Finding f(0) When f(f(x))=x^2-x+1

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[JRasdf]

Only need to notice that Settinh x=1 gives f(1)=1, and setting x=0 gives f(0)=1 or 0. Then do the final check.

[bartolhrg]

Function cannot have multiple results
But a question can have multiple functions as the result

[Mathematicsst]

I try so hard to create math teaching videos for the student in Cambodia, Although there are no more subscriptions or views, I do it from my heart to be an introductory video when they need it at one time. Thank SyberMath for a goodvideo.

[randomz]

Great video and a really clear explanation. I love how you almost walked through the thought process for this, explaining each stage step-by-step with no huge leaps in logic. I guess I'll put it like this, I did not have to pause once in this video because of an unclear explanation. Keep up the great work!

[jucom]

Important to note with functional equations is that multiple values for an input can also just imply multiple functions complying with the equation

[DavidTurchickVEGAN]

Very cool! As for the question at 7:47, “Does that automatically imply f(0)=1?”, the answer should be “Yes, it does”. Unless I misunderstood the problem, its very statement ensures the existence of function f.

[tildarusso]

The solution can be simpler: 
let x=0,  
f(f(0))=0*0-0+1=1,
let f(0)=a, therefore f(a)=1
also, f(f(a))=a*a-a+1 = f(1) =1
then a(a-1)=0, aka a=0 or a=1
if f(0)=0, f(f(0))=0-0+1=1=f(0), contradict!
if f(0)=1, f(f(1)) = 1-1+1=1=f(0), check!
therefore, f(0)=1

[mehrdadmohajer]

Hi,
my approach to find the solution is as follow:

F(x) = x^2 -x +1

F´(x) = f(x) = 2x-1
F´( x) =0 Tangente and per definition @ x of 0, therefor :
f(x of 0 ) is set to Zero, meaning 2x -1 = 0 ----> x of Zero = 1/ 2 which refers to the Point P ( 1/2 ; 0) or point of Tangency to the curve F(x).
Now :
By substitution x=0 and evaluating f(0) = 2( 0) f(0 ) = - 1

Conclusion :
By plotting the graf of F( x) and considering P (1/2 ; 0) as condition for Tangent Line and ONLY in Domain [0 <x<1] do we have f( x ) AT ALL!. At x=1/2, F(x) =1 und constant, but for F(1/2) = 3/4 and < 1= F(x), THEREFOR no solution .
If interested for one Answer, then we Shifted X- axis up by 1/4 = b and f(x) changes into : f(x) = mx + b = 2x +( -1 + 1/4) ----> f(x) = 2x -3/4, more likely F(x)= x^2 -3/4 x + c ( as some constant.


Hope it´s being useful.

[antonyqueen]

The solution as presented in the video (starting by calculating f(1) when f(0) is asked) is based on hindsight.

The correct way is to start by x=0 and then you find out you have to calculate f(1), which is solved by setting x=1 and solving f°f°f(1)=f(1)=f^2(1)-f(1)+1

[MarcelCox]

This problem, like most other functional equations you posted on your channel suffers from one major weakness. You do not specify over what domain the functional equation holds. In your reasoning, while assuming the existenance of a and b, you assume that they might have any value and then limit the possible values at the end. However in the end, your reasoning only holds if there is really a function f, and you would need to know the domain of it. Of course, if you limit the domain to the numebrs 0 and 1, then f exists and f(x)=1 is a function that satisfies the equation for 0 and 1. It would however be much more interessting to find out if there is a function f that is defined either for all real numbers, or at least for a non trivial subset that stisfies the functional equation. And then of course, the problem could also be extended to complex numbers.

[roderickdewar]

I boycott all products advertised by YouTube. Have a nice day.

[florianbasier]

0:22 "to evaluate f(0) I first need to evaluate f(1)" except that is NOT a logical step. Why 1? Why not -1 or 2 or e? The only logical first step is to look at f(f(0))=1 [a]. Now f(f(f(0)))=f(0)^2-f(0)+1 but also f(f(f(0)))=f(1) by injecting [a] and therefore f(0)^2-f(0)+1=f(1) [b]. That's when you want to look at f(1). f(f(1))=1 and f(f(f(1)))=1 but also f(f(f(1)))=f(1) so f(1)=1. You reinject that in [b] and f(0)(f(0)-1)=0 so f(0)=0 or f(0)=1.

[TimCrinion]

This is incredible. Could we come up with a general method of applying any function 0.5 times? The "square root" of any function?

[arpangoswami]

The thing that I like most about you is that you explain the techniques neatly which is really helpful for students.

[douglassouza]

I think in this way:
Define g(x) = x² - x so:
f(f(x)) = g(x) + 1 => f(f(0)) = g(0) + 1 => f(f(0)) = 1.
But, by definition
f(f(0)) = [f(0)]² - f(0) + 1 = 1 => [f(0)]² - f(0) = 0, and, therefore, f(0) = 0 or f(0) = 1.
Suppose f(0) = 0 so
0 = f(0) = f(f(0)) = 0² - 0 + 1 = 1. Contradiction!!!
Therefore f(0) = 1.

[slaviskapyromaner]

My solution:

f(f(x))=x²-x+1
f(f(0))=1 ==> f(0)=f⁻¹(1)
f(f(1))=1 ==> f(1)=f⁻¹(1)
The intersection of a function with its inverse is always on the y=x line, except for when the function is itself its own inverse. We know that f(x) is not its own inverse, because then f(f(x)) would be equal to x, but that is not the case.
Therefore, f(1)=f⁻¹(1)=1.
But f(1)=f⁻¹(1)=f(0).
Thus, f(0)=1.

[tulikavarma]

I did this question with a different approach:-
I first replaced all f(x) by x and all x by f(x), in the given equation, so it became,
f(x) = (f(x))² - f(x) +1
We then get,
(f(x))² + 1 - 2f(x) = 0
Therefore,
(f(x) - 1)² = 0
Square rooting both sides we finally get,
f(x) = 1
Therefore, for all values of x belonging to it's domain, f(x) = 1
Hence, f(0) = 1

[serhiislobodianiuk]

If the problem were like: prove that any function that satisfies this equation satisfies also that f(0) = 1 than your solution is ok. But the question is to find f(0) that means that if there is no such functions than the answer is: the set of possible values for f(0) is empty. So you should find an example of such f.

[a_man]

Arkadaşlar bu ortalama bir yks öğrencisinin çözmesi gereken soru

[petervieren]

I had a shorter solution, not sure if anything is wrong with this?

I called the unknown f(0)=y
f(y) = f(f(0)) = 1 (by putting x=0 in the original equation)
f(f(y)) = y²-y+1 = f(1) (follows from previous)
f^3(y) = f(f(1)) = 1 (by putting x=1 in the original equation)
f^4(y) = f(1) = y²-y+1

So we have
f(y) = 1
f^2(y) = y²-y+1
f^3(y) = 1
f^4(y) = y²-y+1
and so on. It's clear that operation f will always place 1 -> y²-y+1 and y²-y+1 -> 1. Therefore y = y²-y+1 (because f(y)=1), which gives y=1.