Functions

Aug 2018
137
7
România
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:
Jun 2019
493
262
USA
\(\displaystyle f(x) = f^2(\lfloor x \rfloor)-f(x-\lfloor x \rfloor)+1\) 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.
 
Aug 2018
137
7
România
\(\displaystyle f(x) = f^2(\lfloor x \rfloor)-f(x-\lfloor x \rfloor)+1\) 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 a moderator:

skipjack

Forum Staff
Dec 2006
21,479
2,470
Start by finding $f(0)$, then find $f(x)$ when $x$ is a non-zero integer.
 
Aug 2018
137
7
România
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
 

skipjack

Forum Staff
Dec 2006
21,479
2,470
That implies that for non-integer values of $x$, $f(x) = 0\text{ or }1$.
 
Aug 2018
137
7
România
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 a moderator:

skipjack

Forum Staff
Dec 2006
21,479
2,470
$f(x)=1\ \forall x \in \mathbb Z$ implies $f([x]) = 1$, so $f(x)=f^2(\{x\})$.
 
Aug 2018
137
7
România
$f(x)=1\ \forall x \in \mathbb Z$ 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 a moderator:

skipjack

Forum Staff
Dec 2006
21,479
2,470
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 $x$.