IF X IS BLACK . THE GIVEN CONDITION BECOMES PFOR ANY SPHERE S WHICH...

1, if X is black . The given condition becomes P

for any sphere S which passes through at least 4 points of M . For any 3 given points A, B, C in M , denote by

S (A, B, C) the set of all spheres which pass through A, B , C and at least one other point of M and by |S (A, B, C)|

the number of these spheres. Also, denote by P

the sum P

X ∈M f (X).

We have

X

(1)

0 = X

f (X) = (|S (A, B, C)| − 1) (f (A) + f (B ) + f (C)) + X

X∈S

S∈S(A,B,C)

since the values of A, B, C appear |S (A, B, C)| times each and the other values appear only once.

If there are 3 points A, B, C such that |S (A, B, C)| = 1, the proof is finished.

If |S (A, B, C)| > 1 for any distinct points A, B , C in M , we will prove at first that P

= 0.

Assume that P

> 0. From (1) it follows that f (A) + f (B ) + f (C) < 0 and summing by all ⁿ ₃

possible

< 0 (contradicts the starting assumption). The

choices of (A, B, C) we obtain that ⁿ ₃ P

< 0, which means P

same reasoning is applied when assuming P

< 0.

Now, from P

= 0 and (1), it follows that f (A) + f (B) + f (C) = 0 for any distinct points A, B , C in M .

Taking another point D ∈ M, the following equalities take place

f (A) + f (B ) + f (C) = 0

f (A) + f (B ) + f (D) = 0

f (A) + f (C) + f (D) = 0

f (B) + f (C) + f (D) = 0

which easily leads to f (A) = f (B ) = f (C) = f (D) = 0, which contradicts the definition of f .

Problem 5. Let X be a set of ^2k−4 _k−2

+ 1 real numbers, k ≥ 2. Prove that there exists a monotone sequence

{x _i } ^k _i=1 ⊆ X such that

|x _i+1 − x 1 | ≥ 2|x _i − x 1 |

for all i = 2, . . . , k − 1. [20 points]

Solution. We prove a more general statement:

Lemma. Let k, l ≥ 2, let X be a set of ^k+l−4 _k−2

+ 1 real numbers. Then either X contains an increasing sequence

{x _i } ^k _i=1 ⊆ X of length k and

|x _i+1 − x ₁ | ≥ 2|x _i − x ₁ | ∀i = 2, . . . , k − 1,

or X contains a decreasing sequence {x _i } ^l _i=1 ⊆ X of length l and

|x _i+1 − x 1 | ≥ 2|x _i − x 1 | ∀i = 2, . . . , l − 1.

Proof of the lemma. We use induction on k + l. In case k = 2 or l = 2 the lemma is obviously true.

Now let us make the induction step. Let m be the minimal element of X, M be its maximal element. Let

X _m = {x ∈ X : x ≤ m + M

2 }, X _M = {x ∈ X : x > m + M

2 }.

Since ^k+l−4 _k−2

= ^{k+(l−1)−4} _k−2

+ ^{(k−1)+l−4} _(k−1)−2

, we can see that either

k + (l − 1) − 4

(k − 1) + l − 4

|X _m | ≥

+ 1, or |X _M | ≥

+ 1.

(k − 1) − 2

k − 2

In the first case we apply the inductive assumption to X m and either obtain a decreasing sequence of length l

with the required properties (in this case the inductive step is made), or obtain an increasing sequence {x _i } ^k−1 _i=1 ⊆

X m of length k − 1. Then we note that the sequence {x ₁ , x 2 , . . . , x k−1 , M } ⊆ X has length k and all the required

properties.

In the case |X _M | ≥ ^{k+(l−1)−4} _k−2

+ 1 the inductive step is made in a similar way. Thus the lemma is proved.

The reader may check that the number ^k+l−4 _k−2

+ 1 cannot be smaller in the lemma.

Problem 6. For every complex number z / ∈ {0, 1} define

f(z) := X

(log z) ⁻⁴ ,

where the sum is over all branches of the complex logarithm.

a) Show that there are two polynomials P and Q such that f (z) = P (z)/Q(z) for all z ∈ C \ {0, 1}. [10

points]

b) Show that for all z ∈ C \ {0, 1}

f(z) = z z ² + 4z + 1

6(z − 1) ⁴ . [10 points]

Solution 1. It is clear that the left hand side is well defined and independent of the order of summation, because

we have a sum of the type P

n ⁻⁴ , and the branches of the logarithms do not matter because all branches are taken.

It is easy to check that the convergence is locally uniform on C \ {0, 1}; therefore, f is a holomorphic function on

the complex plane, except possibly for isolated singularities at 0 and 1. (We omit the detailed estimates here.)

The function log has its only (simple) zero at z = 1, so f has a quadruple pole at z = 1.

Now we investigate the behavior near infinity. We have Re(log(z)) = log |z|, hence (with c := log |z|)

| log z| ⁻⁴ = X

| X

(log z) ⁻⁴ | ≤ X

(log |z| + 2πin) ⁻⁴ + O(1)

Z ∞

=

(c + 2πix) ⁻⁴ dx + O(1)

−∞

(1 + 2πix/c) ⁻⁴ dx + O(1)

= c ⁻⁴

= c ⁻³

(1 + 2πit) ⁻⁴ dt + O(1)

≤ α(log |z|) ⁻³

for a universal constant α. Therefore, the infinite sum tends to 0 as |z| → ∞. In particular, the isolated singularity

at ∞ is not essential, but rather has (at least a single) zero at ∞.

The remaining singularity is at z = 0. It is readily verified that f (1/z) = f (z) (because log(1/z) = − log(z));

this implies that f has a zero at z = 0.

We conclude that the infinite sum is holomorphic on C with at most one pole and without an essential singularity

at ∞, so it is a rational function, i.e. we can write f (z) = P (z)/Q(z) for some polynomials P and Q which we

may as well assume coprime. This solves the first part.

Since f has a quadruple pole at z = 1 and no other poles, we have Q(z) = (z − 1) ⁴ up to a constant factor

which we can as well set equal to 1, and this determines P uniquely. Since f(z) → 0 as z → ∞, the degree of P

is at most 3, and since P (0) = 0, it follows that P (z) = z(az ² + bz + c) for yet undetermined complex constants

a, b, c.

There are a number of ways to compute the coefficients a, b, c, which turn out to be a = c = 1/6, b = 2/3.

Since f(z) = f (1/z), it follows easily that a = c. Moreover, the fact lim

z→1 (z − 1) ⁴ f (z) = 1 implies a + b + c = 1

(this fact follows from the observation that at z = 1, all summands cancel pairwise, except the principal branch

which contributes a quadruple pole). Finally, we can calculate





n ⁻⁴ = 2π ⁻⁴ X

f (−1) = π ⁻⁴ X

n ⁻⁴ − X

n ⁻⁴

n ⁻⁴ = 2π ⁻⁴



 = 1