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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

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

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

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- Thread starter
- #1

- Jan 25, 2013

- 1,225

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

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

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- Mar 5, 2012

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

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

$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} $

Use full induction going down.

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 $f(84) = 997$.

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- #3

- Jan 25, 2013

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I like Serena :very smart inductionLemma

$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 $f(84) = 997$.

- Mar 31, 2013

- 1,309

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