Nice solution, I solved it a little differently with inverses but I think there might be a small flaw in my solution(assuming f has an inverse on some interval). Anyway, this is what I did: Let g(x) be the inverse of f(x), (f^-1(x) is annoying to type). Then we have f(x)=g(x^2-x+1). Plug in 1 and we get that f(1)=g(1). Since f and g are inverses, they are mirrored over the line y=x in the Cartesian coordinate plane. Since they both have the same y value at x=1, that must mean that f(1)=g(1)=1. If we then plug in 0, we get f(0)=g(1), but we know g(1)=1, so f(0)=1.
Let f(0) = A, let A^2 - A + 1 = B, then f(A) = f(f(0)) = 0 - 0 + 1 = 1, and f(1) = f(f(A)) = A^2 - A + 1 = B, and f(B) = f(f(1)) = 1 - 1 + 1 = 1. Now f(f(B)) = f(1) = B, but also f(f(B)) = B^2 - B + 1 => B^2 - B + 1 = B => (B - 1)^2 = 0 => B = 1 => A*(A-1) = 0. Then check that A=0 leads to a contradiction, and the only answer is A = 1.
I would like to present a solution that does not involve heavy calculation, but instead understanding of the problem and logical reasoning. We can easily calculate by setting x=0 and x=1 that f(f(0)) = f(f(1)) = 1, so if we apply f twice to 0, we get 1 and if we apply f twice to 1, we get 1. The problem is to find out what we get if we apply f only once. 0 -> f(0)=? -> f(f(0)) = 1 It’s easy to verify that if f(0)=1, then f(1)=1 and the equation holds, so f(0)=1 is definitely a possible solution. The problem is to find out if it’s also the only solution. Now, let’s see what sequence we get if we start with 0 and keep applying f. To ease the notation, set f(0) = a. f(0) = a f(a) = f(f(0)) = 1 f(1) = f(f(a)) = a^2-a+1 f(a^2-a+1) = f(f(1)) = 1 f(1) = a^2-a+1 etc. Now if we set a^2-a+1 = b, we see that f(f(b))= a^2-a+1 = b But from the initial equation, it also follows that f(f(b))=b^2-b+1 Solving b = b^2-b+1 leads to b=1 and so a=0 or a=1. a=0 leads to a contradiction and therefore a = f(0) = 1. Finally, let me give a “counterexample”, with f(0)?> 1 that really confused me: f(0)=5 f(5)=1 f(1)=21 f(21)=1 f(1)=21 etc. I didn’t see why such a function is not possible. The problem is that f(f(21))=21, but it must also be much larger, because f(f(21))=21^2-21+2. This leads to a contradiction, so the only possibility is f(0)=1.
You should solve some questions from the exam konkur(کنکور), it’s the university entry exam across nation and it is notorious for it’s difficult questions and little time
Very nicely explained! On further exploration, I was curious to know if there is a way to determine the function itself from knowing that f(f(x))=x^2-x+1?
I don't see how we're allowed to use f(x)=x. Clearly, the question shows that f(x)!=x. So how could the assumption f(x)=x be valid under any situation? Aren't we basically contradicting the question itself by taking f(x)=x?
Since f is defined on the whole reals, it has a value at f(x) for any x, let’s say f(x)=a, so we can also evaluate the function at x=a=f(x). It’s not that we are saying f(x)=x everywhere but we evaluate f at one point, namely f(x).
@@justtimo8638but then he is not talking about the particular value where x is equal to f(x) instead he generalized it for all X (the equation f(f(f(x))) = f(x)²-f(x)+1
@@MathAPlus But if we take f(x)=x, the new equation would be invalid for any x that doesn't meet the f(x)=x condition. So we can no longer input x=0 inside the equation because f(0)!=0 (if f(0)=0 then f(f(0))=f(0)=0 which is wrong because f(f(0))=1) This 'trick' contradicts the question because to use this, we are assuming that x=0 meets our new equation (i.e. we assume f(0)=0)
In function problems like this we always try to find some basic values then based on experience we apply some trick to solve the problem. I will post function problem next week again. Please subscribe my channel for more interesting videos 🙏
Its not luck, you can see the relation by inspection and from the fact that the question asks you to find f(0). So immediately you should know to let f(f(x)) = f(f(0)). There is no one method to solve problems you have to learn to be ingenious
Another way to approach this is the following (it is very similar to his method, but in a different order): Consider f(f(f(x))). This is something that is often helpful to consider in these types of problems, because we can interpret it in two different ways (and they are equal since function composition is associative). On one hand, f(f(f(x)) = (f(x))^2 - f(x) + 1 (where I used the f(f()) relation on the first two f's). On the other hand, f(f(f(x))= f(x^2-x+1) (where I used the f(f()) relation on the last two f's). Hence, we now have the identity f(x^2-x+1)= (f(x))^2-f(x)+1. We wish the work out f(0), so we will plug in x=0 (since then the right hand side is an equation of f(0)) to get f(1) = f(0^2-0+1)= (f(0))^2-f(0)+1. If we can obtain the value for f(1), then we would be able to compute f(0) via the quadratic equation. Similarly then, we will try plugging in x=1. This gives us f(1) = f(1^2-1+1) = (f(1))^2-f(1)+1. Now we officially have the two equations in the video, and can finish it off in the exact same way.
This solution is overcomplicated. Factor out -x to get f(f(x))=-x(-x+1)+1 and extract f(x)=-x+1. Verify the expression for f(f(x)) matches and plug in 0 to get f(0)=1
Unfortunately this is not correct, f(x) cannot be a linear function. In your case f(f(0)) is equal to 0 which is not correct, it should be 1. The reasoning in your solution is that in the first f(x) you are saying it multiplies the number by -1 and then adds 1, however in the outside f(x) it multiplies it by itself and -1, then adding 1. So f(x) is not -x+1
Nice explanation, but slightly disappointed at the fact that these type of problems seem to always involve the constants 0 and 1 AND the answer is Also 0 or 1. Will work on creating at least a double digit integer constant term version of this.
your solution is better I think, because I assumed f has an inverse. If it does f(f(0))=1 and f(f(1))=1 we can see f(0)=f(1) because they both equal f_inv (1). Then from the earlier equation f(f(1))=1 we can see f_inv (1) = f(1) meaning that 1 is it's own inverse and if you graph it, would lie on the line y=x. So f(1)=1. and as we already said f(0)=f(1) that means f(0)=1 as well.
My method Let g be the inverse of f: f(g(x))=g(f(x))=x So: (1) f(g(x))=g^2-g+1=x g^2-g+(1-x)=0 (2) g=(-1+-sqrt(1-4(1-x))/2) According to (1): g(f(0))=0 If we use it in (2): (-1+-sqrt(1-4(1-f(0))/2)=0 1=+-sqrt(1-4(1-f(0)) 1=1-4(1-f(0)) 0=4(1-f(0)) f(0)=1
@@TimeFadesMemoryLasts No its mathematically not correct. Of course he can do it to get an idea what the solution could be but its not correct. Even if a solution exists it does not mean that an inverse exists.
What the hell is going on in the very first point? We had f(f(x))=x^2-x+1, when substituting g(x) for x, we get f(f(g(x)))=f(x)=g(x)^2-g(x)+1, and everything that follows is just nonsense
even though if we assume if f(x) is x the f(f(x)) is also x. then it means f(0) is 0.. but this is not applicable. as it says: x = x^2 -x +1 which means 0=1 and that is totally wrong.
This problem is INVALID. The reason is that a function is called either outside the function (itself) for regular solution or inside the body of function for a recursive solution. you can't call the same function in it's argument. because for a funtion we need a base value which will remain unreachable as to get a base value it asks to solve the function and to solve a function we need a base value of aurgument. and this loop goes on and never stops as both actions asks for each other. By the way i am not a mathematician. I am just a normal programmer. So I might have used a different set of terms. So if you dont get see below: fuction_name (function_arguments) = { function_body_which_leads_to_output }
@@슬픈개구리-i8v that's not right, a linear function cannot generate a non linear one, any linear function in x f(x) = ax + b, f(f(x)) = a(ax+b) + b = a^2x + ab + b, which is still linear in x hence cannot be a generator of x^2 - x + 1, in fact I am struggling to find anything can get a quadratic on iteration, seems impossible.
the "at x = f(x)" step is bothering me because can we really assume a fixed point exists in this function? edit: it's actually (probably?) not wrong take any variable x and evaluate the function at f(x). but writing "x = f(x)" threw me off since it seemed to be claiming that there is a fixed point.
@@brianjjames1 Of course it is possible. X has a continuous representation and so does f(x), so f(x) is only passed to the function as another argument. The confusing thing here might be the way how the substitution was written because it assumes that f(x) = x is indeed just a plain, linear function and that is not true.
How can we assume f(x)=x exists. Let's say f(x) =x+1, than the function g(x)=x anf f(x) would never intersect, hence there is no point x for which f(x)=x. In my opinion he would need to proof that this intersection exists but maybe I am missing a point here?
f(f(x)) = x^2 - x + 1 => f(f(0)) = 1 then inverse both sides f(0) = inverse_f(1) => f(f(1)) = 1 same again but for x = 1 instead f(1) = inverse_f(1) which means f(1) = f(0) now sub in inverse_f(x) into the original equation f(f( inverse_f(x))) = [inverse_f(x)]^2 - inverse_f(x) + 1 => f(x) = [inverse_f(x)]^2 - inverse_f(x) + 1 sub in 1 for x f(1) = [inverse_f(1)]^2 - inverse_f(1) + 1 sub in f(0) for f(1) and f(0) for inverse_f(1) f(0) = [f(0)]^2 - f(0) + 1 => 0 = [f(0)]^2 - f(0) + 1 = [f(0) -1]^2 0 = f(0) - 1 therefore f(0) = 1
@@dan_mirnejhad Ok and how to prove it has an inverse in the range of interest? I think I got it in a simple way: We see f(f(0))=f(f(1))=1 Now let us act f on it once more: (a). f(f(f(0)))=f(f(f(1)))=f(1) The second term now equals f(f(x)) for x=f(1) and therefore: (b). f(f(f(1)))=f(1)^2-f(1)+1 Now insert (b) for the middle term in (a) which yields f(1)=f(1)^2-f(1)+1 and further f(1)=1 We now look at the left term in (a) and See with this result: (c). f(f(f(0)))=1 Again subsitute x=f(0) in the primary equation we get: f(0)^2-f(0)+1=1 from which follows: f(0)=0,1 We can simply see by plugging this into (c) that by recursion the solution f(0)=0 leads to 0=1 but f(0)=1 to 1=1. So the solution is f(0)=1. No inverse, no x=f(x), simple evaluation at fixed points.
The way i did it, i saw that f inverse of 1 was equal to f(1), thus implying that f(1) = 1. And since f(0) = f inverse (1), we get f(0) = 1. Is this approach valid?
Ok it is rather elementary to show that f(0)=f(1)=1. But how do we know that there exist a solution to the equation? And what is this solution. That is an interesting question.
that's interesting indeed. No polynomial function would work : a degree one polynomial composed with itself would still be of degree one, and if we choose a degree two polynomial ax^2+bx+c and compose it with itself, we must have a=0 to not have a 4th degree term which implies that the polynomial is just the degree one polynomial bx+c...
This paper talks about a similar problem : yaroslavvb.com/papers/rice-when.pdf There are no function f defined over the entire complex plane such that f(f(z))=az^2+bz+c for any choice of a,b,c. However it is sometimes possible to have such an f if we just define it over the reals, and in particular for the equation in the video it works (cf. their result in the epilogue). I'm not sure if finding an explicit expression for f is doable.
@@hach1kokothx so much. İnteresting article. İn very few cases it is possible to find explicit solutions of this functional squareroot like problem. Fx squarerootfunction of x^2 is x^(sqrroot 2).
I find this problem very silly, I came to the answer very quickly, although giving it a little more thought, I figured out there can not exist a function of (f o f) in that scenario. But Regardless thank you for you effort.
I got confused by the thumbnail and thought we needed the primitive at 0. Both f(x)=1 and f(x)=|cos(pi*x)| satisfy the requirements for f(0) and f(1). For F(x)=x+C, F(0)=C. For the cosine one, F(0)=C as well. Does this mean anything?
a function by definition can't have two images for any x belonging to its domain. Also if f(0) = 0 it will imply that f(0) = 1 which is a contradiction so f(0) has to be 1
f(fx)=x^2-x+1 put fx= y => x=y^-1 fy= (y^-1)^2-(y^-1)+1 exchange y with x fx= (x^-1)^2-(x^-1)+1 put x^-1 = y again fx=y^2-y+1 put y = fx again fx =fx^2-fx+1 (fx - 1} ^2 =0 fx= 1 f0=1
I tried a bit and noticed that it's possible for f(f(x))=ax, because then f(x)=x√a, same for f(f(x))=x+a, f(x)=x+a/2, and even you can go for f(f(x))=x², f(x)=x^a such that a^a=2, this can be solved with Lambert W. But all at once sounds hard, because even f(f(x))=x²-1 is extreme, (x^a+b)^a+b=x²-1 is it possible to find b?
You can take the inverse only if the function is one one and onto and as you can clearly see f(0) and f(1) are same so the function is not one one so you can't take the inverse of the function
Hi. I got the same solution setting f(x)=ax+b, then determining a and b. But I suppose that this result should be valid whatever function you adopt. Thanks
because f(x) is a function and x is variable, while your suggestion suppose both these being variables. Equation that you'll get will give you f(x) = 1 that's obviously not true :)
bro dont you know that you should say what the domain of f is? the idea is cool, and visible, but bro you need to become more precise... like how can you write x=f(x) and not actually mean it... bro!
Your solution is incorrect : when plugin your solution f(0) = 1 into the given equation, it turns out that f(1) = 1 . Now f(0) = 1 , f(1) = 1. So in your own words, this is none sense!
f(0)=0 creates a contradiction because f(f(0))=1 would imply f(0)=1, and that would mean 0=1, but there's no contradiction for f(0)=1: f(f(0))=1 is still true if f(0)=1 because we've seen that f(1)=1.
what is f(-2) and f(2) for f(x)=x^2 if a given function , gives the same output for two different inputs , that doesn't mean that the function/solution to it is inherently wrong
چون ایرانی هستی پس کامنت فارسی میذارم .فقط پنجاه درصد علم رسیدن به جواب هست ، پنجاه درصد دیگه از قبل همه جوابش رو میدونند ، منتهی در این پنجاه درصد هدف روش رسیدن هست و نه خود جواب .فرض کنیم داریم 2+2 ، خب همه جواب این رو میدونند ، و بنظر سوال احمقانه یی میاد اما اگر یکی نابغه باشه و بخوایم استعدادش رو کشف کنیم به جوابش نگاه نمیکنم بلکه به نحوه ی رسیدنش به جواب رو نگاه می کنیم .نبوغ در روش رسیدن هست نه خود رسیدن .ما الگوی ذهنی دیگران رو تکرار میکنیم ، مثلا شما یه این روش رو یاد دادید حالا دیگه همه الگو برداری میکنند نبوغ از بین میره،الگو برداری ذهنی از علم تا دین و مذهب یه مرض ناعلاج هست.همیشه یکی باید راه بهتری برای رسیدن به جواب 4 پیدا کنه.!! و البته صد در صد همیشه یک راه دگر وجود داره منتهی کسی باید راه رو پیدا کنه و نه جواب رو ، البته من گفتم علم پنجاه درصدش فقط جواب هست ، اما واقعیت رو بخوام از بعد دیگری بگم ، علم هرگز جواب نبوده ، بلکه همیشه راه بوده ، تمام علم در تمام خلقت وجود داره و جواب داده شده ، فقط راه و روش هست که ما امروزه به عنوان علم میشناسیم نه خود هدف یا پاسخ یک معما.
@iro4201you can copy it and paste it in Google Translater and this is an easy way to read Persian language. How ever, the answer to your question is from up to down and from right to left.
ببخشید آقای اهورا مزدا، بنده اهورا تویوتا هستم: حالا اینا را گفتی سادهتر بگو ما بفهمیم. من یه کپه مدرک از داخل و خارج دارم ولی نتونستم بفهمم چی میگی و چه ربطی به حل این مسئله داره.
It cannot be considered as any sort Olympic maths because it is too easy. Even a primary school child can do it if how to solve it were like the way how you mistakenly did it because if all whst you had to do were to plug x = 0 in the inner function, there would be not any thing to do here. The question is f(0) = ? is a misunderstanding or an incorrect notation of f The way how you solved the prob, and the way how you put the question are messy, ambiguous and obviously not aligned because whenever you use a capital or upper -case F , it means the outer function, and a lower case f , it means the inner function. But if you use the lower-case to denote both the outer function and the inner function, it will become ambiguous because f(0) can be interpreted as either the outer function f(0), or the inner function f(0). That messy and ambiguous language is usually used by individuals whose knowledge of both maths and English is poor If you denote the outer function as F(0) , then it means that the inner function f(x) = 0. If you denote f(0), it means x = 0. Here the inner function is f(x) , and the outer function is F(f(x)). In this case , you did not give any explicit inner function, and gave only an outer function which contains the inner function f(x) , the composite function should have been put like this: F(f(x)), or you could have used another letter to denote the outer function to avoid ambiguity in your expression, which has made yourself misunderstand the question If F(0), f(x) = x^2 - x + 1 = 0, because 0 inside the parentheses of the outer function stands for the inner function f(x), but not for x of the inner function If f(0), then x = 0 . That's all what the question asks for , and there will be nothing to do other than plugging x = 0 in x^2 - x + 1 It the question had been f(0), f(1), f(f(x), f(f(f(x))) , then you could evaluate the outer function with the inputs = 0, 1, f(x), f(f(x)). But your question is find f(0) , which has nothing to do with all the other inputs 0, f(x), f(f(x)) It means that you cannot even understand the question correctly. it asks for only f(0) , but not f(1), f(f(x), f(f(f(x))) if F(0), then x^2 - x + 1 = 0 is a false equation because x^2 is always non-negative , and either > or = x , and thus x^2 -x can never be 1. Therefore, no solution You will need to go back to school to learn how to denote functions correctly and clearly
8 min???? These questions come in our exam called "konkoor" that we have only a minute to solve it or we lose our time for 8 question and thats so bad for our grade te get accept in good university thats whole shit system
@@natevanderwit’s defined in the question, stupid. In math olympiads they define the domain, usually saying f(x): R -> R, for the real numbers. Complex number almost never appear
@@Mc3U you have that clown prince, Do it already. You won't get a worthy leader like Khomeini with your non-existent ideology (sorry, i don't think getting naked is counted as one).
I think you are wrong, the topics of combining elements are explained in the 12th level books in schools in Turkey at a superficial level, and I read those materials, you could never solve such questions without having an idea of such a question.
Poland 🇵🇱 Math Olympiad 1993 - Algebra - Functional Problem f(x)?!
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Nice solution, I solved it a little differently with inverses but I think there might be a small flaw in my solution(assuming f has an inverse on some interval). Anyway, this is what I did:
Let g(x) be the inverse of f(x), (f^-1(x) is annoying to type). Then we have f(x)=g(x^2-x+1). Plug in 1 and we get that f(1)=g(1). Since f and g are inverses, they are mirrored over the line y=x in the Cartesian coordinate plane. Since they both have the same y value at x=1, that must mean that f(1)=g(1)=1. If we then plug in 0, we get f(0)=g(1), but we know g(1)=1, so f(0)=1.
Let f(0) = A, let A^2 - A + 1 = B, then
f(A) = f(f(0)) = 0 - 0 + 1 = 1, and
f(1) = f(f(A)) = A^2 - A + 1 = B, and
f(B) = f(f(1)) = 1 - 1 + 1 = 1.
Now f(f(B)) = f(1) = B, but also f(f(B)) = B^2 - B + 1 => B^2 - B + 1 = B => (B - 1)^2 = 0 => B = 1 => A*(A-1) = 0.
Then check that A=0 leads to a contradiction, and the only answer is A = 1.
Simply, we note that:
f(f(0)) = f(f(1)) = 1
From here, it follows that:
f(1) = f(f(f(1))) = f(1)² - f(1) + 1
So (f(1) - 1)² = 0 -> f(1) = 1
Now, 1 = f(1) = f(f(f(0))) = f(0)² -f(0) + 1.
So f(0)² -f(0) = 0 -> either f(0) = 0 or f(0) = 1.
We check manually that f(0) could not be 0, in fact otherwise:
0 = f(0) = f(f(0)) == 0² - 0 + 1 = 1, contraddiction.
So it has to be f(0) = 1
Please watch my new video: USA 🇺🇸 Math Olympiad - Number Theory - Find the max of N to make the term integer!
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Brazil 🇧🇷 Math Olympiad - Algebra - System of Nonlinear Equations
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China 🇨🇳 Geometry problem
China 🇨🇳 Math Olympiad - Geometry - Find the length Y?
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World 🌎 International Math Olympiad 2019 - Algebra - Find f(x)?! Boost Your Confidence 😆
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England 🏴 Math Olympiad 2009 - Algebra - Find f(x)??
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Spain 🇪🇸 Math Olympiad 1981 - Algebra - Find the sum of numbers?!
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I would like to present a solution that does not involve heavy calculation, but instead understanding of the problem and logical reasoning.
We can easily calculate by setting x=0 and x=1 that f(f(0)) = f(f(1)) = 1, so if we apply f twice to 0, we get 1 and if we apply f twice to 1, we get 1. The problem is to find out what we get if we apply f only once.
0 -> f(0)=? -> f(f(0)) = 1
It’s easy to verify that if f(0)=1, then f(1)=1 and the equation holds, so f(0)=1 is definitely a possible solution. The problem is to find out if it’s also the only solution.
Now, let’s see what sequence we get if we start with 0 and keep applying f. To ease the notation, set f(0) = a.
f(0) = a
f(a) = f(f(0)) = 1
f(1) = f(f(a)) = a^2-a+1
f(a^2-a+1) = f(f(1)) = 1
f(1) = a^2-a+1
etc.
Now if we set a^2-a+1 = b, we see that
f(f(b))= a^2-a+1 = b
But from the initial equation, it also follows that
f(f(b))=b^2-b+1
Solving b = b^2-b+1 leads to b=1 and so a=0 or a=1.
a=0 leads to a contradiction and therefore a = f(0) = 1.
Finally, let me give a “counterexample”, with f(0)?> 1 that really confused me:
f(0)=5
f(5)=1
f(1)=21
f(21)=1
f(1)=21
etc.
I didn’t see why such a function is not possible. The problem is that f(f(21))=21, but it must also be much larger, because f(f(21))=21^2-21+2. This leads to a contradiction, so the only possibility is f(0)=1.
Great
Interesting Canadian math Olympiad: Canada 🇨🇦 Math Olympiad - Algebra - Find all real X values
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You should solve some questions from the exam konkur(کنکور), it’s the university entry exam across nation and it is notorious for it’s difficult questions and little time
For sure, I will post university entrance exam in the future weeks 😊
Russia 🇷🇺 Math Olympiad - Algebra - Very Interesting Problem!
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The problem in the thumbnail is unsolvable as we need to find F(0) but we are only given information on f
Very nicely explained! On further exploration, I was curious to know if there is a way to determine the function itself from knowing that f(f(x))=x^2-x+1?
Understand the question as if such a function exists. But does such a function exist?
Nice video. Completely understandable
I don't see how we're allowed to use f(x)=x. Clearly, the question shows that f(x)!=x. So how could the assumption f(x)=x be valid under any situation? Aren't we basically contradicting the question itself by taking f(x)=x?
Since it is R to R, we can use this trick to simplify the functional relation
You can say x = f(a), it does not matter in this case
Since f is defined on the whole reals, it has a value at f(x) for any x, let’s say f(x)=a, so we can also evaluate the function at x=a=f(x). It’s not that we are saying f(x)=x everywhere but we evaluate f at one point, namely f(x).
@@justtimo8638but then he is not talking about the particular value where x is equal to f(x) instead he generalized it for all X (the equation f(f(f(x))) = f(x)²-f(x)+1
@@MathAPlus But if we take f(x)=x, the new equation would be invalid for any x that doesn't meet the f(x)=x condition. So we can no longer input x=0 inside the equation because f(0)!=0 (if f(0)=0 then f(f(0))=f(0)=0 which is wrong because f(f(0))=1)
This 'trick' contradicts the question because to use this, we are assuming that x=0 meets our new equation (i.e. we assume f(0)=0)
I would appreciate if you could tell more about why you approached it like this. Not just trying out values and being "lucky"
In function problems like this we always try to find some basic values then based on experience we apply some trick to solve the problem. I will post function problem next week again. Please subscribe my channel for more interesting videos 🙏
Its not luck, you can see the relation by inspection and from the fact that the question asks you to find f(0). So immediately you should know to let f(f(x)) = f(f(0)).
There is no one method to solve problems you have to learn to be ingenious
Another way to approach this is the following (it is very similar to his method, but in a different order):
Consider f(f(f(x))). This is something that is often helpful to consider in these types of problems, because we can interpret it in two different ways (and they are equal since function composition is associative). On one hand,
f(f(f(x)) = (f(x))^2 - f(x) + 1 (where I used the f(f()) relation on the first two f's). On the other hand,
f(f(f(x))= f(x^2-x+1) (where I used the f(f()) relation on the last two f's). Hence, we now have the identity
f(x^2-x+1)= (f(x))^2-f(x)+1.
We wish the work out f(0), so we will plug in x=0 (since then the right hand side is an equation of f(0)) to get
f(1) = f(0^2-0+1)= (f(0))^2-f(0)+1.
If we can obtain the value for f(1), then we would be able to compute f(0) via the quadratic equation. Similarly then, we will try plugging in x=1. This gives us f(1) = f(1^2-1+1) = (f(1))^2-f(1)+1. Now we officially have the two equations in the video, and can finish it off in the exact same way.
Say f(0)=A
So f(A)=1
f(f(A))=A^2-A+1
This is also f(0) since f(A)=0, So A=A^2-A+1
A=1. If there are more than one root, f would not be a function.
birebir olduğunu nereden biliyorsun
birebirliği kullandığımı nereden çıkardın. Kullandığım şey fonksiyonun bir elemanı yalnızca bir elemana götürmesi.@@yiggiecorleone
This solution is overcomplicated. Factor out -x to get f(f(x))=-x(-x+1)+1 and extract f(x)=-x+1. Verify the expression for f(f(x)) matches and plug in 0 to get f(0)=1
Wow, your answer is really awesome you have completed this problem so easly in a very smart way
Unfortunately this is not correct, f(x) cannot be a linear function. In your case f(f(0)) is equal to 0 which is not correct, it should be 1. The reasoning in your solution is that in the first f(x) you are saying it multiplies the number by -1 and then adds 1, however in the outside f(x) it multiplies it by itself and -1, then adding 1. So f(x) is not -x+1
You brain ded?
@@dorukturkoglu4813 f(f(0)) is equal to 1 u missed the +1 at the end
Nice explanation, but slightly disappointed at the fact that these type of problems seem to always involve the constants 0 and 1 AND the answer is Also 0 or 1.
Will work on creating at least a double digit integer constant term version of this.
thanks i will keep this way of approaching in my mind if i see a similar question
Thanks for following my channel
your solution is better I think, because I assumed f has an inverse. If it does f(f(0))=1 and f(f(1))=1 we can see f(0)=f(1) because they both equal f_inv (1). Then from the earlier equation f(f(1))=1 we can see f_inv (1) = f(1) meaning that 1 is it's own inverse and if you graph it, would lie on the line y=x. So f(1)=1. and as we already said f(0)=f(1) that means f(0)=1 as well.
f doesn't have an inverse because it is not injective, f(0)=f(1)
Nice solution!
My method
Let g be the inverse of f:
f(g(x))=g(f(x))=x
So:
(1) f(g(x))=g^2-g+1=x
g^2-g+(1-x)=0
(2) g=(-1+-sqrt(1-4(1-x))/2)
According to (1):
g(f(0))=0
If we use it in (2):
(-1+-sqrt(1-4(1-f(0))/2)=0
1=+-sqrt(1-4(1-f(0))
1=1-4(1-f(0))
0=4(1-f(0))
f(0)=1
you cant assume that an inverse exists
@@peterluger1400he can because he gets a valid solution. If there was no solution, then you would have a point
@@TimeFadesMemoryLasts No its mathematically not correct. Of course he can do it to get an idea what the solution could be but its not correct. Even if a solution exists it does not mean that an inverse exists.
What the hell is going on in the very first point? We had f(f(x))=x^2-x+1, when substituting g(x) for x, we get f(f(g(x)))=f(x)=g(x)^2-g(x)+1, and everything that follows is just nonsense
@@TimeFadesMemoryLasts He has only the correct answer, it's just that the roots coincided
even though if we assume if f(x) is x the f(f(x)) is also x. then it means f(0) is 0.. but this is not applicable. as it says:
x = x^2 -x +1
which means 0=1 and that is totally wrong.
ff(0)) is x2-x+1
We dont assumed that f(x) is x. It is just a substitution
. You can substitute x=f(a) the same way
You should verify that the second solution is correct too.
This problem is INVALID.
The reason is that a function is called either outside the function (itself) for regular solution or inside the body of function for a recursive solution. you can't call the same function in it's argument. because for a funtion we need a base value which will remain unreachable as to get a base value it asks to solve the function and to solve a function we need a base value of aurgument. and this loop goes on and never stops as both actions asks for each other.
By the way i am not a mathematician. I am just a normal programmer. So I might have used a different set of terms. So if you dont get see below:
fuction_name (function_arguments) = { function_body_which_leads_to_output }
Your argument is invalid. Take the simple example of f(f(x))=x. Notice f(x)=x, clearly.
Great bro ❤❤❤
This problem is invalid if you cannot show a function which satisfies f(f(x)) = x^2-x+1, I don't find any.
f(x) = -x+1
@@슬픈개구리-i8v that's not right, a linear function cannot generate a non linear one, any linear function in x f(x) = ax + b, f(f(x)) = a(ax+b) + b = a^2x + ab + b, which is still linear in x hence cannot be a generator of x^2 - x + 1, in fact I am struggling to find anything can get a quadratic on iteration, seems impossible.
the "at x = f(x)" step is bothering me because can we really assume a fixed point exists in this function?
edit: it's actually (probably?) not wrong take any variable x and evaluate the function at f(x). but writing "x = f(x)" threw me off since it seemed to be claiming that there is a fixed point.
I had this same question
@@brianjjames1 Of course it is possible. X has a continuous representation and so does f(x), so f(x) is only passed to the function as another argument. The confusing thing here might be the way how the substitution was written because it assumes that f(x) = x is indeed just a plain, linear function and that is not true.
yes. it's not x = f(x). it is just another composition with f. (let x=f(t) for some t) it is misleading indeeed.
İf f(f(X))=X ,then X is a fixed point by the theory
@@F1Reyesif for example F was ln(x), you can't chose x such that x = ln(x). This step looks wrong to me.
How can we assume f(x)=x exists.
Let's say f(x) =x+1, than the function g(x)=x anf f(x) would never intersect, hence there is no point x for which f(x)=x. In my opinion he would need to proof that this intersection exists but maybe I am missing a point here?
f(f(x)) = x^2 - x + 1
=> f(f(0)) = 1 then inverse both sides
f(0) = inverse_f(1)
=> f(f(1)) = 1 same again but for x = 1 instead
f(1) = inverse_f(1)
which means f(1) = f(0)
now sub in inverse_f(x) into the original equation
f(f( inverse_f(x))) = [inverse_f(x)]^2 - inverse_f(x) + 1
=> f(x) = [inverse_f(x)]^2 - inverse_f(x) + 1
sub in 1 for x
f(1) = [inverse_f(1)]^2 - inverse_f(1) + 1
sub in f(0) for f(1) and f(0) for inverse_f(1)
f(0) = [f(0)]^2 - f(0) + 1
=> 0 = [f(0)]^2 - f(0) + 1
= [f(0) -1]^2
0 = f(0) - 1 therefore f(0) = 1
@@dan_mirnejhad Ok and how to prove it has an inverse in the range of interest?
I think I got it in a simple way:
We see f(f(0))=f(f(1))=1
Now let us act f on it once more:
(a). f(f(f(0)))=f(f(f(1)))=f(1)
The second term now equals f(f(x)) for x=f(1) and therefore:
(b). f(f(f(1)))=f(1)^2-f(1)+1
Now insert (b) for the middle term in (a) which yields
f(1)=f(1)^2-f(1)+1 and further f(1)=1
We now look at the left term in (a) and See with this result:
(c). f(f(f(0)))=1
Again subsitute x=f(0) in the primary equation we get:
f(0)^2-f(0)+1=1 from which follows:
f(0)=0,1
We can simply see by plugging this into (c) that by recursion the solution f(0)=0 leads to 0=1 but f(0)=1 to 1=1.
So the solution is f(0)=1.
No inverse, no x=f(x), simple evaluation at fixed points.
@@anatolbeck1992 f(f(x)) is polynomial so f(x) is polynomial and polynomials have inverses so we can assume f(x) has an inverse
Great video :)
Very cool video!! Thank you !
Thanks 🙏 stay tuned for more videos and introduce my channel to friends 😆
If we ignore the left part and just set x as 0, we still get 1:
X^2 - X + 1 = 0^2-0+1=1
ok and
ok an
f(x) = x should give a specific answer for x. You can't assume that for any x.
In general case yes! But in this problem we assumed f(f(x)) is available means for x----f(x) the expression is valid
The way i did it, i saw that f inverse of 1 was equal to f(1), thus implying that f(1) = 1. And since f(0) = f inverse (1), we get f(0) = 1. Is this approach valid?
I don't think so as you are assuming that f is invertible.@@suyashmisra7406
New trick up my sleeves
Amazing ❤❤❤❤
Ok it is rather elementary to show that f(0)=f(1)=1. But how do we know that there exist a solution to the equation? And what is this solution. That is an interesting question.
that's interesting indeed. No polynomial function would work : a degree one polynomial composed with itself would still be of degree one, and if we choose a degree two polynomial ax^2+bx+c and compose it with itself, we must have a=0 to not have a 4th degree term which implies that the polynomial is just the degree one polynomial bx+c...
This paper talks about a similar problem : yaroslavvb.com/papers/rice-when.pdf
There are no function f defined over the entire complex plane such that f(f(z))=az^2+bz+c for any choice of a,b,c. However it is sometimes possible to have such an f if we just define it over the reals, and in particular for the equation in the video it works (cf. their result in the epilogue). I'm not sure if finding an explicit expression for f is doable.
@@hach1kokothx so much. İnteresting article. İn very few cases it is possible to find explicit solutions of this functional squareroot like problem. Fx squarerootfunction of x^2 is x^(sqrroot 2).
this should be mental math
Cool 😎
Why not set f(X)=ax^2+bx+c then get it's composition
Try it. On composition you get a term with x^4. This implies a =0. Try composing ax +b and you dont get the required x^2.
@@davidmitchell3881 got it; İ belittled the equation; didn't mind this subtle characteristics
why would f be quadratic function
I find this problem very silly, I came to the answer very quickly, although giving it a little more thought, I figured out there can not exist a function of (f o f) in that scenario. But Regardless thank you for you effort.
I got confused by the thumbnail and thought we needed the primitive at 0. Both f(x)=1 and f(x)=|cos(pi*x)| satisfy the requirements for f(0) and f(1). For F(x)=x+C, F(0)=C. For the cosine one, F(0)=C as well. Does this mean anything?
@iro4201 That'd be |sin(pi*x)|. Plugging in 0 gives sin0 which is 0, not 1. sin(pi) is also 0. Sine doesn't quite work
@iro4201 Also nice pfp
Very instructive task
Taylor expansion?
Nice Solution
I don't get in the end why it is not acceptable to have 2 solutions...
a function by definition can't have two images for any x belonging to its domain. Also if f(0) = 0 it will imply that f(0) = 1 which is a contradiction so f(0) has to be 1
buna olimpiyat sorusu diyolarsa bide türkiyede yks 2021 e baksınlar
Harbiden aq iki tarafında f tersini alıp 30 saniye de falan çözülür
@@con394530 saniyelik çözümü atabilir misin
One is two or two is one? I don't understand
Who just put this question in olympics bro
First level national Olympiad in Iran 😆
This would be the easiest question in Turkey’s university entrance exam
Is it possible to solve for f(x) in terms of like a polynomial or whatever?
I do not think so!
f(x) approaches x^✓2 as x goes to infinity.
I didn't solve it like that but it's a fun problem
Yup, good video
f(0)=0 then
f(0)=1
why need difficult
f(fx)=x^2-x+1
put fx= y => x=y^-1
fy= (y^-1)^2-(y^-1)+1
exchange y with x
fx= (x^-1)^2-(x^-1)+1
put x^-1 = y again
fx=y^2-y+1
put y = fx again
fx =fx^2-fx+1
(fx - 1} ^2 =0
fx= 1
f0=1
you make me remember my jee days
f(x)=1??
If f(x)=1, f(f(x)) is always 1
But f(f(3))=7, not 1
What bull shit
How you take x=y^-1??
سپاس
Are you iranian?
Yes
@@MathAPlus Are you gay?
@@btb2954wtf
سلام داداش فالوت کردیم.
we are watching your career with great interest ❤❤
Palpatine
1 just basic recursion
Let y = f(x)
How to discover f(x)?
I tried a bit and noticed that it's possible for f(f(x))=ax, because then f(x)=x√a, same for f(f(x))=x+a, f(x)=x+a/2, and even you can go for f(f(x))=x², f(x)=x^a such that a^a=2, this can be solved with Lambert W.
But all at once sounds hard, because even f(f(x))=x²-1 is extreme, (x^a+b)^a+b=x²-1 is it possible to find b?
Take inverse of f both sides. Calculate f0 and f1. Then multiply finverse1. F0 will be 1.😊
you need to prove that you can take the inverse
You can take the inverse only if the function is one one and onto and as you can clearly see f(0) and f(1) are same so the function is not one one so you can't take the inverse of the function
@@John-cg1sx see Andrew michel reply to video. What I am saying is the same..
what is wrong with -x+1 ?
?
@@mmmmmmok5292 f(x)= -x+1
@@CUSELİSFANIf f(x) = -x+1,
then f(f(x)) = -f(x) + 1 = -(-x+1) + 1 = x
and not = x^2-x+1.
面白い問題だ！
China 🇨🇳 Math Olympiad - Geometry - Find the length Y?
th-cam.com/video/1N1BybFiw7E/w-d-xo.html
日本語コメ適当にあしらわれてて泣いた
@@ssusp
それな、3日前から涙止まらない
Now, is it possible to find f(x)?
Good question! I believe we can not find f(x) but if you find the solution please share it here
We can solve for a closely related composition: if f[f[x]] = x^2-x+3/4, then f[x] = |x-1/2|^(2^1/2) + 1/2
@@MathAPlusCan you notneolve it without replacing x with f(×) in the equation..
f(x) = -x + 1
but, how to prove: idk haha
Hi. I got the same solution setting f(x)=ax+b, then determining a and b. But I suppose that this result should be valid whatever function you adopt. Thanks
Why cant we just substite f(x)=x and get a quadratic in f(x)
Wtf how can you do that ?
because f(x) is a function and x is variable, while your suggestion suppose both these being variables. Equation that you'll get will give you f(x) = 1 that's obviously not true :)
bro dont you know that you should say what the domain of f is? the idea is cool, and visible, but bro you need to become more precise... like how can you write x=f(x) and not actually mean it... bro!
This is a half functional analysis.
i solved it its f(0) = 1:D
Ans- 1
Why you put x=f(x) ?
To simplify the equation we use this trick since the function is defined in R domain and range. Thanks watching my video Danial 😊
This is also cool:
England 🏴 Math Olympiad 2009 - Algebra - Find f(x)??
th-cam.com/video/_eXmLJkW6W0/w-d-xo.html
Exactly
The solution is wrong because at the same time,
he wrote f(x)=x
and
He wrote x=0
He wrote two Assumption at same time
highschool question
F(1)= 1?
solved it in less than 10 seconds, too easy, but thanks for posting g it anyway.
you did it wrong
@seeneverything
That's a bold accusation with no proof
What was your method
Nice job!
Your solution is incorrect : when plugin your solution f(0) = 1 into the given equation, it turns out that f(1) = 1 . Now f(0) = 1 , f(1) = 1. So in your own words, this is none sense!
f(0)=0 creates a contradiction because f(f(0))=1 would imply f(0)=1, and that would mean 0=1, but there's no contradiction for f(0)=1: f(f(0))=1 is still true if f(0)=1 because we've seen that f(1)=1.
Bro, what would be the problem in a function having the same solution for different values of x?
what is f(-2) and f(2) for f(x)=x^2
if a given function , gives the same output for two different inputs , that doesn't mean that the function/solution to it is inherently wrong
so what? a function can take on the same value for different inputs
Well that's a quadratic function my friend... I hope you've dealt with some of those parabolic graphs.
Please read about "many-one type functions"
چون ایرانی هستی پس کامنت فارسی میذارم .فقط پنجاه درصد علم رسیدن به جواب هست ، پنجاه درصد دیگه از قبل همه جوابش رو میدونند ، منتهی در این پنجاه درصد هدف روش رسیدن هست و نه خود جواب .فرض کنیم داریم 2+2 ، خب همه جواب این رو میدونند ، و بنظر سوال احمقانه یی میاد اما اگر یکی نابغه باشه و بخوایم استعدادش رو کشف کنیم به جوابش نگاه نمیکنم بلکه به نحوه ی رسیدنش به جواب رو نگاه می کنیم .نبوغ در روش رسیدن هست نه خود رسیدن .ما الگوی ذهنی دیگران رو تکرار میکنیم ، مثلا شما یه این روش رو یاد دادید حالا دیگه همه الگو برداری میکنند نبوغ از بین میره،الگو برداری ذهنی از علم تا دین و مذهب یه مرض ناعلاج هست.همیشه یکی باید راه بهتری برای رسیدن به جواب 4 پیدا کنه.!! و البته صد در صد همیشه یک راه دگر وجود داره منتهی کسی باید راه رو پیدا کنه و نه جواب رو ، البته من گفتم علم پنجاه درصدش فقط جواب هست ، اما واقعیت رو بخوام از بعد دیگری بگم ، علم هرگز جواب نبوده ، بلکه همیشه راه بوده ، تمام علم در تمام خلقت وجود داره و جواب داده شده ، فقط راه و روش هست که ما امروزه به عنوان علم میشناسیم نه خود هدف یا پاسخ یک معما.
@iro4201you can copy it and paste it in Google Translater and this is an easy way to read Persian language. How ever, the answer to your question is from up to down and from right to left.
@iro4201you can't read it
@iro4201بهتره نخونی چلغوز
ببخشید آقای اهورا مزدا، بنده اهورا تویوتا هستم: حالا اینا را گفتی سادهتر بگو ما بفهمیم. من یه کپه مدرک از داخل و خارج دارم ولی نتونستم بفهمم چی میگی و چه ربطی به حل این مسئله داره.
من بابا كورنيستيم از دانشكاه اصفهان
It was very very easy
asnwer=1 isit
it's been a day bro
👍
Indians exams would give such a question as level 1 just to help us save some time in exam!!
Cringe
Ok shut up
It cannot be considered as any sort Olympic maths because it is too easy. Even a primary school child can do it if how to solve it were like the way how you mistakenly did it because if all whst you had to do were to plug x = 0 in the inner function, there would be not any thing to do here.
The question is f(0) = ? is a misunderstanding or an incorrect notation of f
The way how you solved the prob, and the way how you put the question are messy, ambiguous and obviously not aligned because whenever you use a capital or upper -case F , it means the outer function, and a lower case f , it means the inner function. But if you use the lower-case to denote both the outer function and the inner function, it will become ambiguous because f(0) can be interpreted as either the outer function f(0), or the inner function f(0). That messy and ambiguous language is usually used by individuals whose knowledge of both maths and English is poor
If you denote the outer function as F(0) , then it means that the inner function f(x) = 0. If you denote f(0), it means x = 0. Here the inner function is f(x) , and the outer function is F(f(x)). In this case , you did not give any explicit inner function, and gave only an outer function which contains the inner function f(x) , the composite function should have been put like this: F(f(x)), or you could have used another letter to denote the outer function to avoid ambiguity in your expression, which has made yourself misunderstand the question
If F(0), f(x) = x^2 - x + 1 = 0, because 0 inside the parentheses of the outer function stands for the inner function f(x), but not for x of the inner function
If f(0), then x = 0 . That's all what the question asks for , and there will be nothing to do other than plugging x = 0 in x^2 - x + 1
It the question had been f(0), f(1), f(f(x), f(f(f(x))) , then you could evaluate the outer function with the inputs = 0, 1, f(x), f(f(x)). But your question is find f(0) , which has nothing to do with all the other inputs 0, f(x), f(f(x))
It means that you cannot even understand the question correctly. it asks for only f(0) , but not f(1), f(f(x), f(f(f(x)))
if F(0), then x^2 - x + 1 = 0 is a false equation because x^2 is always non-negative , and either > or = x , and thus x^2 -x can never be 1. Therefore, no solution
You will need to go back to school to learn how to denote functions correctly and clearly
Incorrect.
I feel you are very angry. The question and the solution were both understandable.
I've no idea what r u typed. But the way he explain is more understandable than yours
take your pills on time
8 min???? These questions come in our exam called "konkoor" that we have only a minute to solve it or we lose our time for 8 question and thats so bad for our grade te get accept in good university thats whole shit system
Easy question
This is nonsense. What about imaginary roots? Such as:( - 1 + square root 3 i)/2.
The domain is defined in real numbers
Why do you say its nonsense?
@@MathAPlus wdym this isn't the 1600's. The domain is always the complex's
@@natevanderwit’s defined in the question, stupid. In math olympiads they define the domain, usually saying f(x): R -> R, for the real numbers. Complex number almost never appear
@@MathAPlusWhere is this stated in the video?
H
Yes ef af sqw
f(x) =1-x
Что ты несёшь
Все правильно вроде несёт
0 or 1
only 1 is the correct answer.
too ez
Why do youbsay that ? How did you solve..i think his method is wrong..
@@leif1075
nonsense
@@wylitebee2648 what's nonsense?
@@wylitebee2648??
@@leif1075 it's pretty easy to solve, his method is right just very verbose
This is haram 😂
Buddy it’s math💀 please be quiet
Muslims invented algebra when west was in dark ages, Pooping each other out in caves.
@@mazedabari6438 I think Indians came up with the constructs of Mathematical notations we use today, but Arabians helped spread them
@@yty1941that's lies told to you by proud indian nationalist bogus
@@ahnaflfc369Indians did invent numbers though.
The flag is wrong
☀️🦁
Least delusional iranian opposition
@@frsteemid This belief is beyond the opposition.
Most of people are opposite of government.
People only need a leader to make a revolution.
@@Mc3U you have that clown prince,
Do it already. You won't get a worthy leader like Khomeini with your non-existent ideology (sorry, i don't think getting naked is counted as one).
بلاخره باید همه جا آبرو ریزی کنید دیگه😂
Wow, youtube hid my reply.
Gotta love the censorship in this so-called free social media😅
Bro this question is very easy question for Turkish students LOL
olm nasıl kolay. düşünebildin mi bu çözümü?
@@lastabroad2601 kanka 12. sinifta de test kitaplarinda bu soru tipinden vardi.
I think you are wrong, the topics of combining elements are explained in the 12th level books in schools in Turkey at a superficial level, and I read those materials, you could never solve such questions without having an idea of such a question.
@@EhsanFromBT yes youre right. But every Turkish students whos in grade 12th solve this type of questions. I graduated one year ago from highschool
Does Iran have mathematicians? They can only beat the heads on the ground in praying
As far as I know they are among top 10 mathematicians
Where tf are you getting that shifty statement from💀 I'm iranian. No one here "beats their heads on the ground in praying"
thats kinda offensive
No we are number 8 of the most medals of olympiad in the world
well we got an ignorant who's fed up with western media trash
Please use Iran's real flag
That's Islamic republics flag
This is the real flag, now cry about it.