Let’s solve the problem systematically.

We are given:

1. $f(1) = 1$
2. $f(n+1) = f(n)^2 - 3$
3. We need to find $f(4)$.

Step-by-step Calculation:

1. Start with $f(1)$: $f(1) = 1$

2. Use the recurrence relation to calculate $f(2)$: $f(2) = f(1)^2 - 3 = 1^2 - 3 = -2$

3. Use the recurrence relation to calculate $f(3)$: $f(3) = f(2)^2 - 3 = (-2)^2 - 3 = 4 - 3 = 1$

4. Use the recurrence relation to calculate $f(4)$: $f(4) = f(3)^2 - 3 = 1^2 - 3 = -2$

Final Answer:

$f(4) = -2$

Let me know if you'd like further clarification or additional details!

Follow-Up Questions:

1. What is the general behavior of $f(n)$ in this recurrence? Does it repeat?
2. Can we prove whether the sequence $f(n)$ is periodic?
3. How would you determine $f(n)$ for larger values of $n$, like $f(10)$?
4. Can you think of other problems where recurrence relations like this are used?
5. Is there a way to write $f(n)$ as an explicit function of $n$?

Tip: When working with recurrence relations, calculate a few terms first to look for patterns or cycles.