cos(n − 1)(n − 1)a cos(n − 1)na

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

Rn→∑nk=1

Rk=========

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 1 · · · 1 1


.... . .

......

cos(n − 2)a cos 2(n − 2)a... cos(n − 2)(n − 1)a cos(n − 2)na

n−1∑k=0

cos(ka)n−1∑k=0

cos(2ka) · · ·n−1∑k=0

cos((n − 1)ka)n−1∑k=0

cos(nka)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

(1)==

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 1 · · · 1 1


.... . .

......

cos(n − 2)a cos 2(n − 2)a · · · cos(n − 2)(n − 1)a cos(n − 2)na

0 0 · · · 0

n−1∑k=0

1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 1 · · · 1 1


.... . .

......

cos(n − 2)a cos 2(n − 2)a · · · cos(n − 2)(n − 1)a cos(n − 2)na0 0 · · · 0 n

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

16

= n

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 1 · · · 1 1


.... . .

......

cos(n − 2)a cos 2(n − 2)a · · · cos(n − 2)(n − 1)a cos(n − 2)na0 0 · · · 0 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

= n

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 1 · · · 1

cosa cos 2a · · · cos(n − 1)acos 2a cos 4a · · · cos 2(n − 1)a...

.... . .

...cos(n − 2)a cos 2(n − 2)a · · · cos(n − 2)(n − 1)a

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣= n |An−1| .

Adding the equations

|An | = n|An−1|

n|An−1| = n(n − 1)|An−2|...

...

n(n − 1) · · · 5 · 4 |A3| = n(n − 1) · · · 4 · 3 |A

2|

n(n − 1) · · · 4 · 3 |A2| = n(n − 1) · · · 3 · 2 |A

1|

we have that

|An | = n! |A1| = n!

So the matrix An is invertible forall n ∈ N.

�

10 Let V be a linear space over R such that dimRV < ∞ and f : V →V be a linear projection such that any non zero vector of V is an

eigenvector of f. Prove that there exists λ ∈ R such that f = λ Id where

Id is the identity endomorphism.

17

Solution. Let dimRV = n. Since any non zero vector is an eigenvectorit follows that every basis of V is also an eigenbasis. Let B =

{−→ei

}ni=1

besuch a basis and λi , i = 1, 2, . . . , n be the respective , not necessarilydistinct , eigenvalues of the eigenvectors of B. For the vector ~y =

n∑i=1

−→ei

which also happens to be eigenvector with eigenvalue λ , it holds that:

f (~y) = λ~y⇒ f

n∑i=1

−→ei

= λn∑i=1

−→ei

f linear=====⇒

n∑i=1

f(−→ei

)= λ

n∑i=1

−→ei

⇒

n∑i=1

λi−→ei = λ

n∑i=1

−→ei

⇒

n∑i=1

(λi − λ)−→ei = 0

ei linearly independent=============⇒ [(∀i = 1, 2, . . . , n) λi − λ = 0]

⇒ [(∀i = 1, 2, . . . , n) λi = λ]

But then for each ~x =n∑i=1

xi−→ei ∈ V it holds that:

f (~x) = f

n∑i=1

xi−→ei

=

n∑i=1

xif(−→ei

)=

n∑i=1

xiλ−→ei = λ

n∑i=1

xi−→ei = λ~x

The result follows.

�


11 Give an example of a matrix A that is not normal but eA is.

Solution. Let P =

(2 1

1 1

)which is clearly invertible and P−1 =

(1 −1

−1 2

).

Let D =

(2πi 0

0 −2πi

). Now consider

A = PDP−1 =

(2 1

1 1

) (2πi 0

0 −2πi

) (1 −1

−1 2

)=

(6πi −8πi4πi −6πi

)


18

It follows easily that the conjugate transpose matrix of A is

A∗ =

(−6πi −4πi8πi 6πi

)Also,

AA∗ =

(6πi −8πi4πi −6πi

) (−6πi −4πi8πi 6πi

)=

(100π2

72π2i72π2

52π2i

)and

A∗A =

(−6πi −4πi8πi 6πi

) (6πi −8πi4πi −6πi

)=

(52π2 −72π2

−72π2100π2

)Hence AA∗ , A∗A and therefore A is not normal. Furthermore,

expA = exp(PDP−1

)= P exp(D)P−1

= P I2×2 P

−1

= PP−1

= I2×2

�


12 If C,D are n×n symmetric real matrices we write C ≥ D if-f the matrix

C − D is non negative definite. Examine if

exp(A + B

2

)≤

expA + expB2

for each pair 2 × 2 real symmetric matrices A, B such that AB = BA.

Solution. Since A, B commute it follows from the properties of the ex-ponential function

expA + B

2

= expA

2

expB

2

(1)

Noting that


19

expA + expB2

− exp(A + B

2

)=

1

2

(exp

A

2

− expB

2

)2(2)

Setting M = expA

2

− expB

2

we must prove that M2 ≥ 0. Since A, B

commute so are expA

2

, expB

2

. Hence M is symmetric. Thus,

x>M2x = x>M>Mx = ‖Mx‖2 (3)

and we’re done.

�


13 Let ‖·‖2

denote ‖A‖2

=

√√n∑i=1

n∑j=1

∣∣∣aij∣∣∣2. Consider A ∈ Mn×n(C) and let

`1, . . . , `n be its eigenvalues. Prove that:

‖A‖22≥

n∑i=1

|`i |2

Solution. Noting that ‖A‖22

= tr(A∗A), it is clear that this norm is in-variant under conjugation by a unitary matrix. Since Schur tells us thatevery matrix is unitarily equivalent to an upper triangular matrix (whereof course the diagonal entries are just the eigenvalues) the claim followsimmediately.

�

Interpretation: The above result has a nice interpretation; It says thatfor a square complex matrix the sum of squares of the singular valuesis greater of equal than the sum of squares of absolute values of theeigenvalues, with equality iff the matrix is normal. Of course if itâ™snormal then the singular values are exactly the absolute values of theeigenvalues, and the above inequality implies that the reverse implicationholds as well.


14 Let n > 2 . Define the group

Q2n = 〈x, y | x2 = y2

n−2, y2

n−1= 1, x−1yx = y−1〉



20

Prove that Q2n/Z (Q

2n ) ' D

2n−1 where D is the dihedral group.

Solution. Using x−1yx = y−1 or equivalently yx = xy−1 we can write eachelement of Q

2n in the form x rys where r, s ∈ N ∪ {0}. Using x2 = y2

n−2

we may assume that r ∈ {0, 1}. Using y2n−1

= 1 we may also assume thats ∈ {0, 1, . . . , 2n−1 − 1}. It is easy to prove inductively that ytx = xy−t .

Let Z = Z(Q2n ). We prove that Z = {1, y2

n−2}. Obviously 1 ∈ Z.

Furthermore, y2n−2∈ Z since

y2n−2

(x rys) = xy−2n−2x r−1ys = · · · = x ry2

n−2ys = (x rys) y2

n−2

If yk ∈ Z (such that 0 ≤ k < 2n−1) then xyk = ykx = xy−k hence y2k = 1

and therefore k = 0 or k = 2n−2. If xyk ∈ Z then xyk+1 = yxyk = xyk−1

hence y2 = 1 which is a contradiction since n ≥ 2.

Therefore,

Q2n/Z = 〈x, y|x2 = y2

n−2= 1, yx = xy−1〉

which is precisely the dihedral group with 2n−1 elements.

�


15 Prove that Q is not a subcartesian product of infinite cyclic group.

Solution. Recall that a group G is subcartesian product of X groupsif and only if G is a residually X -group. We will show that Q is notresidually infinite cyclic group. Assume on the contrary that it is. Thenfor any 0 , m

n ∈ Q there exists Nm/n such that mn < Nn/m and that Q/Nm/n

is infinite cyclic. So for any k ∈ Z we have that kmn < Nn/m . Clearly Q is

not cyclic so there exists 0 , ab ∈ Nm/n . Hence

ma = bma

b∈ Nm/n

Thus, it follows that Q/Nm/n is finite which is a contradiction.

�




2

2PART

Calculus

21

23

1 Let n ∈ N. Prove that ∫ ∞

−∞

(n

x

)dx = 2

n

Solution. First of all,

(n

x

)=

n!x! (n − x)!

=n!

Γ(x + 1)Γ(n + 1 − x)

=n!

Γ(1 + x)(n − x) (n − 1 − x) · · · (1 − x) · Γ(1 − x)

=n!

xΓ(x)(n − x) (n − 1 − x) · · · (1 − x)Γ(1 − x)

=n!π·

sin πx(n − x) (n − 1 − x) · · · (1 − x) · x

due to the well known Γ(x + 1) = xΓ(x) and Γ(x)Γ(1 − x) = π csc πx .

However using the residue theorem we get that

1

(n − x) (n − 1 − x) · · · (1 − x) · x=

n∑k=0

1

x − k·

(−1)k

n!

(n

k

)Thus,

∫ ∞

−∞

(n

x

)dx =

∫ ∞

−∞

n!π·

sin πx(n − x) (n − 1 − x) · · · (1 − x) · x

dx

=

∫ ∞

−∞

n!π·

n∑k=0

sin πxx − k

·(−1)k

n!

(n

k

)dx

=1

π

n∑k=0

(−1)k(n

k

) ∫ ∞

−∞

sin πxx − k

dx

=1

π

n∑k=0

(−1)k(n

k

)(−1)kπ

=

n∑k=0

(n

k

)

24

= 2n

due to the binomial theorem and the well known fact

∫ ∞

−∞

sin πxx − k

dx = (−1)kπ

�


2 Evaluate the infinite product:

Π =

∞∏n=3

(n3 + 3n

)2

n6 − 64

Solution.

∞∏n=3

(n3 + 3n

)2

n6 − 64

=

∞∏n=3

n2

(n2 + 3

)2

n6 − 69

=

∞∏n=3

n2

(n2 + 3

)2(

n3 − 8

) (n3 + 8

)=

∞∏n=3

n2

(n2 + 3

)2

(n − 2)(n2 + 2n + 4

)(n + 2)

(n2 − 2n + 4

)=

∞∏n=3

n

n − 2

·n

n + 2

·n2 + 3

(n − 1)2 + 3

·n2 + 3

(n + 1)2 + 3

= limN→+∞

N (N − 1)1 · 2

·3 · 4

(N + 1) (N + 2)·N2 + 3

22 + 3

·

·3

2 + 3

(N + 1)2 + 3

=72

7

�




25

3 The sinc function is defined as:

sinc x =

1 , x = 0

sin xx

, x , 0

Prove that for any couple (α, �) of real numbers in (0, 1) the following

result holds:

∞∑n=−∞

sinc na sinc n� =π

max{α, �}

Solution. Since

∞∑n=−∞

sinc na sinc n� = 1 +2

α�

∞∑n=1

sinnα sinn�n2

due to the addition formulas for the sine and cosine functions it is enoughto prove the equality

f (θ) =

∞∑n=1

cosnθn2

=π2

6

−θ (2π − θ)

4

forall θ ∈ [0, 2π]

which is an immediate consequence of the Fourier series by integratingthe sawtooth wave function;

∞∑n=1

sinnθn

=π − θ

2

Hence,

∞∑n=−∞

sinc na sinc n� = 1 +f (|α − �|) + f (α + �)

α�

= 1 +1

4α�

((α − �)2 − (α + �)2 +

+ 2π (α − � − |α + �|))

= 1 +1

4α�(−4α� + 4πmin{α, �})

26

=π

max{α, �}

�


4 Let H(s)n denote the n - th harmonic number of weight s. Prove that

∞∑n=1

H2

n −H(2)n

(n + 1)(n + 2)= 2

Solution. Recall the generating function

∞∑n=1

(H2

n −H(2)n

)xn =

ln2(1 − x)1 − x

(1)

Thus integrating (1) from 0 to t we get that

∞∑n=1

(H2

n −H(2)n

) tn+1

n + 1

= −1

3

ln3(1 − t) (2)

Integrating (2) from 0 to 1 the result follows.�


5 Let ζ denote the Riemann zeta function and Lis the polylogarithm of

order s. Let s ≥ 2. Evaluate the limit:

` = limx→1

−(ζ (s) − Lis(x)) ln(1 − x)

Solution. We distinguish cases:

� For s > 2 we have:

` = limx→1

−(ζ (s) − Lis(x)) ln(1 − x)

= limx→1

−

ζ (s) − Lis(x)1 − x

· (1 − x) ln(1 − x)

=

(limx→1

−

ζ (s) − Lis(x)1 − x

)·

(limx→1

−(1 − x) ln(1 − x)

)= Li

′s(1) · 0

= 0



27

� The dilogarithm function satisfies the following identity

Li2(x) + Li

2(1 − x) = ζ (2) − ln x ln(1 − x) , x ∈ (0, 1)

Hence for s = 2 we have:

` = limx→1

−(ζ (2) − Li

2(x)) ln(1 − x)

= limx→1

−

(π2

6

− Li2(x)

)ln(1 − x)

= limx→1

−(Li

2(1 − x) + ln x ln(1 − x)) ln(1 − x)

= limx→0

+(Li

2(x) + ln x ln(1 − x)) ln x = 0

since

limx→0

+Li

2(x) ln x = lim

x→0+

Li2(x)x· x ln x = 1 · 0 = 0

�


6 Prove that

∞∑n=1

arctan(

10n

(3n2 + 2)(9n2 − 1)

)= ln 3 −

π

4

Solution. he key ingredient is the observation arctan x = arg(1 + ix).Then we note that

1 +10in(

3n2 + 2

) (9n2 − 1

) =(n − i) (3n − (1 − i)) (3n + i) (3n + (1 + i))

(3n − 1) (3n + 1)(3n2 + 2

)=

(1 − i

n

) (1 + i

3n−1

) (1 + i

3n+1

) (1 + i

3n

)1 + 2

3n2

Using the arg’s property we get that

arctan(

10n(3n2 + 2

) (9n2 − 1

)) = arctan(

1

3n − 1

)+ arctan

1

3n+


28

+ arctan(

1

3n + 1

)− arctan

1

n

Hence the initial sum telescopes:

∞∑n=1

arctan(

10n(3n2 + 2

) (9n2 − 1

)) = limN→+∞

N∑n=1

[arctan

(1

3n − 1

)+

+ arctan1

3n+ arctan

(1

3n + 1

)−

− arctan1

n

]= lim

N→+∞

3N+1∑n=N+1

arctan1

n− arctan 1

= limN→+∞

3N+1∑n=N+1

[1

n+ O

(1

n3

)]−

− arctan 1

= ln 3 −π

4

since3N+1∑n=N+1

1

n∼ ln

(3N + 1

N + 1

)which explains why ln 3 pops up.

�


7 Evaluate the sum

S =

∞∑n=1

arctan2

n2

Solution. First of all note that:

N∑n=1

arctan2

n2

=

N∑n=1

arctan(n + 1) − arctan(n − 1)

= −π

4

+ arctanN + arctan(N + 1)


29

Hence letting N → +∞ we have that

∞∑n=1

arctan2

n2

=3π

4

�


8 Prove that

limx→1−

√1 − x

∞∑n=0

xn2

=

√π

2

Solution. We are working near 1. It follows from the Integral Compar-ison Test that

∫ ∞

0

x t2

dt ≤∞∑n=0

xn2

≤ 1 +

∫ ∞

0

x t2

dt

However,

∫ ∞

0

x t2

dt =

∫ ∞

0

exp[−

(t√− log x

)2

]dt

=1√− log x

∫ ∞

0

e−u2

du

=1

2

√π

− log x

The result follows from the Sandwich Theorem.

�


9 Evaluate the limit

` = limn→+∞

n∑i=1

1

i−

n2∑i=n+1

1

i



30

Solution. We have that

n∑i=1

1

i−

n2∑i=n+1

1

i= Hn − (−γ +Hn2 + γ −Hn)

= 2Hn −Hn2

= 2

(logn + γ + O

(1

n

))−

(logn2 + γ + O

(1

n2

))= γ + O

(1

n

)

since Hn = logn + γ + O(

1

n

). Thus the limit is γ.

�


10 Evaluate the integral:

J =

∫ ∞

0

x log xe−√xdx

Solution. We are evaluating the Mellin transform of the function f (x) =

e−√x .

M (f ) =

∫ ∞

0

xs−1e−√xdx

u=√x

=====2

∫ ∞

0

u2s−1e−u du

= 2Γ (2s)

where Γ is the Euler’s Gamma function. Hence,

∫ ∞

0

x log xe−√xdx =M′(f )

∣∣∣∣∣s=2

= (2Γ(2s))′∣∣∣∣∣s=2

= 4Γ(2s)ψ(0)(2s)∣∣∣∣∣s=2


31

= 4Γ(4)ψ(0)(4)

= 4 · 6 ·

(11

6

− γ

)= 44 − 24γ

where γ is the Euler - Mascheroni constant and ψ(0) is the digamma.

�



32

3

3PART

Analysis

33

35

1 Prove that there exists no continuous and 1 − 1 map ( depiction ) from

a sphere to a proper subset of it.

Solution. Let x0∈ Sn and consider f : Sn → Sn \ {x0

} that is a continousand 1−1 map . We consider the stereographic projection g : Sn\{x0

} → Rn

that is also 1− 1 and continous. Hence the composition (g ◦ f ) : Sn → Rn

is continuous and 1 − 1 which is an obscurity due to Borsuk-Ulam.

�


2 Let a , b be real numbers. Consider the function:

f (x) =

{a , 0 ≤ x < cb , c ≤ x ≤ 1

as well as the polynomial Bn(x) =

n∑i=0

(n

i

)x i (1 − x)n−i f

( in

). Evaluate

the limit limn→+∞

Bn(c).

Solution. The polynomial Bn is nothing else than a Bernstein one. fis Riemann integrable on [0, 1] and c is a discontinuity of a first kind.Thus,

limn→+∞

n∑i=0

(n

i

)(1 − c)n−i cif

( in

)=f (c−) + f (c+)

2

=a + b

2

where f (c−) = limx→c−

f (x) and f (c+) = limx→c+

f (x).

�


3 Prove that ρ(x, y) = |f (x) − f (y)| where

f (x) =

{arctan x , x ∈ Q

arctan(x + 1) , x ∈ R \ Q

defines a metric on R.

Solution. (i) Obviously ρ(x, y) = |f (x) − f (y)| = |f (y) − f (x)| = ρ(y, x)and



36

(ii) ρ(x, y) = |f (x) − f (y)| ≤ |f (x) − f (z)|+ |f (z) − f (y)| = ρ(x, z)+ρ(z, y)as well as ρ(x, y) ≥ 0.

(iii) Let ρ(x, y) = |f (x) − f (y)| = 0. It can’t one of x, y be rational andthe other irrational since then it would be arctan x = arctan(y + 1),hence x = y + 1 which is an obscurity. Hence both of them areeither rational or irrational. In both cases x = y.

Hence ρ is a metric.�


4 In the metric space (Q, |·|) , find a decreasing sequence {An}n∈N of non

closed subsets of Q such that diam (An)n→+∞−−−−−→ 0 and

∞⋂n=1

An = ∅.

Solution. Let us pick An = Q ∩[√

2 − 1

n ,√2 + 1

n

]. Indeed,

diam(An) =

∣∣∣∣∣∣√2 +1

n−

(√2 −

1

n

)∣∣∣∣∣∣ =2

n

n→+∞−−−−−→ 0

and

∞⋂n=1

An =

∞⋂n=1

(Q ∩

[√2 −

1

n,√2 +

1

n

])=

∞⋂n=1

[√2 −

1

n,√2 +

1

n

] ∩ Q=

{√2

}∩ Q

= ∅

�


5 Let f : R→ R be a continuous function such that f (x) ≥ 0 forall x ∈ R

and

∞∫−∞

f (x) dx = 1 . Let r ≥ 0. Define:

In(r) =

(x2

1+x2

2+···+x2

n≤r

f (x1)f (x

2) · · · f (xn) d(x

1, x

2, . . . , xn)

Evaluate limn→+∞

In(r) for a given r .



37

Solution. If∫ √

r

−√rf (x) dx = a < 1 then

In(r) ≤

∫ √r

−√rf (x) dx

n = ann→+∞−−−−−→ 0

We can assume that∫ √

r

−√rf (x) dx = 1. From the continuity of f it must

hold f (x) = 0 for |x | ≥√r . Again, from the continuity of f we deduce

that f has a maximum value M . Thus,

In(r) ≤ MnVn√rn

where Vn is the volume of the unit sphere in Rn . It is known howeverthat

Vn =πn/2

Γ(n/2 + 1)

and hence In(r)→ 0 as n → +∞.�


6 Let {yn}n∈N be a sequence of real numbers such that for all real se-

quences {xn}n∈N with limn→+∞

xn = 0 the series

∞∑n=1

xnyn converges. Prove

that the series

∞∑n=1

|yn | also converges.

Solution. Suppose on the contrary that the series diverges. Define thesequence of the naturals n

0, n

1, n

2, . . . as follows. We set n

0= 1. Having

defined n0, n

1, . . . , nk we define nk+1

to be the least natural m that isgreater than nk and satisfies

m∑r=nk+1

|yr | > 1. The sequence is well defined

because the series∞∑n=1

|yn | diverges.

For nk < i ≤ nk+1we define xi =

sign(yi)k + 1

. Then we note that xn → 0

and

nk+1∑i=nk+1

xiyi =

nk∑i=nk+1

|yi |

k + 1

>1

k + 1


38

thus the series∞∑n=1

xnyn diverges. This leads to contradiction hence theresult.

�


7 Let Vn(1) be the volume of the sphere centered at 0 and radius 1 in

Rn . Prove that for n ≥ 3 it holds that

Vn(1) =2π

nVn−2(1)

Solution. The volume of the sphere in Rn is given by:

Vn(1) =

(x2

1+x2

2+···+x2

n≤1

1 d (x1, x

2, . . . , xn)

Parametrize the sphere by

x1

= r cos θ1

x2

= r sin θ1cos θ

2

x3

= r sin θ1sin θ

1cos θ

3

...xn−1 = r sin θ

1sin θ

2· · · sin θn−2 cos θn−1

xn = r sin θ1sin θ

2· · · sin θn−1

taking

0 ≤ r ≤ 1

0 ≤ θi ≤ π forall i = 1, 2, . . . , n − 2

0 ≤ θn−1 < 2π

It then follows from the Change of Variables formula that the rectan-gular volume element dV = dx

1dx

2· · · dxn can be written in spherical

coordinates as

dV =

∣∣∣∣∣∣∣∣det

∂x1

∂(r, θj

)∣∣∣∣∣∣∣∣ dr d (θ

1, θ

2, . . . , θn−1)

= rn−1 sinn−2 θ1sinn−3 θ

2· · · sin θn−2dr d (θ

1, θ

2, . . . , θn−1)


39

Thus,

Vn(1) =

(x2

1+x2

2+···+x2

n≤1

1 d (x1, x

2, . . . , xn)

=

∫2π

0

∫ π

0

· · ·

∫1

0

rn−1 sinn−2 θ1· · · sin θn−2dr d (θ

1, . . . , θn−1)

=

∫2π

0

dθn−1

∫1

0

rn−1 dr

∫ π

0

sinn−2 θ dθ · · ·

∫ π

0

sin θ dθ

=2π

n

∫ π

0


∫ π

0

sin θ dθ

Hence, ∫ π

0


∫ π

0

sin θ dθ =n

2πVn(1) (1)

In particular since n − 4 = (n − 2) − 2 we get that:∫ π

0

sinn−4 dθ · · ·

∫ π

0

sin θ dθ =n − 2

2πVn−2(1) (2)

Using (2) as well as Wallis’ integral we are able to prove the result. Letus assume that n is even, then:

Vn(1) =2π

n

∫ π

0

sinn−2 θ dθ

∫ π

0

sinn−3 θ dθ

∫ π

0


· · ·

∫ π

0

sin θ dθ

=2π

n

∫ π

0

sinn−2 θ dθ

∫ π

0

sinn−3 θ dθ ·n − 2

2πVn−2(1)

=2π

n·

(n − 3

n − 2

· · ·1

2

· π

)·

(n − 4

n − 3

· · ·2

3

· 2

)·n − 2

2πVn−2(1)

=2π

n· Vn−2(1)

If n is odd we work similarly.�



40

8 Let ∆ : {(x, y) ∈ R2 : x2 + y2 ≤ 4x}. Evaluate the integral:

J =

"∆

arctan exy dy dx

Solution. The key observation to nail the integral is that the domain ofintegration is symmetric with respect to the x axis.

x

y

1 2 3 40

So,

J =

"∆

arctan exy d(y, x)

=

∫4

0

∫ √4x−x2

−√4x−x2

arctan exy d(y, x)

=

∫4

0

∫0

−√4x−x2

arctan exy d(y, x) +

∫4

0

∫ √4x−x2

0

arctan exy d(y, x)

=

∫4

0

∫ √4x−x2

0

arctan exy d(y, x) +

∫4

0

∫ √4x−x2

0

arctan e−xy d(y, x)

=

∫4

0

∫ √4x−x2

0

π

2

d(y, x)

=π

2

∫4

0

√4x − x2

dx

=π

2

· 2π

= π2

41

since

arctan x + arctan1

x=π

2

forall x > 0

�


9 Let n ≥ 1 be an integer and let f : [0, 1]→ R be a continuous function.

Suppose that

∫1

0

xkf (x) dx = 1 for all 0 ≤ k ≤ n − 1. Show that

∫1

0

f 2(x)dx ≥ n2

Solution. We state a lemma:

Lemma

For an integer n ≥ 1 the n×n Hilbert matrix is defined byHn = [aij]where

aij =1

i + j − 1

, 1 ≤ i, j ≤ n

It is known that Hn is invertible and if H−1n = [bij] then∑i,jbij = n2.

SinceHn , the n×n Hilbert matrix , is invertible there exist real numbersp

0, p

1, . . . , pn−1 such that

n∑i=1

pi−1i + j − 1

= 1 , 1 ≤ j ≤ n.

So the polynomial p(x) =n−1∑k=0

pkxk satisfies the conditions

∫1

0

xkp(x) dx = 1 , 0 ≤ k ≤ n − 1

Clearlyn−1∑k=0

pk is the sum of all the entries of H−1n and son−1∑k=0

pk = n2.

Now let f be a real-valued continuous function on [0, 1] such that


42

∫1

0

xkf (x) dx = 1 , 0 ≤ k ≤ n − 1

Let p(x) be the above polynomial.Then since

f 2(x) − 2f (x)p(x) + p2(x) = (f (x) − p(x))2 ≥ 0

integrating gives:

∫1

0

(f (x))2 dx ≥ 2

∫1

0

f (x)p(x) dx −

∫1

0

(p(x))2 dx

= 2

n−1∑k=0

pk

∫1

0

xkf (x) dx −n−1∑k=0

pk

∫1

0

xkp(x) dx

= 2

n−1∑k=0

pk −n−1∑k=0

pk

=

n−1∑k=0

pk

= n2

and the result follows.�


10 Let {an}n∈N be a decreasing sequence. Prove that

∞∑n=1

an sinnx converges

uniformly if-f nann→+∞−−−−−→ 0.

Solution. Since nan → 0 then for a given ϸ > 0 there exists positiveinteger n

0such that if n ≥ n

0to hold

|nan | − nan <ϸ

2(π + 1)(1)

It suffices to prove the result for x ∈ [0, π] due to symmetry and period-

icity. So, it suffices to prove that for n ≥ n0

it holds

∣∣∣∣∣∣∣n+p∑k=n+1

ak sin kx

∣∣∣∣∣∣∣ < ϸ

for all x ∈ [0, π] and all p ∈ N.We distinguish cases:


43

� x ∈[0, π

n+p

];

∣∣∣∣∣∣∣n+p∑k=n+1

ak sin kx

∣∣∣∣∣∣∣ =

n+p∑k=n+1

ak sin kx

≤

n+p∑k=n+1

akkx (sin kx ≤ kx, x ≥ 0)

=

n+p∑k=n+1

(kak)x

(1)≤

ϸ

2(π + 1)πp

n + p

< ϸ

� x ∈[πn+p , π

];

∣∣∣∣∣∣∣n+p∑k=n+1

ak sin kx

∣∣∣∣∣∣∣ ≤b πx c∑k=n+1

akkx +

∣∣∣∣∣∣∣∣∣n+p∑

k=b πx c+1

ak sin kx

∣∣∣∣∣∣∣∣∣≤

b πx c∑k=n+1

akkx +2am+1

sin x2

(summation by parts)

≤ϸ

2(π + 1)

(⌊πx

⌋− n

)x +

2am+1

sin x2

≤ϸ

2(π + 1)mx + 2am+1

π

x

(since

2x

π≤ sin x

)≤

ϸ

2(π + 1)π + 2am+1

(m + 1)

<ϸ

2

+ 2 ·ϸ

2(π + 1)< ϸ

where m =

⌊πx

⌋.

This completes the proof of the first part. Now the converse is mucheasier. For each n such that k ≤ n ≤ 2k − 1 we have nπ

6k ∈(π6, π

2

). Hence

1

2≤ sin nπ

6k . Therefore, picking x = π6k we have:

44

2k−1∑n=k

an sinnπ

6k≥

1

2

2k−1∑n=k

an

≥1

2

2k−1∑n=k

a2k−1

=1

2

ka2k−1

≥1

4

(2k − 1)a2k−1

≥ 0 (2)

Since the series converges uniformly we have that2k−1∑n=k

an sinnx → 0

uniformly. More specifically,2k−1∑n=k

an sin nπ6k → 0 as k → +∞. Using the

sandwich theorem it follows that (2k − 1)a2k−1 → 0. It only remains to

prove that 2ka2k → 0. This follows from

0 ≤ 2ka2k ≤ 2ka

2k−1 ≤ 4(2k − 1)a2k−1 → 0

�


11 Let f : [0, 1]→ R be a continous function such that f (0) = 0 and

∫1

0

f (x) dx =

∫1

0

xf (x) dx (1)

Prove that there exists a c ∈ (0, 1) such that∫ c

0

xf (x) dx =c

2

∫ c

0

f (x) dx

Solution. Let I(x) =

∫x

0

f (t) dt and G(x) =

∫x

0

I(t) dt. Integrating byparts (1) reveals that

∫1

0

I(t) dt = 0 =⇒ G(1) = 0


45

Now let us consider the function K(x) =G(x)x2

. It holds that K(1) = 0.As we can also see using two consecutive DeL’ Hospital’s Rules , it alsoholds that lim

x→0

K(x) = 0. So, by Rolle’s theorem there exists a c ∈ (0, 1)such that

G (c) =c

2

I(c)

However integration by parts reveals that

G(c) = cI(c) −∫ c

0

xf (x) dx

and thus∫ c

0

xf (x) dx =c

2

∫ c

0

f (x) dx which is the desired output.

�


12 Let f : [0, 1]→ R be a continous function such that

∫1

0

f (x) dx =

∫1

0

xf (x) dx (1)

Prove that there exists a c ∈ (0, 1) such that

cf (c) = 2

∫0

cf (x) dx

Solution. Let F be a primitive of f . Consider the function G(x) =

x2F (x). Trivially G(0) = 0. Now, we note that:

∫1

0

F (x) dx =

∫1

0

(x)′F (x) dx

= [xF (x)]10−

∫1

0

xf (x) dx

= F (1) −∫

1

0

xf (x) dx

= 0


46

because F is of the form F (x) =

∫x

0

f (t) dt. Thus

F (1) =

∫1

0

f (t) dt =

∫1

0

tf (t) dt

due to the initial assumptions. Applying the Integral Mean Value Theo-rem we have that there exists an m ∈ (0, 1) such that

∫1

0

F (x) dx = 0 = F (m)

Thus G(0) = 0 = G(m). The conclusion now follows from Rolle’s theo-rem.

�


13 Let f : [0, 1] → R be a differentiable function with continuous deriva-

tive such that f (0) = f ′(0) = 0. If

∫1

0

f (x) dx =

∫1

0

xf (x) dx (1)

then prove that there exists a c ∈ (0, 1) such that∫ c

0

xf (x) dx =3c

2

∫ c

0

f (x) dx

Solution. We consider the function

F (x) = x2

∫ x

0

tf (t) dt − x3

∫ x

0

f (t) dt , x ∈ [0, 1]

which is continuous in [0, 1] and differentiable in (0, 1). Rolle’s theoremyields the result.

�

14 Let f : R→ R be a continuous function such that

f (x) = f

(x +

1

√n

)where n = 1, 2, . . . . Prove that f is constant.


47

Solution. Since it holds

f (x) = f

(x +

k√n

)where k ∈ Z and n = 1, 2, . . . . The set

{k√n, k ∈ Z, n = 1, 2, . . .

}is

dense in R and since f is continuous and characterised by its values ina dense set we conclude that f is constant since

f (0) = f

(k√n

)�


15 Let n ∈ N and f be an entire function. Prove that for any arbitrary

positive numbers a, b it holds that:∫2π

0

e−intf (z + aeit) dt∫2π

0

e−intf (z + beit) dt=

(ab

)nSolution. Since our function is entire this means that it is holomorphicand can be represented in the form

f (x) =

∞∑m=0

am (x − z)m

This series converges uniformly on [0, 2π] thus we can interchange sum-mation and integral. Hence:

∫2π

0

e−intf(z + aeit

)dt =

∫2π

0

∞∑m=0

amameit(m−n)

dt

=

∞∑m=0

amam

∫2π

0

eit(m−n)dt

= 2π∞∑m=0

amamδmn


48

= 2πanan

where δmn is Kronecker’s delta. Similarly for the denominator. Dividingwe get the result.

�


16 Define

f (z) =1

z·1 − 2z

z − 2

· · ·1 − 10z

z − 10

Evaluate the contour integral �|z|=100

f (z) dz

Solution. We are applying the substitution u = 1

z thus:

�|z|=100

f (z) dz = −

|w|=1/100

f

(1

w

)dw

w2

=

�|w|=1/100

1

w

5∏n=1

w − 2n

1 − 2nwdw

= −2πi3840

since the function g(w) =1

w

5∏n=1

w − 2n

1 − 2nwhas only one pole in the specific

contour , namely w = 0.

�


17 Let f be analytic in the disk |z| < 2. Prove that:

1

2πi

∮|z|=1

f (z)z − α

dz =

f (0) , |α| < 1

f (0) − f(

1

α

), |α| > 1



49

Solution. It follows from Taylor that f (z) =∞∑n=0

cnzn and the conver-

gence is uniform. Hence,

1

2πi

∮|z|=1

f (z)z − α

dz =1

2πi

∮|z|=1

∞∑n=0

cn zn

z − αdz

=

∞∑n=0

cn2πi

∮|z|=1

zn

z − αdz

|z|=1⇒zn= 1

zn==========

∞∑n=0

cn2πi

∮|z|=1

1

zn(z − α)dz

We have that Res

(1

zn(z − α);α

)=

1

αnand for n ≥ 1 we also have that

Res

(1

zn(z − α); 0

)= −

1

αndue to

1

zn(z − α)= −

1

αzn1

1 − z/α= −

1

αzn

(1 +

z

α+z2

α2

+ · · ·

)So if |α| < 1 then α lies within the disk |z| = 1; hence the integral equalsc0

= f (0) whereas if |α| > 1 then α lies outside the disk |z| = 1; hence

1

2πi

∮|z|=1

f (z)z − α

dz = −

∞∑n=1

cnαn

= cn −∞∑n=0

cnαn

= f (0) −

∞∑n=0

cnαn

= f (0) − f

(1

α

)

�



50

18 Prove that

limx→1−

√1 − x

∞∑n=0

xn2

=

√π

2

Solution. Near 1, it follows from the Integral Comparison Test that

∫ ∞

0

x t2

dt ≤∞∑n=0

xn2

≤ 1 +

∫ ∞

0

x t2

dt

However,

∫ ∞

0

x t2

dt =

∫ ∞

0

exp[−

(t√− log x

)2

]dt

=1√− log x

∫ ∞

0

e−u2

du

=1

2

√π

− log x

The result follows from the Sandwich Theorem.

�


19 Let f : R2 → R2. If:

� f (0) = 0

� |f (u) − f (v)| = |u − v| for all u, v

then prove that f is linear.

Solution. For convenience, identify R2 with C here. Then note that forany such function f : C → C, also z

1· f (z) a solution for any point

z1

on the unit circle. Also f (z) is a solution. Note that |f (1)| = 1 andhence we can wlog assume that f (1) = 1. So f (i) is a point on the unitcircle with distance

√2 to 1. Hence f (i) = ±i, so w.l.o.g. assume that

f (i) = i. But then for any z ∈ C, both z and f (z) have the same distanceto 0, 1 and i. So supposing z , f (z), all 0, 1, i lie on the perpendicularbisector between these points and in particular 0, 1 and i are collinearwhich clearly is absurd. Hence f (z) = z for all z which proves the claim.


51

�


20 Let ·! denote the factorial of a real number; that is x! = Γ(x + 1).Evaluate the limit:

` = limx→n

x! − n!x − n

Solution. It holds that

limx→n

x! − n!x − n

= limx→n

Γ(x + 1) − Γ(n + 1)x − n

= Γ′(n + 1)

= Γ(n + 1)ψ(0)(n + 1)= n! (Hn − γ)

whereHn denotes the n-th harmonic number and γ the Euler - Mascheroniconstant.

�




52

4

4PART

General Mathematics -

Inequalities

53

55

1 Let n ∈ N. Prove that the number

n =√

1111 · · · 11︸︷︷︸2n

− 2222 · · · 22︸︷︷︸n

is rational.

Solution. We have successively that

n =√

111 · · · 11︸︷︷︸2n

− 222 · · · 22︸︷︷︸n

=

√√2n−1∑k=0

10k − 2

n−1∑k=0

10k

=

√1

9

(10

2n − 1

)−

2

9

(10n − 1)

=

√(10n − 1)2

9

=10

n − 1

3

= 3

(1 + 10 + 100 + · · · + 10

n−1)

= 333 · · · 33︸︷︷︸n

since every number n ∈ N has an expansion of the form

n−1∑k=0

ak10k, 0 ≤ ak ≤ 9

�


2 Find all positive integers α such that it holds that

1

α= 0.α (1)

where · stands for the period.


56

Solution. We have successively:

1

α= 0.α ⇔

1

α= 0.ααα . . .

⇔1

α=α

10

∞∑i=0

1

10i

⇔1

α=α

9

α>0

⇐==⇒ α = 3

�


3 Let ABC be a given triangle such that A = 2C. Prove that

a2 = c2 + bc

Solution. We are working on the following figure.

B C

A

Dφ

2φφ φ

a

bc

Let AD be the bisector of BAC. Then:

{M ADB ∼M ABC

AD = DC⇒

AD

AC=AB

BCAD = DC

⇒DC

AC=AB

BCDC= a·b

b+c ,AC=B,AB=c,BC=a=================⇒

a·bb+c

b=c

a


57

⇒a

b + c=c

a⇒ a2 = c2 + bc

and the result follows.

�


4 Prove that in any triangle ABC the following equivalence relation holds:

a2 = ab + c2 ⇔ A = 90◦ +

C

2

Solution. We are working on the following figure

Let I be the incenter of the triangle. Thus,

A = 900 +

C

2

⇔ A = AIB ⇔ BAI = AEB =A

2

meaning that AB is tangent to the circumcircle of the triangle AIE. Thus,

c2 = BI · BE (1)

Hence,

BI

BE=

a + c

a + b + c

(1)⇔ c2 =

a + c

a + b + c· BE2 =

a + c

a + b + c

(ac −

b2

(a + c)2

)⇔ c2 =

ac (a + c − b)a + c


58

⇔ a2 = ab + c2

�


5 Let ϕ denote the golden ratio. Prove that:

sin 666◦ = −

ϕ

2

Solution. First of all we note that

sin 666◦ = sin (630◦ + 36

◦)= sin (7 · 90◦ + 36

◦)= − cos 36

◦

We are working on theisosceles triangle.

A

B Γx

E

x1 − x

x

72◦

36◦

36◦

We note that the triangles ABΓ , ΓBE are similar. Thus,

x

1 − x=

1

x⇔ x2 + x − 1 = 0

x>0

⇔ x =

√5 − 1

2

It follows from the law of cosines on the triangle ΓBE that:

BE2 = EΓ2 + BΓ2 − 2EΓ · BΓ · cos 36

◦ ⇔ (1 − x)2 = x2 + x2 − 2x2 cos 36◦

⇔ 1 − 2x + x2 = x2 + x2 − 2x2 cos 36◦

⇔ 1 − 2x = x2 − 2x2 cos 36◦

⇔ 1 − 2x − x2 = −2x2 cos 36◦

⇔ x2 + 2x − 1 = 2x2 cos 36◦

⇔ cos 36◦ =

x2 + 2x − 1

2x2

⇔ cos 36◦ =

x2 + x − 1 + x

2x2

x2+x−1=0

⇐=====⇒ cos 36◦ =

x

2x2


59

⇔ cos 36◦ =

1

2xx=

√5−12

⇐====⇒ cos 36◦ =

√5 + 1

4

⇔ cos 36◦ =

ϕ

2

Thus, sin 666◦ = −

ϕ

2

.

�


6 Evaluate the integral

J =

∫1

0

√4 − x2

dx

using geometric methods.

Solution. We are working on the following figure

x

y

O

Γ

-2 -1 2

A

B

Thus,

∫1

0

√4 − x2

dx =

(4

OAB

)+ (AOΓ)

=

√3

2

+ π · 22 ·30

360

=

√3

2

+4π

12

=π

3

+

√3

2


60

since the red angle is 60◦ due to the triangle OAB since tan O =

√3 (

OB = 1 , AB =√3 ). Therefore , the green angle is 180

◦−90◦−60◦ = 30◦.

Finally, the area of the circular sector is equal to

E = πr2 ·µ

360

where µ = 30 and r = 2.

�


7 Prove that

cosπ

5

+ cos3π

5

=1

2

Solution. We are making use of the formula

cosA + cosB = 2 cosA + B

2

cosA − B

2

thus

cosπ

5

+ cos3π

5

= 2 cos2π

5

cosπ

5

=2 cos π

5sin π

5cos 2π

5

sin π5

=sin 2π

5cos 2π

5

sin π5

=2 sin 2π

5cos 2π

5

2 sin π5

=sin 4π

5

2 sin π5

=sin

(π − π

5

)2 sin π

5

=sin π

5

2 sin π5

=1

2


61

�


8 Prove that

cosπ

7

+ cos3π

7

+ cos5π

7

=1

2

Solution. 1st solution: Consider the tridiagonal matrix A =

1 1 0

1 0 1

0 1 0

.

Its eigenvalues are 2 cosπ

7

, 2 cos3π

7

, 2 cos5π

7

. Hence,

Tr (A) = 1 = 2

(cos

π

7

+ cos3π

7

+ cos5π

7

)and the result follows.

2nd solution: The points(cos

(2n + 1)π7

, sin(2n + 1)π

7

), n ∈ {0, 1, 2, . . . , 6}

are vertices of a regular septagon. It follows from the rotational symme-try of the polygon that:

6∑i=0

cos(2i + 1) π

7

= 0⇔ 2

(cos

π

7

+ cos3π

7

+ cos5π

7

)+ cos π = 0

⇔ cosπ

7

+ cos3π

7

+ cos5π

7

=1

2

�


9 Given a heptagon of side a and diagonals b, c such that b < c

a

b

c



62

prove that:

b2

a2

+c2

b2

+a2

c2

= 5

Solution. Let a be the side of the heptagon and b, c be its diagonalsrespectively. It holds that

b2 − a2 = ac (1)

c2 − b2 = ab (2)

a2 − c2 = −bc (3)

c

a+a

b−b

c= 2 (4)

Equation (4) comes naturally from Vieta’s formulae since ca ,

ab , −

bc are

the roots of the equation t3 − 2t2 − t + 1 = 0. Thus,

b2

a2

+c2

b2

+a2

c2

=b2 − a2 + a2

a2

+c2 − b2 + b2

b2

+a2 − c2 + c2

c2

=

(b2 − a2

a2

+a2

a2

)+

(c2 − b2

b2

+b2

b2

)+

(a2 − c2

c2

+c2

c2

)=

(aca2

+ 1

)+

(abb2

+ 1

)+

(−bc

c2

+ 1

)=c

a+a

b−b

c+ 3

= 2 + 3

= 5

�


10 Let a, b, c be positive real numbers such that a + b + c = π. Prove the

following trigonometric identities:

(i) sina + sin b + sin c = 4 cosa

2

cosb

2

cosc

2


63

(ii) cosa + cos b + cos c = 1 + 4 sina

2

sinb

2

sinc

2

Solution. (i) 1st solution: We have successively:

sina + sin b + sin c = 2 sina + b

2

cosa − b

2

+ sin c

= 2 sin(π2

−c

2

)cos

a − b

2

+ 2 sinc

2

cosc

2

= 2 cosc

2

cosa − b

2

+ 2 sinc

2

cosc

2

= 2 cosc

2

(cos

a − b

2

+ sinc

2

)= 2 cos

c

2

(cos

a − b

2

+ cosa + b

2

)= 2 cos

c

2

· 2 cosa

2

cosb

2

= 4 cosa

2

cosb

2

cosc

2

2nd solution: Since cosA

2

=

√τ(τ − a)bc

and E =abc

4Rwe have:

4 cosA

2

cosB

2

cosC

2

= 4

√τ3(τ − a)(τ − b)(τ − c)

a2b2c2

=4τ√τ(τ − a)(τ − b)(τ − c)

abc

=4τE

abc

=4τ

4R

=a + b + c

2R= sinA + sinB + sinC

where τ denotes the semiperimeter.

(ii) We have successively:

cosa + cos b + cos c = 2 cosa + b

2

cosa − b

2

+ cos c

64

= 2 cos(π2

−c

2

)cos

a − b

2

+ cos c

= 2 sinc

2

cosa − b

2

+

(1 − 2 sin2

c

2

)= 1 + 2 sin

c

2

(cos

a − b

2

− sinc

2

)= 1 + 2 sin

c

2

[cos

a − b

2

− sin(π2

−a + b

2

)]= 1 + 2 sin

c

2

(cos

a − b

2

− cosa + b

2

)= 1 + 2 sin

c

2

· 2 sina

2

sinb

2

= 1 + 4 sina

2

sinb

2

sinc

2

�


11 Let f : R→ R be the Dirichlet function;

f (x) =

{1 , x ∈ Q0 , x ∈ R \ Q

Evaluate the limit

` = limn→+∞

1

n

(f (1) + f

(√2

)+ f

(√3

)+ · · · + f

(√n))

Solution. We simply note that

0 ≤f (1) + f

(√2

)+ f

(√3

)+ · · · + f

(√n)

n=bnc

n≤

1

√n

and the limit follows to be 0. The reason why

f (1) + f(√

2

)+ f

(√3

)+ · · · + f

(√n)

= bnc

is because√m is rational if-f m is a perfect square.

�




65

12 Prove that the following inequality holds in any triangle:

(sin

A

2

+ sinB

2

)2+

(sin

B

2

+ sinC

2

)2+

(sin

C

2

+ sinA

2

)2≤ 3

Solution. Let s denote the semiperimeter of the triangle. Using thecosine theorem we have that

c2 = a2 + b2 − 2ab cosC = (a − b)2 + 4ab sin2C

2

from which it follows that

sin2C

2

=(s − a) (s − b)

ab(1)

sin2A

2

=(s − b) (s − c)

bc(2)

sin2B

2

=(s − a) (s − c)

ac(3)

Thus, by Cauchy’s inequality we have:

(sin

A

2

+ sinB

2

)2=s − c

c

(√s − b ·

1

√b

+√s − a ·

1

√a

)2

≤s − c

c·

(1

a+

1

b

)· (s − a + s − b)

=(a + b − c) (a + b) c

2abc

=

(a2 + b2

)c − c2 (a + b) + 2abc

2abc

= 1 +

(a2 + b2

)c − c2 (a + b)

2abc

Hence,

∑(sin

A

2

+ sinB

2

)2≤ 3 +

∑ (a2 + b2

)c − c2 (a + b)

2abc

66

= 3 +1

2abc

(∑(a2 + b2

)c −

∑c2 (a + b)

)= 3

�


13 Let ABΓ be a triangle. Show that

1 − cos B

sin B

·1 − cos Γ

sin Γ= 1 −

2a

a + b + c

Solution. Let s denote the semiperimeter of the triangle. On account ofthe well known relations,

1 − cos B

sin B

= tanB

2

=

√(s − a) (s − c)s (s − b)

1 − cos Γ

sin Γ= tan

Γ

2

=

√(s − a) (s − b)s (s − c)

we have

1 − cos B

sin B

·1 − cos Γ

sin Γ= tan

B

2

· tanΓ

2

=

√(s − a) (s − c)s (s − b)

·

√(s − a) (s − b)s (s − c)

=s − a

s

= 1 −2a

a + b + c

�


14 Prove that in any acute triangle ABC the following inequality holds:

sinA sinBcosC

+sinB sinC

cosA+

sinC sinAcosB

≥9

2



67

Solution. Since A + B + C = π it holds that

cosC = − cos(A + B) = − cosA cosB + sinA sinB (1)

and thus

sinA sinBcosC

= 1 +cosA cosB

cosC= 1 +

tanCtanA + tanB

(2)

Using Nesbitt’s inequality we see that

∑ sinA sinBcosC

=∑(

1 +tanC

tanA + tanB

)= 3 +

∑ tanCtanA + tanB

≥ 3 +3

2

=9

2

Equality holds if-f A = B = C = π3.

�


15 Show that in any triangle ABC with area A the following holds:

1

2

sinA + sinB + sinC1

a + 1

b + 1

c

≤ ASolution. Let s is the semiperimeter , R the circumradius and r theinradius. From the law of sines we find

sinA + sinB + sinC =s

R(1)

as well as

A = rs (2)

Now,


68

1

2A

∑sinA ≤

∑1

a2

Substitute the preceding equalities into the last inequality and simplifyingwe obtain

1

2rR≤

∑1

a2

From

A = rs =abc

4R

the last inequality becomes

a + b + c

abc=

1

bc+

1

ca+

1

ab≤

1

a2

+1

b2

+1

c2

which is true by the rearrangement inequality.

�


16 Given a triangle ABC let ma , mb, mc denote the median points of the

sides a, b, c respectively. Prove that

(ma +mb +mc)2 + 4r (R − 2r) ≤ (4R + r)2

where R denotes the the circumradius and r the inradius respectively.

Solution. Let s denote the semiperimeter. We have successively

(∑ma

)2

=∑

m2

a + 2

∑mamb

=3

4

∑a2 + 2

∑mamb

mamb≤2c2+ab

4

≤3

4

∑a2 +

1

2

∑(2c2 + ab

)=

3

4

∑a2 +

∑a2 +

1

2

∑ab

=7

4

∑a2 +

1

2

∑ab


69

=1

4

[∑a2 + 2

∑ab + 6

∑a2

]=

4s2 + 12

(s2 − 4Rr − r2

)4

= 4s2 − 12Rr − 3r2

Gerretsen≤ 4

(4R2 + 4Rr + 3r2

)− 12Rr − 3r2

= 16R2 + 8Rr + r2 − 4Rr + 8r2

= (4R + r)2 − 4r (R − 2r)

�


17 Prove that in any triangle ABC it holds that

∑ √sinA

sinB sinC=

√2R

r

∑sinA

where R denotes the circumradius and r the inradius.

Solution. Using the law of sines we have that

a

sinA=

b

sinB=

c

sinC= 2R

and if we denote A the area of the triangle then

r (a + b + c)2

= A =abc

4R

Thus,

∑ √sinA

sinB sinC=

∑ √a

2R·2R

b·2R

c

=∑ √

a

bc· 2R

=∑ √

a

bc·abc

2A

=a + b + c√2A


70

=

√a + b + c

2A

a+b+c

=

√a + b + c

r

=

√1

r

∑a

=

√2R

r

∑ a

2R

=

√2R

r

∑sinA

�


18 Let x, y, z > 0. Prove that:

y3z

x2(xy + z2)+

z3x

y2(zy + x2)+

x3y

z2(xz + y2)≥

3

2

Solution. Due to homogeneity we may assume xyz = 1. Thus thereexist positive a, b, c such that

x =a

b, y =

b

c, z =

c

a

Hence,

∑ y3z

x2(xy + z2)=

a5

bc(b3 + c3)+

b5

ca(c3 + a3)+

+c5

ab(a3 + b3)

=a6

abc(b3 + c3)+

b6

abc(c3 + a3)+

+c6

abc(a3 + b3)

≥(a3 + b3 + c3)2

2abc(a3 + b3 + c3)


71

=1

2

a3 + b3 + c3

abc

≥3

2

�


19 Let x, y, z be positive numbers. Prove that

3xy

xy + x + y+

3yz

yz + y + z+

3zx

zx + z + x≤ 2 +

x2 + y2 + z2

3

Solution. We are invoking the AM - GM inequality

2 +x2 + y2 + z2

3

=∑ x2 + y2 + 4

6

≥∑

6

√x2y2

=∑ xy

3

√(xy)xy

≥∑ xy

xy+x+y3

=∑

3xy

xy + x + y

and we conclude the result.

�


20 Let x, y, z be positive real numbers such that xy+yz+zx = 2016. Prove

that

∑ √yz

x2 + 2016

≤3

2

Solution. Applying AM - GM we have that

∑ √xy

z2 + 2016

=∑ √

xy

z2 + xy + xz + yz



72

=∑ √

xy

(x + z)(y + z)

≤1

2

∑(x

x + z+

y

y + z

)=

1

2

∑( x

x + z+

z

z + x

)=

3

2

�


21 Let x, y, z > 0 satisfying x + y + z = 1. Prove that

1

x+

1

y+

1

z≥

√3

xyz

Solution. Since 1

x ,1

y ,1

z > 0 then the numbers√1

x+

1

y,

√1

x+

1

z,

√1

y+

1

z

could be sides of a triangle. The area of this triangle is

A =1

2

√1

xy+

1

xz+

1

yz=

1

2

√x + z + y

xyz=

1

2

√xyz

However , in any triangle is holds that [Weitzenböck]

a2 + b2 + c2 ≥ 4A√3 (1)

where A is the area of the triangle. Thus

1

x+

1

y+

1

z≥

√3

xyz

22 Let a, b, c be positive real numbers such that a + b + c = 3. Prove that√b

a2 + 3

+

√c

b2 + 3

+

√a

c2 + 3

≤3

2

4

√1

abc


73

Solution. If we apply AM -GM to (a2, 1, 1, 1) we obtain

a2 + 3 ≥ 4

√a (1)

and similarly if we apply AM - GM to (b, b, b, c) we obtain

3b + c

4

≥4√b3c (2)

We have successively,

√b

a2 + 3

+

√c

b2 + 3

+

√a

c2 + 3

(1)≤

√b

4

√a

+

√c

4

√b

+

√a

4

√c

=1

24√abc

(4√b3c +

4√c3a +

4√a3b

)(2)≤

1

24√abc

(3b + c

4

+3c + a

4

+3a + b

4

)=

3

2

4

√1

abc

�


23 Let ABC be a triangle and denote a, b, c the lengths of the sides BC ,

CA and AB respectively. If abc ≥ 1 then prove that√sinA

a3 + b6 + c6

+

√sinB

b3 + c6 + a6

+

√sinC

c3 + a6 + b6

≤4

√27

4

Solution. Applying Cauchy’s inequality to the vectors

u =(√

sinA,√

sinB,√

sinC)

and

v =

√ 1

a3 + b6 + c6

,

√1

b3 + c6 + a6

,

√1

c3 + a6 + b6

we get that


74

∑√

sinAa3 + b6 + c6

2

≤(∑

sinA) (∑

1

a3 + b6 + c6

)≤

3

√3

2

∑1

a3 + b6 + c6

due to the well known fact

sinA + sinB + sinC ≤3

√3

2

(1)

It suffices to prove that∑

1

a3 + b6 + c6

≤ 1. Let x, y be two positive realnumbers. Then,

(x − y)2(x4 + x3y + x2y2 + xy3 + y4

)≥ 0⇔ x6 + y6 ≥ xy

(x4 + y4

)(2)

On the other hand if xyz ≥ 1 then x + y + z ≥ 33

√xyz = 3 and

x4 + y4 + z4 ≥(x + y + z)4

27

(3)

Then it follows that

∑1

a3 + b6 + c6

≤∑

1

ab(a4 + b4

)+ c3

≤∑

1

a4+b4

c + c3

=∑ c

a4 + b4 + c4

=a + b + c

a4 + b4 + c4

≤27

(a + b + c)3

≤ 1

�



75

24 Let a, b, c be three positive real numbers such that

√a +√b +√c = 1.

Prove that

√a

a2 + 2bc+

√b

b2 + 2ca+

√c

c2 + 2ab≤

1

a+

1

b+

1

c

Solution. By AM - GM we have,

∑ √a

a2 + 2bc≤

∑ √a

a2 + 2

(b2+c2

2

)=

1

a2 + b2 + c2

∑ √a

=1

a2 + b2 + c2

However,

1 =(∑ √

a)2

=

(∑ a√a

)2

≤(∑

a2

)·

(∑1

a

)

Hence1

a2 + b2 + c2

≤1

a+

1

b+

1

cand the result follows.

�


25 Let x1, x

2, . . . , xn be n ≥ 2 positive numbers other than 1 such that

x2

1+ x2

2+ · · · + x2

n = n3. Prove that

log4

x1 x2

x1+ x

2

+log4

x2 x3

x2

+ x3

+ · · · +log4

xn x1

xn + x1

≥1

2

Solution. The Engels form of the Cauchy - Schwartz inequality givesus:


76

∑ log4

x1 x2

x1+ x

2

≥

(∑log2

x1 x2

)2∑

(x1+ x

2)

=

(∑log2

x1 x2

)2

2

∑x

1

AM-GM≥

[n

(∏logx1 x2

)2/n

]2

2

∑x

1

C-B-S≥

n2

2n2

=1

2

and the inequality is proven.

�


26 Let x, y, z > 0 such that

√xy +

√yz +

√yz = 1. Prove that

x2

x + y+

y2

y + z+

z2

z + x≥

1

2

Solution. 1st solution: It follows from Cauchy - Schwartz that

x2

x + y+

y2

y + z+

z2

z + x≥

(x + y + z)2

2(x + y + z)

=x + y + z

2

≥

√xy +

√yz +

√yz

2

=1

2

2nd solution: We have successively:

x2

x + y+

y2

y + z+

z2

z + x=

(x −

xy

x + y

)+

(y −

yz

y + z

)+

(z −

zx

z + x

)


77

= (x + y + z) −(xy

x + y+

yz

y + z+

zx

z + x

)≥ (x + y + z) −

( √xy

2

+

√yz

2

+

√zx

2

)≥

1

2

(√xy +

√yz +

√zx

)=

1

2

�


27 Let a, b, c be positive real numbers such that a + b + c = 3. Prove that√b

a2 + 3

+

√c

b2 + 3

+

√a

c2 + 3

≤3

2

4

√1

abc

Solution. Due to the AM - GM we have that

a2 + 3 ≥ 4

√a (1)

and

3b + c ≥ 4

4√b3c (2)

Thus,

∑ √b

a2 + 3

≤

√b

4

√a

+

√c

4

√b

+

√a

4

√c

=1

24√abc

(4√b3c +

4√c3a +

4√a3b

)≤

1

24√abc

(3b + c

4

+3c + a

4

+3a + b

4

)=

3

2

4

√1

abc

�




78

28 The number n ranges over all possible powers with both the base and

the exponent positive integers greater than n, assuming each such value

only once. Prove that:

∑n

1

n − 1

= 1

Solution. Let us denote by M the set of positive integers greater than1 that are not perfect powers ( i.e are not of the form ap , where a is apositive integer and p ≥ 2 ). Since the terms of the series are positive ,we can freely permute them. Thus,

∑n

1

n − 1

=∑m∈M

∞∑k=2

1

mk − 1

=∑m∈M

∞∑k=2

∞∑j=1

1

mkj

=∑m∈M

∞∑j=1

∞∑k=2

1

mkj

=∑m∈M

∞∑j=1

1

m j (m j − 1)

=

∞∑n=2

1

n (n − 1)

=

∞∑n=2

(1

n − 1

−1

n

)= 1

�


29 Let x, y, z, n ∈ N∗ and n ≥ z. Prove that the equation

xn + yn = zn

has no solution.


79

Solution. Without loss of generality , assume that x < y. If xn+yn = zn

held , then it would be zn > yn thus zn ≥ (y + 1)n . It follows fromBernoulliâ™s inequality that,

zn = xn + yn

< 2yn

≤

(1 +

n

z

)yn

≤

(1 +

1

z

)nyn

<

(1 +

1

y

)nyn

= (y + 1)n

≤ zn

which is an obscurity. The result follows.

�


30 Given the points A(0, 0) , B(1.5, 0) , Γ(2.5,−1) , ∆(3.5, 0), E(5, 0) ,

Z(5, 2) , H(3.5, 2) , Θ(2.5, 3), I(1.5, 2) , K(0, 2) we construct a polygon

as seen in the figure below.

A B

Γ

∆ E

ZH

Θ

IK

Evaluate the area of the polygon ABΓ∆EZHΘIK.


80

Solution. We are using Green’s theorem; Green’s Theorem states that,for a "well-behaved" curve C forming the boundary of a region D∮

C

P(x, y) dx + Q(x, y) dy =

"D

∂Q

∂x−∂P

∂ydA (1)

Since the area of D is equal to!D

dA we can use Green’s Theorem to

calculate area by choosing P(x, y) = 0 and Q(x, y) = x that satisfy

∂Q

∂x−∂P

∂y= 1

Thus letting A be the area of region D we have that

A =

∮C

x dy (2)

Now, consider the polygon below, bordered by the piecewise-smoothcurve C = C

0∪ C

1∪ · · · ∪ Cn−1 ∪ Cn where Ck starts at the point (xk, yk)

and ends at the "next" point along the polygon’s edge when proceedingcounter-clockwise .

(xn−1, yn−1)

(xn, yn)

(x0, y

0)

(x1, y

1)

(x2, y

2)

Since line integrals over piecewise-smooth curves are additive over length,we have that:

A =

∮C

xdy =

∫C0

x dy + · · · +

∫Cn

x dy (3)

81

To compute the k-th line integral above, parametrise the segment from(xk, yk) to (xk+1

, yk+1). Hence

Ck :((xk+1

− xk)t + xk, (yk+1− yk)t + yk

), 0 ≤ t ≤ 1 (4)

Substituting this parametrisation into the integral, we find:

∮Ck

x dy =

∫1

0

((xk+1− xk)t + xk) (yk+1

− yk) dt

=(xk+1

+ xk)(yk+1− yk)

2

Summing all of the Ck ’ s we then find the total area:

A =

n∑k=0

(xk+1+ xk)(yk+1

− yk)2

Returning back to the original problem and using the general formula weget that the area of the polygon ABΓ∆EZHΘIK is 12.

�

82

5

5PART

Jom ... Proposes

83

85

1 Prove that

b1∑a1=0

b2∑a2=0

· · ·

bn∑an=0

(b1

a1

)(b2

a2

)· · ·

(bnan

)(b1+b2+···+bna1+a2+···+an

) = b1+ b

2+ · · · + bn + 1

2 Let

f (x) =

arctan

2

x2

, x , 0

π

2

, x = 0

Prove that

∫ ∞

−∞

f (x) dx =

∞∑n=−∞

f (n) = 2π

3 Prove that

(i)

∞∑n=1

arctan2

n2 + n + 4

= arctan 2.

(ii)

∞∑n=1

arctan2n

n4 − n2 + 1

=π

2

.

(iii)

∞∑n=1

arctan(

1

n2 + n + 1

)=π

4

.

4 Let ABC be a triangle. Prove that:

∑1

sin2 A≥

2R

r≥

4

√3s

9r≥ 4

5 Let ABC be a triangle and let us denote as r the inradius and R the

circumradius. Prove that:

(i)

(∑sin

A

2

)2≤

∑cos2

A

2

(ii)

∑sin

A

2

sinB

2

≤1

2

(1 +

r

R

)

86

6 Let x, y > 0 such that x , y. Prove that

√xy ≤

x − y

ln x − ln y≤x + y

2

7 Let θs (q; z) denote the Jacobi Theta functions. Prove that:

(i)

∫1

0

ϑ2(0; q) dq = π tanh π

(ii)

∫1

0

ϑ2(0; q) ln

1

qdq =

π

2

(tanh π −

π

cosh2 π

)8 Prove that

∞∑n=1

1

n3 sin(√

2πn) = −

13π3

360

√2

6

6PART

JoM ... Study

87

89

Author: Tolaso

JoM ... studies a class of trigonometric integrals

The following two integrals look almost the same, and even to those fairly

well-versed in the art, are the same. But alas, as we shall see.

J1

=

∫ π

0

x sin x1 + cos2 x

dx J2

=

∫ π

0

x cos x1 + sin2 x

dx

Let us deal with J1. The substitution x 7→ π − x reveals a whole lot about

this integral. Thus,

J1

=

∫ π

0

x sin x1 + cos2 x

dx

x 7→π−x======

∫ π

0

(π − x) sin (π − x)1 + cos2 (π − x)

dx

=

∫ π

0

π sin x1 + cos2 x

dx −

∫ π

0

x sin x1 + cos2 x

dx

= π

∫ π

0

sin x1 + cos2 x

dx − J1

=π2

2

− J1

and finally

J1

=π2

4

Easy right? Now , let us move on to the second integral. Unfortunately, the

substitution x 7→ π − x reveals nothing; the reason being that the sub

changes the sign of the cosine in the numerator. We will prove that∫ π

0

x cos x1 + sin2 x

dx = ln2

(√2 + 1

)−π2

4

We split the integral as follows:

∫ π

0

x cos x1 + sin2 x

dx =

∫ π/2

0

x cos x1 + sin2 x

dx +

∫ π

π/2

x cos x1 + sin2 x

dx

=

∫ π/2

0

x cos x1 + sin2 x

dx −

∫ π/2

0

(x + π/2) sin x1 + cos2 x

dx

90

= 2

∫ π/2

0

x cos x1 + sin2 x

dx − π

∫ π/2

0

cos x1 + sin2 x

dx

= 2

∫ π/2

0

x cos x1 + sin2 x

dx −π2

4

So the evaluation of the integral boils down to the evaluation of∫ π/2

0

x cos x1 + sin2 x

dx

It is easy to note that∫ π/2

0

x cos x1 + sin2 x

dx =π2

8

−

∫ π/2

0

arctan(sin x) dx

Hence , it suffices to evaluate the integral

J3

=

∫ π/2

0

arctan(sin x) dx

By induction we have ∫ π/2

0

sin2k+1 x dx =2k k!

(2k + 1)!!Thus,

∫ π/2

0

arctan(sin x) dx =

∞∑k=0

(−1)k

2k + 1

∫ π/2

0

sin2k+1 x dx

=

∞∑k=0

(−1)k

2k + 1

2k k!

(2k + 1)!!

=

∞∑k=0

(−1)k

(2k + 1)24k(

2kk

)Using the Beta function, we get the following identity:

1(2nn

) = (2n + 1)∫

1

0

tn(1 − t)ndt

Thus,

∞∑n=0

(−4)nx2n

(2n + 1)(2nn

) =

∫1

0

dt

1 + 4x2t(1 − t)

91

=

∫1

0

dt

1 + x2 − x2(2t − 1)2

=1

1 + x2

∫1

0

dt

1 − x2

1+x2(2t − 1)2

=1

2

·1

1 + x2

∫1

−1

dt

1 − x2

1+x2t2

=1

2x√1 + x2

∫ x/√1+x2

−x/√1+x2

dt

1 − t2

=1

x√1 + x2

arctanh

(x

√1 + x2

)=

arcsinhx

x√1 + x2

Hence,

∞∑n=0

(−4)n

(2n + 1)2(2nn

) =

∫1

0

arcsinhx

x√1 + x2

dx

= −

∫1

0

arcsinhx1√

1 + 1/x2

d

(1

x

)

= −

∫1

0

arcsinhx d

(arcsinh

(1

x

))= − arcsinh

2

1 +

∫1

0

arcsinh

(1

x

)d (arcsinhx)

= − arcsinh2

1 −

∫ ∞

1

arcsinh(x) d(arcsinh

(1

x

))= − arcsinh

2

1 +

∫ ∞

1

arcsinhx

x√1 + x2

dx

= −1

2

arcsinh2

1 +1

2

∫ ∞

0

arcsinhx

x√1 + x2

dx

= −1

2

arcsinh2

1 +1

2

∫ ∞

0

t

sinh tdt

On the other hand expanding1

sinh x as a series we get

∫ ∞

0

t

sinh(t)dt =

∫ ∞

0

∞∑k=0

2t e−(2k+1)tdt

92

=

∞∑k=0

2

(2k + 1)2

=π2

4

Combining all the results we get that

J2

=

∫ π

0

x cos x1 + sin2 x

dx = ln2

(√2 + 1

)−π2

4

Notes:

(a) We can also evaluate the integral

J =

∫ π/2

0

arctan(sin x) , dx

as follows.

Proof. Applying the change of variables u = sin x we are asked to

evaluate the equivalent integral

∫1

0

arctanu√1 − u2

du

Let us consider the function

f (a) =

∫1

0

arctanat√1 − t2

dt

We differentiate with respect to a. Thus,

f ′(a) =

∫1

0

t

(1 + a2t2)√1 − t2

dt

=

∫1

0

u

[1 + a2(1 − u2)]udu

=1

1 + a2

∫1

0

1

1 −(

au√1+a2

)2

du

=1

a√1 + a2

arctanh

(a

√1 + a2

)

93

=ln

((a +

√1 + a2)2

)2a√

1 + a2

=ln

(a +√1 + a2

)a√1 + a2

=arcsinh a

a√1 + a2

Integrating back our integral reappears and thus the result. �

(b) Using the arctan defintion

arctanα =

∫ ∞

1

α

x2 + αdα

the integral

J =

∫ π/2

0

arctan(sin x) dx

can be expressed as

2

∫ π/2

0

arctan (sin x) dx = 2

∫ π/2

0

∫ ∞

1

sin xy2 + sin2 x

dy dx

= 2

∫ ∞

1

∫ π/2

0

sin xy2 + 1 − cos2 x

dx dy

t=cos x=====

∫ ∞

1

∫1

0

dt dy

y2 + 1 − t2

=

∫ ∞

1

1√y2 + 1

ln

√y2 + 1 + 1√y2 + 1 − 1

dy

y=sinhφ======

∫ ∞

arcsinh1

ln(coshφ + 1

coshφ − 1

)dφ

= 2

∫ ∞

arcsinh1

ln(1 + e−φ

1 − e−φ

)dφ

u=e−φ====2

∫ √2−1

0

ln (1 + u) − ln (1 − u)u

= 2

(Li

2

(√2 − 1

)− Li

2

(1 −√2

))

94

It remains to show that

Li2

(√2 − 1

)− Li

2

(1 −√2

)=

ln2

(1 +√2

)2

−π2

8

This can be shown by using the following Dilogarithm function

properties:

Li2(x) + Li

2(1 − x) = ζ (2) − ln x ln(1 − x) , 0 < x < 1

Li2(1 − x) + Li

2

(1 − 1

x

)= − ln2 x

2, 1

2≤ x < 2

Li2(x) + Li

2(−x) = 1

2Li

2

(x2

), −1 ≤ x ≤ 1

95

Author: Tolaso

JoM ... studies particular Poisson integrals

In this article we shall investigate two particular Poisson Integrals.

Evaluate ∫ π

0

dθ

1 − 2a cos θ + a2

, |a| < 1

We present two solutions for this given integral.

Solution. 1st Solution: For |a| < 1 it holds that

1 + 2

∞∑n=1

an cosnθ =1 − a2

1 − 2a cos θ + a2

⇒ (1)

1

1 − a2

+2

1 − a2

∞∑n=1

an cosnθ =1

1 − 2a cos θ + a2

(2)

Integrating (2) we have that

∫ π

0

dθ

1 − 2a cos θ + a2

=1

1 − a2

∫ π

0

dθ +2

1 − a2

∫ π

0

∞∑n=1

an cosnθ dθ

=π

1 − a2

+2

1 − a2

∞∑n=1

an∫ π

0

cosnθ dθ

=π

1 − a2

+2

1 − a2

∞∑n=1

an��

��: 0

sinnπn

=π

1 − a2

2nd Solution: For |a| < 1 we have successively:

∫ π

0

dx

1 − 2a cos x + a2

=1

2

∫ π

−π

dx

1 − 2a cos x + a2

=1

2

(1 − a2

) ∫ π

−π

∞∑n=−∞

a |n|einx dx

=2π

2

(1 − a2

)

96

=π

1 − a2

�

The second integral is trickier. Let a ≥ 0. We will prove that

I(a) =

∫ π

0

ln(1 − 2a cos x + a2

)dx =

{0 , |a| ≤ 1

2π ln |a| , otherwise

Solution. We give a simple solution breaking it into several steps.

1. Making the substitution x 7→ π − x we note that

I(a) = I(−a)

2. We also note that

I(a) + I(−a) =

∫ π

0

log( (

1 − 2a cos x + a2

) (1 + 2a cos x + a2

) )dx

=

∫ π

0

log( (

1 + a2

)2

− (2a cos x)2)dx

Using double angle formulae we get

I(a) + I(−a) =

∫ π

0

log(1 + 2a2 + a4 − 2a2 (1 + cos 2x)

)dx

=

∫ π

0

log(1 − 2a2 cos 2x + a4

)dx

so by setting x 7→ x2

we get

I(a) + I(−a) =1

2

∫2π

0

log(1 − 2a2 cos x + a4

)dx

Splitting the integral at π and setting x 7→ 2π − x at the second integralwe get

97

I(a) + I(−a) =1

2

I(a2

)+

1

2

∫2π

πlog

(1 − 2a2 cos x + a4

)dx

=1

2

I(a2

)+

1

2

∫ π

0

log(1 − 2a2 cos x + a4

)dx

= I(a2)

Hence

I(a) =1

2

I(a2) (1)

It follows from (1) that I(0) = 0 and I(1) = 0.We now distinguish cases:

� 0 ≤ a < 1 Iterating n times (1) we get:

I(a) =1

2nI(a2

n)

Letting n → +∞ we get that I(a) = 0.

� When a > 1 it follows that 0 < 1

a < 1 and consequently I(

1

α

)= 0.

Thus,

I(a) =

∫ π

0

ln(a2

(1

a2

+cos xa

+ 1

))dx

= 2π lna + I

(1

a

)= 2π lna

Extending the result for negative a we get to our conclusion.

�

One can also note that when |a| ≤ 1 the fact that I(a) = 0 follows

immediately from the previously discussed integral.

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