Poland 🇵🇱 Math Olympiad 1993 - Algebra - Functional Problem f(x)?! kzbin.info/www/bejne/sJi4gpaIrdedepo
@andrewmichel25255 ай бұрын
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.
@mamenchisaurus87315 ай бұрын
Nice video. Completely understandable
@a_k66895 ай бұрын
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.
@luigibaronchelli31834 ай бұрын
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
@nathanas645 ай бұрын
Very cool video!! Thank you !
@MathAPlus5 ай бұрын
Thanks 🙏 stay tuned for more videos and introduce my channel to friends 😆
@azizkudaikulov9934 ай бұрын
Nice solution!
@jamesjackson1284 ай бұрын
Great bro ❤❤❤
@huseyin88635 ай бұрын
thanks i will keep this way of approaching in my mind if i see a similar question
@MathAPlus5 ай бұрын
Thanks for following my channel
@QED.4 ай бұрын
Great video :)
@ShayanMohammadian4 ай бұрын
Amazing ❤❤❤❤
@radmehrhakhamanesh68165 ай бұрын
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
@MathAPlus5 ай бұрын
For sure, I will post university entrance exam in the future weeks 😊
@HritikRC4 ай бұрын
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?
@SanderBessels5 ай бұрын
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.
@MathAPlus5 ай бұрын
Great
@math_qz_25 ай бұрын
Very instructive task
@penguin92573 ай бұрын
The problem in the thumbnail is unsolvable as we need to find F(0) but we are only given information on f
@minhhungle74885 ай бұрын
New trick up my sleeves
@feyakut2 ай бұрын
This would be the easiest question in Turkey’s university entrance exam
@abtinm9344 ай бұрын
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?
@MathAPlus4 ай бұрын
Since it is R to R, we can use this trick to simplify the functional relation
@Hotmedal4 ай бұрын
You can say x = f(a), it does not matter in this case
@justtimo86384 ай бұрын
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).
@hououinkyouma43884 ай бұрын
@@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
@user-eh2ec3rn6w5 ай бұрын
Nice Solution
@dnadns84535 ай бұрын
I would appreciate if you could tell more about why you approached it like this. Not just trying out values and being "lucky"
@MathAPlus5 ай бұрын
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 🙏
@james-md1cf5 ай бұрын
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
@everettmeekins15825 ай бұрын
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.
@richardfredlund88465 ай бұрын
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.
@anasshaki75315 ай бұрын
f doesn't have an inverse because it is not injective, f(0)=f(1)
@Set_Get5 ай бұрын
سپاس
@maximju.31845 ай бұрын
Yup, good video
@getanehkassa5 ай бұрын
Understand the question as if such a function exists. But does such a function exist?
@MathAPlus5 ай бұрын
China 🇨🇳 Geometry problem China 🇨🇳 Math Olympiad - Geometry - Find the length Y? kzbin.info/www/bejne/Z3-Uc6yYe86qbac
@MathAPlus5 ай бұрын
Please watch my new video: USA 🇺🇸 Math Olympiad - Number Theory - Find the max of N to make the term integer! kzbin.info/www/bejne/jZC4e6WqnqqcgLM
@bonzinip5 ай бұрын
Why not just write the series for f(f(x)) in terms of the coefficients of the series for f(x)? Set x=0 and you get f(0)=f(f(0))=1
@ScandGeek5 ай бұрын
This implies that the power series is convergent, which we do not know
@bonzinip5 ай бұрын
You do not need to assume that you're able to compute f(1) using the series though, so it only has to converge for x=0. For example 1+x+x^2+3!x^3+4!x^4 only converges for x=0 but the value of the analytic continuation at 0 is indeed 1. So all you have to assume, is that f(x) is analytic.
@ScandGeek5 ай бұрын
@@bonzinip fair enough, but analyticity is a rather strong requirement, or rather analyticity around x=0
@bonzinip5 ай бұрын
It is but for a competition question it's fair to try a couple sufficient conditions and see if one works, especially given how trivial the solution becomes. (Initially I looked at the series just because another comment was asking what f(x) is like).
@ToguMrewuku5 ай бұрын
i dont think analyticity is the problem, but rather how f(0) equals f(f(0)). why would that be?
@karabatmanferhat4 ай бұрын
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.
@yiggiecorleone4 ай бұрын
birebir olduğunu nereden biliyorsun
@karabatmanferhat4 ай бұрын
birebirliği kullandığımı nereden çıkardın. Kullandığım şey fonksiyonun bir elemanı yalnızca bir elemana götürmesi.@@yiggiecorleone
@MyOneFiftiethOfADollar3 ай бұрын
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.
@fnsankaku59205 ай бұрын
面白い問題だ!
@MathAPlus5 ай бұрын
China 🇨🇳 Math Olympiad - Geometry - Find the length Y? kzbin.info/www/bejne/Z3-Uc6yYe86qbac
@ssusp5 ай бұрын
日本語コメ適当にあしらわれてて泣いた
@fnsankaku59205 ай бұрын
@@ssusp それな、3日前から涙止まらない
@pecfexfextus44375 ай бұрын
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.
@brianjjames15 ай бұрын
I had this same question
@F1Reyes5 ай бұрын
@@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.
@Q_205 ай бұрын
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.
@khayalaliyev35195 ай бұрын
İf f(f(X))=X ,then X is a fixed point by the theory
@louison32165 ай бұрын
@@F1Reyesif for example F was ln(x), you can't chose x such that x = ln(x). This step looks wrong to me.
@saleemhafiz194 ай бұрын
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.
@MathAPlus4 ай бұрын
ff(0)) is x2-x+1
@30svich4 ай бұрын
We dont assumed that f(x) is x. It is just a substitution . You can substitute x=f(a) the same way
@saleemhafiz194 ай бұрын
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 }
@greenpewdiepie42074 ай бұрын
Your argument is invalid. Take the simple example of f(f(x))=x. Notice f(x)=x, clearly.
@geostorm81925 ай бұрын
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?
@iro42015 ай бұрын
Vhat about sine?
@geostorm81925 ай бұрын
@@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
@geostorm81925 ай бұрын
@@iro4201 Also nice pfp
@iro42015 ай бұрын
Interesting stuff, tell me more @@geostorm8192
@MathAPlus4 ай бұрын
Spain 🇪🇸 Math Olympiad 1981 - Algebra - Find the sum of numbers?! kzbin.info/www/bejne/d5TCfpt6eLODa5I
@MathAPlus5 ай бұрын
England 🏴 Math Olympiad 2009 - Algebra - Find f(x)?? kzbin.info/www/bejne/lZa7nn-AoLxpjZI
@MathAPlus5 ай бұрын
Interesting Canadian math Olympiad: Canada 🇨🇦 Math Olympiad - Algebra - Find all real X values kzbin.info/www/bejne/pn3QlKp7bbx8ntE
@vik24oct19915 ай бұрын
This problem is invalid if you cannot show a function which satisfies f(f(x)) = x^2-x+1, I don't find any.
@user-pv4ey9pj4g4 ай бұрын
f(x) = -x+1
@vik24oct19914 ай бұрын
@@user-pv4ey9pj4g 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.
@MathAPlus5 ай бұрын
World 🌎 International Math Olympiad 2019 - Algebra - Find f(x)?! Boost Your Confidence 😆 kzbin.info/www/bejne/g2mwfplupdJjfK8
@user-di7em8zz9b4 ай бұрын
One is two or two is one? I don't understand
@henrykim63984 ай бұрын
this should be mental math
@MathAPlus4 ай бұрын
Cool 😎
@anatolbeck19924 ай бұрын
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?
@dan_mirnejhad3 ай бұрын
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
@anatolbeck19923 ай бұрын
@@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.
@dan_mirnejhad3 ай бұрын
@@anatolbeck1992 f(f(x)) is polynomial so f(x) is polynomial and polynomials have inverses so we can assume f(x) has an inverse
@DipterixWang4 ай бұрын
Taylor expansion?
@pixtane74275 ай бұрын
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
@mmmmmmok52925 ай бұрын
ok and
@btb29545 ай бұрын
ok an
@mohammadmahdimovahedfar32455 ай бұрын
f(x) = x should give a specific answer for x. You can't assume that for any x.
@MathAPlus5 ай бұрын
In general case yes! But in this problem we assumed f(f(x)) is available means for x----f(x) the expression is valid
@suyashmisra74065 ай бұрын
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?
@aadhavan71275 ай бұрын
I don't think so as you are assuming that f is invertible.@@suyashmisra7406
@khayalaliyev35195 ай бұрын
Why not set f(X)=ax^2+bx+c then get it's composition
@davidmitchell38815 ай бұрын
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.
@khayalaliyev35195 ай бұрын
@@davidmitchell3881 got it; İ belittled the equation; didn't mind this subtle characteristics
@mishania66785 ай бұрын
why would f be quadratic function
@oakpope5 ай бұрын
You should verify that the second solution is correct too.
@apomount3175 ай бұрын
Who just put this question in olympics bro
@MathAPlus5 ай бұрын
First level national Olympiad in Iran 😆
@MathAPlus3 ай бұрын
Brazil 🇧🇷 Math Olympiad - Algebra - System of Nonlinear Equations kzbin.info/www/bejne/m4ramaavrJ6Zb6M
@klausg18435 ай бұрын
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.
@hach1koko5 ай бұрын
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...
@hach1koko5 ай бұрын
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.
@klausg18435 ай бұрын
@@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).
@pouyanchakameh85945 ай бұрын
سلام داداش فالوت کردیم. we are watching your career with great interest ❤❤
@benyseus63255 ай бұрын
Palpatine
@djigoo4 ай бұрын
I didn't solve it like that but it's a fun problem
@mvirts4 ай бұрын
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
@riccardolaporta70844 ай бұрын
Wow, your answer is really awesome you have completed this problem so easly in a very smart way
@dorukturkoglu48134 ай бұрын
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
@Mehraj_IITKGP4 ай бұрын
You brain ded?
@em985744 ай бұрын
@@dorukturkoglu4813 f(f(0)) is equal to 1 u missed the +1 at the end
@marxcarton38585 ай бұрын
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.
@mohamadserhan95814 ай бұрын
I don't get in the end why it is not acceptable to have 2 solutions...
@jishubharadwaj38174 ай бұрын
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
@schizoframia48744 ай бұрын
Is it possible to solve for f(x) in terms of like a polynomial or whatever?
@MathAPlus4 ай бұрын
I do not think so!
@nenhard4 ай бұрын
f(x) approaches x^✓2 as x goes to infinity.
@ISDL29TV5 ай бұрын
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
@peterluger14005 ай бұрын
you cant assume that an inverse exists
@TimeFadesMemoryLasts5 ай бұрын
@@peterluger1400he can because he gets a valid solution. If there was no solution, then you would have a point
@peterluger14005 ай бұрын
@@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.
@user-es6hc4qk3t5 ай бұрын
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
@user-es6hc4qk3t5 ай бұрын
@@TimeFadesMemoryLasts He has only the correct answer, it's just that the roots coincided
@ibrahimmeteoglu3204 ай бұрын
buna olimpiyat sorusu diyolarsa bide türkiyede yks 2021 e baksınlar
@con39454 ай бұрын
Harbiden aq iki tarafında f tersini alıp 30 saniye de falan çözülür
@ozgurozgurluk.4 ай бұрын
@@con394530 saniyelik çözümü atabilir misin
@miner43745 ай бұрын
Are you iranian?
@MathAPlus5 ай бұрын
Yes
@btb29545 ай бұрын
@@MathAPlus Are you gay?
@mishania66785 ай бұрын
@@btb2954wtf
@emremokoko5 ай бұрын
what is wrong with -x+1 ?
@mmmmmmok52925 ай бұрын
?
@emremokoko5 ай бұрын
@@mmmmmmok5292 f(x)= -x+1
@magicmulder5 ай бұрын
@@emremokokoIf f(x) = -x+1, then f(f(x)) = -f(x) + 1 = -(-x+1) + 1 = x and not = x^2-x+1.
@guillermo34125 ай бұрын
i solved it its f(0) = 1:D
@raphaelnoronha14195 ай бұрын
How to discover f(x)?
@madghostek30265 ай бұрын
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?
@HaniyeHashemi-tp9rc5 ай бұрын
👍
@gerakore89485 ай бұрын
1 just basic recursion
@Narayanabhakt4 ай бұрын
Ans- 1
@megofya4 ай бұрын
highschool question
@danial.amirian5 ай бұрын
Why you put x=f(x) ?
@MathAPlus5 ай бұрын
To simplify the equation we use this trick since the function is defined in R domain and range. Thanks watching my video Danial 😊
@MathAPlus5 ай бұрын
This is also cool: England 🏴 Math Olympiad 2009 - Algebra - Find f(x)?? kzbin.info/www/bejne/lZa7nn-AoLxpjZI
@alireza90064 ай бұрын
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
@ownallamohammed94324 ай бұрын
من بابا كورنيستيم از دانشكاه اصفهان
@ToguMrewuku5 ай бұрын
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!
@boetiemaloneАй бұрын
Let y = f(x)
@justafanofalphabetlore5 ай бұрын
This is a half functional analysis.
@sheelu455 ай бұрын
Take inverse of f both sides. Calculate f0 and f1. Then multiply finverse1. F0 will be 1.😊
@John-cg1sx5 ай бұрын
you need to prove that you can take the inverse
@hardikgarg92185 ай бұрын
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
@sheelu455 ай бұрын
@@John-cg1sx see Andrew michel reply to video. What I am saying is the same..
@triangle25173 ай бұрын
Why cant we just substite f(x)=x and get a quadratic in f(x)
@Theuomr5 ай бұрын
Now, is it possible to find f(x)?
@MathAPlus5 ай бұрын
Good question! I believe we can not find f(x) but if you find the solution please share it here
@matthewarnold88685 ай бұрын
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
@leif10755 ай бұрын
@@MathAPlusCan you notneolve it without replacing x with f(×) in the equation..
@Luciano-nv8kp5 ай бұрын
f(x) = -x + 1 but, how to prove: idk haha
@paolorepeto43685 ай бұрын
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
@MathAPlus4 ай бұрын
Russia 🇷🇺 Math Olympiad - Algebra - Very Interesting Problem! kzbin.info/www/bejne/bZTEeoOgmreVZqs
@sa_ket_roy5 ай бұрын
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
@radicalengineer23315 ай бұрын
you make me remember my jee days
@user-zc8qv9ok7q5 ай бұрын
f(x)=1?? If f(x)=1, f(f(x)) is always 1 But f(f(3))=7, not 1
@albinskariah74185 ай бұрын
What bull shit
@imrozahmad3525 ай бұрын
How you take x=y^-1??
@koktonglan75664 ай бұрын
f(0)=0 then f(0)=1 why need difficult
@dimitrijbender72854 ай бұрын
F(1)= 1?
@ibrahimyeter74785 ай бұрын
It was very very easy
@ahuramazda92025 ай бұрын
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.
@iro42015 ай бұрын
Hov do I read yoiur language, from right to left or from up to dovnside?
@user-ph8po7wv6u5 ай бұрын
@@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.
@iranmaster5 ай бұрын
@@iro4201you can't read it
@Set_Get4 ай бұрын
@@iro4201بهتره نخونی چلغوز
@Set_Get4 ай бұрын
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.
@comdo7775 ай бұрын
asnwer=1 isit
@stickbug3975 ай бұрын
it's been a day bro
@MT-in3tp5 ай бұрын
solved it in less than 10 seconds, too easy, but thanks for posting g it anyway.
@seeneverything51505 ай бұрын
you did it wrong
@itsohaya40965 ай бұрын
@seeneverything That's a bold accusation with no proof
@FishSticker5 ай бұрын
What was your method
@MathAPlus5 ай бұрын
Nice job!
@lemon79995 ай бұрын
Easy question
@michealscofield47227 күн бұрын
H Yes ef af sqw
@user-dy5qg3yl4d4 ай бұрын
0 or 1
@dante58764 ай бұрын
only 1 is the correct answer.
@The1WhoKnocks_5 ай бұрын
Что ты несёшь
@debikk42045 ай бұрын
Все правильно вроде несёт
@wylitebee26485 ай бұрын
too ez
@leif10755 ай бұрын
Why do youbsay that ? How did you solve..i think his method is wrong..
@wylitebee26485 ай бұрын
@@leif1075 nonsense
@leif10755 ай бұрын
@@wylitebee2648 what's nonsense?
@leif10755 ай бұрын
@@wylitebee2648??
@wylitebee26485 ай бұрын
@@leif1075 it's pretty easy to solve, his method is right just very verbose
@NurmemetAbliz5 ай бұрын
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!
@antoniofornari74515 ай бұрын
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.
@tedchirvasiu5 ай бұрын
Bro, what would be the problem in a function having the same solution for different values of x?
@naveen.v47345 ай бұрын
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
@pmcate25 ай бұрын
so what? a function can take on the same value for different inputs
@aryanmishra95145 ай бұрын
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"
@dsikefxx82785 ай бұрын
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
@danielcrespo78245 ай бұрын
f(x) =1-x
@lgtvsmart40785 ай бұрын
This is haram 😂
@ChadNarukamii5 ай бұрын
Buddy it’s math💀 please be quiet
@mazedabari64385 ай бұрын
Muslims invented algebra when west was in dark ages, Pooping each other out in caves.
@yty19415 ай бұрын
@@mazedabari6438 I think Indians came up with the constructs of Mathematical notations we use today, but Arabians helped spread them
@ahnaflfc3695 ай бұрын
@@yty1941that's lies told to you by proud indian nationalist bogus
@BuzzyBee135 ай бұрын
@@ahnaflfc369Indians did invent numbers though.
@chocolatemilkshake91684 ай бұрын
Indians exams would give such a question as level 1 just to help us save some time in exam!!
@hououinkyouma43884 ай бұрын
Cringe
@yuseifudo60752 ай бұрын
Ok shut up
@marcusclarkson55205 ай бұрын
Bro this question is very easy question for Turkish students LOL
@lastabroad26015 ай бұрын
olm nasıl kolay. düşünebildin mi bu çözümü?
@marcusclarkson55205 ай бұрын
@@lastabroad2601 kanka 12. sinifta de test kitaplarinda bu soru tipinden vardi.
@EhsanFromBT4 ай бұрын
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.
@marcusclarkson55204 ай бұрын
@@EhsanFromBT yes youre right. But every Turkish students whos in grade 12th solve this type of questions. I graduated one year ago from highschool
@Mc3U4 ай бұрын
The flag is wrong ☀️🦁
@frsteemid4 ай бұрын
Least delusional iranian opposition
@Mc3U4 ай бұрын
@@frsteemid This belief is beyond the opposition. Most of people are opposite of government. People only need a leader to make a revolution.
@frsteemid4 ай бұрын
@@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).
@user-ls5dq8eu2u4 ай бұрын
بلاخره باید همه جا آبرو ریزی کنید دیگه😂
@frsteemid4 ай бұрын
Wow, youtube hid my reply. Gotta love the censorship in this so-called free social media😅
@vansf34335 ай бұрын
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
@abdc29905 ай бұрын
Incorrect.
@hamidrezaamini86265 ай бұрын
I feel you are very angry. The question and the solution were both understandable.
@rifqy0285 ай бұрын
I've no idea what r u typed. But the way he explain is more understandable than yours
@iranmaster5 ай бұрын
take your pills on time
@user-xw6ky8ob4l5 ай бұрын
This is nonsense. What about imaginary roots? Such as:( - 1 + square root 3 i)/2.
@MathAPlus5 ай бұрын
The domain is defined in real numbers
@leif10755 ай бұрын
Why do you say its nonsense?
@natevanderw5 ай бұрын
@@MathAPlus wdym this isn't the 1600's. The domain is always the complex's
@oliveirafilipe54165 ай бұрын
@@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
@AmosNewcombe5 ай бұрын
@@MathAPlusWhere is this stated in the video?
@user-fd6dx9mm3l4 ай бұрын
Please use Iran's real flag That's Islamic republics flag
@mohamadserhan95814 ай бұрын
This is the real flag, now cry about it.
@robertmonroe97285 ай бұрын
Does Iran have mathematicians? They can only beat the heads on the ground in praying
@MathAPlus5 ай бұрын
As far as I know they are among top 10 mathematicians
@theallmightyowlcat92815 ай бұрын
Where tf are you getting that shifty statement from💀 I'm iranian. No one here "beats their heads on the ground in praying"
@imshiruba5 ай бұрын
thats kinda offensive
@alishahani98925 ай бұрын
No we are number 8 of the most medals of olympiad in the world
@a9majd5 ай бұрын
well we got an ignorant who's fed up with western media trash