6th Swedish Mathematical Society Problems 1966

1.  Let {x} denote the fractional part of x = x - [x]. The sequences x₁, x₂, x₃, ... and y₁, y₂, y₃, ... are such that lim {x_n} = lim {y_n} = 0. Is it true that lim {x_n + y_n} = 0? lim {x_n - y_n} = 0?

2.  a₁ + a₂ + ... + a_n = 0, for some k we have a_j ≤ 0 for j ≤ k and a_j ≥ 0 for j > k. If a_i are not all 0, show that a₁ + 2a₂ + 3a₃ + ... + na_n > 0.

3.  Show that an integer = 7 mod 8 cannot be sum of three squares.

4.  Let f(x) = 1 + 2/x. Put f₁(x) = f(x), f₂(x) = f(f₁(x)), f₃(x) = f(f₂(x)), ... . Find the solutions to x = f_n(x) for n > 0.

5.  Let f(r) be the number of lattice points inside the circle radius r, center the origin. Show that lim_r→∞ f(r)/r² exists and find it. If the limit is k, put g(r) = f(r) - kr². Is it true that lim_r→∞ g(r)/r^h = 0 for any h < 2?

Solutions

Problem 1

Let {x} denote the fractional part of x = x - [x]. The sequences x₁, x₂, x₃, ... and y₁, y₂, y₃, ... are such that lim {x_n} = lim {y_n} = 0. Is it true that lim {x_n + y_n} = 0? lim {x_n - y_n} = 0?

Answer

Yes, no.

Solution

Since {x_n} tends to 0, we must have {x_n} < 1/3 for sufficiently large n. Similarly for y_n. Hence {x_n + y_n} = {x_n} + {y_n} for sufficiently large n, which proves the first part.

A counter-example for the second part is x_n = 0, y_n = 1/n. Then {x_n - y_n} = 1 - 1/n which does not tend to 0.

Problem 2

a₁ + a₂ + ... + a_n = 0, for some k we have a_j ≤ 0 for j ≤ k and a_j ≥ 0 for j > k. If a_i are not all 0, show that a₁ + 2a₂ + 3a₃ + ... + na_n > 0.

Solution

We have a_i ≤ 0 for i ≤ k, so i a_i ≥ k a_i for i ≤ k. Hence ∑₁^k i a_i ≥ k ∑₁^k a_i (*). Similarly, ∑_k+1ⁿ i a_i ≥ k ∑_k+1ⁿ a_i (**). Since not all a_i are 0 we cannot have equality in (*) and (**). Hence ∑ i a_i > k ∑₁^k a_i + k ∑_k+1ⁿ a_i = 0.

Problem 3

Show that an integer = 7 mod 8 cannot be sum of three squares.

Solution

All squares are 0, 1 or 4 mod 8.

Problem 4

Let f(x) = 1 + 2/x. Put f₁(x) = f(x), f₂(x) = f(f₁(x)), f₃(x) = f(f₂(x)), ... . Find the solutions to x = f_n(x) for n > 0.

Answer

x = -1 or 2.

Solution

f(x) = x iff 0 = x² - x - 2 = (x+1)(x-2). Now it is clear by a trivial induction that f_n(x) = linear/linear. So f_n(x) = x is a quadratic and so has at most two roots. But x = -1 and 2 are obviously roots, so they are the only roots.

Problem 5

Let f(r) be the number of lattice points inside the circle radius r, center the origin. Show that lim_r→∞ f(r)/r² exists and find it. If the limit is k, put g(r) = f(r) - kr². Is it true that lim_r→∞ g(r)/r^h = 0 for any h < 2?

Answer

k = π

Yes, any h > 1.

Solution

The area of the circle is πr². We may tile the plane with squares side 1 centered on the lattice points. A circle encloses some complete squares and some incomplete squares. If all the squares were complete the circle would enclose πr², since the incomplete squares have smaller area, it encloses ≥ πr² squares. Any square which intersects the disk radius r-2 must be completely inside the disk radius r, so there are at least π(r-2)² squares completely inside the disk radius r. So the number of lattice points is at least π(r-2)². If the lattice point lies inside the circle then its square lies entirely inside the circle radius r+2, so the number of lattice points is at most π(r+2)². So π(1 - 2/r)² ≤ f(r)/r² ≤ π(1 + 2/r)². Hence lim f(r)/r² = π.

The inequality above gives |g(r)| ≤ 4π(r+1), so lim g(r)/r^h = 0 for h > 1.