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 July 14th, 2019, 07:31 AM #1 Member   Joined: Aug 2018 From: România Posts: 84 Thanks: 6 Functions Hello all, Determine the functions $\displaystyle f:\mathbb R \rightarrow \mathbb R$ that verify the relationship $\displaystyle f(x)=f^2(\{x\})-f([x])+1$ , $\displaystyle \forall x\in \mathbb R$ and where $\displaystyle \{x\}$ and $\displaystyle [x]$ are defined as in the following site: Home Math Notes Algebra II TRIGONOMETRY INTEGER PART OF NUMBERS. FRACTIONAL PART OF NUMBER Suppose, $\displaystyle x$ is the real number. Its integral part is the greatest integral number, that isn′t exceed $\displaystyle x$. The integral part of number $\displaystyle x$ denotes as $\displaystyle [x]$. The fractional number of number $\displaystyle x$ is difference between the number and its integral part, i.e. $\displaystyle x−[x]$ . The fractional part of number $\displaystyle x$ denotes as $\displaystyle \{x\}$. For example, $\displaystyle [3.47]=3;\{3.47\}=0.47;$ $\displaystyle [−2.3]=−3;\{−2.3\}=−2.3−(−3)=0.7;$ $\displaystyle [15]=15;\{15\}=0$ All the best, Integrator Last edited by Integrator; July 14th, 2019 at 07:37 AM.
 July 14th, 2019, 08:54 AM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 120 Thanks: 40 $\displaystyle f(x) = f^2(\lfloor x \rfloor)-f(x-\lfloor x \rfloor)+1$ would be the more universal way to write that. $\displaystyle f(x) = 1$ is a trivial solution. Not sure how you would go about finding all the functions. I would maybe start with taking the derivative and maybe evaluating as Taylor series expansions(?), since $\displaystyle d\lfloor x\rfloor/dx=0$ for x not an integer.
July 14th, 2019, 08:14 PM   #3
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 Originally Posted by DarnItJimImAnEngineer $\displaystyle f(x) = f^2(\lfloor x \rfloor)-f(x-\lfloor x \rfloor)+1$ would be the more universal way to write that. $\displaystyle f(x) = 1$ is a trivial solution. Not sure how you would go about finding all the functions. I would maybe start with taking the derivative and maybe evaluating as Taylor series expansions(?), since $\displaystyle d\lfloor x\rfloor/dx=0$ for x not an integer.
Hello,

The proposed issue is for class IX, and the problem must therefore be resolved at this level of knowledge. Thank you very much!

All the best,

Integrator

Last edited by skipjack; July 14th, 2019 at 08:22 PM.

 July 14th, 2019, 08:25 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 Start by finding $f(0)$, then find $f(x)$ when $x$ is a non-zero integer.
July 14th, 2019, 08:46 PM   #5
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 Originally Posted by skipjack Start by finding $f(0)$, then find $f(x)$ when $x$ is a non-zero integer.
Hello,

$\displaystyle f(x)=1$ $\displaystyle \forall x \in \mathbb Z$ , but we have to find the functions $\displaystyle \forall x\in \mathbb R$.Thank you very much!

All the best,

Integrator

 July 14th, 2019, 11:17 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 That implies that for non-integer values of $x$, $f(x) = 0\text{ or }1$.
July 15th, 2019, 10:00 PM   #7
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 Originally Posted by skipjack That implies that for non-integer values of $x$, $f(x) = 0\text{ or }1$.
Hello,

I do not understand! Please detail! Thank you very much!

All the best,

Integrator

Last edited by skipjack; July 15th, 2019 at 10:44 PM.

 July 15th, 2019, 10:40 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 $f(x)=1\ \forall x \in \mathbb Z$ implies $f([x]) = 1$, so $f(x)=f^2(\{x\})$.
July 16th, 2019, 09:12 PM   #9
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 Originally Posted by skipjack $f(x)=1\ \forall x \in \mathbb Z$ implies $f([x]) = 1$, so $f(x)=f^2(\{x\})$.
Hello,

I do not understand! From $\displaystyle f(x)=1$ and $\displaystyle f(x)=f^2(\{x\}) \forall x \in \mathbb Z$ it results $\displaystyle f(x)=f^2(0)=1$ and no $\displaystyle f(x)=0$ and so it seems to me wrong to say that $\displaystyle f (0) = f ^ 2 (0)$ results in solutions $\displaystyle f (0) = 0$ and $\displaystyle f (0) = 1$, which would imply that we can also have $\displaystyle f (x) = 0$....Thank you very much!

All the best,

Integrator

Last edited by skipjack; July 17th, 2019 at 10:59 AM.

July 17th, 2019, 11:17 AM   #10
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 Originally Posted by Integrator From $\displaystyle f(x)=1$ and $\displaystyle f(x)=f^2(\{x\}) \forall x \in \mathbb Z$ it results $\displaystyle f(x)=f^2(0)=1$
Replacing $\{x\}$ with 0 assumes that $x$ is an integer, in which case it's already assumed that $f(x) = 1$.

I suggested $f(x) = 0$ is possible for non-integer values of $x$.

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