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

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Discussion Overview

The discussion revolves around the functional equation ##f(2x)=f^2(x)-2f(x)-1/2## and the task of finding the value of ##f(3)##. Participants explore various interpretations of the functional equation, propose different forms of the function, and question the validity of certain steps in their reasoning. The conversation includes technical reasoning, mathematical exploration, and challenges to the proposed solutions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose a functional equation ##f(x+y)=f(x)f(y)-[f(x)+f(y)]-1/2## and derive values for ##f(2)## and ##f(3)##.
  • Others challenge the use of the same notation for different functions and question the derivation of the proposed functional equation from the original statement.
  • There is a discussion about whether ##f^2(x)## refers to ##(f(x))^2## or ##f(f(x))##, with implications for the validity of the derived equations.
  • Some participants suggest that there may be multiple functions that satisfy the given conditions, leading to different values for ##f(3)##.
  • A later reply presents a specific function ##f(x) = 2^x + 1## and calculates ##f(6) = 65##, while also noting that other forms of ##g(x)## could lead to different results.
  • There are corrections and refinements of earlier claims, particularly regarding the interpretation of the functional equation and the values of ##f(2)## and ##f(4)##.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the functional equation and the validity of various derived forms. There is no consensus on a single solution for ##f(3)##, as multiple competing interpretations and potential solutions are presented.

Contextual Notes

Participants note that the function is defined over positive reals, which affects the interpretation of certain values. There are unresolved questions regarding the assumptions made in deriving the functional equations and the implications of different interpretations of ##f^2(x)##.

Who May Find This Useful

This discussion may be of interest to those studying functional equations, mathematical reasoning, and the exploration of different functional forms in the context of problem-solving in mathematics.

littlemathquark
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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)##
 
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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))##?
 
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fresh_42 said:
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##
 
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?
 
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.
 
fresh_42 said:
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.
 
littlemathquark said:
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.
 
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
 
littlemathquark said:
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.
 
  • #10
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.
 
  • #11
littlemathquark said:
##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##.

littlemathquark said:
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)]##.
 
  • #12
fresh_42 said:
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.
 
  • #13
pasmith said:
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##
 
  • #14
littlemathquark said:
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.
 
  • #15
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##
 
  • #16
littlemathquark said:
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##.
 
  • #17
D H said:
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##.
 
  • #18
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?
 

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