Solving the functional equation f[x+f(y)]=x+y+1

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[yusufjamal]

Putting x=x-f(y) in the original equation cancels out f(y) in LHS. Putting y=x then results in f(x)=x+(1/2)

f(x-f(y)+f(y))=x-f(y)+y+1
f(x)=x-f(y)+y+1

put y=x
f(x)=x-f(x)+x+1
2f(x)=2x+1
f(x)=x+1/2

[robertgerbicz]

One liner put x=y-f(y) then f(y)=2*y-f(y)+1, so f(y)=y+1/2, what is really solution.

[snejpu]

Again checking what happens for x=0 and y=0 solves the problem almost until the end. Good problem, as usual. : )

[Eknoma]

This took me about... let's be generous 10 seconds.

f(x+f(y)) = x + y + 1

f obviously has to be linear, with slope 1, so the f(y) becomes y + something.

What this "something" is is really easy to figure out, you see that it is added once in the f(y), and then again, and that should then equal 1.

Thus "something" has to be 1/2

f(x) = x + 1/2

[madhavgopakumar]

Note that f is injective and bijective(proving this is not that hard) so f-1 exists. Plug x = -y to get f(f(y) - y) = 1 so f(y) = y + f-1(1). Hence, f-1(1) = f(x + f(y)) - (x + f(y)) = y - f(y) + 1 = - f-1(1) + 1. Hence, f-1(1) = 0.5 and f(y) = y + 0.5

[josephwong]

Differentiate f from first principle:
Lim (f(x+f(y)+h)-f(x+f(y)))/h
=lim (x+h+y+1-(x+y+1))/h (Apply the functional equation to both terms)
=lim h/h
=1

So the function f is differentiable and the derivative is constant 1, meaning f is a linear polynomial and f(x)=x+c for some constant c.

Then the rest is simple:
f(f(y))=y+1 (from the functional equation) and f(f(y))=y+2c (from that f is linear with derivative 1, which is easily solved that c=1/2.

[angelmendez-rivera]

So, because we are merely dealing with polynomials and compositions, and no division, we can safely postulate the problem as looking for solutions R —> R. Specifying the domain is important for these problems to not be ill-formed. The video does say f should be assumed to be continuous, and that f is a nice function, but still, I can definitely see how this keeps the problem ambiguous for many people.

With that out of the way, what you will notice is that the main source of difficulty for this equation is that pesky f(y) on the inside. Generally, composition of functions is much more annoying to deal with than multiplication or addition, so we would like to get rid of that first. Here is what you do: you set x = x' – f(y), y = y'. This substitution results in f(x') = x' – f(y') + y' + 1. No more compositions! In fact, there is no multiplication either: we only have addition to deal with. So we should being every f to one side, to simplify out thought process: we have f(x') + f(y') = x' + y' + 1. Now, you can set x' = t, y' = t. The advantage with this is that it takes this two-variable problem and turns it into a one-variable problem. Hence 2·f(t) = 2·t + 1. Now this is just a basic algebra problem: the solution is f(t) = t + 1/2. And indeed, f has no singularities, as expected.

I am aware other comments have already mentioned this approach, but I wrote this because I wanted to be a bit more careful with the maths, and I also wanted to explain why you would let x = x' – f(y). After all, to many other people, this substitution would not be obvious. It becomes obvious once you think through the lens of "yeah, I don't like nested functions or functional composition", but for someone jumping into this problem without any hints, it may be hard for them to think of that right off bat, especially if they are not experienced with functional equations.

Also, an important bonus of this approach is that there is actually no need to assume f is continuous from the start of the problem, which definitely makes the solution more convincing.

[jamesstrickland]

I like that there are so many ways to solve this problem. my way,
let x=0
f(f(y))=y+1
y=x
f(f(x))=x+1
f(f(x))=(x+c)+c
so 2c = 1
so c = 1/2
f(x)=x+1/2

[tetramur]

x = 0
f(f(y)) = y+1
This is easy (for example, Taylor's series helps), result is f(y) = y+1/2. So the solution is f(x) = x+1/2

[mohamedfarouk]

f(0) = c
f(x) = x - c + 1
Let x = 0
f(0) = - c + 1
But we know f(0) = c
So c = - c + 1
c = ½

[benjaminvatovez]

Thank you for the solution. y=-1 gives x=f(something) for all x then f is surjective then there is at least one y' such that f(y')=0. We get f(x)=x+y'+1=x+c as you got and we get c=0.5 when replaced in the equation.

[aleksszukovskis]

Hi SyberMath, i'm a beginner to functional equations. can you link some textbook or an introductory video to this field?

[josephsilver]

Nice problem. You used a clever approach that made the solution simple!

[davidseed]

at 1:19 you have f(x+c)=x+1.
from here you should substitute x:= -c, this gives
f(-c+c) = -c+1, hence
f(0)=c=-c+1 hence c=1/2
then substitute x:=x-c we have f(x)=x-c+1= x+1/2

[Packerfan]

f(x + f(y)) = x + y + 1

For y = 0, we have f(x + f(0)) = x + 1.
Set t = x + f(0). Then
f(t) = (t - f(0)) + 1
f(t) = t + 1 - f(0)

f(x + f(y)) = x + y + 1
f(x + (y + 1 - f(0)) = x + y + 1
f(x + y + 1 - f(0)) = x + y + 1
(x + y + 1 - f(0)) + 1 - f(0) = x + y + 1
x + y + 2 - 2 f(0) = x + y + 1
2 f(0) = 1
f(0) = 1/2

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

[fix]

i did it quite similar: substituting x=0 and y=f(y) gives f(f(f(y)))=f(y)+1 which implies f(y+1)-f(y)= 1, thus f is of the form x+n, substituting back gives n=1/2

[ramaprasadghosh]

Linearity of the functional value as evident from f(x + f(y) ) = x + y + 1, essentially implies f(x) = a*x + b
Hereby f(x + f(y) ) = a(x + f(y) ) + b
= a(x + ay + b) + b
Hereby
a (x+y) + b(a+1) = x + y + 1
or a = 1, b = 1/ (a+1) = 1/2

[thierryvieville]

my half a penny + x variant, not better, just showing another way ...
1/ introducing z = x + f(y), yields f(z) = x + y + 1
2/ considering z = 0 and eliminating x, yields f(y) = -x = y + 1 - f(0)
3/ considering y = 0 in this last equation yields 2 f(0) = 1 so that finally
we must have f(y) = y + 1/2
and are left to verify that
f(x+f(y)) = (x + (y + 1/2) + 1/2) = x + y + 1
works, and ``this brings us to this end of this ´´ nice problem 😁
P.S.: next time i will write f(0) = b, so that we may consider 2 b (or not) … 😂😎

Thanks again for this nice problem, with several interesting way through. ..

[djridoo]

I replaced y with -1 and x with x - f(-1) and it goes like the same, but faster ;)

[hogehoge]

for y, choose x := -f(y); then you'll have f(y) = y - f(0) + 1, and hence f(x) = x - f(0) + 1. Substituting 0 for x, you can calculate f(0).