will'o'wisp2
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I can't think of an example to prove that if f(f(x))=g(g(x)) then g(x)=f(x) also note that g,f \mathbb R \Rightarrow \mathbb R


the problem is that no matter what i can think of it never works. I can sub g for f and say that f(g(x)) or g(f(x)) but as long as the question says that f and g are the same then i can't see how there'd ever be an example to prove this wrong.


I thought maybe that the statement was in the form if P then Q so i thought maybe that P does not imply Q but however it is Q implies P but i dont' quite fully understand the implies stuff so i don't even know if that's the right way to go.
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RDKGames
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(Original post by will'o'wisp2)
I can't think of an example to prove that if f(f(x))=g(g(x)) then g(x)=f(x) also note that g,f \mathbb R \Rightarrow \mathbb R


the problem is that no matter what i can think of it never works. I can sub g for f and say that f(g(x)) or g(f(x)) but as long as the question says that f and g are the same then i can't see how there'd ever be an example to prove this wrong.


I thought maybe that the statement was in the form if P then Q so i thought maybe that P does not imply Q but however it is Q implies P but i dont' quite fully understand the implies stuff so i don't even know if that's the right way to go.
Prove the contrapositive statement:

If f(x) \neq g(x) then ff(x) \neq gg(x)
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will'o'wisp2
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(Original post by RDKGames)
Prove the contrapositive statement:

If f(x) \neq g(x) then ff(x) \neq gg(x)
sorry my bad, i think it and don't write it out correctly, lol if only i could just contrapositive this and it'd be all right, sorry about the mess.


I was supposed to prove this is an incorrect statement.
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RDKGames
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(Original post by will'o'wisp2)
sorry my bad, i think it and don't write it out correctly, lol if only i could just contrapositive this and it'd be all right, sorry about the mess.


I was supposed to prove this is an incorrect statement.
Just pick a counter example.

f(x)=x and g(x)=\frac{1}{x} if x \neq 0 and g(x)=0 if x=0 (dunno how to write g piecewise atm on this, but you get the function) then both give ff(x)=gg(x) and both are \mathbb{R} \rightarrow \mathbb{R}. Reversing this order, clearly we do not get an implication like the question says.
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MR1999
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(Original post by RDKGames)
Prove the contrapositive statement:

If f(x) \neq g(x) then ff(x) \neq gg(x)
Would ff(x) not be the product of the function f and the (real) number f(x)?
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will'o'wisp2
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(Original post by RDKGames)
Just pick a counter example.

f(x)=x and g(x)=\frac{1}{x} if x \neq 0 and g(x)=0 if x=0 (dunno how to write g piecewise atm on this, but you get the function) then both give ff(x)=gg(x) and both are \mathbb{R} \rightarrow \mathbb{R}. Reversing this, clearly we do not get an implication like the question says.
ye but f and g must be the same

f(x)=g(x)

i think you use the code called cases to produce them big squigly {} and write it kinda like a matrix


But ye f and g gotta be the same
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RDKGames
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(Original post by will'o'wisp2)
ye but f and g must be the same

f(x)=g(x)

i think you use the code called cases to produce them big squigly {} and write it kinda like a matrix


But ye f and g gotta be the same
Do they? Where does it say that? The statement says that IF ff(x)=gg(x) THEN f(x)=g(x) but if you use my example above, we get the 'IF' condition, but it doesn't lead us to the 'THEN' condition, hence the statement is false as the 'IF' doesn't imply the 'THEN'
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RDKGames
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(Original post by Desmos)
Would ff(x) not be the product of the function f and the (real) number f(x)?
Product? No, the title says composite so you stuff one into itself.
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MR1999
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(Original post by RDKGames)
Product? No, the title says composite so you stuff one into itself.
I dunno, but saying  ff(x) = f(f(x)) seems a to 'go against' function notation, where, as far as I'm aware, the argument of a function is written next to function inside brackets, eg.,  f(x), g(y), h(z) etc.

Personally, I wouldn't agree the above assertion is true because it allows ambiguous notation.
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_gcx
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(Original post by Desmos)
I dunno, but saying  ff(x) = f(f(x)) seems a to 'go against' function notation, where, as far as I'm aware, the argument of a function is written next to function inside brackets, eg.,  f(x), g(y), h(z) etc.

Personally, I wouldn't agree the above assertion is true because it allows ambiguous notation.
ff(x) is a common alternative to (f \circ f)(x) or f(f(x)). I wouldn't call it ambiguous.
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RDKGames
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(Original post by Desmos)
I dunno, but saying  ff(x) = f(f(x)) seems a to 'go against' function notation, where, as far as I'm aware, the argument of a function is written next to function inside brackets, eg.,  f(x), g(y), h(z) etc.

Personally, I wouldn't agree the above assertion is true because it allows ambiguous notation.
The standard composite function notation shortens it from f(g(x)) to fg(x), the same for ff(x). If it's anything other than this, the question's context would clarify on it.
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MR1999
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(Original post by RDKGames)
The standard composite function notation shortens it from f(g(x)) to fg(x), the same for ff(x). If it's anything other than this, the question's context would clarify on it.
Is is then okay to say that  gf = g \circ f?
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MR1999
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(Original post by _gcx)
ff(x) is a common alternative to (f \circ f)(x) or f(f(x)). I wouldn't call it ambiguous.
So if I write two functions without any gaps, eg., fg, then I'm referring to f\circ g not  f \cdot g?
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will'o'wisp2
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(Original post by RDKGames)
Do they? Where does it say that? The statement says that IF ff(x)=gg(x) THEN f(x)=g(x) but if you use my example above, we get the 'IF' condition, but it doesn't lead us to the 'THEN' condition, hence the statement is false as the 'IF' doesn't imply the 'THEN'
oh rite i see so as long as the function equal each other but f and g aren't the same then the statement is false, thanks so much man, i was struggling so much .__.
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_gcx
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(Original post by Desmos)
So if I write two functions without any gaps, eg., fg, then I'm referring to f\circ g not  f \cdot g?
depends on the context. If it was in shorthand and we were omitting arguments anyway I'd normally read fg as f(x) \cdot g(x). But written as fg(x) I'd interpret it as (f \circ g)(x). I'm personally not a fan of it though.
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MR1999
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(Original post by _gcx)
depends on the context. If it was in shorthand and we were omitting arguments anyway I'd normally read fg as f(x) \cdot g(x). But written as fg(x) I'd interpret it as (f \circ g)(x). I'm personally not a fan of it though.
Is there any other notation where the context can change what it means?
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DFranklin
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(Original post by Desmos)
Is there any other notation where the context can change what it means?
Yes, lots.

f^* might mean the complex conjugate of f, the adjoint function, the dual function, the function with removable singularities removed, the function f restricted to the interior of a given domain, the function f induces on a quotient space, etc...

That said, I agree with your concern in this case; I really don't like ff(x) here. All the same, I don't think it's worth derailing the thread over.
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MR1999
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(Original post by DFranklin)
All the same, I don't think it's worth derailing the thread over.
Sorry about that. It's a bit easy to get carried away sometimes. :/ Next time I'll try to reel it in a bit.

Btw, if the question has been answered satisfactorily, is it okay to then derail the thread?
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DFranklin
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(Original post by Desmos)
Sorry about that. It's a bit easy to get carried away sometimes. :/ Next time I'll try to reel it in a bit.

Btw, if the question has been answered satisfactorily, is it okay to then derail the thread?
If it's clear the OP isn't going to be confused as a consequence, then I think it's OK (but to be clear, I don't make/enforce the rules here, although I have no hesitation in making my own views known).
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