##f(2x)=f^2(x)-2f(x)-1/2## then find ##f(3)##

[littlemathquark]

TL;DR

Let $f$ be defined over positive reals. $f(2x)=f^2(x)-2f(x)-1/2$ and $f(1)=2$ then find $f(3)$

My solution:
Let $f(x+y)=f(x)f(y)-[f(x)+f(y)]-1/2$
İf $y=x$ we find functional equation that given us. So for $x=y=1$ then $f(2)=-1/2$

İf we evaluate $x=1, y=2$ at above equation $f(3)=-3$

My question is: What is the solution of that functional equation; I mean are there other solutions? İf so it must be other values of $f(3)$

 

[fresh_42]

You shouldn't use the same name for at prior two different functions. And you should show us what you did to arrive at your conclusions. Can you derive your equation for $f(x+y)$ from the given functional equation? Since otherwise, you cannot use it. And is $f^2(x)=(f(x))^2$ or is it $f^2(x)=f(f(x))$?

 

[littlemathquark]

You shouldn't use the same name for at prior two different functions. And you should show us what you did to arrive at your conclusions. Can you derive your equation for $f(x+y)$ from the given functional equation? Since otherwise, you cannot use it. And is $f^2(x)=(f(x))^2$ or is it $f^2(x)=f(f(x))$?

$f^2(x)=(f(x))^2$

 

[fresh_42]

I have another question. If I write down the formulas for $x=1$ then I get
\begin{align*}
f(2x)&=f^2(x)-2f(x)- 1/2\, , \,f(1)=2\\[8pt]
f(2)&=f^2(1)-2f(1)-1/2=\begin{cases}-1/2 &\text{ if }f^2(x)=(f(x))^2\\f(2)-4-1/2 &\text{ if }f^2(x)=f(f(x))\end{cases}
\end{align*}
The first interpretation isn't in $\mathbb{Z}$ anymore and the second is impossible. Did I make a mistake?

 

[littlemathquark]

Once upon times I saw my equation at solution of that question $f(1)=3$ and $(f (x))^2=f (2x)+2f (x)-2$ then find $f (6)=?$
But I don't know origin of the equation.

 

[littlemathquark]

I have another question. If I write down the formulas for $x=1$ then I get
\begin{align*}
f(2x)&=f^2(x)-2f(x)- 1/2\, , \,f(1)=2\\[8pt]
f(2)&=f^2(1)-2f(1)-1/2=\begin{cases}-1/2 &\text{ if }f^2(x)=(f(x))^2\\f(2)-4-1/2 &\text{ if }f^2(x)=f(f(x))\end{cases}
\end{align*}
The first interpretation isn't in $\mathbb{Z}$ anymore and the second is impossible. Did I make a mistake?

Sorry, my mistake. $f$ defined over positive reals not integers.

 

[fresh_42]

Once upon times I saw my equation at solution of that question $f(1)=3$ and $(f (x))^2=f (2x)+2f (x)-2$ then find $f (6)=?$
But I don't know origin of the equation.

$-1/2$ and $-2$ makes a significant difference.

 

[littlemathquark]

Solution to my second question:
f(x+y) = f(x).f(y) - [f(x) + f(y)] + 2 ...(1)
İf y = x
f(2x) = [f(x)]^2 - 2.f(x) + 2 ... (2)
İf there is a function Y = f(x) that satisfies (1), it also satisfies (2)
f(1) = 3

f(2) = f(1+1) = 5

f(3) = f(1+2) = 9

f(4) = f(2+2) = 17 [or; f(4) = f(1+3) = 17 ]

f(5) = f(1+4) = 33

f(x) = 2^x + 1
So f(6)=65

 

[fresh_42]

Solution to my second question:
f(x+y) = f(x).f(y) - [f(x) + f(y)] + 2 ...(1)
İf y = x
f(2x) = [f(x)]^2 - 2.f(x) + 2 ... (2)
İf there is a function Y = f(x) that satisfies (1), it also satisfies (2)
f(1) = 3

f(2) = f(1+1) = 5

f(3) = f(1+2) = 9

f(4) = f(2+2) = 17 [or; f(4) = f(1+3) = 17 ]

f(5) = f(1+4) = 33

f(x) = 2^x + 1
So f(6)=65

Yes, but this is the wrong direction. We are given (2) and have no control whether its solution satisfies (1).

And this is meanwhile the seventh function that you call $f$!

1. $f$ that satisfies (2).
2. $f$ that satisfies (1).
3. $f$ with constant term $-1/2.$
4. $f$ with constant term $-2.$
5. $f$ with constant term $+2.$
6. $f$ with $f(1)=2.$
7. $f$ with $f(1)=3.$

Sorry, but this creates a lot of confusion and it does not make sense to discuss this on the internet.

 

[littlemathquark]

Sorry for mass and trouble. I belive that there are a lot of f(3) values for my frst question but I could'nt show. Thank you.

 

[Mark44]

$f^2(x)=(f(x))^2$

Which is how $f^2(x)$ would normally be defined. Compare this to $\sin^2(x)$ which in a more verbose form would be written as $(\sin(x))^2$.

My solution:
Let $f(x+y)=f(x)f(y)-[f(x)+f(y)]-1/2$

I don't see that this is valid. From the problem statement $f(x+y)= f^2(\frac {x + y}2) - 2f(\frac {x + y}2) - 1/2$.
Clearly you can do this: $f(2x) = f^2(x) - 2f(x) - 1/2$, but I don't see how you can turn the first and second terms on the right side above to $f(x)f(y)$ and $-2[f(x) + f(y)]$.

 

[pasmith]

And is $f^2(x)=(f(x))^2$ or is it $f^2(x)=f(f(x))$?

If f^2(x) = f(f(x)) then putting x = 1 in the functional equation yields \begin{split}<br /> f(2) &amp;= f(f(1)) - 2f(1) - \tfrac12 \\<br /> &amp;= f(2) - 2f(1) - \tfrac12 \\<br /> &amp;= f(2) -\tfrac{9}{2} \end{split} which is clearly false. So f^2(x) = (f(x))^2, and we can rewrite the functional equation as f(2x) = (f(x) - 1)^2 - \frac32. We also have <br /> f\left( \frac x2 \right) = 1 \pm \sqrt{\frac{3 + 2f(x)}{2}} and I don't think there is a way to fix the sign of the root in general, since <br /> f(1) = 1 + \sqrt{\frac{3 + 2f(2)}{2}} \quad\mbox{but}\quad f(2) = 1 - \sqrt{\frac{3 + 2f(4)}2}. Putting x = 0 in the functional equation yields f(0) = \frac{3 \pm \sqrt{11}}2.

I don't think that it is possible to determine f(3) from this.

 

[littlemathquark]

If f^2(x) = f(f(x)) then putting x = 1 in the functional equation yields \begin{split}<br /> f(2) &amp;= f(f(1)) - 2f(1) - \tfrac12 \\<br /> &amp;= f(2) - 2f(1) - \tfrac12 \\<br /> &amp;= f(2) -\tfrac{9}{2} \end{split} which is clearly false. So f^2(x) = (f(x))^2, and we can rewrite the functional equation as f(2x) = (f(x) - 1)^2 - \frac32. We also have <br /> f\left( \frac x2 \right) = 1 \pm \sqrt{\frac{3 + 2f(x)}{2}} and I don't think there is a way to fix the sign of the root in general, since <br /> f(1) = 1 + \sqrt{\frac{3 + 2f(2)}{2}} \quad\mbox{but}\quad f(2) = 1 - \sqrt{\frac{3 + 2f(4)}2}. Putting x = 0 in the functional equation yields f(0) = \frac{3 \pm \sqrt{11}}2.

I don't think that it is possible to determine f(3) from this.

So Let $f(x) - 1=g(x)$ then $g(2x)=(g(x))^2-5/2$

 

[littlemathquark]

So Let $f(x) - 1=g(x)$ then $g(2x)=(g(x))^2-5/2$

Since $f$ is positive defined $f(x/2)>0$ and f(2x),f(4x),... can be find.

 

[littlemathquark]

My second question was $f(1)=3$ and $(f (x))^2=f (2x)+2f (x)-2$ find $f (6)=?$

İn this case $g(x)=f(x)-1$ then it would be $g(2x)=(g(x))^2$ then it would be $g(x)=a^x$

 

[D H]

TL;DR Summary: Let $f$ be defined over positive reals. $f(2x)=f^2(x)-2f(x)-1/2$ and $f(1)=2$ then find $f(3)$

My solution:
Let $f(x+y)=f(x)f(y)-[f(x)+f(y)]-1/2$
İf $y=x$ we find functional equation that given us. So for $x=y=1$ then $f(2)=-1/2$

İf we evaluate $x=1, y=2$ at above equation $f(3)=-3$

My question is: What is the solution of that functional equation; I mean are there other solutions? İf so it must be other values of $f(3)$

Your $f(x+y)=f(x)f(y)-[f(x)+f(y)]-1/2$ isn't consistent with the problem statement. If one takes $x=0, y=1$, then $f(0+1)=f(1)=f(0)f(1)-(f(0)+f(1))-\frac12$, and since $f(1)=2$ is a given, this means $f(0)=\frac92$. However, the problem statement says $f(2*0) = f(0)= f^2(0) - 2f(0)-\frac12$, from which $f(0)=\frac{3\pm\sqrt{11}}{2}$, neither of which is consistent with $f(0)=\frac92$.

 

[littlemathquark]

Your $f(x+y)=f(x)f(y)-[f(x)+f(y)]-1/2$ isn't consistent with the problem statement. If one takes $x=0, y=1$, then $f(0+1)=f(1)=f(0)f(1)-(f(0)+f(1))-\frac12$, and since $f(1)=2$ is a given, this means $f(0)=\frac92$. However, the problem statement says $f(2*0) = f(0)= f^2(0) - 2f(0)-\frac12$, from which $f(0)=\frac{3\pm\sqrt{11}}{2}$, neither of which is consistent with $f(0)=\frac92$.

Yes, thank you. İt's the same for $x=4$. We can find two different image of $x=4$; I mean $f(4)=-11/2$ and $f(4)=3/4$.

 

[littlemathquark]

My other question was this:
$(f (x))^2=f (2x)+2f (x)-2$ and $f(1)=3$ then find $f(6)$
We can find $g(2x)=(g(x))^2$ where $f(x)=g(x)+1$ and one of the solution is $g(x)=2^x$ and $f(x)=2^x+1$ and $f(6)=65$
But if
$g(x)= \left\{\begin{array}{cc} 1, & x= 3\cdot 2^n , n\in \mathbb Z & \\ 2^x , & & \mbox{ other values} \end{array} \right.$ so $g(6)=(g(3))^2=1$ and $f(6)=g(6)+1=2$

But my main question is $(f (x))^2=f (2x)+2f (x)+1/2$ and $f(1)=2$ then find $f(3)$
For this question $g(2x)=(g(x))^2-\frac52$. Can we define $g(x)$ function that similar to above $g(x)$ partial function?