f(x+1) = f(x+2) +1

In this video, I solved a functional equation using the f(x) =kx assumption

[honestadministrator]

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)

[malvoliosf]

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.

[Risuchan]

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

[andrewfischer-garbutt]

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

[zakarialatrache-sw]

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

[michaelmcgruder]

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)!

[chaosredefined]

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.

[mitrofankostikov]

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

[АлександрГуленко-ен]

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

[徐瑞斌-io]

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).

[Hanible]

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

[massimograzzini]

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.

[cameronspalding]

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

[MCentral]

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

[rainerzufall]

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.

[netravelplus]

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.

[lukaskamin]

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

[zviman]

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

[abitpalgyawali]

Loving the videos! I wanna add my two cent to this video: For this problem (if it was in a timed test or a competition), I think the trick would be to zoom out a little and look at it intuitively. f(x) = f(x+1) + 1 or f(x) - 1 = f(x+1) --> The right hand equation makes it much easier to spot the linear relationship between x and f(x)! And, with a little bit of logic, you can figure out that f(x) = -x. Also, you can even expand the solution so something like f(x) = -x + c (where c is a constant). Regardless, I really enjoyed your video - it would've definitely taken me much longer to find a mathematical proof to the problem rather than an intuitive one. Thanks so much!

[tonystoneK]

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!!