Animated Solution for Mathematics - Functions: The number of functions $f:\{1,2,3,4\}\to \{a\in \mathbb{Z}:|a|\le 8\}$ satisfying $f(n)+\frac{1}{n}f(n+1)=1,\forall n\in \{1,2,3\}$ is

Have you ever watched a row of dominoes fall? One push at the start dictates the entire sequence. In mathematics, recurrence relations are exactly like that.

Today, we are going to explore a problem that feels like a puzzle, where a single starting value, $f(1)$, triggers a chain reaction that determines the entire function. Our goal is to find how many such functions exist within a strictly defined boundary.

The Engine of the Recurrence

We are given the relation $f(n)+\frac{1}{n}f(n+1)=1$. This is our engine. To see how it works, let's isolate the term that pushes us forward, $f(n+1)$.

By rearranging, we get:

This formula is our generator. It tells us that to find the next value, we take the current value, subtract it from one, and multiply by the index $n$. It is elegant, simple, and incredibly powerful.

The Domino Effect

Now, let's watch the chain reaction unfold. We start with $f(1)$.

For $n=1$:

For $n=2$:

For $n=3$:

Look at that! We have successfully expressed every single output—$f(1)$, $f(2)$, $f(3)$, and $f(4)$—solely in terms of $f(1)$. The entire function is now a slave to the value of $f(1)$.

The Boundary Patrol

But we aren't free to choose any value for $f(1)$. The problem imposes a strict codomain: all outputs must be integers between $-8$ and $8$. This is our 'Boundary Patrol'.

We must ensure that for every $n$, $|f(n)|\le 8$.

1. For $f(1)$: $|f(1)|\le 8\Rightarrow -8\le f(1)\le 8$.

2. For $f(2)$: $|1-f(1)|\le 8\Rightarrow -8\le 1-f(1)\le 8$. Solving this, we find $-7\le f(1)\le 9$.

3. For $f(3)$: $|2f(1)|\le 8\Rightarrow -8\le 2f(1)\le 8$. Dividing by $2$, we get $-4\le f(1)\le 4$.

4. For $f(4)$: $|3-6f(1)|\le 8\Rightarrow -8\le 3-6f(1)\le 8$. Subtracting $3$ gives $-11\le -6f(1)\le 5$.

Dividing by $-6$ (and flipping the signs!) gives:

The Final Verdict

We have four windows of opportunity for $f(1)$. To satisfy the function, $f(1)$ must fall into the intersection of all these intervals. The most restrictive window is the last one: $-\frac{5}{6}\le f(1)\le \frac{11}{6}$.

Since $f(1)$ must be an integer, the only candidates are $0$ and $1$. Let's test them:

- If $f(1)=0$: The outputs are $\{0,1,0,3\}$. All are within $[-8,8]$. Valid!

- If $f(1)=1$: The outputs are $\{1,0,2,-3\}$. All are within $[-8,8]$. Valid!

We have found exactly two functions. This problem teaches us that even when a system seems complex, finding the right 'base' variable can turn a daunting functional equation into a simple, manageable set of constraints. Keep this strategy in your toolkit—it is a classic JEE winner.