# Find f(84)

#### Albert

##### Well-known member
$\text{given } :x \in\mathbb{Z}$

$f(x)= \begin{cases}x-3 & x \geq 1000 \\f\big [f(x+5)\big ]& x<1000 \end{cases}$

$\text{find } :\,\, f(84)$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
$\text{given } :x \in\mathbb{Z}$

$f(x)= \begin{cases}x-3 & x \geq 1000 \\f\big [f(x+5)\big ]& x<1000 \end{cases}$

$\text{find } :\,\, f(84)$
Lemma
$f(x)= \begin{cases} x-3 & x \geq 1000 \\ 997 & x<1000 \text{ and$x$even} \\ 998 & x<1000 \text{ and$x$odd} \\ \end{cases}$

Proof
Use full induction going down.
Initial condition: we can verify that it is true for any $x \ge 997$.
Induction step: suppose it is true for any $x$ and some $y$ with $y < x$ and $y< 997$.
Then we need to distinguish the cases that $y$ is even or $y$ is odd.
When we fill in what we already have for $f(y)$ it follows that the given formula is also true for $f(y)$, which completes the proof.

Using the lemma we find that $f(84) = 997$.

#### Albert

##### Well-known member
Lemma
$f(x)= \begin{cases} x-3 & x \geq 1000 \\ 997 & x<1000 \text{ and$x$even} \\ 998 & x<1000 \text{ and$x$odd} \\ \end{cases}$

Proof
Use full induction going down.
Initial condition: we can verify that it is true for any $x \ge 997$.
Induction step: suppose it is true for any $x$ and some $y$ with $y < x$ and $y< 997$.
Then we need to distinguish the cases that $y$ is even or $y$ is odd.
When we fill in what we already have for $f(y)$ it follows that the given formula is also true for $f(y)$, which completes the proof.

Using the lemma we find that $f(84) = 997$.
I like Serena :very smart induction

##### Well-known member
My approach is different from I like serena

we evaluate f(999) through f(995)

f(999) = 998
f(998) = 997
f(997) = 998
f(996) = 997
f(995) = 998

now f(85)= f^183 (998) notation for f is applied 183

applying f twice gives 998 and so on applying 182 time gives 998 and then once more gives 997 which is the ans