Le the be a real valued functions satisfying `f(x+1) + f(x-1) = 2 f(x)` for all `x, y in R` and `f(0) = 0`, then for any `n in N`, f(n) =

Options

A. n f(1)
B. `[ f(1)]^(n)`
C. 0
D. none of these

Explanation

To solve the problem, we need to analyze the functional equation given and derive a general formula for \( f(n) \) where \( n \) is a natural number.

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
The functional equation provided is:
\[ f(x+1) + f(x-1) = 2f(x) \]
This resembles a second-order linear recurrence relation.

2. **Initial Condition**:
We are given that \( f(0) = 0 \).

3. **Finding \( f(1) \)**:
Let's substitute \( x = 0 \) into the functional equation:
\[ f(1) + f(-1) = 2f(0) \]
Since \( f(0) = 0 \), we have:
\[ f(1) + f(-1) = 0 \quad \text{(1)} \]
This implies \( f(-1) = -f(1) \).

4. **Finding \( f(2) \)**:
Now, substitute \( x = 1 \):
\[ f(2) + f(0) = 2f(1) \]
Again, since \( f(0) = 0 \), we get:
\[ f(2) = 2f(1) \quad \text{(2)} \]

5. **Finding \( f(3) \)**:
Substitute \( x = 2 \):
\[ f(3) + f(1) = 2f(2) \]
Using equation (2) where \( f(2) = 2f(1) \):
\[ f(3) + f(1) = 2(2f(1)) = 4f(1) \]
Thus, we can express \( f(3) \) as:
\[ f(3) = 4f(1) - f(1) = 3f(1) \quad \text{(3)} \]

6. **Finding \( f(4) \)**:
Substitute \( x = 3 \):
\[ f(4) + f(2) = 2f(3) \]
Using \( f(2) = 2f(1) \) and \( f(3) = 3f(1) \):
\[ f(4) + 2f(1) = 2(3f(1)) = 6f(1) \]
Therefore:
\[ f(4) = 6f(1) - 2f(1) = 4f(1) \quad \text{(4)} \]

7. **Finding a Pattern**:
From the values we have computed:
- \( f(1) = 1f(1) \)
- \( f(2) = 2f(1) \)
- \( f(3) = 3f(1) \)
- \( f(4) = 4f(1) \)

We can see a pattern emerging:
\[ f(n) = nf(1) \quad \text{for } n \in \mathbb{N} \]

8. **Conclusion**:
Thus, for any natural number \( n \):
\[ f(n) = nf(1) \]

### Final Answer:
For any \( n \in \mathbb{N} \), \( f(n) = nf(1) \).