@@PrimeNewtons it actually gets worse, without an extra condition like requiring f to be continuous, you cannot extend the generality of the solution f(x) = f(0)-x outside the integers. In general, you can have a different "line" of -x for any element of the interval [0,1). f(x) = f(v)-x, where v is an element of [0,1) chosen such that x-v is an integer. This covers all possible x and is unique. Each v gives an arbitrary choice for f(v). Analysis can be hairy sometimes. To your credit though, the original question only asks to find A solution, not all solutions.
@@Dc4nt absolutely. f(x) being defined for the series (x, x +1, ..) for each of the fractional value 0 < frac < 1 f ( N + frac) = - ( N + frac) + ( frac + f ( frac) ) f ( z) = - z + A( frac) Here in N = integral part ( z) frac = z - N A ( z - N) = z - N + f ( z - N) More precisely f ( z) = - z + A ( z - integral part ( z))
I think my solution is the same but less mathematically expressed. I was thinking: it drops by one if we go one step further therefore it has a gradient of -1 so the formula must be f(x) = -x + a edit: to make things clear: i pictured a graph in my head and thought about what would happen to the y coordinate if i went one step further in the x direction. So it was sort of a graphical way to solve it
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
@@TechToppers Certainly. We define p(x) = f(x)+x thus f(x) = p(x)-x The original equation: f(x+1) = f(x+2) + 1 becomes: p(x+1)-(x+1) = p(x+2)-(x+2)+1 p(x+1)-(x+1)+(x+2)-1 = p(x+2) p(x+1)-x-1+x+2-1 = p(x+2) p(x+1) = p(x+2) As it is valid for all numbers x, we can shift x+1 -> x p(x) = p(x+1)
@@TechToppers f'(x+1)=f'(x+2) It is enough to take the derivative and we get a periodic function. However, the period does not have to equal one. 1/2 will also work. Two periods fit into one.
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
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)!
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
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.
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
Even that answer is incomplete as it is only true for differentiable functions. You can have a function which looks like a staircase which satisfies the condition.
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
Same idea here, solved “graphically”. Based on the formula, f(x) goes down 1 unit at each step so it had a downward slope of -1 and any f(x) = -x + constant works. We only had to find a solution, not all of them so I stopped there. You can also use the standard formula slope = (f(b)-f(a))/(b-a) , use b=x+2 and a=x+1 in this case, substitute and get the same slope.
f(x) = -x is just one of the solutions. In fact f(x) = g(x) - x, where g(x) is ANY SYMMETRICAL FUNCTION from x = 0. So, for example, and just for example, f(x) = *x² - x* is allowed (because x² is SYMMETRICAL from x = 0)
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.
It is not the most general solution. There are other functions that can satisfy it. Even if you know that f(0) = 0, that's insufficient information to determine what f(0.01) is equal to for instance (or what any non-integer value would give), because you can never get to 0.01 by adding 1 to 0 repeatedly.
@@asdfqwerty14587 He said nothing about x so I assumed x is a real number. And from that it follows that f(x) too must be a real number, that’s I assumed f: R->R. So the solution f(x) = -x + a is a simple first grade polynomial and it is a continuous function. Once you fix the value of ‘a’ the function is determined for every x of R. For example, if you set f(0)=0 then a=0. The function is f(x)=x so f(0.01) = 0.01. If you consider a function f:Z->Z the solution is the same but is defined on Z, so there is no sense to ask for f(0.01).
@@christianfunintuscany1147 That is not true. Once you've set the value for f(0), then you can derive any integer value of f(x), but not any non-integer value. You can never get to 0.01 by adding or subtracting 1 from 0. For instance, a function like f(x) = sin(2*pi*x) - x also satisfies that condition without being a linear function (and sin(2*pi*x) could be replaced by any periodic function with a period of 1).
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
Claim: Every solution f to the equation f(x)=f(x+1)+1 is of the form f(x)=g(x)-x for some 1-periodic function g Proof: Let f be a solution for the equation. Let g be the function that on [0,1) equals f(x)+x and is continued with period 1 across the reals. recursive claim: for all natural numbers n and for all real numbers r on the interval [0,1) g(n+r)=f(n+r)+n+r recursive proof: Let 0≤r
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
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.
LOL, this was not even near to complete. You could for example set N = N(z), where N(z) is any arbitrary function with integer values at integer parameters! And A and C may be periodical with a period of 1, or even just constant over IZ. For example: f(z) = (1 - cos(π z))² exp(2 π i (z³ - z) + (-1)^(2 z)) - z [Edit: This example is incorrect!]
Okay, I must confess, that was too optimistic! My bad! The given rule has also be obeyed between the integer parameters, e.g. f(0,5) = f(1.5) + 1. Therefore, it must be always the same additional g(x) : [0, 1) -> |C function, also between the integer nodes. Thus A has to be a constant or 1-periodical and C(z) = C(z + n) and N(z) = N(z + n) for all n € IZ and z € IC. This reduces the fun, that I had with the example, but this still holds (as an example): f(z) = |cos(π Re(z))| exp(20 π i + sin(Re(2 π z) + Im(z)^7) - z
I wish we had teachers with a fraction of your attitude over here (educational system is broken beyond repair here. Germany, that is). youtube helps my kids more than school does. it‘s said, but thanks, mate!
The most general solution would be as follows: Let's associate each number p on a semi-interval [0, 1) with some constant C_p. Then for each number x such that x = p + n, where n is a whole number, we define f(x) = -x + C_p. Example: f(x) = 34 - x for whole numbers and f(x) = -3 - x for any other number
What we can do is assume our function is differentiable, and see that f’(x+1)=f’(x+2). Let’s notice that this can be true if f’(x) is some constant m. Ok, since this question only asks us to find a function f, this will be our function that we will choose, in which f’=m Therefore f(x) is some linear function mx+c. If f(x+1) = f(x+2)+1, then f(x+1)-f(x+2)=1. This means that m(x+1)-m(x+2)= -m=1, therefore m=-1. We therefore have f(x)=-x+c, where c is any real number. By the way, this is quite often a good way to approach these problems. When we differentiate we quite often simplify our problem
For function equations in general and for this one in particular, it is always important to specify the domain on which the problem must be solved. If you go for integers, the solution to the problem is relatively simple: F(x)=f(0)-x where f(0) can take any value. However, once you get to real functions, it quickly becomes an “abomination”. In fact, if you solve on real numbers, you take x0 in the interval [0,1) and you can assign any possible real to f(x0). So the function becomes f(x)=f(x0)-x where x0 is the fractional part of x. Most of these functions are not even continuous.
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.
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))
Thats the answer...you can not make assumptions without support them mathematically ( like prime newtons done). You need to prove them then use them like your answer
@@omaraladib2165 it won't work if you try to assume f(x) = ax^2 + bx + c ax^2 + bx + c = a(x+1)^2 + b(x+1) + c + 1 = ax^2 + 2ax + a + bx + b + c + 1 everything on the left cancels with the right 0 = 2ax + a + b + 1 there is no x term so a = 0 0 = b + 1 b = -1 we can't make any assumptions on c so leave it alone So the polynomial is 0x^2 - x + c it's forced to be linear
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)
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
Functional equations aren't something I studied as part of my math education, and usually I'm stumped when I see one. I actually had your answer to this one in just a few seconds from glancing at the original equation, though, and considering what the effect on the input was. Each time you add one to the parameter, there's a compensation of adding exactly one to the expression on the outside required. What function would have that property? The additive inverse.
The functions of the form c-x aren't the only solutions by the way! For example, you also have the function -[x], where [x] is the floor (rounding down) of x. This may be advanced for some viewers here, but you can use the axiom of choice to get a much larger family of solutions. For example, you can define f(x) to equal -x if x is rational, and 1-x if x is irrational. More generally, for every subset of the real numbers where its elements are equally-spaced by 1 (so like the integers), you can define f on it to be c-x, for whatever value of c you want for that subset, and then another subset could have a different c, and so on. (this is where the axiom of choice is used) Since all the numbers that are 1 apart have the same fractional part, you can also define f(x) = {x}-x, where {x} is the fractional part of x. So here f equals -[x] from above.
Wow, I felt intuitively somewhat uncomfortable with the implicit change of integer increment to rearrange one, now I see why. So it could be inferred that the conditions of the problem are not quite correct, cause it allows 3 different types of functions: continuous, piecewise, the one that is not differentiable at any point( I guess that is some type if Dirichlet function). So obviously there should be some condition that only a continuous function is to be searched
@@lukaskamin755 Why "3" types? You could have a function that is not differentiable on a countably infinite set of points, for example; or whatever. Chose *any* function that is defined on the interval [0,1) and extend it using the recurrence to obtain a valid solution.
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.
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!
Take literally any function (continuous or not, any of the aleph^aleph functions), copy it periodically for the entire number line, reducing it by one for every copy to the right and increase by one for every copy to the left. Alternate (completely equivalent) definition: take any periodic function with a period of 1 (again, aleph^aleph options), and add to it -x
The functional equation f(x + 1) = f(x + 2) + 1 implies a constant decrement by 1 of f(x) for each increment of 1 in x. Assuming f(x) is differentiable, its derivative f'(x) represents this rate of change. The equation f(x) - f(x + 1) = -1 is equivalent to the original condition, highlighting the decrement as x increases. This decrement can be seen as the differential df for a delta x of 1, leading to f'(x) = df/dx =(f(x) - f(x + 1))/1 = -1. Integrating this derivative gives us the original function f(x), i.e., Integral of f'(x) dx = Integral of -1 dx = -x + C, where C is the constant of integration. This process confirms that f(x) = -x + C satisfies the given functional equation. And let assume C = 0 to match with given by teacher solution.
I can do it visually with translation where translating from f(x) into f(x-a)+b means moving to the right by a units and up by b units. If we are use a point (a, b) in the function f(x), then it would be (a-1, b+1) if we use f(x+1)+1 instead from the equation. To make it a bit clearer (not really needed), I can also use the equation f(x)=f(x-1)-1, which is the other way around. I will have the point (a+1, b-1) If we "draw a line" that connects the 3 points we got so far, we will have the gradient (-1)/1 from rise/run, giving us the gradient -1. This gives us: f(x)=mx+c f(x)=-x+c How do we find the c? There's no need. What we've done applies for any function with the gradient of -1 because if we use b+1 or b+2 instead of just b in our original point (a, b), it will still lead to our gradient being -1. Therefore, we can conclude that f(x)=-x+c, where c is an arbitrary real number.
Oh and one more method, I can also try use the concept of arithmetic progression by assuming f(x) as a1, f(x+1) as a2, f(x+2) as a3, etc. a1=a2+1 d=a2-a1=-1 Now, we can say that as we go forward in the arithmetic progression, we subtract from the previous term by 1, showing that the gradient is -1. Moreover, because I can use any number for a1, it shows that it works for all functions with the gradient of -1, even with different y-intercepts (a.k.a. the c in y=mx+c) We can conclude that f(x)=-x+c
. f(x+1)=f(x+2)+1 ⇒ f(x)=f(x+1)+1 (put x=x-1) ⇒ f(x)-f(x+1)=1 ⇒ f'(x)-f'(x+1)=0 (By taking derivatives on both side) ⇒ f'(x)=f'(x+1) ∴ By observation, Slope of f(x) is the same on all points of f(x) where x∈ℝ ∴ f(x) is a linear function To find f'(x), f'(x)=[f(x)-f(x+1)]/[x-(x+1)]=1/-1=-1 ∴ By using slope-intercept form: f(x)=mx+c (where m is slope (= -1) and c is any y-intercept) ╔════════╗ ║ ∴ f(x)=-x+c ║ ╚════════╝
F doesn't need to be differentiable In fact in doesn't need to be continuous, so it's determined by giving values on [0, 1) And saying f'(x) = f'(x+1) doesn't implies that slope at 0 is the same as at 1/2
f'(x) = f'(x+1) doesnt imply slope is constant e.g counterexample sin(2πx) i think a true exhaustive solution would be f(x) - floor(x) where f(x) is any periodic function with period 1/k where k is an integer >0
f(x)-f(x+1)=1 f(x+1)-f(x)=-1 The sequence f(n) for natural numbers is, by definition, an arithmetic sequence so, the general formula takes the form of is f(n)=(-1)*n+c. But what if x is not a natural number? It's easy to check that the constant is not necessarily the same for every number in the interval (0, 1]. For example, the function f(x)=−x+5f(x)=−x+5 for rational numbers, f(x)=−xf(x)=−x for irrational numbers, satisfying the functional equation. i ts becuse the group of rational numbers is a normal subgroup of the group of real numbers, so if x-y is not a rational k can be diffrent.
By iteration we get: f(x) = f(x+k) + k f(x) - f(x+k) = k f(x) - f(x+k) = k ------ - x - (x+k) x - (x+k) f(x) - f(x+k) = k ------ - x - (x+k) -k f(x) - f(x+k) = -1 ------ x - (x+k) lim f(x) - f(x+k) = lim -1 k->0 ------ k->0 x - (x+k) f’(x) = -1 By integration we get: f(x) = -x + c Where c is a constant As we don’t have an initial value for the function so for each c belongs to R the function is valid.
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
I ve been scrolling 3 min to find a comment with the correct solution. This should be pinned and this guy shouldnt be posting problems if he cant solve them.
f(x+2)-f(x+1)=-1 It can be written as f(x+1)-f(x)=-1 It clearly looks like a straight line equation As you move +1 units in X axis the Y decreases by 1 Correct for any x And as you move +2 units, the change becomes -2 That means the slope at each point is constant tan(theta)=Increase in Y/Increase in X = -2/2=-1/1=-N/N=-1 That means f(x)=-x +C C is the Y intercept While C=f(1)+1
Some of the steps were wrong. You can only make that generalization for adding or subtracting integer values from x - if you're adding or subtracting any non-integer value then there's no guarantee that it will be linear anymore. f(0) = 0, f(1) = -1, f(2) = -2, but f(0.01) = 1000, f(1.01) = 999, f(2.01) = 998 etc. - that pattern won't contradict the equation, but it's definitely not a linear function at that point.
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
You can easily solve it by inserting a Taylorexpansion of f, then collecting all terms with x on one side. As the equation must hold for all x all higher order coefficients must be 0. However this only gives you the class of continuous solutions. Generally, taking any function on [0,1] and tiling it shifting it by one down each time as you go along the x-axis will be a solution as well meaning for any f defined on 0 to 1 g(x) = f(x - floor(x)) - floor(x) is a solution
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.)
i'm curious how to show the solution (f(x)=-x+a) is unique if you assume f is differentiable over its domain, we can get that f'(x)=f'(x+1), which has to mean the derivative is a constant, so then we get that f is linear, but I assumed it's differentiable
You can't show that it's unique because it isn't unique. Even if you assume that it's differentiable, it still wouldn't be unique. For instance, something like f(x) = sin(2*pi*x) - x also satisfies that condition.
@@asdfqwerty14587 well, I understand, you showed that even within that assumption, my proof is wrong (when I said that f'(x)=f'(x+1) means that f'(x) is a constant) great example btw
If you write it as f(x+1) = f(x)-1 then you can view it as a recurrence relationship from an arithmetic series which is very easy to find the nth term for, and by a simple test you can see that it works for non integer values too
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.
Actually, the strategy I used in my head was a bit different ... by substitution (u=x+1) I came to the same f(x) = f(x+1) +1 .... but the I sort of rewrote this as ( f(x+1) - f(x) ) / (x+1 - x) = -1 the left hand is now an expession describing the first derivative of f(x) ... so integrate both sides and you get f(x) = -x + c ..... and that is why I believe your solution (c=0) is incomplete, because the expression f(x) = -x +c fulfills the requirements for any given c.
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
f(x)=f(x+1)+1 f(x-1)=f(x)+1 f(x)=f(x-1)-1 f(1)=f(0)-1 f(2)=f(1)-1=f(0)-2 it seems that f(x)=f(0)-x we have a base case already and let the assumption be the inductive hypothesis, thus we have to show that f(x+1)=f(0)-x-1 LHS=f(x+1) =f(x)-1 =f(0)-x-1 RHS so, f(x)=f(0)-x let c=f(0) f(x)=c-x ; c is an arbitrary constant.
1/First of all, f(x)is a linear function. see graphically as follows: 2/At point (x+1) on x axis, vertical value is f(x+2) +1 3/between (x+1) and (x+2), horizontal value is 1 and vertical value is 1 therefore slope is 1 on 1 but negative. Therefore, f(x)= mx + c m=-1 , f (x) =-x + c
It doesn't have to be linear though. It could be a stepwise function, where f(x) is the integral part of -x. It could even be something like f(x) = -sin(2πx) - x
my solution [ edited/reconsidered ] f(x + 1) = f(x + 2) + 1 so *f(x) = f(x + 1) + 1* and f(x - 1) = f(x) + 1 = f(x + 1) + 2 and f(x - 2) = f(x - 1) + 1 = f(x + 1) + 3 and f(x - 3) = f(x - 2) + 1 = f(x + 1) + 4 and so on so f(x - a) = f(x + 1) + a + 1 = f(x) + a or *f(x + a) = f(x) - a (I)* Looks like something polinomial.. So let - by "instinct" - f(x) = g(x) - x (II) From I and II we have g(x + a) - (x + a) = g(x) - x - a So g(x + a) = g(x), and at this point I can only see g(x) as a constant function. So, finally, *f(x) = -x + C* (C is a constant value)
excuse me friends but I think I made some mistake(s). Maybe f(x) can only be *-x + C* (C as a constant). I must reconsider my solution ASAP . (I think the main mistake was consider, at some point, "a" as a variable, and not as a constant)
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.
Thanks for the tips, Sir. Love your videos. I guess f(x) = (-x+c) since when put in slope form [(f(x+1)-f(x)]/[(x+1)-x], the slope is always = -1. So f(x) = - x+c where c is any constant, would be the answer and because the information when y=y1, x= x1 is not given, the constant c remains unknown. Further, found it also interesting that f(x) when taken as = (coshx) whole squared and f(x+1) taken as =(sinhx) whole squared , and then f(x+1) - f(x) = -1 . Requesting you to also explain whether the function can be periodic also and how
@@piyushaspiretoachieve3179 Its there in the comment. I don't have much knowledge of mathematics that is why I am asking to explain whether the function can be periodic also and how.
Why not f(x)=Sum(An*sin(2*pi*n*x+phi_n))-x+B where n is integer(-inf,inf) while An,B and phi_n are constants. This would cover all periodic functions (All hail Fourier).
This linear function is definitely a solution, but there are so many others! I fact for any function on the interval [0,1), you can extend it to a solution by taking the condition as a defining property. Even if you want continuous functions only, just use functions on [0,1] where f(1)=f(0)-1 so the intervals glue neatly together. For differentiability... I think it'd be enough to make sure f is n-differentiable on [0,1] with dⁿf/dxⁿ being the same at both ends.
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.
It doesn't have to be a constant. This can be any periodic function with period = 1 or harmonic (period = 1/2, 1/3, etc.). For example f(x) = - x + sin(4πx).
The way I saw it: Let f(x)=y f(x + 1) = f(x + 2) + 1 implies Δy = -1 when Δx = 1 for all x The slope of a line is defined as Δy/Δx, so here the slope is -1 Thus, any equation of the form f(x) = -x + k will satisfy the initial conditions
As mentioned f(x) =-x+a works also ∀ a ∈ ℝ. But we haven't established that this family of linear functions are the sole solutions of the problem! Are there non-linear solutions? Ideas of demonstration? Actually, any function like f(x)= -x + g(x) with g(x+1)=g(x) (periodic function) is a solution as well, for example f(x)=-x + cos(2πx).
Fabulous way solutions are approached - almost "discovering" the answer live. That's most interesting abt this channel What's a good textbook for functional equations like these to get practice solving them (without getting into heavy math definitions)? Thanks
Assuming that x+1=X. It it possible to write the equation like this f(X)=f(X+1)+1 Then (f(X+1)-f(X))/1=-1 It means that for any X the slope between (X;f(X)) and (X+1;(f(X+1)) is -1. The family of function satisfying this regularity of slope for any X is f(X)= -X+a (a real constant)
Cool problem! I haven't watched the video in its entirety yet, but I would like to point out you do not need to make the assumption that f(x) is a line, you can actually show it. If we use the definition of the derivative, we see that f'(0)=f'(1)=f'(2)...=f'(n), meaning the first derivative is constant. So the function has to be something in the form of a line f(x)=kx+m. Thereafter you can determine that k=-1 and that m can be some arbitrary real number!
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!
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 ...
f(x+1)=f(x)-1, it means every time that x increases by 1, the function value is gonna decrease by 1, so it implies it’s a linear function with a slope being -1, therefore f(x)=-x+C, where C can be any real number
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.
This is a meta-comment. A) Prime Newtons is the sort of man a thinking human being wants to hang out with. B) The level of the comments is very high and PN's manner of exposition is the reason why.
As pointed out, there are infinitely many solutions since if you replace f(x) by f(x)+g(x) where g is any periodic function with period 1, then it is still a solution.
Since - 1 is the first difference and constant f is a linear polynomial of the form f(x) =mx+c and m= [f(x+1)-f(x)]/1=-1 This gives a family of functions of the form f(x)=-x +c.
f(x+1) = f(x+2) +1 substitue x-1 for x f(x-1+1) = f(x-1 +2) +1 f(x) = f(x+1) +1 f(x+1) -f(x) = -1 left side is the defention of derivative for an increas of 1 in x direction, thus d(f(x))/dx = -1 thus f(x) = -x
C1 is unnecessary as g(x) is already any 1-periodic function and -x part isn't general enough. Others have pointed out that sin(2pix)-floor(x) +k also fits
At least f(x) = -0x +floor(-x) and f(x) = -x and f(x) = -2x + floor(x) all fit the equation So the more general form is f(x) = g(x) -kx +floor( (k-1)x ) + c But I suspect more can be done
floor( (k-1)x )-kx = floor( (k-1)x )-(k-1)x - x = floor( (k-1)x )-floor((k-1)x)-frac((k-1)x) - x = -x - frac((k-1)x) Does frac((k-1)x) a periodic function? I'm not sure that this one also meets the requirements
I wonder why you decided to take partial type of linear dependence, namely direct proportionality ( I'm not sure what it's called in English, but that's the y = kx dependence), instead of general type y = kx+ b of linear dependence. From that recurrent equation we can easily derive ( e.g. through math induction method) that f(x) = f(x+n) + n, or otherwise f(x+n) - f(x)=-n. So it's very much like linear differential equation, just with finite difference ( as soon as for linear function the growth is constant on equal intervals for x), thus we can't derive additive constant from this equation, so basically the function we are looking for is y= -x + C, where C is an arbitrary rearrange constant.
A simple way to realize that the solution has to be linear in x is to take the derivative of the original recursive realationship wrt x. What you find is that the derivative wrt to x is a constant for all x, which means that f(x) is linear in x. Your solution is one solution, but any solution of the form f(x) = -x + constant will also work. Your particular solution is the one for which the constant is zero.
Let f(x) = ax + b So, we can get: f(x + 1) = a(x + 1) + b = ax + a + b f(x + 2) = a(x + 2) + b = ax + 2a + b We can input to the equation: ax + a + b = ax + 2a + b + 1 a = 2a + 1 a = -1 After we get the value of a = -1 ; we can get the value of b: -x - 1 + b = -x - 2 + b + 1 -1 + b = -1 + b b = 0 So, finally we can get that f(x) = -x 🎉🎉🎉
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.
My reasoning was this (not proof): f(x+1) just moves the function to the left by 1 f(x)+1 just moves the function up by 1 f(x+1)+1 will move the function 1 to the left and 1 up If you can imagine those two transformations together (assuming that the f is continuous) it's easy to imagine that the slope must be -1 So f'(x) = -1 f(x) = -x+c For non continuous functions you have can something like this: f(x) = -x+1 -> for x in (a, a+1] f(x) = -x-100 -> for x in (a-1, a] for a being equal to 2*k were k in integers or you could even have discrete functions
In my experience, if we take an arbitrary real number called alpha, we have same vertical and horizontal displacements if this alpha is in an argument of the function or it is out of that. So, another way to solve is graphically.
Take ANY function g(x) that is defined on [0, 1) define f(x) = g(x-n)-n for x in [n, n+1) Congradulation! This function corresponds to the property that you wrote. Also - this set of such functions f are the ONLY functions that satisfy the given property.
My strategy was to suppose f is an arithmetic progression, so f(x) = ax + b. Then substitute f(x) in the functional equation and we get a = -1. Ergo, f(x) = -x + b for any b chosen.
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.
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)
Oh my, this was my first solution! 😆
@@PrimeNewtons it actually gets worse, without an extra condition like requiring f to be continuous, you cannot extend the generality of the solution f(x) = f(0)-x outside the integers.
In general, you can have a different "line" of -x for any element of the interval [0,1).
f(x) = f(v)-x, where v is an element of [0,1) chosen such that x-v is an integer. This covers all possible x and is unique. Each v gives an arbitrary choice for f(v).
Analysis can be hairy sometimes.
To your credit though, the original question only asks to find A solution, not all solutions.
@@Dc4nt absolutely. f(x) being defined for the series (x, x +1, ..)
for each of the fractional value
0 < frac < 1
f ( N + frac) = - ( N + frac)
+ ( frac + f ( frac) )
f ( z) = - z + A( frac)
Here in N = integral part ( z)
frac = z - N
A ( z - N) = z - N + f ( z - N)
More precisely
f ( z) = - z + A ( z - integral part ( z))
@@PrimeNewtons I think it would be more didactic and complete than the "guessing" what is in the video.
I think my solution is the same but less mathematically expressed. I was thinking: it drops by one if we go one step further therefore it has a gradient of -1 so the formula must be f(x) = -x + a
edit: to make things clear: i pictured a graph in my head and thought about what would happen to the y coordinate if i went one step further in the x direction. So it was sort of a graphical way to solve it
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.
i was about to say the same
In fact you can replace "a" for ANY SYMMETRICAL FUNCTION (from x = 0)
Yes i was stressing finding the a when it can be true for all value 😂😂😂
@@SidneiMV Symmetrical isn't the required property. Periodic with period 1 is the required property.
www.youtube.com/@user_math2023
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
Can I get a little bit of insight how you came up with the substitution?😅
@@TechToppers Certainly.
We define p(x) = f(x)+x
thus f(x) = p(x)-x
The original equation:
f(x+1) = f(x+2) + 1
becomes:
p(x+1)-(x+1) = p(x+2)-(x+2)+1
p(x+1)-(x+1)+(x+2)-1 = p(x+2)
p(x+1)-x-1+x+2-1 = p(x+2)
p(x+1) = p(x+2)
As it is valid for all numbers x, we can shift x+1 -> x
p(x) = p(x+1)
best answer! I was sure it had to just be a constant but any 1-periodic function would've also worked!
Excellent! The global solutions.
@@TechToppers f'(x+1)=f'(x+2) It is enough to take the derivative and we get a periodic function. However, the period does not have to equal one. 1/2 will also work. Two periods fit into one.
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
The solutions are beyond -x + C (being C a constant)
For example, and just for example, *f(x) = x² - x* is also allowed
@@SidneiMV
f(x) = x² - x
f(0) = 0
f(1) = 0
This function does not satisfy the equation
www.youtube.com/@user_math2023
@@yoavmor9002 yes. You are right mate. Because this is not a matter of symmetry, this is about periods/intervals ( period = 1 in this case)
f(x) = f(0) - x => this represents a set of parallel straight lines \, each with a different intercept f(0) at the y axis
f(x + 2) = f(0) - (x + 2)
f(x + 1) = f(0) - (x + 1) = f(0) - (x + 1 + 1) - 1 = [f(0) - (x + 2)] + 1 = f(x + 2) + 1
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)!
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
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.
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
You really dug deep. I've taken some notes from your exposition. Thank you.
right answer is f(x)=-x+C, not just -x
Even that answer is incomplete as it is only true for differentiable functions. You can have a function which looks like a staircase which satisfies the condition.
the question was to find a function satisfying this property, not all. Thus the given answer in the video suffices
No. It is not the case. f(x+1) = f(x+2) + 1 holds for some x in a
staircase-like function but not for all x.
f (1.1)=f (2.1)+1
@@cosmosapien597do you mean functions like f(x) = -[x]+C?
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).
Excellent. The best approach.
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
❤👍🙏🙏🙏
Same idea here, solved “graphically”. Based on the formula, f(x) goes down 1 unit at each step so it had a downward slope of -1 and any f(x) = -x + constant works. We only had to find a solution, not all of them so I stopped there.
You can also use the standard formula slope = (f(b)-f(a))/(b-a) , use b=x+2 and a=x+1 in this case, substitute and get the same slope.
Why you decided f(0)=0???😮
Nothing requires f(0) to be zero!
f(x) = -x is just one of the solutions.
In fact f(x) = g(x) - x, where g(x) is ANY SYMMETRICAL FUNCTION from x = 0.
So, for example, and just for example, f(x) = *x² - x* is allowed (because x² is SYMMETRICAL from x = 0)
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.
It is not the most general solution. There are other functions that can satisfy it. Even if you know that f(0) = 0, that's insufficient information to determine what f(0.01) is equal to for instance (or what any non-integer value would give), because you can never get to 0.01 by adding 1 to 0 repeatedly.
@@asdfqwerty14587can you show me one of these functions ?
@@asdfqwerty14587 He said nothing about x so I assumed x is a real number. And from that it follows that f(x) too must be a real number, that’s I assumed f: R->R. So the solution f(x) = -x + a is a simple first grade polynomial and it is a continuous function. Once you fix the value of ‘a’ the function is determined for every x of R. For example, if you set f(0)=0 then a=0. The function is f(x)=x so f(0.01) = 0.01. If you consider a function f:Z->Z the solution is the same but is defined on Z, so there is no sense to ask for f(0.01).
@@christianfunintuscany1147 That is not true. Once you've set the value for f(0), then you can derive any integer value of f(x), but not any non-integer value. You can never get to 0.01 by adding or subtracting 1 from 0.
For instance, a function like f(x) = sin(2*pi*x) - x also satisfies that condition without being a linear function (and sin(2*pi*x) could be replaced by any periodic function with a period of 1).
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
great answer！
My answer is that f(x)= g(x) - x where g(x) is a 1-periodic function.
Claim: Every solution f to the equation f(x)=f(x+1)+1 is of the form f(x)=g(x)-x for some 1-periodic function g
Proof:
Let f be a solution for the equation.
Let g be the function that on [0,1) equals f(x)+x and is continued with period 1 across the reals.
recursive claim: for all natural numbers n and for all real numbers r on the interval [0,1) g(n+r)=f(n+r)+n+r
recursive proof:
Let 0≤r
Can be 1/N periodic
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
yes but curiously that assumption is wrong.
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.
f(z) = A exp(2 π i N z + C) - z, with some arbitrary N € IZ, A € IC, and C € IC, would make a wonderful example for complex functions!
LOL, this was not even near to complete. You could for example set N = N(z), where N(z) is any arbitrary function with integer values at integer parameters! And A and C may be periodical with a period of 1, or even just constant over IZ. For example:
f(z) = (1 - cos(π z))² exp(2 π i (z³ - z) + (-1)^(2 z)) - z [Edit: This example is incorrect!]
Okay, I must confess, that was too optimistic! My bad! The given rule has also be obeyed between the integer parameters, e.g. f(0,5) = f(1.5) + 1. Therefore, it must be always the same additional g(x) : [0, 1) -> |C function, also between the integer nodes. Thus A has to be a constant or 1-periodical and C(z) = C(z + n) and N(z) = N(z + n) for all n € IZ and z € IC. This reduces the fun, that I had with the example, but this still holds (as an example): f(z) = |cos(π Re(z))| exp(20 π i + sin(Re(2 π z) + Im(z)^7) - z
Your teaching is soo good
I wish we had teachers with a fraction of your attitude over here (educational system is broken beyond repair here. Germany, that is).
youtube helps my kids more than school does.
it‘s said, but thanks, mate!
The most general solution would be as follows:
Let's associate each number p on a semi-interval [0, 1) with some constant C_p. Then for each number x such that x = p + n, where n is a whole number, we define f(x) = -x + C_p.
Example: f(x) = 34 - x for whole numbers and f(x) = -3 - x for any other number
What we can do is assume our function is differentiable, and see that f’(x+1)=f’(x+2). Let’s notice that this can be true if f’(x) is some constant m. Ok, since this question only asks us to find a function f, this will be our function that we will choose, in which f’=m
Therefore f(x) is some linear function mx+c. If f(x+1) = f(x+2)+1, then f(x+1)-f(x+2)=1. This means that m(x+1)-m(x+2)= -m=1, therefore m=-1. We therefore have f(x)=-x+c, where c is any real number.
By the way, this is quite often a good way to approach these problems. When we differentiate we quite often simplify our problem
This is my favourite solution but again the problem didn't give us the necessary information
For function equations in general and for this one in particular, it is always important to specify the domain on which the problem must be solved.
If you go for integers, the solution to the problem is relatively simple:
F(x)=f(0)-x where f(0) can take any value.
However, once you get to real functions, it quickly becomes an “abomination”. In fact, if you solve on real numbers, you take x0 in the interval [0,1) and you can assign any possible real to f(x0). So the function becomes f(x)=f(x0)-x where x0 is the fractional part of x. Most of these functions are not even continuous.
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.
www.youtube.com/@user_math2023
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))
لماذا فرضت إن الدالة من الدرجة الأولى وليست مثلا من الدرجة الثانية؟؟
Thats the answer...you can not make assumptions without support them mathematically ( like prime newtons done). You need to prove them then use them like your answer
@@huseinali4415 prime newtons perfectly justified his use of assumption, i didn't even prove the recursion step
@@omaraladib2165 it won't work if you try to assume f(x) = ax^2 + bx + c
ax^2 + bx + c = a(x+1)^2 + b(x+1) + c + 1
= ax^2 + 2ax + a + bx + b + c + 1
everything on the left cancels with the right
0 = 2ax + a + b + 1
there is no x term so a = 0
0 = b + 1
b = -1
we can't make any assumptions on c so leave it alone
So the polynomial is 0x^2 - x + c
it's forced to be linear
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)
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!!
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
Functional equations aren't something I studied as part of my math education, and usually I'm stumped when I see one. I actually had your answer to this one in just a few seconds from glancing at the original equation, though, and considering what the effect on the input was. Each time you add one to the parameter, there's a compensation of adding exactly one to the expression on the outside required. What function would have that property? The additive inverse.
The functions of the form c-x aren't the only solutions by the way!
For example, you also have the function -[x], where [x] is the floor (rounding down) of x.
This may be advanced for some viewers here, but you can use the axiom of choice to get a much larger family of solutions.
For example, you can define f(x) to equal -x if x is rational, and 1-x if x is irrational.
More generally, for every subset of the real numbers where its elements are equally-spaced by 1 (so like the integers), you can define f on it to be c-x, for whatever value of c you want for that subset, and then another subset could have a different c, and so on. (this is where the axiom of choice is used)
Since all the numbers that are 1 apart have the same fractional part, you can also define f(x) = {x}-x, where {x} is the fractional part of x. So here f equals -[x] from above.
So the general solution should be -x +g({x}) where g(x) is any function
Wow, I felt intuitively somewhat uncomfortable with the implicit change of integer increment to rearrange one, now I see why. So it could be inferred that the conditions of the problem are not quite correct, cause it allows 3 different types of functions: continuous, piecewise, the one that is not differentiable at any point( I guess that is some type if Dirichlet function). So obviously there should be some condition that only a continuous function is to be searched
@@lukaskamin755 Why "3" types? You could have a function that is not differentiable on a countably infinite set of points, for example; or whatever.
Chose *any* function that is defined on the interval [0,1) and extend it using the recurrence to obtain a valid solution.
@@lukaskamin755 Even for only continuous or even differentiable functions there are many solutions to the equation
@@satindra.r Any function defined on [0,1), yes. Doesn't have to be continuous/differentiable/etc.
Just anything defined on [0,1).
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.
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!
There are solutions not on the form f(x)=-x+c. Consider f(x)=floor of -x
It's true! f(x) = -x + C (C is a constant)
Or f(x) = -x + sin(2pi x) and many many more...
@@landsgevaer what?? sin(2πx) is NOT a constant!
@@SidneiMV No, but it is a solution.
Try it.
Take literally any function (continuous or not, any of the aleph^aleph functions), copy it periodically for the entire number line, reducing it by one for every copy to the right and increase by one for every copy to the left.
Alternate (completely equivalent) definition: take any periodic function with a period of 1 (again, aleph^aleph options), and add to it -x
The functional equation f(x + 1) = f(x + 2) + 1 implies a constant decrement by 1 of f(x) for each increment of 1 in x. Assuming f(x) is differentiable, its derivative f'(x) represents this rate of change. The equation f(x) - f(x + 1) = -1 is equivalent to the original condition, highlighting the decrement as x increases. This decrement can be seen as the differential df for a delta x of 1, leading to f'(x) = df/dx =(f(x) - f(x + 1))/1 = -1. Integrating this derivative gives us the original function f(x), i.e., Integral of f'(x) dx = Integral of -1 dx = -x + C, where C is the constant of integration. This process confirms that f(x) = -x + C satisfies the given functional equation.
And let assume C = 0 to match with given by teacher solution.
I can do it visually with translation where translating from f(x) into f(x-a)+b means moving to the right by a units and up by b units.
If we are use a point (a, b) in the function f(x), then it would be (a-1, b+1) if we use f(x+1)+1 instead from the equation.
To make it a bit clearer (not really needed), I can also use the equation f(x)=f(x-1)-1, which is the other way around. I will have the point (a+1, b-1)
If we "draw a line" that connects the 3 points we got so far, we will have the gradient (-1)/1 from rise/run, giving us the gradient -1. This gives us:
f(x)=mx+c
f(x)=-x+c
How do we find the c? There's no need. What we've done applies for any function with the gradient of -1 because if we use b+1 or b+2 instead of just b in our original point (a, b), it will still lead to our gradient being -1. Therefore, we can conclude that f(x)=-x+c, where c is an arbitrary real number.
Oh and one more method, I can also try use the concept of arithmetic progression by assuming f(x) as a1, f(x+1) as a2, f(x+2) as a3, etc.
a1=a2+1
d=a2-a1=-1
Now, we can say that as we go forward in the arithmetic progression, we subtract from the previous term by 1, showing that the gradient is -1.
Moreover, because I can use any number for a1, it shows that it works for all functions with the gradient of -1, even with different y-intercepts (a.k.a. the c in y=mx+c)
We can conclude that f(x)=-x+c
. f(x+1)=f(x+2)+1
⇒ f(x)=f(x+1)+1 (put x=x-1)
⇒ f(x)-f(x+1)=1
⇒ f'(x)-f'(x+1)=0 (By taking derivatives on both side)
⇒ f'(x)=f'(x+1)
∴ By observation, Slope of f(x) is the same on all points of f(x) where x∈ℝ
∴ f(x) is a linear function
To find f'(x), f'(x)=[f(x)-f(x+1)]/[x-(x+1)]=1/-1=-1
∴ By using slope-intercept form: f(x)=mx+c (where m is slope (= -1) and c is any y-intercept)
╔════════╗
║ ∴ f(x)=-x+c ║
╚════════╝
F doesn't need to be differentiable
In fact in doesn't need to be continuous, so it's determined by giving values on [0, 1)
And saying f'(x) = f'(x+1) doesn't implies that slope at 0 is the same as at 1/2
f'(x) = f'(x+1) doesnt imply slope is constant e.g counterexample sin(2πx)
i think a true exhaustive solution would be f(x) - floor(x) where f(x) is any periodic function with period 1/k where k is an integer >0
f(x)-f(x+1)=1
f(x+1)-f(x)=-1
The sequence f(n) for natural numbers is, by definition, an arithmetic sequence so, the general formula takes the form of is
f(n)=(-1)*n+c.
But what if x is not a natural number? It's easy to check that the constant is not necessarily the same for every number in the interval (0, 1]. For example, the function
f(x)=−x+5f(x)=−x+5 for rational numbers,
f(x)=−xf(x)=−x for irrational numbers,
satisfying the functional equation.
i ts becuse the group of rational numbers is a normal subgroup of the group of real numbers, so if x-y is not a rational k can be diffrent.
By iteration we get:
f(x) = f(x+k) + k
f(x) - f(x+k) = k
f(x) - f(x+k) = k
------ -
x - (x+k) x - (x+k)
f(x) - f(x+k) = k
------ -
x - (x+k) -k
f(x) - f(x+k) = -1
------
x - (x+k)
lim f(x) - f(x+k) = lim -1
k->0 ------ k->0
x - (x+k)
f’(x) = -1
By integration we get:
f(x) = -x + c
Where c is a constant
As we don’t have an initial value for the function so
for each c belongs to R the function is valid.
Makes perfect sense. You should look at other comments where a periodic function is also a solution
Loving your videos, I really like your style
Why impose a first-degree function and not, for example, a second-degree function?
exactly! For example, and just for example, *f(x) = x² - x* , and many others , are allowed too .
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
I ve been scrolling 3 min to find a comment with the correct solution. This should be pinned and this guy shouldnt be posting problems if he cant solve them.
f(x+2)-f(x+1)=-1
It can be written as
f(x+1)-f(x)=-1
It clearly looks like a straight line equation
As you move +1 units in X axis the Y decreases by 1
Correct for any x
And as you move +2 units, the change becomes -2
That means the slope at each point is constant
tan(theta)=Increase in Y/Increase in X = -2/2=-1/1=-N/N=-1
That means f(x)=-x +C
C is the Y intercept
While C=f(1)+1
Some of the steps were wrong. You can only make that generalization for adding or subtracting integer values from x - if you're adding or subtracting any non-integer value then there's no guarantee that it will be linear anymore. f(0) = 0, f(1) = -1, f(2) = -2, but f(0.01) = 1000, f(1.01) = 999, f(2.01) = 998 etc. - that pattern won't contradict the equation, but it's definitely not a linear function at that point.
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
You can easily solve it by inserting a Taylorexpansion of f, then collecting all terms with x on one side. As the equation must hold for all x all higher order coefficients must be 0.
However this only gives you the class of continuous solutions. Generally, taking any function on [0,1] and tiling it shifting it by one down each time as you go along the x-axis will be a solution as well meaning for any f defined on 0 to 1 g(x) = f(x - floor(x)) - floor(x) is a solution
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.)
i'm curious how to show the solution (f(x)=-x+a) is unique
if you assume f is differentiable over its domain, we can get that f'(x)=f'(x+1), which has to mean the derivative is a constant, so then we get that f is linear, but I assumed it's differentiable
or maybe I'm wrong with f'(x)=f'(x+1)=>f' is a constant
You can't show that it's unique because it isn't unique. Even if you assume that it's differentiable, it still wouldn't be unique.
For instance, something like f(x) = sin(2*pi*x) - x also satisfies that condition.
@@asdfqwerty14587 well, I understand, you showed that even within that assumption, my proof is wrong (when I said that f'(x)=f'(x+1) means that f'(x) is a constant)
great example btw
If you write it as f(x+1) = f(x)-1 then you can view it as a recurrence relationship from an arithmetic series which is very easy to find the nth term for, and by a simple test you can see that it works for non integer values too
Nicely presented and explained. Cool teacher
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.
Actually, the strategy I used in my head was a bit different ... by substitution (u=x+1) I came to the same
f(x) = f(x+1) +1 ....
but the I sort of rewrote this as
( f(x+1) - f(x) ) / (x+1 - x) = -1
the left hand is now an expession describing the first derivative of f(x) ...
so integrate both sides and you get
f(x) = -x + c
..... and that is why I believe your solution (c=0) is incomplete, because
the expression f(x) = -x +c fulfills the requirements for any given c.
I agree. I fixed my c as zero since I only needed any function that fulfills the condition. Your answer is really cool.
Oh got it its really cool
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
I agree. Pay attention to the question. We are looking for a solution. Any solution.
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
f(x)=f(x+1)+1
f(x-1)=f(x)+1
f(x)=f(x-1)-1
f(1)=f(0)-1
f(2)=f(1)-1=f(0)-2
it seems that f(x)=f(0)-x
we have a base case already and let the assumption be the inductive hypothesis, thus we have to show that f(x+1)=f(0)-x-1
LHS=f(x+1)
=f(x)-1
=f(0)-x-1
RHS
so, f(x)=f(0)-x
let c=f(0)
f(x)=c-x ; c is an arbitrary constant.
1/First of all, f(x)is a linear function. see graphically as follows:
2/At point (x+1) on x axis, vertical value is f(x+2) +1
3/between (x+1) and (x+2), horizontal value is 1 and vertical value is 1 therefore slope is 1 on 1 but negative.
Therefore, f(x)= mx + c
m=-1 , f (x) =-x + c
It doesn't have to be linear though. It could be a stepwise function, where f(x) is the integral part of -x.
It could even be something like f(x) = -sin(2πx) - x
I really think that the complete solution should be: f(x)=k(x), if 0≤x
my solution [ edited/reconsidered ]
f(x + 1) = f(x + 2) + 1
so *f(x) = f(x + 1) + 1*
and f(x - 1) = f(x) + 1 = f(x + 1) + 2
and f(x - 2) = f(x - 1) + 1 = f(x + 1) + 3
and f(x - 3) = f(x - 2) + 1 = f(x + 1) + 4
and so on
so f(x - a) = f(x + 1) + a + 1 = f(x) + a
or *f(x + a) = f(x) - a (I)*
Looks like something polinomial..
So let - by "instinct" - f(x) = g(x) - x (II)
From I and II we have
g(x + a) - (x + a) = g(x) - x - a
So g(x + a) = g(x), and at this point I can only see g(x) as a constant function.
So, finally, *f(x) = -x + C* (C is a constant value)
excuse me friends but I think I made some mistake(s). Maybe f(x) can only be *-x + C* (C as a constant). I must reconsider my solution ASAP .
(I think the main mistake was consider, at some point, "a" as a variable, and not as a constant)
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.
Thanks for the tips, Sir. Love your videos. I guess f(x) = (-x+c) since when put in slope form [(f(x+1)-f(x)]/[(x+1)-x], the slope is always = -1. So f(x) = - x+c where c is any constant, would be the answer and because the information when y=y1, x= x1 is not given, the constant c remains unknown.
Further, found it also interesting that f(x) when taken as = (coshx) whole squared and f(x+1) taken as =(sinhx) whole squared , and then f(x+1) - f(x) = -1 . Requesting you to also explain whether the function can be periodic also and how
I didn’t get your second logic. If you can explain it will be great
@@piyushaspiretoachieve3179 I have requested to explain whether the function can be periodic also and how
I am asking about your cosine and sine function logic what exactly are your trying to say@@AzmiTabish
@@piyushaspiretoachieve3179 Its there in the comment. I don't have much knowledge of mathematics that is why I am asking to explain whether the function can be periodic also and how.
f(x) = A*sin(2pi*x) - x + B, where A, B are unknown constants is most probably the complete answer@@piyushaspiretoachieve3179
f(x) = A*sin(2pi*x) - x + B, where A, B are unknown constants
Why not
f(x)=Sum(An*sin(2*pi*n*x+phi_n))-x+B
where n is integer(-inf,inf) while An,B and phi_n are constants.
This would cover all periodic functions (All hail Fourier).
Yes. Many thanks@@amareshsingh1460
Thanks teacher. Your video arrived in Brazil (São Paulo).
Glad it was helpful!
This linear function is definitely a solution, but there are so many others! I fact for any function on the interval [0,1), you can extend it to a solution by taking the condition as a defining property. Even if you want continuous functions only, just use functions on [0,1] where f(1)=f(0)-1 so the intervals glue neatly together. For differentiability... I think it'd be enough to make sure f is n-differentiable on [0,1] with dⁿf/dxⁿ being the same at both ends.
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.
It doesn't have to be a constant. This can be any periodic function with period = 1 or harmonic (period = 1/2, 1/3, etc.). For example f(x) = - x + sin(4πx).
The way I saw it:
Let f(x)=y
f(x + 1) = f(x + 2) + 1 implies Δy = -1 when Δx = 1 for all x
The slope of a line is defined as Δy/Δx, so here the slope is -1
Thus, any equation of the form f(x) = -x + k will satisfy the initial conditions
this guy teaches in 10 mins, but skl teachers take atleast 2 days to teach us this, salute my guy
W videos, I really enjoy your style of 'teaching' or 'resolving equations' :)
As mentioned f(x) =-x+a works also ∀ a ∈ ℝ. But we haven't established that this family of linear functions are the sole solutions of the problem! Are there non-linear solutions? Ideas of demonstration? Actually, any function like f(x)= -x + g(x) with g(x+1)=g(x) (periodic function) is a solution as well, for example f(x)=-x + cos(2πx).
f(x) = -x + SUM_n [a_n cos(2pi n x) + b_n sin(2pi n x)]
And that is just a tiny part of the solutions...
You know too much math
@@PrimeNewtons One can never know too much math!
@@landsgevaer Only a mathematician says that.
@@PrimeNewtons At least one physicist wrote it too. 😉
Fabulous way solutions are approached - almost "discovering" the answer live. That's most interesting abt this channel What's a good textbook for functional equations like these to get practice solving them (without getting into heavy math definitions)? Thanks
Assuming that x+1=X. It it possible to write the equation like this f(X)=f(X+1)+1
Then (f(X+1)-f(X))/1=-1
It means that for any X the slope between (X;f(X)) and (X+1;(f(X+1)) is -1.
The family of function satisfying this regularity of slope for any X is f(X)= -X+a (a real constant)
use coordinate，y=-x+a
Where się you get this amazing comma sign grom? It is bigger than usual.
Hey guys, f(x+2)-f(x+1)=-1, the difference between successive terms are constant and equal to -1, so f(x)=mx+c=-x+c,
Oh no! My teacher has never taught me about function equation before😮so amazing🎉
Cool problem! I haven't watched the video in its entirety yet, but I would like to point out you do not need to make the assumption that f(x) is a line, you can actually show it. If we use the definition of the derivative, we see that f'(0)=f'(1)=f'(2)...=f'(n), meaning the first derivative is constant. So the function has to be something in the form of a line f(x)=kx+m. Thereafter you can determine that k=-1 and that m can be some arbitrary real number!
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!
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 ...
f(x+1)=f(x)-1, it means every time that x increases by 1, the function value is gonna decrease by 1, so it implies it’s a linear function with a slope being -1, therefore f(x)=-x+C, where C can be any real number
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.
This is a meta-comment.
A) Prime Newtons is the sort of man a thinking human being wants to hang out with.
B) The level of the comments is very high and PN's manner of exposition is the reason why.
As pointed out, there are infinitely many solutions since if you replace f(x) by f(x)+g(x) where g is any periodic function with period 1, then it is still a solution.
Since - 1 is the first difference and constant f is a linear polynomial of the form f(x) =mx+c and
m= [f(x+1)-f(x)]/1=-1
This gives a family of functions of the form f(x)=-x +c.
f(x+1) = f(x+2) +1
substitue x-1 for x
f(x-1+1) = f(x-1 +2) +1
f(x) = f(x+1) +1
f(x+1) -f(x) = -1
left side is the defention of derivative for an increas of 1 in x direction, thus
d(f(x))/dx = -1
thus
f(x) = -x
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
I'm just thinking. Does f'(x) = f'(x+1) mean that it is a linear function?
Ah. It's not. f(x) = C*sin(2Pi*x) - x + k is also a solution as f(x+1) = C*sin(2Pi*x + 2Pi) - x - 1+ k = C*sin(2Pi*x) - x + k - 1= f(x) - 1
so the full anser is f(x) = C1*g(x)-x+C2, where C1 and C2 any constants and g(x) any periodic function with period equals to 1 (g(x+1) = g(x))
C1 is unnecessary as g(x) is already any 1-periodic function and -x part isn't general enough. Others have pointed out that sin(2pix)-floor(x) +k also fits
At least f(x) = -0x +floor(-x) and f(x) = -x and f(x) = -2x + floor(x) all fit the equation
So the more general form is
f(x) = g(x) -kx +floor( (k-1)x ) + c
But I suspect more can be done
floor( (k-1)x )-kx = floor( (k-1)x )-(k-1)x - x = floor( (k-1)x )-floor((k-1)x)-frac((k-1)x) - x = -x - frac((k-1)x)
Does frac((k-1)x) a periodic function? I'm not sure that this one also meets the requirements
I approached that problem as follows:
Any x ∈ ℝ has the following form:
x = floor(x) + fraction(x), floor(x) ∈ ℤ, fraction(x) ∈ [0,1).
f(x+1) = f(x+2)+1
==> f(x) = f(x+1)+1
f(x+1) = f(x)-1
f(x+n) = f(x)-n, n ∈ ℤ
f(floor(x)+fraction(x)) = f(fraction(x))-floor(x)
==> g(x) := f(x)+floor(x) is periodic on distance 1
Let h be any function from [0,1) to ℝ and g(x) := h(x), or
let h be any function from (0,1] to ℝ and g(x) := h(1-x);
f(x) := g(fraction(x)) - floor(x).
==> f(x+1) = g(fraction(x+1)) - floor(x+1)
= g(fraction(x)) - floor(x+1)
= g(fraction(x)) - (floor(x)+1)
= g(fraction(x)) - floor(x) - 1 = f(x) - 1
Examples:
1)
h(x) := e^x,
g(x) := h(x),
f(x) := g(fraction(x)) - floor(x)
= h(fraction(x)) - floor(x)
= e^(1-fraction(x)) - floor(x)
f(2.5) = e^(1-fraction(2.5)) - floor(2.5) = e^(1-0.5) - 2 = e^0.5 - 2
f(1.5) = e^(1-fraction(1.5)) - floor(1.5) = e^(1-0.5) - 2 = e^0.5 - 1 = e^0.5 - 2 + 1 = f(2.5) + 1
2)
h(x) := 1/x,
g(x) := h(1-x),
f(x) := g(fraction(x)) - floor(x)
= h(1-fraction(x)) - floor(x)
= 1/(1-fraction(x)) - floor(x)
f(2.5) = 1/(1-fraction(2.5)) - floor(2.5) = 1/(1-0.5) - 2 = 1/0.5 - 2
f(1.5) = 1/(1-fraction(1.5)) - floor(1.5) = 1/(1-0.5) - 2 = 1/0.5 - 1 = 1/0.5 - 2 + 1 = f(2.5) + 1
I wonder why you decided to take partial type of linear dependence, namely direct proportionality ( I'm not sure what it's called in English, but that's the y = kx dependence), instead of general type y = kx+ b of linear dependence. From that recurrent equation we can easily derive ( e.g. through math induction method) that f(x) = f(x+n) + n, or otherwise f(x+n) - f(x)=-n. So it's very much like linear differential equation, just with finite difference ( as soon as for linear function the growth is constant on equal intervals for x), thus we can't derive additive constant from this equation, so basically the function we are looking for is y= -x + C, where C is an arbitrary rearrange constant.
Because delta f(x) is constant
f(x)=c-x
@@prashantgujar9159 I am not sure which question were you answering
A simple way to realize that the solution has to be linear in x is to take the derivative of the original recursive realationship wrt x. What you find is that the derivative wrt to x is a constant for all x, which means that f(x) is linear in x. Your solution is one solution, but any solution of the form f(x) = -x + constant will also work. Your particular solution is the one for which the constant is zero.
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?
Let f(x) = ax + b
So, we can get:
f(x + 1) = a(x + 1) + b = ax + a + b
f(x + 2) = a(x + 2) + b = ax + 2a + b
We can input to the equation:
ax + a + b = ax + 2a + b + 1
a = 2a + 1
a = -1
After we get the value of a = -1 ; we can get the value of b:
-x - 1 + b = -x - 2 + b + 1
-1 + b = -1 + b
b = 0
So, finally we can get that f(x) = -x 🎉🎉🎉
Well, you can rearrange to f(x+1)-f(x)=-1.
This dictates a linear equation with common difference -1, i.e. f(x) = -x + a
if x=n+y where 0
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.
My reasoning was this (not proof):
f(x+1) just moves the function to the left by 1
f(x)+1 just moves the function up by 1
f(x+1)+1 will move the function 1 to the left and 1 up
If you can imagine those two transformations together (assuming that the f is continuous) it's easy to imagine that the slope must be -1
So f'(x) = -1 f(x) = -x+c
For non continuous functions you have can something like this:
f(x) = -x+1 -> for x in (a, a+1]
f(x) = -x-100 -> for x in (a-1, a]
for a being equal to 2*k were k in integers
or you could even have discrete functions
Question, Can you take the Laplace transform on both side and solve for F(s) and take the inverse Laplace transform
This is great thanks a lot for sharing. Its very usefull for developing a problem solving mindset. Thank you
In my experience, if we take an arbitrary real number called alpha, we have same vertical and horizontal displacements if this alpha is in an argument of the function or it is out of that. So, another way to solve is graphically.
Take ANY function g(x) that is defined on [0, 1)
define f(x) = g(x-n)-n for x in [n, n+1)
Congradulation! This function corresponds to the property that you wrote.
Also - this set of such functions f are the ONLY functions that satisfy the given property.
My strategy was to suppose f is an arithmetic progression, so f(x) = ax + b. Then substitute f(x) in the functional equation and we get a = -1. Ergo, f(x) = -x + b for any b chosen.
Btw, can be solved by definition of the derivative of a function! 👍
f(x)=g(x)-x, where g(x)=g(x+T), T=1 (for example g(x)=sin(2πx) )
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.
In your solution k does not need to be real. It works for any unreal constant number as well.
really nice chalk board writing man!
Please sir, can you recommend textbooks to solve or practice more functional equations