Finding $f(84)$ with the Defined Function $f$

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Discussion Overview

The discussion revolves around the function \( f \) defined on the set of integers, specifically focusing on finding \( f(84) \). The function is defined piecewise, with different behaviors depending on the value of \( n \) relative to 1000. The scope includes mathematical reasoning and exploration of the function's properties.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • Post 1 presents a definition of the function \( f \) with the condition \( n \geq 1000 \) leading to \( f(n) = n - 3 \) and \( f(n) = f(f(n + 5)) \) for \( n < 1000 \).
  • Post 2 offers a different definition where the condition is \( n \leq 1000 \) for the first case, which leads to a disagreement about the correct formulation of the function.
  • Post 3 acknowledges the typo in Post 1 and expresses gratitude for the correction, indicating a recognition of the error in the condition.
  • Post 4 reiterates the acknowledgment of the typo and emphasizes the importance of checking for errors before posting, while also reflecting on the nature of making mistakes in contributions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct definition of the function \( f \), as there are competing views regarding the conditions for \( n \). The discussion remains unresolved regarding which definition should be accepted.

Contextual Notes

The discussion highlights the ambiguity in the function's definition due to the differing conditions presented in the posts. There is a lack of clarity on how these definitions affect the calculation of \( f(84) \).

anemone
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The function $f$ is defined on the set of integers and satisfies

\[ f(n)=\begin{cases}
n-3, & \text{if} \,\,n\geq 1000 \\
f(f(n+5)), & \text{if}\,\, n< 1000
\end{cases}
\]

Find $f(84)$.
 
Last edited:
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anemone said:
The function $f$ is defined on the set of integers and satisfies

\[ f(n)=\begin{cases}
n-3, & \text{if} \,\,n\leq 1000 \\
f(f(n+5)), & \text{if}\,\, n< 1000
\end{cases}
\]

Find $f(84)$.

I think 1st condition should be $>=$
 
Ah, thanks to kaliprasad for pointing it out! Sorry all! I made a blunder, I think for those who know me well, you can tell this wasn't the first time I made a typo in my posting, hehehe...but, I apologize. This shouldn't happen in the first place, I should have checked both before and after posting too!
 
anemone said:
Ah, thanks to kaliprasad for pointing it out! Sorry all! I made a blunder, I think for those who know me well, you can tell this wasn't the first time I made a typo in my posting, hehehe...but, I apologize. This shouldn't happen in the first place, I should have checked both before and after posting too!

No one is perfect, and only those who do things put themselves at the risk of making a mistake. When I see that someone contributing to a site has possibly made a typo or other error in a post, I tend to send them a PM so they can correct it with minimal embarrassment. :)
 
As for the value of f(n) for n less than 1000 can be computed if we now for n+5 so let us compute f(n) for n = 1004 to 995 downwards
$f(1004) = 1004- 3 = 1001$
$f(1003) = 1003 - 3 = 1000$
$f(1002) = 1002-3 = 999$
$f(1001) = 1001 - 3 = 998$
$f(1000) = 1000-3 = 997$
$f(999) = f(f(1004)) = f(1001) = 998$
$f(998) = f(f(1003)) = f(1000) = 997$
$f(997) = f(f(1002)) = f(999) = 998$
$f(996) = f(f(1001)) = f(998) = 997$
$f(995) = f(f(1000)) = f(997) = 998$

Now as 84 + 915 = 999 or 84 + 183 * 5 = 999 So let us evaluate f(999-5k) for some k

we have $f(994) = f(f(999)) = f(998) = 997$
$f(989) = f(f(994)) = f(997) = 998$
$f(984) = f(f(989)) = f(998) = 997$
from the above we see that $f(999-kn)$ is $998$ if k is even and is $997$ if k is odd

as $999- 84 = 183 * 5$ so we get $f(84) = 997$
 

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