Unlocking The Sum Of Set Sizes: |S₁| + ... + |Sₘ| Formula

Hey everyone! Today, we're diving into a fascinating problem involving sets, fractional parts, and a neat little formula. We're going to explore the sum of the sizes of some special sets, and along the way, we'll flex our mathematical muscles with ceiling and floor functions. Buckle up, because it's going to be a fun ride!

Unveiling the Problem

Let's kick things off by understanding the problem we're tackling. For every positive integer n (that's what $\mathbb{N}^*$ means, by the way, positive integers), we define a set Sₙ. This set contains all the numbers x between 0 (inclusive) and 1 (exclusive) that satisfy a particular condition. This condition involves the fractional part of nx, denoted as {nx}. Specifically, x belongs to Sₙ if x multiplied by the fractional part of nx equals 1/2. Our goal is to show this intriguing equality:

|S₁| + |S₂| + ... + |Sₘ| = [ (m+1) / 2 ] * [ (m+2) / 2 ]

Where |Sₙ| represents the number of elements in the set Sₙ, and the square brackets denote the floor function (which gives us the greatest integer less than or equal to the expression inside).

This problem beautifully combines set theory, number theory (specifically fractional parts), and the world of ceiling and floor functions. Let's break it down step by step to truly grasp its essence and find a solution.

Deconstructing the Sets Sₙ

The heart of this problem lies in understanding the sets Sₙ. Remember, each Sₙ contains values of x in the interval [0, 1) that satisfy the condition x{nx} = 1/2. Let's dissect what this condition really means. The fractional part of a number, {nx}, is the decimal part left after you subtract the integer part. For example, {3.14} = 0.14 and {5} = 0. So, we're looking for x values where multiplying x by the fractional part of nx gives us exactly 1/2.

To get a handle on this, it’s helpful to express nx in terms of its integer part (let's call it k) and its fractional part: nx = k + {nx}. Here, k is an integer. Now we can rewrite the condition x{nx} = 1/2 as {nx} = 1/(2x). Substituting this back into our expression for nx, we get nx = k + 1/(2x). A little bit of algebraic manipulation gives us a quadratic equation in terms of x: 2nx² - 2kx - 1 = 0. Solving this quadratic equation will give us potential candidates for elements in Sₙ.

However, we need to be careful! The solutions x must lie in the interval [0, 1), and their corresponding fractional parts {nx} must also be in the interval [0, 1). This means we'll need to consider the constraints imposed by these intervals when we analyze the solutions of our quadratic equation. We're essentially looking for the roots of this quadratic that fall within specific boundaries, dictated by the nature of fractional parts and the interval [0, 1).

Delving into the Quadratic Equation and its Roots

Let's revisit that quadratic equation we derived: 2nx² - 2kx - 1 = 0. This equation holds the key to finding the elements of our sets Sₙ. Using the quadratic formula, we can find the roots (solutions) for x: x = (2k ± √(4k² + 8n)) / (4n). Simplifying this, we get x = (k ± √(k² + 2n)) / (2n).

Now, remember those constraints we talked about? Since x must be non-negative (because it lies in [0, 1)), we only need to consider the positive square root. This gives us x = (k + √(k² + 2n)) / (2n). Furthermore, x must be less than 1. This leads to the inequality (k + √(k² + 2n)) / (2n) < 1. Manipulating this inequality, we get √(k² + 2n) < 2n - k. Squaring both sides (since both sides are positive), we have k² + 2n < 4n² - 4nk + k². This simplifies to 2n < 4n² - 4nk, which further reduces to k < n - 1/2. Since k is an integer, this inequality tells us that k must be less than or equal to n - 1.

We also need to ensure that the fractional part {nx} is less than 1, which it always will be since 0 <= x < 1. However, we also need to remember that {nx} = 1/(2x), so we must have 1/(2x) < 1, which implies x > 1/2. Substituting our expression for x, we have (k + √(k² + 2n)) / (2n) > 1/2. This leads to k + √(k² + 2n) > n, and subsequently, √(k² + 2n) > n - k. Squaring both sides (again, both sides are positive), we get k² + 2n > n² - 2nk + k², which simplifies to 2n > n² - 2nk. This gives us 2 > n - 2k, or 2k > n - 2, so k > (n-2)/2.

Therefore, for each n, the possible integer values of k that give us solutions x in the set Sₙ lie in the range (n-2)/2 < k <= n-1. The number of integers k in this range determines the number of elements in Sₙ, which is the cardinality |Sₙ|.

Calculating |Sₙ| and the Sum

Now that we've narrowed down the possible values of k, let's figure out how many elements are in each set Sₙ. The number of integers k satisfying (n-2)/2 < k <= n-1 is given by (n - 1) - [ (n-2)/2 ], where the square brackets denote the floor function. This is because we take the largest possible integer k (which is n-1) and subtract the largest integer that is less than or equal to (n-2)/2.

So, |Sₙ| = (n - 1) - [ (n-2)/2 ]. Now comes the crucial part: summing these cardinalities from n = 1 to m. We want to find the value of Σ|Sₙ| (where Σ denotes summation) from n=1 to m. This is the left-hand side of the equation we're trying to prove.

Σ|Sₙ| = Σ (n - 1) - [ (n-2)/2 ] (from n=1 to m)

Let's break this sum into two parts: Σ(n - 1) and -Σ[ (n-2)/2 ]. The first part is a simple arithmetic series: 0 + 1 + 2 + ... + (m - 1) = m(m - 1)/2. The second part requires a bit more finesse. We need to carefully consider the floor function and how it behaves as n changes.

To evaluate Σ[ (n-2)/2 ], let's write out the first few terms: [-1/2] + [0/2] + [1/2] + [2/2] + [3/2] + ... This simplifies to -1 + 0 + 0 + 1 + 1 + 2 + 2 + ... We notice a pattern: each integer appears twice, except for the initial -1 and 0. We need to figure out how many pairs of integers we have in the sum up to m. This depends on whether m is even or odd.

Let's consider the sum Σ[ (n-2)/2 ] from n=1 to m. It can be rewritten as Σ[n/2] from n=-1 to m-2. Breaking the summation, we have Σ[n/2] from n=1 to m-2. Now, if m is even, let m = 2p. Then the sum becomes Σ[n/2] from n=1 to 2p-2, which is equal to 2(1 + 2 + ... + (p-1)) = 2 * (p-1)p / 2 = p(p-1) = (m/2)(m/2 - 1) = m(m-2)/4. If m is odd, let m = 2p+1. Then the sum becomes Σ[n/2] from n=1 to 2p-1, which is equal to 2(1 + 2 + ... + (p-1)) + p = p(p-1) + p = p^2 = ((m-1)/2)^2 = (m-1)^2 / 4.

Therefore, the second part of our original sum can be expressed as:

-Σ[ (n-2)/2 ] =

-m(m-2)/4, if m is even

-(m-1)^2 / 4, if m is odd

The Grand Finale: Proving the Formula

Now we have all the pieces of the puzzle. Let's put them together and prove the formula:

|S₁| + |S₂| + ... + |Sₘ| = [ (m+1) / 2 ] * [ (m+2) / 2 ]

We found that Σ|Sₙ| = m(m - 1)/2 - Σ[ (n-2)/2 ]. We also derived expressions for Σ[ (n-2)/2 ] for both even and odd m. Let's substitute these expressions back into our equation.

Case 1: m is even (m = 2p)

Σ|Sₙ| = m(m - 1)/2 - m(m - 2)/4 = (2m(m - 1) - m(m - 2)) / 4 = (2m² - 2m - m² + 2m) / 4 = m² / 4 = (2p)² / 4 = p². On the other hand, [ (m+1) / 2 ] * [ (m+2) / 2 ] = [ (2p+1) / 2 ] * [ (2p+2) / 2 ] = p * (p+1) = p². Oops! There seems to be a slight discrepancy here. Let's revisit our calculations to find the potential error.

Case 2: m is odd (m = 2p + 1)

Σ|Sₙ| = m(m - 1)/2 - (m-1)^2 / 4 = (2m(m - 1) - (m - 1)²) / 4 = (2m² - 2m - (m² - 2m + 1)) / 4 = (m² - 1) / 4 = ((2p + 1)² - 1) / 4 = (4p² + 4p) / 4 = p(p + 1). And [ (m+1) / 2 ] * [ (m+2) / 2 ] = [ (2p+2) / 2 ] * [ (2p+3) / 2 ] = (p + 1) * [ p + 3/2 ] = (p+1) * p

Let's revisit the even case. It appears there's an error in calculating [ (m+1) / 2 ] * [ (m+2) / 2 ]. When m = 2p, then [ (m+1) / 2 ] * [ (m+2) / 2 ] = [ (2p+1) / 2 ] * [ (2p+2) / 2 ] = p(p+1).

Therefore, revisiting the cases again,

Case 1: m is even (m = 2p), Σ|Sₙ| = p^2 and [ (m+1) / 2 ] * [ (m+2) / 2 ] = p * (p+1) = (m/2)(m/2 +1/2) and it equals p(p+1) = p^2 + p

Case 2: m is odd (m=2p+1), Σ|Sₙ| = p(p + 1) and [ (m+1) / 2 ] * [ (m+2) / 2 ] = (p+1)^2

Oops. After careful re-examination of the calculations, an error is found. It was calculated |Sn| = (n - 1) - [ (n-2)/2 ]. Σ[ (n-2)/2 ] from n = 1 to m is 0 if m = 1, -1 + 0 = -1, -1 + 0 + 0 = -1, -1 + 0 + 0 + 1 = 0 for m = 4. Hence the formula derived seems to be inaccurate. Let's attempt to find the exact formula by observing pattern.

|S1| = 0, |S2| = 1 - [0/2] = 1, |S3| = 2 - [1/2] = 2, |S4| = 3 - [2/2] = 2, |S5| = 4 - [3/2] = 4-1 = 3, |S6| = 5 - [4/2] = 5 - 2 = 3. |S7| = 6 - [5/2] = 6 - 2 = 4, |S8| = 7 - [6/2] = 7 - 3 = 4

|S1| + |S2| + ... + |Sm|

S1 = 0

S1 + S2 = 1

S1 + S2 + S3 = 1 + 2 = 3

S1 + S2 + S3 + S4 = 3 + 2 = 5

S1 + S2 + S3 + S4 + S5 = 5 + 3 = 8

S1 + S2 + S3 + S4 + S5 + S6 = 8 + 3 = 11

S1 + S2 + S3 + S4 + S5 + S6 + S7 = 11 + 4 = 15

S1 + S2 + S3 + S4 + S5 + S6 + S7 + S8 = 15 + 4 = 19

Also, [ (m+1) / 2 ] * [ (m+2) / 2 ] for m = 1 is [1][3/2] = 1, for m = 2 it is [3/2][4/2] = 12 = 2. [4/2][5/2] = 22 = 4. [5/2][6/2] = 23 = 6. [6/2][7/2] = 3 * 3 = 9. [7/2][8/2] = 3 * 4 = 12. [8/2][9/2] = 4 * 4 = 16. [9/2][10/2] = 4 * 5 = 20. The original answer seems to be correct.

Back to Σ|Sₙ| = m(m - 1)/2 - Σ[ (n-2)/2 ], this is equivalent to m(m-1)/2 - sum([ (n-2)/2 ]) from n=1 to m.

Case 1: m = 2p. m(m-1)/2 - sum([ (n-2)/2 ]) from n=1 to 2p. = 2p(2p-1)/2 - p(p-1) = p(2p - 1) - p(p - 1) = 2p^2 - p - p^2 + p = p^2. And [(2p+1)/2] * [(2p+2)/2] = p(p+1).

Case 2: m = 2p + 1, m(m-1)/2 - sum([ (n-2)/2 ]) = (2p+1)(2p)/2 - p^2 = p(2p+1) - p^2 = 2p^2 + p - p^2 = p^2 + p. and [(2p+2)/2] * [(2p+3)/2] = (p+1)[(2p+3)/2] = (p+1) p, an approximate formula exists to be correct

Reflections on the Journey

This problem has been quite a journey! We started with a seemingly simple set definition and ended up exploring quadratic equations, floor functions, and summations. While we encountered a few bumps along the road (those pesky error checks are crucial!), we've gained a much deeper appreciation for the interplay between different areas of mathematics. Math, guys, is about the voyage, not merely the destination! Even though we didn't arrive at the perfect proof just yet, the process of exploring and grappling with the concepts has been incredibly valuable. Sometimes, the most important discoveries happen when we're willing to get our hands dirty and wrestle with the details. Keep exploring, keep questioning, and keep the math magic alive!