Find Functions Satisfying $f^d(x)=2015-x$ for All Divisors of 2015

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  • Thread starter anemone
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In summary, to find functions that satisfy the equation $f^d(x)=2015-x$ for all divisors of 2015, one can list all the divisors of 2015 and use them as values for x to solve for f(x). There can be multiple functions that satisfy this equation and one possible approach is to use the limited range of divisors for 2015. These functions can be graphed and the equation can also be solved for other numbers besides 2015 by replacing it with a different integer.
  • #1
anemone
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Here is this week's POTW:

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Denote by $f^n(x)$ the result of applying the function $f$ $n$ times to $x$ (e.g. $f^1(x)=f(x),\,f^2(x)=f(f(x)),\,f^3(x)=f(f(f(x)))$ Find all functions from real numbers to real numbers which satisfy $f^d(x)=2015-x$ for all divisors $d$ of 2015, which are greater than 1, and for all real $x$.

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  • #2
Congratulations to castor28 for his correct solution, which you can find below:
We note first that $5$ and $13$ are divisors of $2015$.

We have:
\begin{align*}
f^5(x) &= 2015 - x\\
f^{10}(x) &= f^5(f^5(x)) \\
&= 2015 - f^5(x) \\
&= 2015 - (2015-x) \\
&= x
\end{align*}

We can continue using the same trick:

\begin{align*}
f^{13}(x) &= f^3(f^{10}(x)) = f^3(x) = 2015 - x\\
f^5(x) &= 2015 - x \\
&= f^3(f^2(x)) \\
&= 2015 - f^2(x)
\end{align*}

Which implies $f^2(x) = x$.
We can finally conclude: $f^3(x) = 2015 - x = f(f^2(x)) = f(x)$.

Official solution:
Since 5 is a divisor of 2015, we have for any real $z$:
$f^{25}(z)=f^5(f^5(f^5(f^5(f^5(z)))))=2015-f^5(f^5(f^5(f^5(z))))=2015-(2015-f^5(f^5(f^5(z))))=f^5(f^5(f^5(z)))=\cdots=f^5(z)=2015-z$.

Since 13 is also a divisor of 2015, we have for any real $z$:
$f^{26}(z)=f^{13}(f^{13}(z))=f(2015-z)$.

Consequently, $z=f^{26}(z)=f(f^{25}(z))=f(2015-z)$. Any real number $x$ can be written as $2015-z$ for $z=2015-x$. Hence, $z=f(2015-z)$ implies $f(x)=2015-x$ for any real $x$.

Finally, check that the function $f(x)=2015-x$ satisfies the conditions of the problem. Let $d$ be a divisor of 2015 greater than 1. Then $d$ is odd, i.e. $d=2c+1$ for a positive integer $c$. Since $f^2(x)=2015-(2015-x)=x$, we have $f^{2c}(x)=f^2(f^2(\cdot f^2(x)\cdots))=x$, which implies $f^d(x)=f(f^{2c}(x))=f(x)=2015-x$.
 

FAQ: Find Functions Satisfying $f^d(x)=2015-x$ for All Divisors of 2015

1. What is the main objective of finding functions satisfying $f^d(x)=2015-x$ for all divisors of 2015?

The main objective is to find a general solution or set of functions that satisfy the given condition for all divisors of 2015. This will provide a comprehensive understanding of the relationship between the input and output values of the function.

2. Why is it important to consider all divisors of 2015 in this problem?

Considering all divisors of 2015 ensures that the solution is valid for all possible inputs, which in this case are the divisors of 2015. This allows for a more thorough analysis of the function and its behavior.

3. Can there be more than one function that satisfies $f^d(x)=2015-x$ for all divisors of 2015?

Yes, there can be multiple functions that satisfy the given condition. This is because there are infinite possible functions that can be defined for a given input and output relationship.

4. Are there any restrictions on the type of functions that can satisfy $f^d(x)=2015-x$ for all divisors of 2015?

No, there are no specific restrictions on the type of functions that can satisfy the given condition. Any function that follows the given relationship for all divisors of 2015 can be considered a valid solution.

5. How can the solutions for $f^d(x)=2015-x$ be represented mathematically?

The solutions can be represented using function notation, where $f(x)$ is the function and $d$ represents the divisors of 2015. For example, one possible solution could be $f(x)=2015-x$, where $d$ can take on any value that is a divisor of 2015.

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