Functional equation

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sourav das
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Functional equation

Unread post by sourav das » Mon Jan 30, 2012 9:30 pm

Let $f$ be a real-valued function defined for all real numbers $x$ such that,
the equation

$f(x+1)=\frac{1}{2}+\sqrt{f(x)-(f(x))^2}$

holds for all real $x$.

Find such a function (non-constant)...
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Zzzz
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Re: Functional equation

Unread post by Zzzz » Tue Jan 31, 2012 10:33 am

such 'a' function or 'all' functions...?
Every logical solution to a problem has its own beauty.
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sourav das
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Re: Functional equation

Unread post by sourav das » Tue Jan 31, 2012 10:47 am

Such "$a$" function.
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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Re: Functional equation

Unread post by Zzzz » Tue Jan 31, 2012 11:04 am

Am I missing something? :?

What about this one? $f(x)= \frac{1}{2} + \frac{\sqrt{2}}{4},\ \ \forall x$
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Re: Functional equation

Unread post by sourav das » Tue Jan 31, 2012 11:16 am

Oppsss. Sorry Zzzzz. Edited....
You spin my head right round right round,
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SANZEED
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Re: Functional equation

Unread post by SANZEED » Sat Mar 03, 2012 11:49 pm

I doubt if such a non-constant function really exists.Plugging in $-1,1,0$ for $f(x)$ says me that I am right,but I think you do have solutions in hand....will assure me so that I can retry ?
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Phlembac Adib Hasan
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Re: Functional equation

Unread post by Phlembac Adib Hasan » Sun Mar 04, 2012 9:33 am

sourav vaia wrote:Let $f$ be a real-valued function defined for all real numbers $x$ such that,
the equation

$f(x+1)=\frac{1}{2}+\sqrt{f(x)-(f(x))^2}$

holds for all real $x$.

Find such a function (non-constant)...
I also agree with SANZEED.First notice that $f(x)$ can't be negative as $(f(x))^2$ is always positive.So \[f(x)[1-f(x)]\ge 0\]\[\Rightarrow 1-f(x)\ge 0\]\[\Rightarrow 1\ge f(x)\] Hence $1\ge f(x)\ge 0$.But $f(x)$ increases at least by $\frac {1}{2}$ for increasing $x$ by $1$.Which implies sush a non-constant function can not exist.
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*Mahi*
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Re: Functional equation

Unread post by *Mahi* » Sun Mar 04, 2012 10:08 am

Phlembac Adib Hasan wrote: But $f(x)$ increases at least by $\frac {1}{2}$ for increasing $x$ by $1$.
It doesn't imply anything. $f(x)$ increases at least $\frac 12$... from what? What you said just implies that $\sqrt{f(x)-f^2(x)} \leq \frac 12 \forall x \in \mathbb {R}$.
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Phlembac Adib Hasan
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Re: Functional equation

Unread post by Phlembac Adib Hasan » Sun Mar 04, 2012 10:19 am

*Mahi* vaia wrote:
Phlembac Adib Hasan wrote: But $f(x)$ increases at least by $\frac {1}{2}$ for increasing $x$ by $1$.
It doesn't imply anything. $f(x)$ increases at least $\frac 12$... from what?
Oh, sorry.Got the mistake.
Last edited by Phlembac Adib Hasan on Sun Mar 04, 2012 10:31 am, edited 1 time in total.
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*Mahi*
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Re: Functional equation

Unread post by *Mahi* » Sun Mar 04, 2012 10:26 am

Phlembac Adib Hasan wrote:
*Mahi* vaia wrote:
Phlembac Adib Hasan wrote: But $f(x)$ increases at least by $\frac {1}{2}$ for increasing $x$ by $1$.
It doesn't imply anything. $f(x)$ increases at least $\frac 12$... from what?
As in the problem:
sourav vaia wrote:$f(x+1)=$$\frac{1}{2}$$+\sqrt{f(x)-(f(x))^2}$

holds for all real $x$.
Yes, and what I'm pointing to is $f(x)$ increases by $\frac 12$ for increament of $x$ by $1$ from $\sqrt{f(x)-(f(x))^2}$ , which is (not necessarily) greater than $\frac 12$ for some $x \in \mathbb R$.
Hint:
Consider $g(x)=1-f(x)$
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