f(x+1) = f(x+2) +1 

f(x) = f ( x + 1) + 1 f( x + 1) = f (x + 2) + 1 f ( x ) = f ( x + N) + N Hereby f ( N) = - N + f(0)

Don’t forget: f(x) = f(x + 1) + 1 is true for all f of the form f(x) = -x + a for any value of a.

All solutions: f(x) = g(x) - floor(x), where g(x) is an arbitrary function with periodicity of 1, and floor(x) is the function that returns the greatest integer N fulfilling N

this guy teaches in 10 mins, but skl teachers take atleast 2 days to teach us this, salute my guy

f(x+1)=f((x+1)+1)+1 change of variables gives f(x)=f(x+1)+1 we need to find any function that goes down by 1 periodically each time x increases by 1. assuming that the function is linear, it has to be f(x)=x-a. might be interesting to find all the continuous functions and analytical ones. you can graphically construct random solutions. draw any weird function on an clopen(one closed point and one open point) interval of length 1 then translate everything moving down. any such function is a solution to the functional equation.

Good Morning，Sir， We can prove this function by means of slope which is more logical than the assumption of y=mx eg y0=f(x0)=f(x0 + 1)+1=y1+1 therefore，((y1-y0)÷(x1-x0))=-1 then we have the same result ((y2-y1)÷(x2-x1))=-1 hence. m=-1 ⇒ y = f(x) = -x + c 03-23-2024

really nice chalk board writing man!

right answer is f(x)=-x+C, not just -x

Defining p(x) = f(x)+x, with a bit of manipulation we get p(x+1) = p(x). Any function p that respects this property is good. In fact this is the defining property of a periodic function (of period 1). Therefore, the general solution is the set of all the functions f(x) = p(x) - x where p is 1-periodic. A simple example? f(x) = 10 - x. A wild example? f(x) = |¼ + sin(2πx+√3)| - x

You can add an arbitrary function g(x) which is periodic with period 1. Of course, since your question was to find a function with this property what you did is correct. I just wanted to point this out

f(x)=-x+b ; b is arbitrary real number so there exists an infinite family of functions that are solution of the equation

My answer is that f(x)= g(x) - x where g(x) is a 1-periodic function.

That's not the full score. You've proven, that f(x) = -x holds, but that's not at all the full solution. For discrete functions, I'd at least expect "f(x) = f(0) - x" as a solution (assumption f(x) = k x + c) for all x € IZ. For continuous functions, f(x) = g(x) - x is a solution for all x € IR, where g(x) is an arbitrary 1-periodic function. For example: f(x) = A sin(2 π N x + C) - x, with some arbitrary N € IZ, A € IR, and C € IR.

come on, guys, you need to attest that this equation doesn't define any single function, nor does it define even a single family of functions. So y=-x is a solution, but it's tragically not the only solution of this equation, which actually describes families of functions with various properties. 1)First I thought of the obvious - the arbitrary additive constant (like in indefinite integrals), but then I felt uncomfortable about the transition from f(x+n)-f(x)=-n (where n is an integer), which is obviously true, to the real numbers (if we conjecture that n can be any real number, which is true for linear function and even for stepwise function like y=-floor(x) (the integer part of x), that also satisfies this equation). And obviously not in vain, because it's not true for every function that satisfies this equation. 2)Even not leaving the class of continuous functions we can find a broad family of functions, that satisfy the equation but deny the transition to the property with arbitrary real offset n. That would be if we add to -x any periodic function with the period of 1, say y=-x+A*sin(2*pi*{x}), where {x}=x-floor(x), the fractional part of x (for negative numbers those work counterintuitively like [-1.2]=floor(-1.2)=-2, and {-1.2}=0.8). Or even simpler y=-x+A*sin(2*pi*x) + C, where A and С are arbitrary real numbers. So that's only if we consider continuous functions we can add any( absolutely any!) periodic function with a period of 1 (or even 1/m, where m is ANY natural number!). 3)If we allow our function to be piecewise, that is, it consists of continuous fragments the variety grows drastically, 'cause now you can additionally add or subtract any number of ANY functions dependent on {x} (for it has a period of 1 and a saw-shape ) and defined on [0,1) interval either to -x or to -[x]=-floor(x) (simplest example: y=-[x] +A*ln({x})). 4)But even that's not all, there's a crazy class of functions, that are discontinuous at every point of their domain (or nowhere continuous). A great example is Dirichlet function, which is 1 at every rational point and zero at every irrational one. Using it we can take any two of the discussed above functions and define a new function D(x)*f1(x)+(1-D(x))*f2(x), which will be nowhere continuous but still fit our equation (because adding a unit doesn't change rationality/irrationality of the number). Graphically it would look as if we see both graphs of f1(x) and f2(x) simultaneously, while each won't be a continuous line, but consist of single points placed infinitely close one to each other, while for each value there will be a point either on f1 or on f2 graph. At the end, I need to mention that ideas about all this variety came to me after reading other comments, so I just decided to sum up and I took my time to check most of them (except with Dirichlet, which is kinda obvious) in Desmos, so I'm 100% sure they are correct and valid solutions for the discussed functional equation. It totally complies, while it is an Olympiad problem, implying out-of-the-box thinking and not relying just on the most obvious solution. Anyhow, I'm incredible greatful to Prime Newtons (not sure if it's the person's name or just a channel name))) for such an interesting problem that made me dig really deep

What's interesting is that if you substitute f(x) = g(x) -x, you can simplify the equation to g(x) = g(x+1). So that means that a general solution would be -x + (any function of period 1)!

Hey man, I have an easier way!!! f(x)=f(x+1) + 1 f(x+1) - f(x) = -1 [f(x+1) - f(x)] / 1 = -1 This is the derivative of f(x) !!!! f'(x) = -1 So f(x) = -x !!!!!!! ( u can rigorously say f(x) = -x + c then prove c = 0)

f(x) = f(x+1) + 1 = f(x+2) + 2 = f(x+3) + 3 = ... and so on. So f(x) = f(x+k) + k. Put the x=0: f(0) = f(k) + k or f(k) = f(0) - k Replace k with x again: f(x) = f(0) - x, where f(0) is any constant number

Your teaching is soo good

Another way to describe the general solution form for f(x) is f(x) = p( x-[x] ) - [ x ], where p(t) is an arbitrary function defined in t ∈ [0, 1), and [ ] denotes the Gauss brackets, i.e. [ x ] = floor(x).

I resolved it graphically in my head : f(x+c) moves f(x) to the left and f(x)+c moves it up by the same amount. So doing both of those things at the same time you are moving f diagonally (up and left). So for f(x) = f(x+c)+c, all points have to be on that diagonal, which is y = -x. therefore f(x) = -x

Put x = x-1 and you get after a while that f(x) = (f(x-1) + f(x+1)) / 2 , so it is arithmetical average and proof that f(x) must be linear function. A after a next while you get f(x) = -x +b. Then b is any real number.

Please consider that your answer is not complete. The correct and complete answere is: f(x)= -x+r, which "r" can be any real number. You can have both signs "+" as well as "-" for "r" in the above equation

As you note, f(x) = f(x + 1) + 1 This is true for x = 0, x = 1, x = 2, etc... up to x = N-1. So, let's add those up. f(0) + f(1) + f(2) + ... + f(N-1) = [f(1) + 1] + [f(2) + 1] + ... + [f(N-1) + 1] + [f(N) + 1] Rearrange the brackets on the RHS, and I'm going to suggestively add some brackets on the LHS f(0) + [f(1) + f(2) + ... + f(N-1)] = [f(1) + f(2) + f(3) + ... + f(N-1)] + f(N) + [1 + 1 + 1 + ... + 1] Note that there are N 1's added together. Also, I'm going to subtract [f(1) + f(2) + ... + f(N-1)] from both sides f(0) = F(N) + N Finally, rearranging, and swapping in x for N f(x) = f(0) - x Note, this is only true for positive integers. You can do a similar argument for negative integers. We are not given enough information to confirm that this isn't something like f(x) = f(0) - x + A sin(k pi x) for some integer value of k.

f(x) = A*sin(2pi*x) - x + B, where A, B are unknown constants

It sounds really you are fantastic keep up

I initially thought that this was an easy one, but I was only getting a recursive function. Thanks for explaining how to get rid of the recursive function.

Why impose a first-degree function and not, for example, a second-degree function?

All functions of the form f(x) = -x + k (for each real k) are certainly solutions of the propesed equation. Moreover, it is easy verify that such functions are the unique linear solutions of the proposed equations. Do other solutions there exist? Yes, of course. For example, also the map f(x) = - floor(x) is a solution. This hints a way to obtain the general solution of the proposed equation. It may be as follows: f(x) = h({x}) - floor(x), where h: [0; 1[ --> R is generic and {x} is the fractional part of x. Indeed, we have: {x+2} = {x+1}; floor(x+2) =floor(x+1+1) = floor(x+1) + 1. Thus, we obtain f(x+1) = h({x+1}) - floor(x+1) = h({x+1}) - (floor(x+1) + 1) + 1 = h({x+2}) - floor(x+2) + 1 = f(x+2) +1.

Ill stated problem because it doesn't say a word about the domain of f.

Once you pick the right guess namely f(x)=-x you want to see if an extension of this solution may still be a solution, and it is easy to verify that indeed f(x)= -x + a, is the most general solution, as already pointed out in a previous comment. In the cartesian XY plane these equation represent parallel lines. The value of “a” can be fixed by giving the value of f(x) at a specific x.

f(x)=g(x)-x, where g(x)=g(x+T), T=1 (for example g(x)=sin(2πx) )

I say f(x) = -x + 10e10000 prove me wrong

Not sure if this is how you originally got to the solution, but it may be this: f(x+1) = f(x+2) +1 f(x+2) -f(x+1) = -1 (f(x+2)-f(x+1))/(x+2-(x+1)) = -1 / (x+2-(x+1)) But (x+2-(x+1)) = is just 1 and (f(x+2)-f(x+1))/(x+2-(x+1)) = d(f(x)/dx = -1 integrating: f(x) = -x +a.

Perdón, pero una solución más general sería f(x) = a - x, con a arbitrario. Eso es obvio. Sería mucho más interesante demostrar que no existen ninguna solución distinta, cosa que al parecer no es cierto, o mejor aun, encontrar la solución general. Yo se la doy, es la siguiente: Si definimos una función arbitraria h del intervalo [0,1) en R (en los reales), 0 incluido, 1 no incluido (h en realidad queda definido arbitrariamente a través del axioma de elección), me parece que la solución general para f(x) sería de la forma: f(x)=h(x-[x]) - x En que [x] es la parte entera de x (por ejemplo [5,564] =5) mientras que la función h es total y completamente arbitraria. Esa es la solución que yo quería ver cuando intrigado me di tiempo para escuchar su análisis, y me decepcionó Lamentablemente Ud. se conformo con la solución más simple de todas (h=0), y evitó el verdadero problema que no deja de ser interesante. Aquí en Chile diríamos que Ud. "le saco el poto a la jeringa"

Ok f(x)=f(x+1)+1 Lets asume f(1)=k k is a constant f(x)=f(x+1)+1 so f(x+1)=f(x)-1 f(2)=k-1 f(3)=k-2 So now its a Succession k,k-1,k-2... What is the general rule? De difference between them are -1 k-1-k=-1 k-2-(k-1)=-1 So the general rule ar -x+c si g(x)=-x+c g(1)=-1+c=k c=k+1 So f(x)=-x+k+1 k+1 its also a constant so I will call it a f(x)=-x+a Verification : f(x+1)=-(x+1)+a=-x-1+a f(x+2)+1=-(x+2)+a+1=-x-2+a+1=-x-1+a f(x+1)=f(x+2)+1

I think the easier method was F(x+2)-F(x+1)=-1, which is the same as: (F(x+2)-F(x+1))/(x+2-(x+1))=-1 dF/dx=-1 Then F(x) = -x+C

The assumption is unnecessary. Define g(x)=f(x)+x. Then f(x+1)=f(x+2)+1 can be written as g(x+1)=g(x+2) which by induction means that g(x)=g(0) for all integers x. So f(x)=C-x for some constant C and all integers x. For non-integral x, this result does not hold.

This is the wrong solution. You have given only the trivial part of the solution to the differential equation, which must be added to any periodic function. f(x+1)=f(x+2)+1 f'(x+1)=f'(x+2) case 1: if f'(x)=c => f(x)=ax+b f(x+1)= a(x+1)+b f(x+2)+1= a(x+2)+b+1 a(x+1)+b = a(x+2)+b+1 a(x+1) = a(x+2)+1 ax+a=ax+2a+1 a=2a+1 0=a+1 a=-1 f(x)=-x+b case 2: if f'(x+1)!=c then f'(x) periodic function. so solution f(x) = P(x)-x+b, where P(x) periodic function, T may be 1 and other periods that are comparable to 1 may also be suitable. For example: f(x) = sin(2pix)-x f(0) = 0 f(1) = -1 f(2) = -2 f(1/4) = 1-1/4 = 3/4 f(5/4) = 1-5/4 = -1/4

I know what you originally did: f(x+1) = f(x+2) +1 | -1 f(x+1)-1 = f(x+2) | flip the sides to make it clearer f(x+2) = f(x+1)-1 Imagine the graph: Everytime you take on step to the right on the x-axis the y gets reduced by 1 Obviously: y=-x True?

You can rearrange it to F(x+2)-F(x+1) = -1 . So I think it wouldn't be far to assume the function is similar to form of the real function (DY/DX) = - 1, and go from there to Y = -X + C

defining f(x) as a linear funtion, then f(x)=ax+b and so f(x+1) = a(x+1)+b =( ax+b) +a = f(x)+a and so (f(x+2) = a(x+2)+b= (ax+b)+2a = f(x)+2a now, given f(x+1) = f(x+2)+1 let's use the above, and so f(x)+a = f(x)+2a+1 then a=-1 it can be shown that b does not have any effect for any x we choose...(b can be any arbirary number) and so f(x)=-x+b

I knew that it would be -x bc if the number in the function is bigger than the one in the other side and you still need to add a number that means it's negative and bc the number in function is bigger by 1 and you add 1 that means that the function is -1

Oh no! My teacher has never taught me about function equation before😮so amazing🎉

We can also solve this by making it a telescoping series f(x+1)-f(x+2) = 1 => f(0) - f(1) = 1 f(1) - f(2) = 1 f(2) - f(3) = 1 ... f(x-2) - f(x-1) = 1 f(x-1) - f(x) = 1 Adding all: f(0) - f(x) = x f(x) = -x + f(0) f(0) can be any constant So f(x) = -x + a

I really think that the complete solution should be: f(x)=k(x), if 0≤x

One question if there is anyone that knows the answer...¿How we can be sure that f(x)=-x is the only solution of the problem? ¿Is there any other f(x) that satisfy the property? Thank you very much

Nice videos! Btw when introducing a functional equation, the function's domain and codomain should be defined, as well as the domain of the equation variables. This can drastically change the problem solution (olympiad problems always define this). Normally I wouldn't nitpick this but it seems to be a widespread bad habit of many question askers, and even tutors ...

I don't think you've given a good reason to guess. First you should prove f(n)=f(0)-n and then assume f(x)=c-x, for arbitrary constant c. Anyway I like your teaching and follow you. Good luck.

This is not the general solution. More general is −x + c of course, but c is an arbitrary function of [0,1[, the fractional part of x. For example −x + cos(2πx) is another solution.

f(x+1) - f(x) = -1 notice that's just the formula for slope of a secant line. assuming f(x) is continuous, and x is real, then f(x) = -x + c is an answer.

The most general solution is f(x)= p(x)-x, where p is any 1-periodic function on the real line. Why you shrink it to the integers?

This is certainly good, but it is not a complete solution. for example, this function is also suitable: f(x)=sin(2*pi*x)-x in the general case, the solution is any function for which the following rules are valid: 1) on the interval [0;1) the function can take any random values 2) the values of the function on the interval [a;a+1) are obtained by displacement function values by 1 downward from the interval [a-1;a) or by shifting function values by 1 upward from the interval [a+1;a+2)

Obviously linear. Obviously negative. Three seconds. Love your enthusiasm though.

Isnt this an incredibly easy problem? Just rewrite it in recursive form A_n = A_(n-1) - 1 which is trivial. If you need a f(x) form that's again trivially f(x)=-x+C I did this within like 10s from the thumbnail and I really only clicked to see how this was stretched out for ten minutes. It seems it's for teaching people who are new to functions.

i actually used the recursivity of the funtion being that f(0)=f(n)+n for any n so i put n=x in as n is any input assuming function is defined on the reals only. then i took cases. when f(-1)=f(0)+1 for odd and even case i found out f cannot be even so i got f(1)=-1 then put it into my original equation for f(o) and got f(x)=f(o)-x which leaves you with f(x)=-x

F(x)=-x+k Let f(x)=ax+k ax+k=a(x+1)+k+1 ax+k=ax+a+k+1 a=-1 So It’s true for any k So F(x)=-x+k For any k

Thank you for the video. My solution is: we must find ONE such function. f(x)=-x works. Over. 🙂

this equation defines an arithmatic progression with a common difference of -1. so f(x) = f(1) - N

Annoyingly, I solved this in my head from the thumbnail, but didn't really have a strategy. It just came to me as "obvious", which isn't really helpful when trying to develop skills (and certainly isn't helpful when trying to teach it). I like your method here.

you assume too must. first => f(x+n) = f(x) - n second => on the domain [0..1[ define a free function g, can be discontinue on any point. f( x + n ) = g( x ) - n with x on the domain [0..1] you have many many (infinite ) solution

Did you know minus is a Arabic word? It derived from the word menha which means 'from it'. Good to know im Persian and i hate all Arabic words in Farsi.

f(x) = -x is actually just one of the solutions to the equation f(x + 1) = f(x + 2) + 1. I think the problem here is whtether it is also the unique solution given continuity condiion and initial value f(0) = 0

f(x) should have been equal to kx+p, not kx. After the same steps f(x)=-x can be obtained formally.

He did a mistake. He forget the constant. The equation f(x) = kx was wrong. a linear equation means f(x) = kx+c.

How about f(x+2)-f(x+1)=-1 Which is slope. Think about. Then f(x)=-x Done.

Would not any function of the type f(x)=-x+c (c real contsant) fit the bill? F(x+1)=-x-1+b. F(x+2)+1=-x-2+b+1=-x-1+b

Wrong assumption that f(x) is linear function otherwise f(x+1) = k(x+1) will not equal to kx+k

Solution: f(x + 1) = f(x + 2) + 1 Substitute (x + 1) with y: f(y) = f(y + 1) + 1 |-1 f(y + 1) = f(y) - 1 So if you add 1 to the input, the output is reduced by 1 Only f(z) = -z satisfies this: f(x + 1) = f(x + 2) + 1 -(x + 1) = -(x + 2) + 1 -x - 1 = -x - 2 + 1 -x - 1 = -x - 1

f(x)=f(x+1)+1 is the simplest form, it's unnecessary to find out a pure X formulation .

I like math

As I have seen uploader has already pinned a comment that f(x)=-x+f(0). Let us take it further. Assume f(x)=-x+f(0)+ any periodic function which repeats. i.e. f(x)=-x+f(0)+g(x) such that g(x+1)=g(x) Then f(x+1)=-x-1+f(0)+g(x+1)= (-x+f(0)+g(x))-1=f(x)-1 Then f(x)=f(x+1)+1 Thanks to work of Fourier Sir, the general form for g(x) will be g(x)=sum(Cn(sin(2*pi*n*x+phi_n)) where Cn and phi_n are constants and n is integer (-inf,inf). So the exhaustive answer would be f(x)=-x+f(0)+sum(Cn(sin(2*pi*n*x+phi_n)) If we take all Cn and f(0) equal to 0 then f(x)=-x.

I never did functionnal equation but by simple logic from the video's picture I could figure out -x works, it's a bit too easy to find one answer to this problem

As a student of a simple school, I only solved part of it, although I did not complete the task completely, but I am excited to watch this video and learn something new, thanks for the video!

Bad Solution: you cannot assume a solution and show it's true - if you search ALL functions equalizing the starting Equation

But it is necessary to make the hypothesis that f(x) exists for every x. Furthermore f(x)=-x+h with constant h is a solution.

I don't know how I ran into this video, but I immediately subscribed when I saw how you solved this. Sooo good! Thank you!!

f(x)= - x if x is an integer, otherwise f(x) = - x +1. This tricky function satisfies the equation?

I think it was better to say f(x)=kx+b And then by input some numbers find the value of b

I know I'm 3 months late to this, but I think a nifty way to approach this problem is graphically. Start with f(x)=f(x+1)+1 This means that a given point (x,f(x)) on the graph is one unit higher than the point on the graph one unit to the right. Thus, we have a line with slope -1. Plugging this into the slope-intercept form of a linear function, this means that we have a family of functions f(x)=-x+b, where b is an arbitrary constant. (In other words, any choice of b would satisfy the functional equation.)

you can rearrange the order of the equation to (f(x+2)-f(x+1)) / ((x+2)-(x+1)) = -1 Since the Δy/Δx is a constant, assuming x is continuous, the function would be the linear equation y = -x plus some constant c. - hence, f(x)= -x + c

Hurrah for Prime Newtons! 😊

i tried to find a function f(x)=ax+b to satisfy the equation and i arrived at a general function f(x)= -x+b

For sure you can find another answer for this question as f(x)=x+1 That funny thanks for this video

Isn’t f(x)=family of parallel lines with slope -1 and seperated by y=1?

ответ за 5 секунд угадывается, а тут на 11 минут развели демагогию. я так не умею, научите лить воду и говорить ни о чем.

Kind of.. very easy, I litteraly found the answer in 5 to six seconds.. but not that boring

Why you always smile? 😂😂

Very cool! Additionally solving an equation means, however, also to show, that there are no other solutions.

f(x)=-x+c is also a solution but could we proof that these forme of solutions are exhaustive?

That is just an example of a solution. It isn't all the solutions. To solve the equations, you have to get all the solutions (for this one there are infinitely many)

I would say, that the exact Solution has to be f(x) = - x + x0, where x0 is any real or integer

As many others have pointed out, it's only one of many possible equations you have here.

Great job bro! The solution is actually f(x) = -x + c

I think a more general solution is f(x)=c-x, where c is any constant

I believe that f(x) = - floor(x) satisfies the condition too, am I right ?

I would have done it like this: f(x+1) = f(x+2) + 1 f(x) = f(x+1) + 1 = f(x+2) + 2 By recursion, f(x) = f(x+N) + N for all natural numbers N let f(0) = C f(0) = f(N) + N -> f(N) = C - N -> f(x) = C - x Because N is restricted to natural numbers, we can allow to mess with f(x) by any function with a period of 1 so f(x) = C - x + R(x) where R is any function with periodicity 1 (such as 8sin(2pix) or x - floor(x))

if the question is just asking for a random function that satisfies the condition, my first reaction is f(x)=-x

This is only true if f(0)=0. The correct solution is f(x) = -x + f(0)

The answer is true but the solution isn't true you don't prove that the function hasn't another answer

Why not: f(x+1) - f(x+2) = 1 y1 = f(x1); x1=x+1 y2 = f(x2); x2=x+2 (y2-y1)/(x2-x1) = -1 / 1 = -1 => f(x) = -x + C_