This looks reasonable...(5)
If f ( x ) = x 2 , then f ( f ( x ) ) = x 4 .
Are there any other functions f ( x ) that satisfy f ( f ( x ) ) = x 4 ?
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18 solutions
Can we have a non-complex and non-piecewise function?
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A real continuous function other than x 2 ?
I can at least say you can't build an easy one !
If it is possible, it must be a nightmare function.
Some facts that might help you in your thinking :
it can't cross identity x->x line at any other point than 0 and 1, otherwise it would create another fixed point for x -> f(f(x)) ... But 0 and 1 are the only fixed point for x 4
because x->f(f(x)) is injective on [ 0 , + ∞ [ and ] − ∞ , 0 ] , x -> f(x) as to be injective as well on these intervals . As a continuous function, it must therefore be monotonic.
One strategy could be to follow the x 2 curve but to add small accelerations interspersed with small decelerations, themselves interspersed with small accelerations, etc. In a fractal fashion. But my guess is that you simply can't.
Good luck !
Oh, I only found the complex function with the cube root. No idea how to find them all! I hope someone can post a solution that solves all possible functions (with prove that there are no more).
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I guess there are an infinite number of solutions. Let a and b be numbers. We can define f by : f(x) = x 2 except for the following numbers : f ( a ) = b 2 , f ( b ) = a 2 , f ( a 2 ) = b 4 , f ( b 2 ) = a 4 , f ( a 4 ) = b 8 , f ( b 4 ) = a 8 , f ( a 8 ) = b 1 6 ,
and so on. This define one new solution per couple (a,b).
(and complex numbers were not excluded)- Thus anything is possible? I made an assumption that they were and thus my answer was wrong. Me bad.
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Well, I also gave a real-valued solution...
There are, I think, other solutions, but they are not going to be easy to write down. They exist, but don't have "nice" equations.
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So, there is no continuous, real function that satisfies it?
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@Blan Morrison – I think there is, but it does not have an easy formula.
In fact, you can calculate this. It is not so farfetched to try functions of the shape f ( x ) = x a . In that case, we have x 4 = f ( f ( x ) ) = ( x a ) a = x a 2 . Thus, a 2 = 4 yields solutions. That is, a = 2 or a = − 2 .
(Of course, you still need to define f appropriately in 0.)
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As already discussed, the solution f ( x ) = x − 2 is not defined everywhere. You have to modify the definition at x = 0 .
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Sure, but you still calculate it, rather than guess it.
f(x) = 1/(x^2) works............
Are there other more?
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I suggested this one : f(0) = 1, f(1) = 0, everywhere else f ( x ) = x 2
@X X , yes there are infinitely many of them! (Uncountably to be more precise), for more details look at that : example link (my solution and Calvin Lin's comment bellow it). The same method can be applied.
Except at x = 0
Basic problem, basic solution!
f ( x ) = ( − x ) 2 works. This would also give f ( f ( x ) ) = x 4 .
Note: This is equivalent to f ( x ) = x 2 , so it is not another solution.
Thanks to Jordan and Nathan for spotting the earlier error: f ( x ) = − x 2 , which was wrong, and now corrected. But an easy mistake to make if you work with MS Excel a lot! See https://en.wikipedia.org/wiki/Order of operations#Exceptions . It says: -
"Unary minus sign
There are differing conventions concerning the unary operator − (usually read "minus"). In written or printed mathematics, the expression − 3 2 is interpreted to mean 0 − 3 2 = − 9
Some applications and programming languages, notably Microsoft Excel (and other spreadsheet applications) and the programming language bc, unary operators have a higher priority than binary operators, that is, the unary minus has higher precedence than exponentiation, so in those languages − 3 2 will be interpreted as ( − 3 ) 2 = 9."
So, yes, it should have been written f ( x ) = ( − x ) 2 .
f ( f ( x ) ) = − ( − x 2 ) 2 = − x 4 = x 4
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OK, I concede that that was sloppy. Evidence of too much time working with MS Excel! Please see https://en.wikipedia.org/wiki/Order of operations#Exceptions . Here it says: -
"Unary minus sign
There are differing conventions concerning the unary operator − (usually read "minus"). In written or printed mathematics, the expression −3^2 is interpreted to mean 0 − (3^2) = − 9
Some applications and programming languages, notably Microsoft Excel (and other spreadsheet applications) and the programming language bc, unary operators have a higher priority than binary operators, that is, the unary minus has higher precedence than exponentiation, so in those languages −3^2 will be interpreted as (−3)^2 = 9."
So, yes, for this context, it should be written (-x)^2.
Well, technically, this function is the same as f ( x ) = x 2 . It suffices to simplify the equation as f ( x ) = ( − x ) ⋅ ( − x ) = x 2 which still holds true in C .
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yes I agree
Indeed, this is a bad solution, as it does not offer a function that is different from f(x)=x^2 for any complex value x. Thus far Mark Henning's 3 solutions seem to be the only ones.
I was thinking f ( x ) = x − 2 It works because ( ( x − 2 ) ) − 2 ) is just x^(-2 • -2) which gives us x^4.
We can define f(x) = -(x^2 * abs(x)/x)
You could simplify that to f ( x ) = − x ∣ x ∣ . Then f ( f ( x ) ) = − ( − x ∣ x ∣ ) ⋅ ∣ − x ∣ x ∣ ∣ = x ∣ x ∣ ⋅ ∣ x 2 ∣ = x ∣ x ∣ 3
Note that, for x < 0 , x ∣ x ∣ 3 = − x 4 = x 4 . Therefore I don't believe your solution works.
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Yes, you're right. Thanks for pointing that out!
I don't think this one is correct? f(f(x)) = f(1 - x^4) = 1 - (1 - x^4)^4 = x^16 + ...
Which is identical to f(x)=x^2 for all complex values of x, thus not a different solution.
Actually since there is no such thing as a “fuction” the answer would have to be No
Sorry, the staff editted it already.
Although it's trivial, a solution is f ( x ) = ( − x ) 2
This is equivalent to f ( x ) = x 2 (for purposes here), therefore also satisfies f ( f ( x ) ) = x 4
It is not a different/new function.
You need a function f(x) which is different from x^2 for some (complex) input value x, and yours doesn't match this requirement, so I assume your answer is "I don't have a clue if the answer is Yes or No".
f(x)=x^2 then f(f(x)=x^4 same as f(x)^2=x^2^2
Sorry, but The solution below are wrong. Easy solution is -x2
That's what I thought, f(x) = -(x)^2 f(f(x)) = - (-x^2)^2 = x^4 There can be other solutions as others pointed out, but this is what i found easier
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f ( f ( x ) ) is actually − x 4 in this case, which is not correct.
Are you saying that other people's solutions are wrong, or is your solution wrong. Because − x 2 is not a correct solution.
f ( f ( x ) ) = − ( − x 2 ) 2 = − x 4
f ( f ( x ) ) should be x 4 , not 1. Hence, f ( x ) = 1 is wrong.
This might be cheating but there's always:
f(x) = (-x)^2
The given function and yours are identical
- f(0) = 1
- f(1) = 0
- everywhere else, f ( x ) = x 2
I was doing the same thing, but this function has an issue at − 1 . With the above definition, f ( − 1 ) = ( − 1 ) 2 = 1 , so that f ( f ( − 1 ) ) = f ( 1 ) = 0 , which is not the same as ( − 1 ) 4 = 1 .
But you can patch this up by adding in f ( − 1 ) = 0 .
- f ( 0 ) = 1
- f ( 1 ) = 0
- f ( − 1 ) = 0
- everywhere else, f ( x ) = x 2
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It is a solution in real numbers, but it doesn't fit in the complex numbers. f ( f ( i ) ) = f ( − 1 ) = 0
Do you have a way to patch up so it fits all of the complex numbers?
Interesting (and sorry for the mistake), thanks !
The easiest one is f(x) = -x^2
It should be: f(x) = (-x)^2
F(x)=x^n/x^(n-2) will work as domain is not same as the given function....
I think maybe sqrtx^4 counts.
This function is identical to x 2
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i know that but the question is if it counts
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No, because you have not given a different solution.
Yes. I agree with the previous two comments below. The function is identical and the question asks if there are any other solutions, not the same solutions.
f ( x ) = − x 2 seemed to work without thinking too deeply. But as pointed out in the replies, it does not.
No it doesnt. That would give ff(x)= − x 4
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Ah, I see you are correct.
I assumed the minus would cancel after being applied twice, but of course the square cancels the inner minus first. Thanks for the correction.
Not trying to be mean, but it would probably be more efficient to get rid of an incorrect solution.
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No, show the corrected incorrect solution so that others, like me, learn something.
The problem with f ( x ) = x − 2 is that it is not defined everywhere. To be precise, we need f ( x ) = { x − 2 0 x = 0 x = 0 Alternatively (and complex numbers were not excluded) we could go with f ( x ) = ω x 2 where ω is a primitive cube root of unity.