Understanding the Functional Equation

The problem gives us a function \(f(x)\) defined for all positive integers \(x, y \in \mathbb{N}\) that satisfies the property \(f(x + y) = f(x) f(y)\). This is a common type of functional equation. We are also given a specific value: \(f(1) = 2\).

Let's use the given information to find the form of the function \(f(x)\). We can compute the values of \(f(x)\) for the first few positive integers:

• For \(x=1\), we are given \(f(1) = 2\).
• For \(x=2\), we can write \(2 = 1 + 1\). Using the functional equation:
  \(f(2) = f(1 + 1) = f(1) f(1)\)
  Since \(f(1) = 2\), we have \(f(2) = 2 \times 2 = 4\).
• For \(x=3\), we can write \(3 = 2 + 1\). Using the functional equation:
  \(f(3) = f(2 + 1) = f(2) f(1)\)
  Since \(f(2) = 4\) and \(f(1) = 2\), we have \(f(3) = 4 \times 2 = 8\).
• For \(x=4\), we can write \(4 = 3 + 1\). Using the functional equation:
  \(f(4) = f(3 + 1) = f(3) f(1)\)
  Since \(f(3) = 8\) and \(f(1) = 2\), we have \(f(4) = 8 \times 2 = 16\).

Looking at the values \(f(1)=2, f(2)=4, f(3)=8, f(4)=16\), we can observe a pattern. These values are powers of 2:

• \(f(1) = 2 = 2^1\)
• \(f(2) = 4 = 2^2\)
• \(f(3) = 8 = 2^3\)
• \(f(4) = 16 = 2^4\)

It appears that \(f(x) = 2^x\) for all \(x \in \mathbb{N}\). Let's verify if this form satisfies the original functional equation \(f(x + y) = f(x) f(y)\):
If \(f(x) = 2^x\), then \(f(x + y) = 2^{x+y}\).
And \(f(x) f(y) = 2^x \times 2^y\).
Using the rule of exponents, \(2^x \times 2^y = 2^{x+y}\).
So, \(f(x + y) = f(x) f(y)\) is satisfied. Also, \(f(1) = 2^1 = 2\), which matches the given condition.

Therefore, the function is indeed \(f(x) = 2^x\).

Solving the Summation Problem

We are given the summation \(\displaystyle\sum_{x=2}^n\) f(x) = 2044. Now that we know \(f(x) = 2^x\), we can rewrite the summation:

\(\displaystyle\sum_{x=2}^n\) \(2^x\) = 2044

This summation represents the sum of powers of 2 starting from \(x=2\) up to \(x=n\):

\(2^2 + 2^3 + 2^4 + \dots + 2^n = 2044\)

This is a finite geometric series. A geometric series is a series where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

In this series:

• The first term, \(a\), is \(2^2 = 4\).
• The common ratio, \(r\), is \(2^3 / 2^2 = 2\).
• The terms are \(2^2, 2^3, \dots, 2^n\). The exponents are 2, 3, ..., n. The number of terms is \(n - 2 + 1 = n - 1\). Let the number of terms be \(k = n-1\).

The formula for the sum of a finite geometric series is \(S_k = a \frac{r^k - 1}{r - 1}\), where \(a\) is the first term, \(r\) is the common ratio, and \(k\) is the number of terms.

Plugging in our values: \(a=4\), \(r=2\), and \(k=n-1\):

\(S_{n-1} = 4 \frac{2^{n-1} - 1}{2 - 1}\)

\(S_{n-1} = 4 \frac{2^{n-1} - 1}{1}\)

\(S_{n-1} = 4 (2^{n-1} - 1)\)

We are given that the sum is 2044. So:

\(4 (2^{n-1} - 1) = 2044\)

Now, we need to solve this equation for \(n\). Divide both sides by 4:

\(2^{n-1} - 1 = \frac{2044}{4}\)

\(2^{n-1} - 1 = 511\)

Add 1 to both sides:

\(2^{n-1} = 511 + 1\)

\(2^{n-1} = 512\)

We need to find the power of 2 that equals 512. Let's list powers of 2:

• \(2^1 = 2\)
• \(2^2 = 4\)
• \(2^3 = 8\)
• \(2^4 = 16\)
• \(2^5 = 32\)
• \(2^6 = 64\)
• \(2^7 = 128\)
• \(2^8 = 256\)
• \(2^9 = 512\)

So, we have \(2^{n-1} = 2^9\). Since the bases are the same, the exponents must be equal:

\(n - 1 = 9\)

Add 1 to both sides:

\(n = 9 + 1\)

\(n = 10\)

Verification of the Result

Let's check if the sum from \(x=2\) to \(n=10\) of \(f(x) = 2^x\) is indeed 2044.

\(\displaystyle\sum_{x=2}^{10}\) \(2^x\) = \(2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + 2^9 + 2^{10}\)

= \(4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024\)

Alternatively, using the geometric series formula with \(a=4\), \(r=2\), and \(k=10-2+1=9\) terms:

Sum = \(4 \frac{2^9 - 1}{2 - 1} = 4 \frac{512 - 1}{1} = 4 \times 511 = 2044\)

The sum is 2044, which matches the given condition. Thus, the value of \(n\) is 10.

Revision Table

Additional Information on Functional Equations and Series

Functional equations like \(f(x + y) = f(x) f(y)\) are fundamental in mathematics. If the domain was real numbers (\(\mathbb{R}\)) and the function was continuous, the general solution would be \(f(x) = a^x\) for some positive constant \(a\). Given \(f(1)=2\), we would get \(f(x)=2^x\). Since the domain here is restricted to natural numbers (\(\mathbb{N}\)), we showed step-by-step that \(f(x)=2^x\) holds.

The summation of a geometric series is a powerful tool. The formula \(S_k = a \frac{r^k - 1}{r - 1}\) is derived by writing out the sum \(S_k = a + ar + ar^2 + \dots + ar^{k-1}\), multiplying by \(r\) to get \(rS_k = ar + ar^2 + \dots + ar^k\), and then subtracting the first equation from the second: \(rS_k - S_k = (ar + \dots + ar^k) - (a + ar + \dots + ar^{k-1})\). This simplifies to \(S_k(r-1) = ar^k - a = a(r^k - 1)\). Dividing by \((r-1)\) (assuming \(r \ne 1\)) gives the formula. If \(r=1\), the sum is simply \(k \times a\).