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July 14th, 2019, 07:31 AM   #1
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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.
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July 14th, 2019, 08:54 AM   #2
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$\displaystyle f(x) = f^2(\lfloor x \rfloor)-f(x-\lfloor x \rfloor)+1$ 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.
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July 14th, 2019, 08:14 PM   #3
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Quote:
Originally Posted by DarnItJimImAnEngineer View Post
$\displaystyle f(x) = f^2(\lfloor x \rfloor)-f(x-\lfloor x \rfloor)+1$ 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.
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July 14th, 2019, 08:25 PM   #4
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Start by finding $f(0)$, then find $f(x)$ when $x$ is a non-zero integer.
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July 14th, 2019, 08:46 PM   #5
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Quote:
Originally Posted by skipjack View Post
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
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July 14th, 2019, 11:17 PM   #6
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That implies that for non-integer values of $x$, $f(x) = 0\text{ or }1$.
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July 15th, 2019, 10:00 PM   #7
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Quote:
Originally Posted by skipjack View Post
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.
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July 15th, 2019, 10:40 PM   #8
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$f(x)=1\ \forall x \in \mathbb Z$ implies $f([x]) = 1$, so $f(x)=f^2(\{x\})$.
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July 16th, 2019, 09:12 PM   #9
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Quote:
Originally Posted by skipjack View Post
$f(x)=1\ \forall x \in \mathbb Z$ 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.
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July 17th, 2019, 11:17 AM   #10
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Quote:
Originally Posted by Integrator View Post
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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